https://www.glc.us.es/~jalonso/RA2016/api.php?action=feedcontributions&feedformat=atom&user=LucnovdosRazonamiento automático (2016-17) - Contribuciones del usuario [es]2026-07-18T05:48:34ZContribuciones del usuarioMediaWiki 1.31.14https://www.glc.us.es/~jalonso/RA2016/index.php?title=Relaci%C3%B3n_7&diff=1205Relación 72016-12-21T20:41:10Z<p>Lucnovdos: </p>
<hr />
<div><source lang="isar"><br />
chapter {* R7: Árboles binarios completos *}<br />
<br />
theory R7_Arboles_binarios_completos<br />
imports Main <br />
begin <br />
<br />
text {* <br />
En esta relación se piden demostraciones automáticas (lo más cortas<br />
posibles). Para ello, en algunos casos es necesario incluir lemas<br />
auxiliares (que se demuestran automáticamente) y usar ejercicios<br />
anteriores. <br />
<br />
--------------------------------------------------------------------- <br />
Ejercicio 1. Definir el tipo de datos arbol para representar los<br />
árboles binarios que no tienen información ni en los nodos y ni en las<br />
hojas. Por ejemplo, el árbol<br />
·<br />
/ \<br />
/ \<br />
· ·<br />
/ \ / \<br />
· · · · <br />
se representa por "N (N H H) (N H H)".<br />
--------------------------------------------------------------------- <br />
*}<br />
<br />
datatype arbol = H | N arbol arbol<br />
<br />
value "N (N H H) (N H H) = (N (N H H) (N H H) :: arbol)"<br />
<br />
text {* <br />
--------------------------------------------------------------------- <br />
Ejercicio 2. Definir la función<br />
hojas :: "arbol => nat" <br />
tal que (hojas a) es el número de hojas del árbol a. Por ejemplo,<br />
hojas (N (N H H) (N H H)) = 4<br />
--------------------------------------------------------------------- <br />
*}<br />
<br />
(* fraortmoy marcarmor13 josgarsan fracorjim1 *)<br />
fun hojas :: "arbol => nat" where<br />
"hojas H = Suc 0"<br />
| "hojas (N a b) = hojas a + hojas b"<br />
<br />
value "hojas (N (N H H) (N H H)) = 4"<br />
<br />
(* marpoldia1 anaprarod paupeddeg migtermor wilmorort pablucoto <br />
ivamenjim serrodcal crigomgom rubgonmar danrodcha ferrenseg manmorjim1 juacabsou lucnovdos <br />
dancorgar bowma *)<br />
(* Es muy parecida a la definición anterior *)<br />
fun hojas2 :: "arbol => nat" where<br />
"hojas2 H = 1" |<br />
"hojas2 (N i d) = hojas2 i + hojas2 d"<br />
<br />
value "hojas2 (N (N H H) (N H H)) = 4"<br />
<br />
<br />
(* anaprarod *)<br />
(* Equivalencia de las definiciones *)<br />
lemma "hojas a = hojas2 a"<br />
by (induct a) simp_all<br />
<br />
text {* <br />
--------------------------------------------------------------------- <br />
Ejercicio 4. Definir la función<br />
profundidad :: "arbol => nat" <br />
tal que (profundidad a) es la profundidad del árbol a. Por ejemplo,<br />
profundidad (N (N H H) (N H H)) = 2<br />
--------------------------------------------------------------------- <br />
*}<br />
<br />
(* fraortmoy marpoldia1 anaprarod migtermor wilmorort marcarmor13 dancorgar *)<br />
fun profundidad :: "arbol => nat" where<br />
"profundidad H = 0"<br />
| "profundidad (N a b) = (if profundidad a > profundidad b<br />
then 1 + profundidad a <br />
else 1 + profundidad b)"<br />
<br />
value "profundidad (N (N H H) (N H H)) = 2"<br />
<br />
(* anaprarod wilmorort pablucoto ivamenjim serrodcal crigomgom rubgonmar <br />
danrodcha ferrenseg josgarsan juacabsou lucnovdos bowma *)<br />
fun profundidad2 :: "arbol => nat" where<br />
"profundidad2 H = 0"<br />
|"profundidad2 (N i d) = 1 + (max (profundidad2 i) (profundidad2 d))"<br />
<br />
value "profundidad2 (N (N H H) (N H H)) = 2"<br />
<br />
(* anaprarod *)<br />
(* Equivalencia de las definiciones *)<br />
lemma "profundidad a= profundidad2 a"<br />
by (induct a) auto<br />
<br />
<br />
(* paupeddeg *)<br />
fun maximo :: "nat × nat => nat" where<br />
"maximo (a,b) = (if a > b <br />
then a else b)"<br />
<br />
fun profundidad :: "arbol => nat" where<br />
"profundidad H = 0"<br />
| "profundidad (N i d) = 1 + maximo(profundidad i, profundidad d)"<br />
<br />
<br />
(* ivamenjim: llamando a la función anterior profundidad3 *)<br />
<br />
lemma "profundidad a = profundidad3 a"<br />
by (induct a) auto<br />
<br />
<br />
(* ivamenjim manmorjim1 fracorjim1 *)<br />
fun profundidad4 :: "arbol => nat" where<br />
"profundidad4 H = 0"<br />
| "profundidad4 (N i d) = Suc (max (profundidad4 i)(profundidad4 d))"<br />
<br />
lemma "profundidad a = profundidad4 a"<br />
by (induct a) auto<br />
<br />
text {* <br />
--------------------------------------------------------------------- <br />
Ejercicio 5. Definir la función<br />
abc :: "nat ⇒ arbol" <br />
tal que (abc n) es el árbol binario completo de profundidad n. Por<br />
ejemplo, <br />
abc 3 = N (N (N H H) (N H H)) (N (N H H) (N H H))<br />
--------------------------------------------------------------------- <br />
*}<br />
(* fraortmoy marpoldia1 anaprarod paupeddeg migtermor wilmorort <br />
serrodcal crigomgom rubgonmar danrodcha ferrenseg josgarsan manmorjim1 juacabsou fracorjim1 lucnovdos <br />
dancorgar bowma *)<br />
fun abc :: "nat ⇒ arbol" where<br />
"abc 0 = H"<br />
| "abc (Suc n) = (N (abc n) (abc n))"<br />
<br />
value "abc 3 = N (N (N H H) (N H H)) (N (N H H) (N H H))"<br />
<br />
(* ivamenjim pablucoto marcarmor13*)<br />
fun abc2 :: "nat ⇒ arbol" where<br />
"abc2 0 = H"<br />
| "abc2 t = N (abc2 (t-1)) (abc2 (t-1))"<br />
<br />
value "abc2 3 = N (N (N H H) (N H H)) (N (N H H) (N H H))"<br />
<br />
(* ivamenjim: Metaejercicio de demostración *)<br />
lemma "abc t = abc2 t"<br />
by (induct t) auto<br />
<br />
text {* <br />
--------------------------------------------------------------------- <br />
Ejercicio 6. Un árbol binario a es completo respecto de la medida f si<br />
a es una hoja o bien a es de la forma (N i d) y se cumple que tanto i<br />
como d son árboles binarios completos respecto de f y, además, <br />
f(i) = f(r).<br />
<br />
Definir la función<br />
es_abc :: "(arbol => 'a) => arbol => bool<br />
tal que (es_abc f a) se verifica si a es un árbol binario completo<br />
respecto de f.<br />
--------------------------------------------------------------------- <br />
*}<br />
<br />
(* fraortmoy anaprarod migtermor serrodcal crigomgom rubgonmar <br />
danrodcha ferrenseg juacabsou josgarsan fracorjim1 lucnovdos bowma *)<br />
fun es_abc :: "(arbol => 'a) => arbol => bool" where<br />
"es_abc _ H = True"<br />
| "es_abc f (N a b) = (es_abc f a ∧ es_abc f b ∧ (f a = f b))"<br />
<br />
<br />
(* marpoldia1 paupeddeg ivamenjim pablucoto marcarmor13 manmorjim1 dancorgar *)<br />
fun es_abc2 :: "(arbol => 'a) => arbol => bool" where<br />
"es_abc2 f H = True" |<br />
"es_abc2 f (N i d) = ((f i = f d) ∧ (es_abc2 f i) ∧ (es_abc2 f d))"<br />
<br />
(* anaprarod *)<br />
(* Equivalencia de las definiciones *)<br />
lemma "es_abc f a = es_abc2 f a"<br />
by (induct a) auto<br />
<br />
text {* <br />
--------------------------------------------------------------------- <br />
Nota. (size a) es el número de nodos del árbol a. Por ejemplo,<br />
size (N (N H H) (N H H)) = 3<br />
--------------------------------------------------------------------- <br />
*}<br />
<br />
value "size (N (N H H) (N H H)) = 3"<br />
value "size (N (N (N H H) (N H H)) (N (N H H) (N H H))) = 7"<br />
<br />
text {* <br />
--------------------------------------------------------------------- <br />
Nota. Tenemos 3 funciones de medida sobre los árboles: número de<br />
hojas, número de nodos y profundidad. A cada una le corresponde un<br />
concepto de completitud. En los siguientes ejercicios demostraremos<br />
que los tres conceptos de completitud son iguales.<br />
--------------------------------------------------------------------- <br />
*}<br />
<br />
text {* <br />
--------------------------------------------------------------------- <br />
Ejercicio 7. Demostrar que un árbol binario a es completo respecto de<br />
la profundidad syss es completo respecto del número de hojas.<br />
--------------------------------------------------------------------- <br />
*}<br />
<br />
(* fraortmoy marpoldia1 migtermor wilmorort pablucoto serrodcal fracorjim1 josgarsan lucnovdos bowma *)<br />
<br />
lemma abc_prof_num_hojas:<br />
assumes "es_abc profundidad a"<br />
shows "hojas a = 2^(profundidad a)"<br />
using assms<br />
by (induct a) auto<br />
<br />
(* Otra forma de escribir lo mismo *)<br />
(* anaprarod crigomgom ivamenjim paupeddeg juacabsou *)<br />
lemma AUX7: "es_abc profundidad a ⟶ (hojas a = 2^(profundidad a))"<br />
by (induct a) simp_all<br />
<br />
(* Otra forma de escribir lo mismo *)<br />
(* danrodcha *)<br />
lemma aux1: "es_abc profundidad (a::arbol) ⟹ (hojas a = 2^ (profundidad a))"<br />
by (induct a) simp_all<br />
<br />
(* fraortmoy marpoldia1 migtermor anaprarod wilmorort serrodcal crigomgom <br />
rubgonmar ivamenjim danrodcha marcarmor13 paupeddeg juacabsou bowma *)<br />
(* También funciona con AUX7 *)<br />
<br />
lemma lej7: "es_abc profundidad a = es_abc hojas a"<br />
by (induct a) (auto simp add: abc_prof_num_hojas)<br />
<br />
(* danrodcha *)<br />
lemma 7: "es_abc profundidad a = es_abc hojas a"<br />
apply (induct a) <br />
apply simp<br />
apply (auto simp add: aux1)<br />
done<br />
<br />
(* ferrenseg *)<br />
lemma [simp]: "es_abc profundidad a ⟶ hojas a = 2 ^ (profundidad a)"<br />
by (induct a) auto<br />
<br />
theorem es_abc_profundidad_hojas: "es_abc profundidad a = es_abc hojas a"<br />
by (induct a) auto<br />
<br />
(* dancorgar *)<br />
lemma rel_hojas_prof: "es_abc hojas a ∧ es_abc profundidad a<br />
⟹ hojas a = 2 ^ profundidad a"<br />
by (induct a) (auto)<br />
<br />
theorem "es_abc hojas a = es_abc profundidad a"<br />
by (induct a) (auto simp add: rel_hojas_prof)<br />
<br />
<br />
text {* <br />
--------------------------------------------------------------------- <br />
Ejercicio 8. Demostrar que un árbol binario a es completo respecto del<br />
número de hojas syss es completo respecto del número de nodos.<br />
--------------------------------------------------------------------- <br />
*}<br />
<br />
(* fraortmoy marpoldia1 migtermor wilmorort pablucoto serrodcal marcarmor13 josgarsan lucnovdos bowma*)<br />
<br />
lemma abc_hojas_num_nodos:<br />
assumes "es_abc hojas a"<br />
shows "Suc(size a) = hojas a"<br />
using assms<br />
by (induct a) auto<br />
<br />
<br />
(* Otra forma de escribir lo mismo *)<br />
(* anaprarod crigomgom paupeddeg juacabsou rubgonmar *)<br />
lemma AUX8: "es_abc hojas a ⟶ (hojas a = (Suc (size a)))"<br />
by (induct a) auto<br />
<br />
<br />
(* fraortmoy marpoldia1 migtermor anaprarod wilmorort pablucoto serrodcal *)<br />
<br />
lemma lej8: "es_abc hojas a = es_abc size a"<br />
by (induct a) (auto simp add:abc_hojas_num_nodos [symmetric])<br />
<br />
(* anaprarod crigomgom paupeddeg juacabsou rubgonmar *)<br />
(* Usando AUX8 *)<br />
lemma L8: "es_abc hojas a= es_abc size a"<br />
by (induct a) (auto simp add: AUX8)<br />
<br />
<br />
(* ivamenjim *)<br />
(* Teorema auxiliar *)<br />
lemma auxEj8: "hojas a = size a + 1"<br />
by (induct a) auto<br />
<br />
(* ivamenjim *)<br />
lemma lej8: "es_abc hojas a = es_abc size a"<br />
by (induct a) (auto simp add: auxEj8)<br />
<br />
(* danrodcha *)<br />
lemma aux3: "es_abc hojas a ⟹ (hojas a = 1 + size a)"<br />
by (induct a) simp_all<br />
<br />
(* danrodcha *)<br />
lemma 8: "es_abc hojas a = es_abc size a"<br />
apply (induct a) <br />
apply simp<br />
apply (auto simp add: aux3)<br />
done<br />
<br />
(* ferrenseg *)<br />
lemma [simp]: "hojas a = size a + 1"<br />
by (induct a) auto<br />
<br />
theorem es_abc_hojas_size: "es_abc hojas a = es_abc size a"<br />
by (induct a) auto<br />
<br />
(* fracorjim1 *)<br />
<br />
lemma aux10 : "hojas (a::arbol) = Suc(size a)"<br />
apply (induct a)<br />
apply auto<br />
done<br />
<br />
lemma es_abc_hojas_size_10 : "es_abc hojas (a::arbol) = es_abc size a"<br />
apply (induct a)<br />
apply (auto simp add: aux10)<br />
done<br />
<br />
(* dancorgar *)<br />
lemma rel_hojas_size: "es_abc hojas a ∧ es_abc size a<br />
⟹ hojas a = (size a) + 1"<br />
by (induct a) (auto)<br />
<br />
theorem "es_abc hojas a = es_abc size a"<br />
by (induct a) (auto simp add: rel_hojas_size)<br />
<br />
text {* <br />
--------------------------------------------------------------------- <br />
Ejercicio 9. Demostrar que un árbol binario a es completo respecto de<br />
la profundidad syss es completo respecto del número de nodos.<br />
--------------------------------------------------------------------- <br />
*}<br />
<br />
(* fraortmoy marpoldia1 migtermor anaprarod wilmorort pablucoto <br />
serrodcal crigomgom rubgonmar danrodcha ivamenjim marcarmor13 paupeddeg juacabsou<br />
josgarsan bowma lucnovdos *)<br />
<br />
lemma lej9: "es_abc profundidad a = es_abc size a"<br />
by (simp add: lej7 lej8)<br />
<br />
(* ferrenseg *)<br />
<br />
corollary es_abc_size_profundidad: "es_abc size a = es_abc profundidad a"<br />
by (simp add: es_abc_profundidad_hojas es_abc_hojas_size)<br />
<br />
text {* <br />
--------------------------------------------------------------------- <br />
Ejercicio 10. Demostrar que (abc n) es un árbol binario completo.<br />
--------------------------------------------------------------------- <br />
*}<br />
<br />
(* fraortmoy marpoldia1 migtermor wilmorort pablucoto serrodcal crigomgom marcarmor13 paupeddeg juacabsou<br />
dancorgar lucnovdos *)<br />
<br />
lemma lej10: "es_abc profundidad (abc n)"<br />
by (induct n) auto<br />
<br />
<br />
(* anaprarod rubgonmar danrodcha ferrenseg ivamenjim paupeddeg juacabsou bowma*)<br />
(* con un demostrador más débil *)<br />
(* y en general para cualquier medida *)<br />
lemma L10: "es_abc f (abc a)"<br />
by (induct a) simp_all<br />
<br />
(* ivamenjim *)<br />
(* Igual que el anterior pero usando auto *)<br />
lemma lej10: "es_abc f (abc n)"<br />
by (induct n) auto<br />
<br />
text {* <br />
--------------------------------------------------------------------- <br />
Ejercicio 11. Demostrar que si a es un árbolo binario completo<br />
respecto de la profundidad, entonces a es igual a<br />
(abc (profundidad a)).<br />
--------------------------------------------------------------------- <br />
*}<br />
<br />
(* fraortmoy marpoldia1 migtermor wilmorort pablucoto serrodcal marcarmor13 *)<br />
<br />
lemma lej11: <br />
assumes " es_abc profundidad a"<br />
shows "a = (abc (profundidad a))"<br />
using assms<br />
by (induct a) auto<br />
<br />
<br />
(* Otra forma de escribir lo mismo *)<br />
(* anaprarod crigomgom rubgonmar ferrenseg ivamenjim paupeddeg juacabsou dancorgar bowma lucnovdos *)<br />
lemma "es_abc profundidad a ⟶ (a = (abc (profundidad a)))"<br />
by (induct a) auto<br />
<br />
(* danrodcha*)<br />
lemma 11:"es_abc profundidad a ⟹ (a = (abc (profundidad a)))"<br />
by (induct a) simp_all<br />
<br />
<br />
text {* <br />
--------------------------------------------------------------------- <br />
Ejercicio 12. Encontrar una medida f tal que (es_abc f) es distinto de <br />
(es_abc size).<br />
--------------------------------------------------------------------- <br />
*}<br />
<br />
(* migtermor *)<br />
fun medida_nula :: "arbol => nat" where<br />
"medida_nula H = 0"<br />
| "medida_nula (N i d) = 0"<br />
<br />
lemma "es_abc medida_nula a = es_abc size a"<br />
quickcheck<br />
oops<br />
(* Quickcheck encuentra el siguiente contraejemplo:<br />
a= N H (N H H) <br />
Tras evaluar:<br />
es_abc medida_nula a = True<br />
es_abc size a = False*)<br />
<br />
<br />
(* anaprarod wilmorort pablucoto serrodcal danrodcha marcarmor13 ferrenseg ivamenjim paupeddeg juacabsou rubgonmar bowma lucnovdos *)<br />
lemma "es_abc f a = es_abc size a"<br />
quickcheck<br />
(* Quickcheck found a counterexample:<br />
f = λx. a⇩1 <br />
a = N H (N H H)<br />
Evaluated terms:<br />
es_abc f a = True<br />
es_abc size a = False *)<br />
oops<br />
(* el contraejemplo que encuentra es la medida constante a1 *)<br />
<br />
(*crigomgom *)<br />
(* Como en la primera de las soluciones he usado la función constante 0 pero he usado una expresión lambda*)<br />
lemma "es_abc (λx. 0::nat) a = es_abc size a"<br />
quickcheck<br />
oops<br />
<br />
(* dancorgar *)<br />
fun medida :: "arbol => bool" where<br />
"medida H = True"<br />
| "medida (N i d) = ((profundidad i) = (size d))"<br />
<br />
theorem "es_abc size a = es_abc medida a"<br />
quickcheck<br />
oops<br />
<br />
<br />
end<br />
</source></div>Lucnovdoshttps://www.glc.us.es/~jalonso/RA2016/index.php?title=Relaci%C3%B3n_7&diff=1191Relación 72016-12-21T13:10:58Z<p>Lucnovdos: </p>
<hr />
<div><source lang="isar"><br />
chapter {* R7: Árboles binarios completos *}<br />
<br />
theory R7_Arboles_binarios_completos<br />
imports Main <br />
begin <br />
<br />
text {* <br />
En esta relación se piden demostraciones automáticas (lo más cortas<br />
posibles). Para ello, en algunos casos es necesario incluir lemas<br />
auxiliares (que se demuestran automáticamente) y usar ejercicios<br />
anteriores. <br />
<br />
--------------------------------------------------------------------- <br />
Ejercicio 1. Definir el tipo de datos arbol para representar los<br />
árboles binarios que no tienen información ni en los nodos y ni en las<br />
hojas. Por ejemplo, el árbol<br />
·<br />
/ \<br />
/ \<br />
· ·<br />
/ \ / \<br />
· · · · <br />
se representa por "N (N H H) (N H H)".<br />
--------------------------------------------------------------------- <br />
*}<br />
<br />
datatype arbol = H | N arbol arbol<br />
<br />
value "N (N H H) (N H H) = (N (N H H) (N H H) :: arbol)"<br />
<br />
text {* <br />
--------------------------------------------------------------------- <br />
Ejercicio 2. Definir la función<br />
hojas :: "arbol => nat" <br />
tal que (hojas a) es el número de hojas del árbol a. Por ejemplo,<br />
hojas (N (N H H) (N H H)) = 4<br />
--------------------------------------------------------------------- <br />
*}<br />
<br />
(* fraortmoy marcarmor13 josgarsan fracorjim1 *)<br />
fun hojas :: "arbol => nat" where<br />
"hojas H = Suc 0"<br />
| "hojas (N a b) = hojas a + hojas b"<br />
<br />
value "hojas (N (N H H) (N H H)) = 4"<br />
<br />
(* marpoldia1 anaprarod paupeddeg migtermor wilmorort pablucoto <br />
ivamenjim serrodcal crigomgom rubgonmar danrodcha ferrenseg manmorjim1 juacabsou lucnovdos *)<br />
(* Es muy parecida a la definición anterior *)<br />
fun hojas2 :: "arbol => nat" where<br />
"hojas2 H = 1" |<br />
"hojas2 (N i d) = hojas2 i + hojas2 d"<br />
<br />
value "hojas2 (N (N H H) (N H H)) = 4"<br />
<br />
<br />
(* anaprarod *)<br />
(* Equivalencia de las definiciones *)<br />
lemma "hojas a = hojas2 a"<br />
by (induct a) simp_all<br />
<br />
text {* <br />
--------------------------------------------------------------------- <br />
Ejercicio 4. Definir la función<br />
profundidad :: "arbol => nat" <br />
tal que (profundidad a) es la profundidad del árbol a. Por ejemplo,<br />
profundidad (N (N H H) (N H H)) = 2<br />
--------------------------------------------------------------------- <br />
*}<br />
<br />
(* fraortmoy marpoldia1 anaprarod migtermor wilmorort marcarmor13*)<br />
fun profundidad :: "arbol => nat" where<br />
"profundidad H = 0"<br />
| "profundidad (N a b) = (if profundidad a > profundidad b<br />
then 1 + profundidad a <br />
else 1 + profundidad b)"<br />
<br />
value "profundidad (N (N H H) (N H H)) = 2"<br />
<br />
(* anaprarod wilmorort pablucoto ivamenjim serrodcal crigomgom rubgonmar <br />
danrodcha ferrenseg josgarsan juacabsou lucnovdos*)<br />
fun profundidad2 :: "arbol => nat" where<br />
"profundidad2 H = 0"<br />
|"profundidad2 (N i d) = 1 + (max (profundidad2 i) (profundidad2 d))"<br />
<br />
value "profundidad2 (N (N H H) (N H H)) = 2"<br />
<br />
(* anaprarod *)<br />
(* Equivalencia de las definiciones *)<br />
lemma "profundidad a= profundidad2 a"<br />
by (induct a) auto<br />
<br />
<br />
(* paupeddeg *)<br />
fun maximo :: "nat × nat => nat" where<br />
"maximo (a,b) = (if a > b <br />
then a else b)"<br />
<br />
fun profundidad :: "arbol => nat" where<br />
"profundidad H = 0"<br />
| "profundidad (N i d) = 1 + maximo(profundidad i, profundidad d)"<br />
<br />
<br />
(* ivamenjim: llamando a la función anterior profundidad3 *)<br />
<br />
lemma "profundidad a = profundidad3 a"<br />
by (induct a) auto<br />
<br />
<br />
(* ivamenjim manmorjim1 fracorjim1 *)<br />
fun profundidad4 :: "arbol => nat" where<br />
"profundidad4 H = 0"<br />
| "profundidad4 (N i d) = Suc (max (profundidad4 i)(profundidad4 d))"<br />
<br />
lemma "profundidad a = profundidad4 a"<br />
by (induct a) auto<br />
<br />
text {* <br />
--------------------------------------------------------------------- <br />
Ejercicio 5. Definir la función<br />
abc :: "nat ⇒ arbol" <br />
tal que (abc n) es el árbol binario completo de profundidad n. Por<br />
ejemplo, <br />
abc 3 = N (N (N H H) (N H H)) (N (N H H) (N H H))<br />
--------------------------------------------------------------------- <br />
*}<br />
(* fraortmoy marpoldia1 anaprarod paupeddeg migtermor wilmorort <br />
serrodcal crigomgom rubgonmar danrodcha ferrenseg josgarsan manmorjim1 juacabsou fracorjim1 lucnovdos *)<br />
fun abc :: "nat ⇒ arbol" where<br />
"abc 0 = H"<br />
| "abc (Suc n) = (N (abc n) (abc n))"<br />
<br />
value "abc 3 = N (N (N H H) (N H H)) (N (N H H) (N H H))"<br />
<br />
(* ivamenjim pablucoto marcarmor13*)<br />
fun abc2 :: "nat ⇒ arbol" where<br />
"abc2 0 = H"<br />
| "abc2 t = N (abc2 (t-1)) (abc2 (t-1))"<br />
<br />
value "abc2 3 = N (N (N H H) (N H H)) (N (N H H) (N H H))"<br />
<br />
(* ivamenjim: Metaejercicio de demostración *)<br />
lemma "abc t = abc2 t"<br />
by (induct t) auto<br />
<br />
text {* <br />
--------------------------------------------------------------------- <br />
Ejercicio 6. Un árbol binario a es completo respecto de la medida f si<br />
a es una hoja o bien a es de la forma (N i d) y se cumple que tanto i<br />
como d son árboles binarios completos respecto de f y, además, <br />
f(i) = f(r).<br />
<br />
Definir la función<br />
es_abc :: "(arbol => 'a) => arbol => bool<br />
tal que (es_abc f a) se verifica si a es un árbol binario completo<br />
respecto de f.<br />
--------------------------------------------------------------------- <br />
*}<br />
<br />
(* fraortmoy anaprarod migtermor serrodcal crigomgom rubgonmar <br />
danrodcha ferrenseg juacabsou josgarsan fracorjim1 lucnovdos *)<br />
fun es_abc :: "(arbol => 'a) => arbol => bool" where<br />
"es_abc _ H = True"<br />
| "es_abc f (N a b) = (es_abc f a ∧ es_abc f b ∧ (f a = f b))"<br />
<br />
<br />
(* marpoldia1 paupeddeg ivamenjim pablucoto marcarmor13 manmorjim1 *)<br />
fun es_abc2 :: "(arbol => 'a) => arbol => bool" where<br />
"es_abc2 f H = True" |<br />
"es_abc2 f (N i d) = ((f i = f d) ∧ (es_abc2 f i) ∧ (es_abc2 f d))"<br />
<br />
(* anaprarod *)<br />
(* Equivalencia de las definiciones *)<br />
lemma "es_abc f a = es_abc2 f a"<br />
by (induct a) auto<br />
<br />
text {* <br />
--------------------------------------------------------------------- <br />
Nota. (size a) es el número de nodos del árbol a. Por ejemplo,<br />
size (N (N H H) (N H H)) = 3<br />
--------------------------------------------------------------------- <br />
*}<br />
<br />
value "size (N (N H H) (N H H)) = 3"<br />
value "size (N (N (N H H) (N H H)) (N (N H H) (N H H))) = 7"<br />
<br />
text {* <br />
--------------------------------------------------------------------- <br />
Nota. Tenemos 3 funciones de medida sobre los árboles: número de<br />
hojas, número de nodos y profundidad. A cada una le corresponde un<br />
concepto de completitud. En los siguientes ejercicios demostraremos<br />
que los tres conceptos de completitud son iguales.<br />
--------------------------------------------------------------------- <br />
*}<br />
<br />
text {* <br />
--------------------------------------------------------------------- <br />
Ejercicio 7. Demostrar que un árbol binario a es completo respecto de<br />
la profundidad syss es completo respecto del número de hojas.<br />
--------------------------------------------------------------------- <br />
*}<br />
<br />
(* fraortmoy marpoldia1 migtermor wilmorort pablucoto serrodcal fracorjim1 josgarsan lucnovdos *)<br />
<br />
lemma abc_prof_num_hojas:<br />
assumes "es_abc profundidad a"<br />
shows "hojas a = 2^(profundidad a)"<br />
using assms<br />
by (induct a) auto<br />
<br />
(* Otra forma de escribir lo mismo *)<br />
(* anaprarod crigomgom ivamenjim paupeddeg juacabsou *)<br />
lemma AUX7: "es_abc profundidad a ⟶ (hojas a = 2^(profundidad a))"<br />
by (induct a) simp_all<br />
<br />
(* Otra forma de escribir lo mismo *)<br />
(* danrodcha *)<br />
lemma aux1: "es_abc profundidad (a::arbol) ⟹ (hojas a = 2^ (profundidad a))"<br />
by (induct a) simp_all<br />
<br />
(* fraortmoy marpoldia1 migtermor anaprarod wilmorort serrodcal crigomgom <br />
rubgonmar ivamenjim danrodcha marcarmor13 paupeddeg juacabsou*)<br />
(* También funciona con AUX7 *)<br />
<br />
lemma lej7: "es_abc profundidad a = es_abc hojas a"<br />
by (induct a) (auto simp add: abc_prof_num_hojas)<br />
<br />
(* danrodcha *)<br />
lemma 7: "es_abc profundidad a = es_abc hojas a"<br />
apply (induct a) <br />
apply simp<br />
apply (auto simp add: aux1)<br />
done<br />
<br />
(* ferrenseg *)<br />
lemma [simp]: "es_abc profundidad a ⟶ hojas a = 2 ^ (profundidad a)"<br />
by (induct a) auto<br />
<br />
theorem es_abc_profundidad_hojas: "es_abc profundidad a = es_abc hojas a"<br />
by (induct a) auto<br />
<br />
<br />
text {* <br />
--------------------------------------------------------------------- <br />
Ejercicio 8. Demostrar que un árbol binario a es completo respecto del<br />
número de hojas syss es completo respecto del número de nodos.<br />
--------------------------------------------------------------------- <br />
*}<br />
<br />
(* fraortmoy marpoldia1 migtermor wilmorort pablucoto serrodcal marcarmor13 josgarsan lucnovdos *)<br />
<br />
lemma abc_hojas_num_nodos:<br />
assumes "es_abc hojas a"<br />
shows "Suc(size a) = hojas a"<br />
using assms<br />
by (induct a) auto<br />
<br />
<br />
(* Otra forma de escribir lo mismo *)<br />
(* anaprarod crigomgom paupeddeg juacabsou rubgonmar *)<br />
lemma AUX8: "es_abc hojas a ⟶ (hojas a = (Suc (size a)))"<br />
by (induct a) auto<br />
<br />
<br />
(* fraortmoy marpoldia1 migtermor anaprarod wilmorort pablucoto serrodcal *)<br />
<br />
lemma lej8: "es_abc hojas a = es_abc size a"<br />
by (induct a) (auto simp add:abc_hojas_num_nodos [symmetric])<br />
<br />
(* anaprarod crigomgom paupeddeg juacabsou rubgonmar *)<br />
(* Usando AUX8 *)<br />
lemma L8: "es_abc hojas a= es_abc size a"<br />
by (induct a) (auto simp add: AUX8)<br />
<br />
<br />
(* ivamenjim *)<br />
(* Teorema auxiliar *)<br />
lemma auxEj8: "hojas a = size a + 1"<br />
by (induct a) auto<br />
<br />
(* ivamenjim *)<br />
lemma lej8: "es_abc hojas a = es_abc size a"<br />
by (induct a) (auto simp add: auxEj8)<br />
<br />
(* danrodcha *)<br />
lemma aux3: "es_abc hojas a ⟹ (hojas a = 1 + size a)"<br />
by (induct a) simp_all<br />
<br />
(* danrodcha *)<br />
lemma 8: "es_abc hojas a = es_abc size a"<br />
apply (induct a) <br />
apply simp<br />
apply (auto simp add: aux3)<br />
done<br />
<br />
(* ferrenseg *)<br />
lemma [simp]: "hojas a = size a + 1"<br />
by (induct a) auto<br />
<br />
theorem es_abc_hojas_size: "es_abc hojas a = es_abc size a"<br />
by (induct a) auto<br />
<br />
(* fracorjim1 *)<br />
<br />
lemma aux10 : "hojas (a::arbol) = Suc(size a)"<br />
apply (induct a)<br />
apply auto<br />
done<br />
<br />
lemma es_abc_hojas_size_10 : "es_abc hojas (a::arbol) = es_abc size a"<br />
apply (induct a)<br />
apply (auto simp add: aux10)<br />
done<br />
<br />
text {* <br />
--------------------------------------------------------------------- <br />
Ejercicio 9. Demostrar que un árbol binario a es completo respecto de<br />
la profundidad syss es completo respecto del número de nodos.<br />
--------------------------------------------------------------------- <br />
*}<br />
<br />
(* fraortmoy marpoldia1 migtermor anaprarod wilmorort pablucoto <br />
serrodcal crigomgom rubgonmar danrodcha ivamenjim marcarmor13 paupeddeg juacabsou<br />
josgarsan<br />
*)<br />
<br />
lemma lej9: "es_abc profundidad a = es_abc size a"<br />
by (simp add: lej7 lej8)<br />
<br />
(* ferrenseg *)<br />
<br />
corollary es_abc_size_profundidad: "es_abc size a = es_abc profundidad a"<br />
by (simp add: es_abc_profundidad_hojas es_abc_hojas_size)<br />
<br />
text {* <br />
--------------------------------------------------------------------- <br />
Ejercicio 10. Demostrar que (abc n) es un árbol binario completo.<br />
--------------------------------------------------------------------- <br />
*}<br />
<br />
(* fraortmoy marpoldia1 migtermor wilmorort pablucoto serrodcal crigomgom marcarmor13 paupeddeg juacabsou*)<br />
<br />
lemma lej10: "es_abc profundidad (abc n)"<br />
by (induct n) auto<br />
<br />
<br />
(* anaprarod rubgonmar danrodcha ferrenseg ivamenjim paupeddeg juacabsou*)<br />
(* con un demostrador más débil *)<br />
(* y en general para cualquier medida *)<br />
lemma L10: "es_abc f (abc a)"<br />
by (induct a) simp_all<br />
<br />
(* ivamenjim *)<br />
(* Igual que el anterior pero usando auto *)<br />
lemma lej10: "es_abc f (abc n)"<br />
by (induct n) auto<br />
<br />
text {* <br />
--------------------------------------------------------------------- <br />
Ejercicio 11. Demostrar que si a es un árbolo binario completo<br />
respecto de la profundidad, entonces a es igual a<br />
(abc (profundidad a)).<br />
--------------------------------------------------------------------- <br />
*}<br />
<br />
(* fraortmoy marpoldia1 migtermor wilmorort pablucoto serrodcal marcarmor13 *)<br />
<br />
lemma lej11: <br />
assumes " es_abc profundidad a"<br />
shows "a = (abc (profundidad a))"<br />
using assms<br />
by (induct a) auto<br />
<br />
<br />
(* Otra forma de escribir lo mismo *)<br />
(* anaprarod crigomgom rubgonmar ferrenseg ivamenjim paupeddeg juacabsou *)<br />
lemma "es_abc profundidad a ⟶ (a = (abc (profundidad a)))"<br />
by (induct a) auto<br />
<br />
(* danrodcha*)<br />
lemma 11:"es_abc profundidad a ⟹ (a = (abc (profundidad a)))"<br />
by (induct a) simp_all<br />
<br />
<br />
text {* <br />
--------------------------------------------------------------------- <br />
Ejercicio 12. Encontrar una medida f tal que (es_abc f) es distinto de <br />
(es_abc size).<br />
--------------------------------------------------------------------- <br />
*}<br />
<br />
(* migtermor *)<br />
fun medida_nula :: "arbol => nat" where<br />
"medida_nula H = 0"<br />
| "medida_nula (N i d) = 0"<br />
<br />
lemma "es_abc medida_nula a = es_abc size a"<br />
quickcheck<br />
oops<br />
(* Quickcheck encuentra el siguiente contraejemplo:<br />
a= N H (N H H) <br />
Tras evaluar:<br />
es_abc medida_nula a = True<br />
es_abc size a = False*)<br />
<br />
<br />
(* anaprarod wilmorort pablucoto serrodcal danrodcha marcarmor13 ferrenseg ivamenjim paupeddeg juacabsou rubgonmar *)<br />
lemma "es_abc f a = es_abc size a"<br />
quickcheck<br />
(* Quickcheck found a counterexample:<br />
f = λx. a⇩1 <br />
a = N H (N H H)<br />
Evaluated terms:<br />
es_abc f a = True<br />
es_abc size a = False *)<br />
oops<br />
(* el contraejemplo que encuentra es la medida constante a1 *)<br />
<br />
(*crigomgom *)<br />
(* Como en la primera de las soluciones he usado la función constante 0 pero he usado una expresión lambda*)<br />
lemma "es_abc (λx. 0::nat) a = es_abc size a"<br />
quickcheck<br />
oops<br />
<br />
<br />
end<br />
</source></div>Lucnovdoshttps://www.glc.us.es/~jalonso/RA2016/index.php?title=Relaci%C3%B3n_6&diff=970Relación 62016-12-06T04:18:17Z<p>Lucnovdos: </p>
<hr />
<div><source lang="isar"><br />
chapter {* R6: Recorridos de árboles *}<br />
<br />
theory R6_Recorridos_de_arboles<br />
imports Main <br />
begin <br />
<br />
text {* <br />
--------------------------------------------------------------------- <br />
Ejercicio 1. Definir el tipo de datos arbol para representar los<br />
árboles binarios que tiene información en los nodos y en las hojas. <br />
Por ejemplo, el árbol<br />
e<br />
/ \<br />
/ \<br />
c g<br />
/ \ / \<br />
a d f h <br />
se representa por "N e (N c (H a) (H d)) (N g (H f) (H h))".<br />
--------------------------------------------------------------------- <br />
*}<br />
<br />
(* ivamenjim marpoldia1 manmorjim1 bowma migtermor wilmorort *)<br />
<br />
datatype 'a arbol = H "'a" | N "'a" "'a arbol" "'a arbol"<br />
<br />
value "N e (N c (H a) (H d)) (N g (H f) (H h))" <br />
<br />
text {* <br />
--------------------------------------------------------------------- <br />
Ejercicio 2. Definir la función <br />
preOrden :: "'a arbol ⇒ 'a list"<br />
tal que (preOrden a) es el recorrido pre orden del árbol a. Por<br />
ejemplo, <br />
preOrden (N e (N c (H a) (H d)) (N g (H f) (H h)))<br />
= [e,c,a,d,g,f,h] <br />
--------------------------------------------------------------------- <br />
*}<br />
<br />
(* ivamenjim marpoldia1 pablucoto bowma fraortmoy migtermor wilmorort lucnovdos *)<br />
<br />
fun preOrden :: "'a arbol ⇒ 'a list" where<br />
"preOrden (H t) = [t]"<br />
| "preOrden (N t i d) = [t] @ (preOrden i) @ (preOrden d)"<br />
<br />
(* danrodcha crigomgom manmorjim1 bowma *)<br />
fun preOrden1 :: "'a arbol ⇒ 'a list" where<br />
"preOrden1 (H x) = [x]"<br />
| "preOrden1 (N x i d) = x#preOrden1 i @ preOrden1 d"<br />
<br />
value "preOrden (N e (N c (H a) (H d)) (N g (H f) (H h))) <br />
= [e,c,a,d,g,f,h]" <br />
value "preOrden1 (N e (N c (H a) (H d)) (N g (H f) (H h))) <br />
= [e,c,a,d,g,f,h]" <br />
<br />
(* danrodcha *)<br />
lemma "preOrden a = preOrden1 a"<br />
by (induct a) simp_all<br />
<br />
text {* <br />
--------------------------------------------------------------------- <br />
Ejercicio 3. Definir la función <br />
postOrden :: "'a arbol ⇒ 'a list"<br />
tal que (postOrden a) es el recorrido post orden del árbol a. Por<br />
ejemplo, <br />
postOrden (N e (N c (H a) (H d)) (N g (H f) (H h)))<br />
= [e,c,a,d,g,f,h] <br />
--------------------------------------------------------------------- <br />
*}<br />
<br />
(* ivamenjim danrodcha crigomgom marpoldia1 manmorjim1 pablucoto bowma fraortmoy migtermor wilmorort lucnovdos *)<br />
<br />
fun postOrden :: "'a arbol ⇒ 'a list" where<br />
"postOrden (H t) = [t]"<br />
| "postOrden (N t i d) = (postOrden i) @ (postOrden d) @ [t]"<br />
<br />
value "postOrden (N e (N c (H a) (H d)) (N g (H f) (H h)))<br />
= [a,d,c,f,h,g,e]"<br />
<br />
text {* <br />
--------------------------------------------------------------------- <br />
Ejercicio 4. Definir la función <br />
inOrden :: "'a arbol ⇒ 'a list"<br />
tal que (inOrden a) es el recorrido in orden del árbol a. Por<br />
ejemplo, <br />
inOrden (N e (N c (H a) (H d)) (N g (H f) (H h)))<br />
= [a,c,d,e,f,g,h]<br />
--------------------------------------------------------------------- <br />
*}<br />
<br />
(* ivamenjim crigomgom marpoldia1 pablucoto bowma fraortmoy migtermor wilmorort lucnovdos*)<br />
<br />
fun inOrden :: "'a arbol ⇒ 'a list" where<br />
"inOrden (H t) = [t]"<br />
| "inOrden (N t i d) = (inOrden i) @ [t] @ (inOrden d)"<br />
<br />
<br />
(* danrodcha manmorjim1 *)<br />
fun inOrden1 :: "'a arbol ⇒ 'a list" where<br />
"inOrden1 (H t) = [t]"<br />
| "inOrden1 (N t i d) = inOrden1 i @ t#inOrden1 d"<br />
<br />
value "inOrden (N e (N c (H a) (H d)) (N g (H f) (H h)))<br />
= [a,c,d,e,f,g,h]"<br />
value "inOrden1 (N e (N c (H a) (H d)) (N g (H f) (H h)))<br />
= [a,c,d,e,f,g,h]"<br />
<br />
(* manmorjim1 *)<br />
lemma "inOrden t = inOrden1 t"<br />
apply (induct t)<br />
apply auto<br />
done<br />
<br />
text {* <br />
--------------------------------------------------------------------- <br />
Ejercicio 5. Definir la función <br />
espejo :: "'a arbol ⇒ 'a arbol"<br />
tal que (espejo a) es la imagen especular del árbol a. Por ejemplo, <br />
espejo (N e (N c (H a) (H d)) (N g (H f) (H h)))<br />
= N e (N g (H h) (H f)) (N c (H d) (H a))<br />
--------------------------------------------------------------------- <br />
*}<br />
<br />
(* ivamenjim danrodcha crigomgom marpoldia1 manmorjim1 pablucoto bowma fraortmoy migtermor wilmorort lucnovdos *)<br />
<br />
fun espejo :: "'a arbol ⇒ 'a arbol" where<br />
"espejo (H t) = H t"<br />
| "espejo (N t i d) = N t (espejo d) (espejo i)"<br />
<br />
value "espejo (N e (N c (H a) (H d)) (N g (H f) (H h))) <br />
= N e (N g (H h) (H f)) (N c (H d) (H a))"<br />
<br />
<br />
text {* <br />
--------------------------------------------------------------------- <br />
Ejercicio 6. Demostrar que<br />
preOrden (espejo a) = rev (postOrden a)<br />
--------------------------------------------------------------------- <br />
*}<br />
<br />
(* ivamenjim migtermor wilmorort *)<br />
<br />
lemma "preOrden (espejo a) = rev (postOrden a)" (is "?P a")<br />
proof (induct a)<br />
fix x <br />
show "?P (H x)" by simp <br />
next<br />
fix x <br />
fix i assume h1: "?P i"<br />
fix d assume h2: "?P d"<br />
have "preOrden (espejo (N x i d)) = preOrden (N x (espejo d) (espejo i))" by simp<br />
also have "... = [x] @ (preOrden (espejo d)) @ (preOrden (espejo i))" by simp<br />
also have "... = [x] @ rev (postOrden d) @ rev (postOrden i)" using h1 h2 by simp <br />
finally show "preOrden (espejo (N x i d)) = rev (postOrden (N x i d))" by simp<br />
qed<br />
<br />
(* danrodcha *)<br />
lemma "preOrden (espejo a) = rev (postOrden a)"<br />
by (induct a) simp_all<br />
<br />
(* danrodcha crigomgom*)<br />
lemma "preOrden (espejo a) = rev (postOrden a)" (is "?P a")<br />
proof (induct a)<br />
fix x<br />
show "?P (H x)" by simp<br />
next<br />
fix x i d <br />
assume HIi: "?P i"<br />
assume HId: "?P d"<br />
have "preOrden (espejo (N x i d)) = preOrden (N x (espejo d) (espejo i))"<br />
by (simp only: espejo.simps(2))<br />
also have "… = x#preOrden (espejo d) @ preOrden (espejo i)"<br />
by (simp only: preOrden.simps(2))<br />
also have"… = x#rev (postOrden d) @ rev (postOrden i)" <br />
using HIi HId by simp<br />
also have "… = rev (postOrden (N x i d))" by simp<br />
finally show "?P (N x i d)" by simp<br />
qed<br />
<br />
(* danrodcha fraortmoy *)<br />
lemma "preOrden (espejo a) = rev (postOrden a)" (is "?P a")<br />
proof (induct a)<br />
fix x<br />
show "?P (H x)" by simp<br />
next<br />
fix x i d <br />
assume HIi: "?P i"<br />
assume HId: "?P d"<br />
show "?P (N x i d)" using HIi HId by simp<br />
qed<br />
<br />
(* bowma *)<br />
lemma "preOrden (espejo a) = rev (postOrden a)"<br />
apply (induct a)<br />
apply simp_all<br />
done<br />
<br />
(* pablucoto marpoldia1*)<br />
<br />
lemma "preOrden (espejo a) = rev (postOrden a)" (is "?P a")<br />
proof (induct a)<br />
fix x <br />
show "?P (H x)" by simp <br />
next<br />
fix x i d <br />
assume h1: "?P i"<br />
assume h2: "?P d"<br />
have "preOrden (espejo (N x i d)) = preOrden (N x (espejo d) (espejo i))" by simp<br />
also have "... = [x] @ (preOrden (espejo d)) @ (preOrden (espejo i))" by simp<br />
also have "... = [x] @ rev (postOrden d) @ rev (postOrden i)" using h1 h2 by simp<br />
also have "... = [x] @ rev (postOrden i @ postOrden d)" by simp <br />
also have "... = rev ( postOrden i @ postOrden d @ [x] ) " by simp<br />
also have "... = rev (postOrden (N x i d)) " by simp<br />
finally show "?P (N x i d)" by simp <br />
qed<br />
<br />
(* bowma *)<br />
lemma "preOrden (espejo a) = rev (postOrden a)" (is "?p a")<br />
proof (induct a)<br />
fix t<br />
show "?p (H t)" by simp<br />
(* Aquí si le diga "preOrden (espejo (H t)) = rev (postOrden (H t))",isabelle dice: <br />
proof (prove)<br />
goal (1 subgoal):<br />
1. preOrden (espejo (H t)) = rev (postOrden (H t)) <br />
Introduced fixed type variable(s): 'b in "t__" <br />
No entiendo porqué *)<br />
next <br />
fix t i d<br />
assume H1: "?p i"<br />
assume H2: "?p d"<br />
have "preOrden (espejo (N t i d)) = preOrden (N t (espejo d) (espejo i))" by simp<br />
also have "... = [t] @ (preOrden (espejo d)) @ (preOrden (espejo i))" by simp<br />
also have "... = [t] @ rev (postOrden d) @ rev (postOrden i)" using H1 H2 by simp<br />
finally show "?p (N t i d)" by simp<br />
qed <br />
<br />
(* fraortmoy lucnovdos *)<br />
<br />
lemma "preOrden (espejo a) = rev (postOrden a)"<br />
by (induct a) auto<br />
<br />
text {* <br />
--------------------------------------------------------------------- <br />
Ejercicio 7. Demostrar que<br />
postOrden (espejo a) = rev (preOrden a)<br />
--------------------------------------------------------------------- <br />
*}<br />
<br />
(* ivamenjim crigomgom bowma migtermor wilmorort*)<br />
<br />
lemma "postOrden (espejo a) = rev (preOrden a)" (is "?P a")<br />
proof (induct a)<br />
fix x <br />
show "?P (H x)" by simp <br />
next<br />
fix x <br />
fix i assume h1: "?P i"<br />
fix d assume h2: "?P d"<br />
have "postOrden (espejo (N x i d)) = postOrden (N x (espejo d) (espejo i))" by simp<br />
also have "... = (postOrden (espejo d)) @ (postOrden (espejo i)) @ [x]" by simp<br />
also have "... = rev (preOrden d) @ rev (preOrden i) @ [x]" using h1 h2 by simp <br />
finally show "postOrden (espejo (N x i d)) = rev (preOrden (N x i d))" by simp<br />
(* "?p (N x i d)" más corto *)<br />
qed<br />
<br />
(* danrodcha fraortmoy *)<br />
lemma "postOrden (espejo a) = rev (preOrden a)" (is "?P a")<br />
proof (induct a)<br />
fix x<br />
show "?P (H x)" by simp<br />
next<br />
fix x i d <br />
assume HIi: "?P i"<br />
assume HId: "?P d"<br />
show "?P (N x i d)" using HIi HId by simp<br />
qed<br />
<br />
(* pablucoto marpoldia1*)<br />
<br />
lemma "postOrden (espejo a) = rev (preOrden a)" (is "?P a")<br />
proof (induct a)<br />
fix x<br />
show "?P (H x)" by simp <br />
next <br />
fix x i d<br />
assume H1: "?P i"<br />
assume H2: "?P d"<br />
have " postOrden (espejo (N x i d)) = postOrden ( N x (espejo d) (espejo i)) " by simp<br />
also have "... = postOrden (espejo d) @ postOrden (espejo i) @ [x] " by simp<br />
also have "... = rev (preOrden d) @ rev (preOrden i) @ [x] " using H1 H2 by simp<br />
also have "... = rev (preOrden d) @ rev (x # preOrden i)" by simp<br />
also have "... = rev (x # preOrden i @ preOrden d)" by simp<br />
also have "... = rev (preOrden (N x i d)) " by simp<br />
finally show "?P (N x i d)" by simp<br />
qed<br />
<br />
(* fraortmoy lucnovdos *)<br />
lemma "postOrden (espejo a) = rev (preOrden a)"<br />
by (induct a) auto<br />
<br />
text {* <br />
--------------------------------------------------------------------- <br />
Ejercicio 8. Demostrar que<br />
inOrden (espejo a) = rev (inOrden a)<br />
--------------------------------------------------------------------- <br />
*}<br />
<br />
(* ivamenjim crigomgom bowma migtermor wilmorort*)<br />
<br />
theorem "inOrden (espejo a) = rev (inOrden a)" (is "?P a")<br />
proof (induct a)<br />
fix x <br />
show "?P (H x)" by simp <br />
next<br />
fix x <br />
fix i assume h1: "?P i"<br />
fix d assume h2: "?P d"<br />
have "inOrden (espejo (N x i d)) = inOrden (N x (espejo d) (espejo i))" by simp<br />
also have "... = (inOrden (espejo d)) @ [x] @ (inOrden (espejo i))" by simp<br />
also have "... = rev (inOrden d) @ [x] @ rev (inOrden i)" using h1 h2 by simp <br />
finally show "inOrden (espejo (N x i d)) = rev (inOrden (N x i d))" by simp<br />
qed<br />
<br />
(* danrodcha *)<br />
theorem "inOrden (espejo a) = rev (inOrden a)"<br />
by (induct a) simp_all<br />
<br />
(* danrodcha *)<br />
theorem "inOrden (espejo a) = rev (inOrden a)" (is "?P a")<br />
proof (induct a)<br />
fix x<br />
show "?P (H x)" by simp<br />
next<br />
fix x i d <br />
assume HIi: "?P i"<br />
assume HId: "?P d"<br />
show "?P (N x i d)" using HIi HId by simp<br />
qed<br />
<br />
(* pablucoto marpoldia1 *)<br />
theorem "inOrden (espejo a) = rev (inOrden a)" (is "?P a")<br />
proof (induct a)<br />
fix x<br />
show "?P (H x) " by simp<br />
next <br />
fix x i d <br />
assume HI1: "?P i"<br />
assume HI2: "?P d"<br />
have " inOrden (espejo (N x i d)) = inOrden ( N x (espejo d) (espejo i) )" by simp<br />
also have "... = inOrden (espejo d) @ [x] @ inOrden (espejo i) " by simp<br />
also have "... = rev (inOrden d) @ [x] @ rev (inOrden i)" using HI1 HI2 by simp<br />
also have "... = rev (x # inOrden d ) @ rev (inOrden i)" by simp<br />
also have "... = rev ( inOrden i @ x # inOrden d) " by simp<br />
also have "... = rev (inOrden (N x i d))" by simp<br />
finally show "?P (N x i d)" by simp<br />
qed<br />
<br />
(* lucnovdos *)<br />
theorem "inOrden (espejo a) = rev (inOrden a)"<br />
by (induct a) auto<br />
<br />
text {* <br />
--------------------------------------------------------------------- <br />
Ejercicio 9. Definir la función <br />
raiz :: "'a arbol ⇒ 'a"<br />
tal que (raiz a) es la raiz del árbol a. Por ejemplo, <br />
raiz (N e (N c (H a) (H d)) (N g (H f) (H h))) = e<br />
--------------------------------------------------------------------- <br />
*}<br />
<br />
(* danrodcha crigomgom manmorjim1 ivamenjim bowma pablucoto migtermor marpoldia1 wilmorort lucnovdos *)<br />
fun raiz :: "'a arbol ⇒ 'a" where<br />
"raiz (H x) = x"<br />
| "raiz (N x i d) = x"<br />
<br />
value "raiz (N e (N c (H a) (H d)) (N g (H f) (H h))) = e"<br />
<br />
text {* <br />
--------------------------------------------------------------------- <br />
Ejercicio 10. Definir la función <br />
extremo_izquierda :: "'a arbol ⇒ 'a"<br />
tal que (extremo_izquierda a) es el nodo más a la izquierda del árbol<br />
a. Por ejemplo, <br />
extremo_izquierda (N e (N c (H a) (H d)) (N g (H f) (H h))) = a<br />
--------------------------------------------------------------------- <br />
*}<br />
<br />
(* danrodcha crigomgom manmorjim1 ivamenjim bowma pablucoto migtermor marpoldia1 wilmorort lucnovdos *)<br />
fun extremo_izquierda :: "'a arbol ⇒ 'a" where<br />
"extremo_izquierda (H x) = x"<br />
| "extremo_izquierda (N x i d) = extremo_izquierda i"<br />
<br />
value "extremo_izquierda (N e (N c (H a) (H d)) (N g (H f) (H h))) = a"<br />
<br />
(* bowma *)<br />
fun extremo_izquierda_1 :: "'a arbol ⇒ 'a" where<br />
"extremo_izquierda_1 (H t) = t"<br />
| "extremo_izquierda_1 (N t i d) = hd (inOrden (N t i d))"<br />
<br />
(* ivamenjim *)<br />
(* Metaejercicio de demostración. Llamando teorema_13 al teorema del ejercicio 13 *)<br />
<br />
lemma "extremo_izquierda a = extremo_izquierda_1 a"<br />
by (induct a, simp_all add: aux_ej12_1 teorema_13)<br />
<br />
text {* <br />
--------------------------------------------------------------------- <br />
Ejercicio 11. Definir la función <br />
extremo_derecha :: "'a arbol ⇒ 'a"<br />
tal que (extremo_derecha a) es el nodo más a la derecha del árbol<br />
a. Por ejemplo, <br />
extremo_derecha (N e (N c (H a) (H d)) (N g (H f) (H h))) = h<br />
--------------------------------------------------------------------- <br />
*}<br />
<br />
(* danrodcha crigomgom manmorjim1 ivamenjim bowma pablucoto migtermor marpoldia1 wilmorort lucnovdos *)<br />
fun extremo_derecha :: "'a arbol ⇒ 'a" where<br />
"extremo_derecha (H x) = x"<br />
| "extremo_derecha (N x i d) = extremo_derecha d"<br />
<br />
value "extremo_derecha (N e (N c (H a) (H d)) (N g (H f) (H h))) = h"<br />
<br />
(* bowma *)<br />
fun extremo_derecha_1 :: "'a arbol ⇒ 'a" where<br />
"extremo_derecha_1 (H t) = t"<br />
| "extremo_derecha_1 (N t i d) = last (inOrden (N t i d))"<br />
<br />
(* ivamenjim *)<br />
(* Metaejercicio de demostración. Llamando teorema_12 al teorema del ejercicio 12 *)<br />
<br />
lemma "extremo_derecha a = extremo_derecha_1 a"<br />
by (induct a, simp_all add: aux_ej12_1 teorema_12)<br />
<br />
text {* <br />
--------------------------------------------------------------------- <br />
Ejercicio 12. Demostrar o refutar<br />
last (inOrden a) = extremo_derecha a<br />
--------------------------------------------------------------------- <br />
*}<br />
<br />
(* danrodcha *)<br />
lemma aux_ej12: "inOrden a ≠ []"<br />
apply (induct a) <br />
apply simp<br />
apply simp<br />
done<br />
<br />
(* danrodcha pablucoto crigomgom wilmorort*)<br />
theorem "last (inOrden a) = extremo_derecha a" (is "?P a")<br />
proof (induct a)<br />
fix x<br />
show "?P (H x)" by simp<br />
next<br />
fix x i d <br />
assume HIi: "?P i"<br />
assume HId: "?P d"<br />
have "last (inOrden (N x i d)) = last (inOrden i @ [x] @ inOrden d)" <br />
by (simp only: inOrden.simps(2))<br />
also have "… = last (inOrden d)" by (simp add: aux_ej12)<br />
also have "… = extremo_derecha d" using HId by simp<br />
also have "… = extremo_derecha (N x i d)" by simp<br />
finally show "?P (N x i d)" by simp<br />
qed<br />
<br />
(* ivamenjim *)<br />
<br />
lemma aux_ej12_1: "inOrden a ≠ []"<br />
by (induct a) simp_all <br />
<br />
(* ivamenjim marpoldia1 *)<br />
(* Igual que la anterior, pero poniendo solo by simp en el primer have *)<br />
<br />
theorem "last (inOrden a) = extremo_derecha a" (is "?P a")<br />
proof (induct a)<br />
fix x <br />
show "?P (H x)" by simp <br />
next<br />
fix x <br />
fix i assume h1: "?P i"<br />
fix d assume h2: "?P d"<br />
have "last (inOrden (N x i d)) = last ((inOrden i) @ [x] @ (inOrden d))" by simp <br />
also have "... = last (inOrden d)" by (simp add: aux_ej12_1)<br />
also have "... = extremo_derecha d" using h2 by simp <br />
finally show "last (inOrden (N x i d)) = extremo_derecha (N x i d)" by simp<br />
qed<br />
<br />
(* bowma *)<br />
(* Casi lo mismo que el anterior,pero no hace falta suponer "?p i" *)<br />
theorem "last (inOrden a) = extremo_derecha a" (is "?p a")<br />
proof (induct a)<br />
fix t<br />
show "?p (H t)" by simp<br />
next <br />
fix t i d<br />
assume HI: "?p d"<br />
have "last (inOrden (N t i d)) = last (inOrden i @ [t] @ inOrden d)" by simp<br />
also have "... = last (inOrden d)" by (simp add:aux_ej12)<br />
also have "... = extremo_derecha d" using HI by simp<br />
finally show "?p (N t i d)" by simp<br />
qed<br />
<br />
(* lucnovdos*)<br />
(* El mismo que el anterior,pero sin usar padrones *)<br />
theorem "last (inOrden a) = extremo_derecha a"<br />
proof (induct a)<br />
fix x ::"'a"<br />
show "last (inOrden (H x)) = extremo_derecha (H x)" by simp<br />
next<br />
fix x ::"'a"<br />
fix i d ::"'a arbol"<br />
assume HI: "last (inOrden d) = extremo_derecha d"<br />
have "last (inOrden (N x i d)) = last ((inOrden i) @ [x] @ (inOrden d))" by simp <br />
also have "… = last (inOrden d)" by (simp add: aux_ej12)<br />
also have "… = extremo_derecha d" using HI by simp<br />
also have "… = extremo_derecha (N x i d)" by simp<br />
finally show "last (inOrden (N x i d)) = extremo_derecha (N x i d)" by simp<br />
qed<br />
<br />
(* migtermor *)<br />
theorem "last (inOrden a) = extremo_derecha a" (is "?P a")<br />
proof (induct a)<br />
fix h<br />
show "?P (H h)" by simp<br />
next<br />
fix n i<br />
fix d assume HId: "?P d"<br />
have AUX: "¬ (inOrden d = [])" (is "?Q d")<br />
proof (induct d)<br />
fix hd<br />
show "?Q (H hd)" by simp<br />
next<br />
fix nd<br />
fix id assume HIid: "?Q id"<br />
fix dd assume HIdd: "?Q dd"<br />
show "?Q (N nd id dd)" using HIid HIdd by simp<br />
qed<br />
have "last (inOrden (N n i d)) = last (inOrden i @[n]@inOrden d)" by simp<br />
also have "… = last (inOrden d)" using AUX by simp<br />
also have "… = extremo_derecha d" using HId by simp<br />
finally show "?P (N n i d)" by simp<br />
qed<br />
<br />
text {* <br />
--------------------------------------------------------------------- <br />
Ejercicio 13. Demostrar o refutar<br />
hd (inOrden a) = extremo_izquierda a<br />
--------------------------------------------------------------------- <br />
*}<br />
<br />
(* danrodcha pablucoto crigomgom *)<br />
theorem "hd (inOrden a) = extremo_izquierda a" (is "?P a")<br />
proof (induct a)<br />
fix x<br />
show "?P (H x)" by simp<br />
next<br />
fix x i d <br />
assume HIi: "?P i"<br />
assume HId: "?P d"<br />
have "hd (inOrden (N x i d)) = hd (inOrden i @ [x] @ inOrden d)" <br />
by (simp only: inOrden.simps(2))<br />
also have "… = hd (inOrden i)" by (simp add: aux_ej12)<br />
also have "… = extremo_izquierda i" using HIi by simp<br />
also have "… = extremo_izquierda (N x i d)" by simp<br />
finally show "?P (N x i d)" by simp<br />
qed<br />
<br />
(* bowma lucnovdos *)<br />
theorem "hd (inOrden a) = extremo_izquierda a" (is "?p a")<br />
proof (induct a)<br />
fix t<br />
show "?p (H t)" by simp<br />
next<br />
fix t i d <br />
assume HI: "?p i"<br />
have "hd (inOrden (N t i d)) = hd (inOrden i @ [t] @ inOrden d)" by simp<br />
also have "… = hd (inOrden i)" by (simp add: aux_ej12)<br />
also have "… = extremo_izquierda i" using HI by simp<br />
finally show "?p (N t i d)" by simp<br />
qed<br />
<br />
(* migtermor *)<br />
theorem "hd (inOrden a) = extremo_izquierda a" (is "?P a")<br />
proof (induct a)<br />
fix h<br />
show "?P (H h)" by simp<br />
next<br />
fix n d<br />
fix i assume HId: "?P i"<br />
have AUX: "¬ (inOrden i = [])" (is "?Q i")<br />
proof (induct i)<br />
fix hi<br />
show "?Q (H hi)" by simp<br />
next<br />
fix ni<br />
fix ii assume HIid: "?Q ii"<br />
fix di assume HIdd: "?Q di"<br />
show "?Q (N ni ii di)" using HIid HIdd by simp<br />
qed<br />
have "hd (inOrden (N n i d)) = hd (inOrden i @[n]@inOrden d)" by simp<br />
also have "… = hd (inOrden i)" using AUX by simp<br />
also have "… = extremo_izquierda i" using HId by simp<br />
finally show "?P (N n i d)" by simp<br />
qed<br />
<br />
(* ivamenjim marpoldia1 wilmorort *)<br />
<br />
theorem "hd (inOrden a) = extremo_izquierda a" (is "?P a")<br />
proof (induct a)<br />
fix x <br />
show "?P (H x)" by simp <br />
next<br />
fix x <br />
fix i assume h1: "?P i"<br />
fix d assume h2: "?P d"<br />
have "hd (inOrden (N x i d)) = hd ((inOrden i) @ [x] @ (inOrden d))" by simp<br />
also have "... = hd (inOrden i)" by (simp add: aux_ej12_1)<br />
also have "... = extremo_izquierda i" using h1 by simp <br />
finally show "hd (inOrden (N x i d)) = extremo_izquierda (N x i d)" by simp<br />
qed<br />
<br />
text {* <br />
--------------------------------------------------------------------- <br />
Ejercicio 14. Demostrar o refutar<br />
hd (preOrden a) = last (postOrden a)<br />
--------------------------------------------------------------------- <br />
*}<br />
<br />
(* danrodcha *)<br />
theorem "hd (preOrden a) = last (postOrden a)" (is "?P a")<br />
proof (induct a)<br />
fix x<br />
show "?P (H x)" by simp<br />
fix x i d<br />
show "?P (N x i d)" by simp<br />
qed<br />
<br />
(* danrodcha *)<br />
theorem "hd (preOrden a) = last (postOrden a)" (is "?P a")<br />
proof (induct a)<br />
fix x<br />
show "?P (H x)" by simp<br />
fix x i d<br />
assume HIi: "?P i"<br />
assume HId: "?P d"<br />
have "hd (preOrden (N x i d)) = hd (x#preOrden i @ preOrden d)"<br />
by (simp only: preOrden.simps(2))<br />
also have "… = x" by simp<br />
also have "… = last (postOrden i @ postOrden d @ [x])" by simp<br />
also have "… = last (postOrden (N x i d))" <br />
by (simp only: postOrden.simps(2))<br />
finally show "?P (N x i d)" by simp<br />
qed<br />
<br />
(* pablucoto crigomgom bowma marpoldia1 wilmorort lucnovdos *) (*Similar al anterior*)<br />
theorem "hd (preOrden a) = last (postOrden a)" (is "?P a")<br />
proof (induct a)<br />
fix x<br />
show "?P (H x)" by simp<br />
next <br />
fix x i d<br />
assume HI1: "?P i"<br />
assume HI2: "?P d"<br />
have " hd (preOrden (N x i d)) = hd ([x] @ preOrden i @ preOrden d)" by simp<br />
also have "... = x" by simp<br />
also have "... = last ( postOrden i @ postOrden d @ [x]) " by simp<br />
also have "... = last ( postOrden (N x i d) )" by simp <br />
finally show "?P (N x i d)" by simp<br />
qed<br />
<br />
(* migtermor *)<br />
theorem "hd (preOrden a) = last (postOrden a)" (is "?P a")<br />
proof (induct a)<br />
fix h<br />
show "?P (H h)" by simp<br />
next<br />
fix n i d<br />
have "hd (preOrden (N n (i :: 'a arbol) (d :: 'a arbol))) = hd ([n]@preOrden i@preOrden d)" <br />
by simp<br />
(* Si no especifico que i y d son árboles, salta un error de tipo. Supongo que será por<br />
no haber asumido hipótesis sobre ellos *)<br />
also have "… = last (postOrden (N n i d))" by simp<br />
show "?P (N n i d)" by simp<br />
qed<br />
<br />
(* ivamenjim *)<br />
<br />
theorem "hd (preOrden a) = last (postOrden a)" (is "?P a")<br />
proof (induct a)<br />
fix x <br />
show "?P (H x)" by simp <br />
next<br />
fix x <br />
fix i assume h1: "?P i"<br />
fix d assume h2: "?P d"<br />
have "hd (preOrden (N x i d)) = hd ([x] @ (preOrden i) @ (preOrden d))" by simp<br />
also have "... = hd ([x])" by simp<br />
finally show "hd (preOrden (N x i d)) = last (postOrden (N x i d))" by simp<br />
qed<br />
<br />
text {* <br />
--------------------------------------------------------------------- <br />
Ejercicio 15. Demostrar o refutar<br />
hd (preOrden a) = raiz a<br />
--------------------------------------------------------------------- <br />
*}<br />
<br />
(* danrodcha *)<br />
theorem "hd (preOrden a) = raiz a" (is "?P a")<br />
proof (induct a)<br />
fix x<br />
show "?P (H x)" by simp<br />
fix x i d<br />
assume HIi: "?P i"<br />
assume HId: "?P d"<br />
have "hd (preOrden (N x i d)) = hd (x#preOrden i @ preOrden d)"<br />
by (simp only: preOrden.simps(2))<br />
also have "… = x" by simp<br />
also have "… = raiz (N x i d)" by simp<br />
finally show "?P (N x i d)" by simp<br />
qed<br />
<br />
(* danrodcha *)<br />
theorem "hd (preOrden a) = raiz a" (is "?P a")<br />
proof (induct a)<br />
fix x<br />
show "?P (H x)" by simp<br />
fix x i d<br />
show "?P (N x i d)" by simp<br />
qed<br />
<br />
(* pablucoto crigomgom ivamenjim marpoldia1 wilmorort lucnovdos *)<br />
theorem "hd (preOrden a) = raiz a" (is "?P a")<br />
proof (induct a) <br />
fix x<br />
show "?P (H x)" by simp<br />
next<br />
fix x i d<br />
assume HI1: " ?P i"<br />
assume HI2: " ?P d"<br />
have " hd (preOrden (N x i d)) = hd ([x] @ preOrden i @ preOrden d) " by simp<br />
also have "... = x" by simp<br />
also have "... = raiz (N x i d)" by simp<br />
finally show " ?P (N x i d)" by simp<br />
qed<br />
<br />
(* bowma *)<br />
(* similar al anterior pero sin suponer "?p d" *)<br />
theorem "hd (preOrden a) = last (postOrden a)" (is "?p a")<br />
proof (induct a)<br />
fix t<br />
show "?p (H t)" by simp<br />
next<br />
fix t i d<br />
assume HI: "?p i"<br />
have "hd (preOrden (N t i d)) = hd ([t] @ preOrden i @ preOrden d)" by simp<br />
also have "… = t" by simp<br />
also have "… = last (postOrden i @ postOrden d @ [t])" by simp<br />
also have "… = last (postOrden (N t i d))" by simp<br />
finally show "?p (N t i d)" by simp<br />
qed<br />
<br />
(* migtermor *)<br />
theorem "hd (preOrden a) = raiz a" (is "?P a")<br />
proof (induct a)<br />
fix h<br />
show "?P (H h)" by simp<br />
next<br />
fix n i d<br />
have "hd (preOrden (N n (i :: 'a arbol) (d :: 'a arbol))) = hd ([n]@preOrden i@preOrden d)"<br />
by simp<br />
also have "… = raiz (N n i d)" by simp<br />
finally show "?P (N n i d)" by simp<br />
qed<br />
<br />
(* ivamenjim: sin usar patrones *)<br />
<br />
theorem "hd (preOrden a) = raiz a" <br />
proof (induct a)<br />
fix x ::"'a"<br />
show "hd (preOrden (H x)) = raiz (H x)" by simp <br />
next<br />
fix x ::"'a"<br />
fix i ::"'a arbol" assume h1: "hd (preOrden i) = raiz i"<br />
fix d ::"'a arbol" assume h2: "hd (preOrden d) = raiz d"<br />
have "hd (preOrden (N x i d)) = hd ([x] @ (preOrden i) @ (preOrden d))" by simp<br />
also have "... = hd ([x])" by simp<br />
finally show "hd (preOrden (N x i d)) = raiz (N x i d)" by simp<br />
qed<br />
<br />
text {* <br />
--------------------------------------------------------------------- <br />
Ejercicio 16. Demostrar o refutar<br />
hd (inOrden a) = raiz a<br />
--------------------------------------------------------------------- <br />
*}<br />
<br />
(*crigomgom pablucoto bowma migtermor ivamenjim wilmorort lucnovdos *)<br />
theorem "hd (inOrden a) = raiz a"<br />
quickcheck<br />
oops<br />
<br />
(* danrodcha:<br />
Auto Quickcheck found a counterexample:<br />
a = N a⇩1 (H a⇩2) (H a⇩1)<br />
Evaluated terms:<br />
hd (inOrden a) = a⇩2<br />
raiz a = a⇩1 *)<br />
<br />
(* ivamenjim marpoldia1 *)<br />
<br />
theorem "hd (inOrden a) = raiz a" (is "?P a")<br />
proof (induct a)<br />
fix x <br />
show "?P (H x)" by simp <br />
next<br />
fix x <br />
fix i assume h1: "?P i"<br />
fix d assume h2: "?P d"<br />
have "hd (inOrden (N x i d)) = hd ((inOrden i) @ [x] @ (inOrden d))" by simp<br />
also have "... = hd (inOrden i)" by (simp add: aux_ej12_1) <br />
(* Perdemos la x, luego se refuta el enunciado del teorema *)<br />
qed<br />
<br />
text {* <br />
--------------------------------------------------------------------- <br />
Ejercicio 17. Demostrar o refutar<br />
last (postOrden a) = raiz a<br />
--------------------------------------------------------------------- <br />
*}<br />
<br />
(* danrodcha *)<br />
theorem "last (postOrden a) = raiz a" (is "?P a")<br />
proof (induct a)<br />
fix x<br />
show "?P (H x)" by simp<br />
fix x i d<br />
show "?P (N x i d)" by simp<br />
qed<br />
<br />
(* danrodcha *)<br />
theorem "last (postOrden a) = raiz a" (is "?P a")<br />
proof (induct a)<br />
fix x<br />
show "?P (H x)" by simp<br />
fix x i d<br />
assume HIi: "?P i"<br />
assume HId: "?P d"<br />
have "last (postOrden (N x i d)) = last (postOrden i @ postOrden d @ [x])"<br />
by (simp only: postOrden.simps(2))<br />
also have "… = x" by simp<br />
also have "… = raiz (N x i d)" by (simp only: raiz.simps(2))<br />
finally show "?P (N x i d)" by simp<br />
qed<br />
<br />
(* pablucoto crigomgom ivamenjim marpoldia1 wilmorort lucnovdos *) (*Similar al anterior*)<br />
<br />
theorem "last (postOrden a) = raiz a" (is "?P a")<br />
proof (induct a )<br />
fix x<br />
show "?P (H x)" by simp<br />
next <br />
fix x i d<br />
assume HI1: "?P i"<br />
assume HI2: "?P d"<br />
have "last (postOrden (N x i d)) = last ( postOrden i @ postOrden d @ [x])" by simp<br />
also have "... = x" by simp<br />
also have "... = raiz (N x i d) " by simp<br />
finally show " ?P (N x i d)" by simp<br />
qed<br />
<br />
(* bowma *)<br />
(* También sin usar el supuesto "?p d" *)<br />
theorem "last (postOrden a) = raiz a" (is "?p a")<br />
proof (induct a)<br />
fix t<br />
show "?p (H t)" by simp<br />
next<br />
fix t i d<br />
assume "?p i"<br />
(* si quito este supuesto, hay error pero no sé dónde se lo está usando *)<br />
have "last (postOrden (N t i d)) = last (postOrden i @ postOrden d @ [t])" by simp<br />
also have "... = t" by simp<br />
also have "... = raiz (N t i d)" by simp<br />
finally show "?p (N t i d)" by simp<br />
qed<br />
<br />
(* migtermor *)<br />
theorem "last (postOrden a) = raiz a" (is "?P a")<br />
proof (induct a)<br />
fix h<br />
show "?P (H h)" by simp<br />
next<br />
fix n i d<br />
have "last (postOrden (N n (i :: 'a arbol) (d :: 'a arbol))) = <br />
last (postOrden i@postOrden d@[n])" by simp<br />
also have "… = raiz (N n i d)" by simp<br />
finally show "?P (N n i d)" by simp<br />
qed<br />
<br />
(* ivamenjim: sin usar patrones *)<br />
<br />
theorem "last (postOrden a) = raiz a" <br />
proof (induct a)<br />
fix x::"'a" <br />
show "last (postOrden (H x)) = raiz (H x)" by simp <br />
next<br />
fix x::"'a" <br />
fix i::"'a arbol" assume h1: "last (postOrden i) = raiz i"<br />
fix d::"'a arbol" assume h2: "last (postOrden d) = raiz d"<br />
have "last (postOrden (N x i d)) = last ((postOrden i) @ (postOrden d) @ [x])" by simp<br />
also have "... = last ([x])" by simp <br />
finally show "last (postOrden (N x i d)) = raiz (N x i d)" by simp<br />
qed<br />
<br />
end<br />
</source></div>Lucnovdoshttps://www.glc.us.es/~jalonso/RA2016/index.php?title=Relaci%C3%B3n_6&diff=969Relación 62016-12-06T03:24:05Z<p>Lucnovdos: </p>
<hr />
<div><source lang="isar"><br />
chapter {* R6: Recorridos de árboles *}<br />
<br />
theory R6_Recorridos_de_arboles<br />
imports Main <br />
begin <br />
<br />
text {* <br />
--------------------------------------------------------------------- <br />
Ejercicio 1. Definir el tipo de datos arbol para representar los<br />
árboles binarios que tiene información en los nodos y en las hojas. <br />
Por ejemplo, el árbol<br />
e<br />
/ \<br />
/ \<br />
c g<br />
/ \ / \<br />
a d f h <br />
se representa por "N e (N c (H a) (H d)) (N g (H f) (H h))".<br />
--------------------------------------------------------------------- <br />
*}<br />
<br />
(* ivamenjim marpoldia1 manmorjim1 bowma migtermor wilmorort *)<br />
<br />
datatype 'a arbol = H "'a" | N "'a" "'a arbol" "'a arbol"<br />
<br />
value "N e (N c (H a) (H d)) (N g (H f) (H h))" <br />
<br />
text {* <br />
--------------------------------------------------------------------- <br />
Ejercicio 2. Definir la función <br />
preOrden :: "'a arbol ⇒ 'a list"<br />
tal que (preOrden a) es el recorrido pre orden del árbol a. Por<br />
ejemplo, <br />
preOrden (N e (N c (H a) (H d)) (N g (H f) (H h)))<br />
= [e,c,a,d,g,f,h] <br />
--------------------------------------------------------------------- <br />
*}<br />
<br />
(* ivamenjim marpoldia1 pablucoto bowma fraortmoy migtermor wilmorort lucnovdos *)<br />
<br />
fun preOrden :: "'a arbol ⇒ 'a list" where<br />
"preOrden (H t) = [t]"<br />
| "preOrden (N t i d) = [t] @ (preOrden i) @ (preOrden d)"<br />
<br />
(* danrodcha crigomgom manmorjim1 bowma *)<br />
fun preOrden1 :: "'a arbol ⇒ 'a list" where<br />
"preOrden1 (H x) = [x]"<br />
| "preOrden1 (N x i d) = x#preOrden1 i @ preOrden1 d"<br />
<br />
value "preOrden (N e (N c (H a) (H d)) (N g (H f) (H h))) <br />
= [e,c,a,d,g,f,h]" <br />
value "preOrden1 (N e (N c (H a) (H d)) (N g (H f) (H h))) <br />
= [e,c,a,d,g,f,h]" <br />
<br />
(* danrodcha *)<br />
lemma "preOrden a = preOrden1 a"<br />
by (induct a) simp_all<br />
<br />
text {* <br />
--------------------------------------------------------------------- <br />
Ejercicio 3. Definir la función <br />
postOrden :: "'a arbol ⇒ 'a list"<br />
tal que (postOrden a) es el recorrido post orden del árbol a. Por<br />
ejemplo, <br />
postOrden (N e (N c (H a) (H d)) (N g (H f) (H h)))<br />
= [e,c,a,d,g,f,h] <br />
--------------------------------------------------------------------- <br />
*}<br />
<br />
(* ivamenjim danrodcha crigomgom marpoldia1 manmorjim1 pablucoto bowma fraortmoy migtermor wilmorort lucnovdos *)<br />
<br />
fun postOrden :: "'a arbol ⇒ 'a list" where<br />
"postOrden (H t) = [t]"<br />
| "postOrden (N t i d) = (postOrden i) @ (postOrden d) @ [t]"<br />
<br />
value "postOrden (N e (N c (H a) (H d)) (N g (H f) (H h)))<br />
= [a,d,c,f,h,g,e]"<br />
<br />
text {* <br />
--------------------------------------------------------------------- <br />
Ejercicio 4. Definir la función <br />
inOrden :: "'a arbol ⇒ 'a list"<br />
tal que (inOrden a) es el recorrido in orden del árbol a. Por<br />
ejemplo, <br />
inOrden (N e (N c (H a) (H d)) (N g (H f) (H h)))<br />
= [a,c,d,e,f,g,h]<br />
--------------------------------------------------------------------- <br />
*}<br />
<br />
(* ivamenjim crigomgom marpoldia1 pablucoto bowma fraortmoy migtermor wilmorort lucnovdos*)<br />
<br />
fun inOrden :: "'a arbol ⇒ 'a list" where<br />
"inOrden (H t) = [t]"<br />
| "inOrden (N t i d) = (inOrden i) @ [t] @ (inOrden d)"<br />
<br />
<br />
(* danrodcha manmorjim1 *)<br />
fun inOrden1 :: "'a arbol ⇒ 'a list" where<br />
"inOrden1 (H t) = [t]"<br />
| "inOrden1 (N t i d) = inOrden1 i @ t#inOrden1 d"<br />
<br />
value "inOrden (N e (N c (H a) (H d)) (N g (H f) (H h)))<br />
= [a,c,d,e,f,g,h]"<br />
value "inOrden1 (N e (N c (H a) (H d)) (N g (H f) (H h)))<br />
= [a,c,d,e,f,g,h]"<br />
<br />
(* manmorjim1 *)<br />
lemma "inOrden t = inOrden1 t"<br />
apply (induct t)<br />
apply auto<br />
done<br />
<br />
text {* <br />
--------------------------------------------------------------------- <br />
Ejercicio 5. Definir la función <br />
espejo :: "'a arbol ⇒ 'a arbol"<br />
tal que (espejo a) es la imagen especular del árbol a. Por ejemplo, <br />
espejo (N e (N c (H a) (H d)) (N g (H f) (H h)))<br />
= N e (N g (H h) (H f)) (N c (H d) (H a))<br />
--------------------------------------------------------------------- <br />
*}<br />
<br />
(* ivamenjim danrodcha crigomgom marpoldia1 manmorjim1 pablucoto bowma fraortmoy migtermor wilmorort lucnovdos *)<br />
<br />
fun espejo :: "'a arbol ⇒ 'a arbol" where<br />
"espejo (H t) = H t"<br />
| "espejo (N t i d) = N t (espejo d) (espejo i)"<br />
<br />
value "espejo (N e (N c (H a) (H d)) (N g (H f) (H h))) <br />
= N e (N g (H h) (H f)) (N c (H d) (H a))"<br />
<br />
<br />
text {* <br />
--------------------------------------------------------------------- <br />
Ejercicio 6. Demostrar que<br />
preOrden (espejo a) = rev (postOrden a)<br />
--------------------------------------------------------------------- <br />
*}<br />
<br />
(* ivamenjim migtermor wilmorort *)<br />
<br />
lemma "preOrden (espejo a) = rev (postOrden a)" (is "?P a")<br />
proof (induct a)<br />
fix x <br />
show "?P (H x)" by simp <br />
next<br />
fix x <br />
fix i assume h1: "?P i"<br />
fix d assume h2: "?P d"<br />
have "preOrden (espejo (N x i d)) = preOrden (N x (espejo d) (espejo i))" by simp<br />
also have "... = [x] @ (preOrden (espejo d)) @ (preOrden (espejo i))" by simp<br />
also have "... = [x] @ rev (postOrden d) @ rev (postOrden i)" using h1 h2 by simp <br />
finally show "preOrden (espejo (N x i d)) = rev (postOrden (N x i d))" by simp<br />
qed<br />
<br />
(* danrodcha *)<br />
lemma "preOrden (espejo a) = rev (postOrden a)"<br />
by (induct a) simp_all<br />
<br />
(* danrodcha crigomgom*)<br />
lemma "preOrden (espejo a) = rev (postOrden a)" (is "?P a")<br />
proof (induct a)<br />
fix x<br />
show "?P (H x)" by simp<br />
next<br />
fix x i d <br />
assume HIi: "?P i"<br />
assume HId: "?P d"<br />
have "preOrden (espejo (N x i d)) = preOrden (N x (espejo d) (espejo i))"<br />
by (simp only: espejo.simps(2))<br />
also have "… = x#preOrden (espejo d) @ preOrden (espejo i)"<br />
by (simp only: preOrden.simps(2))<br />
also have"… = x#rev (postOrden d) @ rev (postOrden i)" <br />
using HIi HId by simp<br />
also have "… = rev (postOrden (N x i d))" by simp<br />
finally show "?P (N x i d)" by simp<br />
qed<br />
<br />
(* danrodcha fraortmoy *)<br />
lemma "preOrden (espejo a) = rev (postOrden a)" (is "?P a")<br />
proof (induct a)<br />
fix x<br />
show "?P (H x)" by simp<br />
next<br />
fix x i d <br />
assume HIi: "?P i"<br />
assume HId: "?P d"<br />
show "?P (N x i d)" using HIi HId by simp<br />
qed<br />
<br />
(* bowma *)<br />
lemma "preOrden (espejo a) = rev (postOrden a)"<br />
apply (induct a)<br />
apply simp_all<br />
done<br />
<br />
(* pablucoto marpoldia1*)<br />
<br />
lemma "preOrden (espejo a) = rev (postOrden a)" (is "?P a")<br />
proof (induct a)<br />
fix x <br />
show "?P (H x)" by simp <br />
next<br />
fix x i d <br />
assume h1: "?P i"<br />
assume h2: "?P d"<br />
have "preOrden (espejo (N x i d)) = preOrden (N x (espejo d) (espejo i))" by simp<br />
also have "... = [x] @ (preOrden (espejo d)) @ (preOrden (espejo i))" by simp<br />
also have "... = [x] @ rev (postOrden d) @ rev (postOrden i)" using h1 h2 by simp<br />
also have "... = [x] @ rev (postOrden i @ postOrden d)" by simp <br />
also have "... = rev ( postOrden i @ postOrden d @ [x] ) " by simp<br />
also have "... = rev (postOrden (N x i d)) " by simp<br />
finally show "?P (N x i d)" by simp <br />
qed<br />
<br />
(* bowma *)<br />
lemma "preOrden (espejo a) = rev (postOrden a)" (is "?p a")<br />
proof (induct a)<br />
fix t<br />
show "?p (H t)" by simp<br />
(* Aquí si le diga "preOrden (espejo (H t)) = rev (postOrden (H t))",isabelle dice: <br />
proof (prove)<br />
goal (1 subgoal):<br />
1. preOrden (espejo (H t)) = rev (postOrden (H t)) <br />
Introduced fixed type variable(s): 'b in "t__" <br />
No entiendo porqué *)<br />
next <br />
fix t i d<br />
assume H1: "?p i"<br />
assume H2: "?p d"<br />
have "preOrden (espejo (N t i d)) = preOrden (N t (espejo d) (espejo i))" by simp<br />
also have "... = [t] @ (preOrden (espejo d)) @ (preOrden (espejo i))" by simp<br />
also have "... = [t] @ rev (postOrden d) @ rev (postOrden i)" using H1 H2 by simp<br />
finally show "?p (N t i d)" by simp<br />
qed <br />
<br />
(* fraortmoy lucnovdos *)<br />
<br />
lemma "preOrden (espejo a) = rev (postOrden a)"<br />
by (induct a) auto<br />
<br />
text {* <br />
--------------------------------------------------------------------- <br />
Ejercicio 7. Demostrar que<br />
postOrden (espejo a) = rev (preOrden a)<br />
--------------------------------------------------------------------- <br />
*}<br />
<br />
(* ivamenjim crigomgom bowma migtermor wilmorort*)<br />
<br />
lemma "postOrden (espejo a) = rev (preOrden a)" (is "?P a")<br />
proof (induct a)<br />
fix x <br />
show "?P (H x)" by simp <br />
next<br />
fix x <br />
fix i assume h1: "?P i"<br />
fix d assume h2: "?P d"<br />
have "postOrden (espejo (N x i d)) = postOrden (N x (espejo d) (espejo i))" by simp<br />
also have "... = (postOrden (espejo d)) @ (postOrden (espejo i)) @ [x]" by simp<br />
also have "... = rev (preOrden d) @ rev (preOrden i) @ [x]" using h1 h2 by simp <br />
finally show "postOrden (espejo (N x i d)) = rev (preOrden (N x i d))" by simp<br />
(* "?p (N x i d)" más corto *)<br />
qed<br />
<br />
(* danrodcha fraortmoy *)<br />
lemma "postOrden (espejo a) = rev (preOrden a)" (is "?P a")<br />
proof (induct a)<br />
fix x<br />
show "?P (H x)" by simp<br />
next<br />
fix x i d <br />
assume HIi: "?P i"<br />
assume HId: "?P d"<br />
show "?P (N x i d)" using HIi HId by simp<br />
qed<br />
<br />
(* pablucoto marpoldia1*)<br />
<br />
lemma "postOrden (espejo a) = rev (preOrden a)" (is "?P a")<br />
proof (induct a)<br />
fix x<br />
show "?P (H x)" by simp <br />
next <br />
fix x i d<br />
assume H1: "?P i"<br />
assume H2: "?P d"<br />
have " postOrden (espejo (N x i d)) = postOrden ( N x (espejo d) (espejo i)) " by simp<br />
also have "... = postOrden (espejo d) @ postOrden (espejo i) @ [x] " by simp<br />
also have "... = rev (preOrden d) @ rev (preOrden i) @ [x] " using H1 H2 by simp<br />
also have "... = rev (preOrden d) @ rev (x # preOrden i)" by simp<br />
also have "... = rev (x # preOrden i @ preOrden d)" by simp<br />
also have "... = rev (preOrden (N x i d)) " by simp<br />
finally show "?P (N x i d)" by simp<br />
qed<br />
<br />
(* fraortmoy lucnovdos *)<br />
lemma "postOrden (espejo a) = rev (preOrden a)"<br />
by (induct a) auto<br />
<br />
text {* <br />
--------------------------------------------------------------------- <br />
Ejercicio 8. Demostrar que<br />
inOrden (espejo a) = rev (inOrden a)<br />
--------------------------------------------------------------------- <br />
*}<br />
<br />
(* ivamenjim crigomgom bowma migtermor wilmorort*)<br />
<br />
theorem "inOrden (espejo a) = rev (inOrden a)" (is "?P a")<br />
proof (induct a)<br />
fix x <br />
show "?P (H x)" by simp <br />
next<br />
fix x <br />
fix i assume h1: "?P i"<br />
fix d assume h2: "?P d"<br />
have "inOrden (espejo (N x i d)) = inOrden (N x (espejo d) (espejo i))" by simp<br />
also have "... = (inOrden (espejo d)) @ [x] @ (inOrden (espejo i))" by simp<br />
also have "... = rev (inOrden d) @ [x] @ rev (inOrden i)" using h1 h2 by simp <br />
finally show "inOrden (espejo (N x i d)) = rev (inOrden (N x i d))" by simp<br />
qed<br />
<br />
(* danrodcha *)<br />
theorem "inOrden (espejo a) = rev (inOrden a)"<br />
by (induct a) simp_all<br />
<br />
(* danrodcha *)<br />
theorem "inOrden (espejo a) = rev (inOrden a)" (is "?P a")<br />
proof (induct a)<br />
fix x<br />
show "?P (H x)" by simp<br />
next<br />
fix x i d <br />
assume HIi: "?P i"<br />
assume HId: "?P d"<br />
show "?P (N x i d)" using HIi HId by simp<br />
qed<br />
<br />
(* pablucoto marpoldia1 *)<br />
theorem "inOrden (espejo a) = rev (inOrden a)" (is "?P a")<br />
proof (induct a)<br />
fix x<br />
show "?P (H x) " by simp<br />
next <br />
fix x i d <br />
assume HI1: "?P i"<br />
assume HI2: "?P d"<br />
have " inOrden (espejo (N x i d)) = inOrden ( N x (espejo d) (espejo i) )" by simp<br />
also have "... = inOrden (espejo d) @ [x] @ inOrden (espejo i) " by simp<br />
also have "... = rev (inOrden d) @ [x] @ rev (inOrden i)" using HI1 HI2 by simp<br />
also have "... = rev (x # inOrden d ) @ rev (inOrden i)" by simp<br />
also have "... = rev ( inOrden i @ x # inOrden d) " by simp<br />
also have "... = rev (inOrden (N x i d))" by simp<br />
finally show "?P (N x i d)" by simp<br />
qed<br />
<br />
(* lucnovdos *)<br />
theorem "inOrden (espejo a) = rev (inOrden a)"<br />
by (induct a) auto<br />
<br />
text {* <br />
--------------------------------------------------------------------- <br />
Ejercicio 9. Definir la función <br />
raiz :: "'a arbol ⇒ 'a"<br />
tal que (raiz a) es la raiz del árbol a. Por ejemplo, <br />
raiz (N e (N c (H a) (H d)) (N g (H f) (H h))) = e<br />
--------------------------------------------------------------------- <br />
*}<br />
<br />
(* danrodcha crigomgom manmorjim1 ivamenjim bowma pablucoto migtermor marpoldia1 wilmorort lucnovdos *)<br />
fun raiz :: "'a arbol ⇒ 'a" where<br />
"raiz (H x) = x"<br />
| "raiz (N x i d) = x"<br />
<br />
value "raiz (N e (N c (H a) (H d)) (N g (H f) (H h))) = e"<br />
<br />
text {* <br />
--------------------------------------------------------------------- <br />
Ejercicio 10. Definir la función <br />
extremo_izquierda :: "'a arbol ⇒ 'a"<br />
tal que (extremo_izquierda a) es el nodo más a la izquierda del árbol<br />
a. Por ejemplo, <br />
extremo_izquierda (N e (N c (H a) (H d)) (N g (H f) (H h))) = a<br />
--------------------------------------------------------------------- <br />
*}<br />
<br />
(* danrodcha crigomgom manmorjim1 ivamenjim bowma pablucoto migtermor marpoldia1 wilmorort lucnovdos *)<br />
fun extremo_izquierda :: "'a arbol ⇒ 'a" where<br />
"extremo_izquierda (H x) = x"<br />
| "extremo_izquierda (N x i d) = extremo_izquierda i"<br />
<br />
value "extremo_izquierda (N e (N c (H a) (H d)) (N g (H f) (H h))) = a"<br />
<br />
(* bowma *)<br />
fun extremo_izquierda_1 :: "'a arbol ⇒ 'a" where<br />
"extremo_izquierda_1 (H t) = t"<br />
| "extremo_izquierda_1 (N t i d) = hd (inOrden (N t i d))"<br />
<br />
(* ivamenjim *)<br />
(* Metaejercicio de demostración. Llamando teorema_13 al teorema del ejercicio 13 *)<br />
<br />
lemma "extremo_izquierda a = extremo_izquierda_1 a"<br />
by (induct a, simp_all add: aux_ej12_1 teorema_13)<br />
<br />
text {* <br />
--------------------------------------------------------------------- <br />
Ejercicio 11. Definir la función <br />
extremo_derecha :: "'a arbol ⇒ 'a"<br />
tal que (extremo_derecha a) es el nodo más a la derecha del árbol<br />
a. Por ejemplo, <br />
extremo_derecha (N e (N c (H a) (H d)) (N g (H f) (H h))) = h<br />
--------------------------------------------------------------------- <br />
*}<br />
<br />
(* danrodcha crigomgom manmorjim1 ivamenjim bowma pablucoto migtermor marpoldia1 wilmorort lucnovdos *)<br />
fun extremo_derecha :: "'a arbol ⇒ 'a" where<br />
"extremo_derecha (H x) = x"<br />
| "extremo_derecha (N x i d) = extremo_derecha d"<br />
<br />
value "extremo_derecha (N e (N c (H a) (H d)) (N g (H f) (H h))) = h"<br />
<br />
(* bowma *)<br />
fun extremo_derecha_1 :: "'a arbol ⇒ 'a" where<br />
"extremo_derecha_1 (H t) = t"<br />
| "extremo_derecha_1 (N t i d) = last (inOrden (N t i d))"<br />
<br />
(* ivamenjim *)<br />
(* Metaejercicio de demostración. Llamando teorema_12 al teorema del ejercicio 12 *)<br />
<br />
lemma "extremo_derecha a = extremo_derecha_1 a"<br />
by (induct a, simp_all add: aux_ej12_1 teorema_12)<br />
<br />
text {* <br />
--------------------------------------------------------------------- <br />
Ejercicio 12. Demostrar o refutar<br />
last (inOrden a) = extremo_derecha a<br />
--------------------------------------------------------------------- <br />
*}<br />
<br />
(* danrodcha *)<br />
lemma aux_ej12: "inOrden a ≠ []"<br />
apply (induct a) <br />
apply simp<br />
apply simp<br />
done<br />
<br />
(* danrodcha pablucoto crigomgom wilmorort*)<br />
theorem "last (inOrden a) = extremo_derecha a" (is "?P a")<br />
proof (induct a)<br />
fix x<br />
show "?P (H x)" by simp<br />
next<br />
fix x i d <br />
assume HIi: "?P i"<br />
assume HId: "?P d"<br />
have "last (inOrden (N x i d)) = last (inOrden i @ [x] @ inOrden d)" <br />
by (simp only: inOrden.simps(2))<br />
also have "… = last (inOrden d)" by (simp add: aux_ej12)<br />
also have "… = extremo_derecha d" using HId by simp<br />
also have "… = extremo_derecha (N x i d)" by simp<br />
finally show "?P (N x i d)" by simp<br />
qed<br />
<br />
(* ivamenjim *)<br />
<br />
lemma aux_ej12_1: "inOrden a ≠ []"<br />
by (induct a) simp_all <br />
<br />
(* ivamenjim marpoldia1 *)<br />
(* Igual que la anterior, pero poniendo solo by simp en el primer have *)<br />
<br />
theorem "last (inOrden a) = extremo_derecha a" (is "?P a")<br />
proof (induct a)<br />
fix x <br />
show "?P (H x)" by simp <br />
next<br />
fix x <br />
fix i assume h1: "?P i"<br />
fix d assume h2: "?P d"<br />
have "last (inOrden (N x i d)) = last ((inOrden i) @ [x] @ (inOrden d))" by simp <br />
also have "... = last (inOrden d)" by (simp add: aux_ej12_1)<br />
also have "... = extremo_derecha d" using h2 by simp <br />
finally show "last (inOrden (N x i d)) = extremo_derecha (N x i d)" by simp<br />
qed<br />
<br />
(* bowma *)<br />
(* Casi lo mismo que el anterior,pero no hace falta suponer "?p i" *)<br />
theorem "last (inOrden a) = extremo_derecha a" (is "?p a")<br />
proof (induct a)<br />
fix t<br />
show "?p (H t)" by simp<br />
next <br />
fix t i d<br />
assume HI: "?p d"<br />
have "last (inOrden (N t i d)) = last (inOrden i @ [t] @ inOrden d)" by simp<br />
also have "... = last (inOrden d)" by (simp add:aux_ej12)<br />
also have "... = extremo_derecha d" using HI by simp<br />
finally show "?p (N t i d)" by simp<br />
qed<br />
<br />
(* lucnovdos*)<br />
(* El mismo que el anterior,pero sin usar padrones *)<br />
theorem "last (inOrden a) = extremo_derecha a"<br />
proof (induct a)<br />
fix x ::"'a"<br />
show "last (inOrden (H x)) = extremo_derecha (H x)" by simp<br />
next<br />
fix x ::"'a"<br />
fix i d ::"'a arbol"<br />
assume HI: "last (inOrden d) = extremo_derecha d"<br />
have "last (inOrden (N x i d)) = last ((inOrden i) @ [x] @ (inOrden d))" by simp <br />
also have "… = last (inOrden d)" by (simp add: aux_ej12)<br />
also have "… = extremo_derecha d" using HI by simp<br />
also have "… = extremo_derecha (N x i d)" by simp<br />
finally show "last (inOrden (N x i d)) = extremo_derecha (N x i d)" by simp<br />
qed<br />
<br />
(* migtermor *)<br />
theorem "last (inOrden a) = extremo_derecha a" (is "?P a")<br />
proof (induct a)<br />
fix h<br />
show "?P (H h)" by simp<br />
next<br />
fix n i<br />
fix d assume HId: "?P d"<br />
have AUX: "¬ (inOrden d = [])" (is "?Q d")<br />
proof (induct d)<br />
fix hd<br />
show "?Q (H hd)" by simp<br />
next<br />
fix nd<br />
fix id assume HIid: "?Q id"<br />
fix dd assume HIdd: "?Q dd"<br />
show "?Q (N nd id dd)" using HIid HIdd by simp<br />
qed<br />
have "last (inOrden (N n i d)) = last (inOrden i @[n]@inOrden d)" by simp<br />
also have "… = last (inOrden d)" using AUX by simp<br />
also have "… = extremo_derecha d" using HId by simp<br />
finally show "?P (N n i d)" by simp<br />
qed<br />
<br />
text {* <br />
--------------------------------------------------------------------- <br />
Ejercicio 13. Demostrar o refutar<br />
hd (inOrden a) = extremo_izquierda a<br />
--------------------------------------------------------------------- <br />
*}<br />
<br />
(* danrodcha pablucoto crigomgom *)<br />
theorem "hd (inOrden a) = extremo_izquierda a" (is "?P a")<br />
proof (induct a)<br />
fix x<br />
show "?P (H x)" by simp<br />
next<br />
fix x i d <br />
assume HIi: "?P i"<br />
assume HId: "?P d"<br />
have "hd (inOrden (N x i d)) = hd (inOrden i @ [x] @ inOrden d)" <br />
by (simp only: inOrden.simps(2))<br />
also have "… = hd (inOrden i)" by (simp add: aux_ej12)<br />
also have "… = extremo_izquierda i" using HIi by simp<br />
also have "… = extremo_izquierda (N x i d)" by simp<br />
finally show "?P (N x i d)" by simp<br />
qed<br />
<br />
(* bowma lucnovdos *)<br />
theorem "hd (inOrden a) = extremo_izquierda a" (is "?p a")<br />
proof (induct a)<br />
fix t<br />
show "?p (H t)" by simp<br />
next<br />
fix t i d <br />
assume HI: "?p i"<br />
have "hd (inOrden (N t i d)) = hd (inOrden i @ [t] @ inOrden d)" by simp<br />
also have "… = hd (inOrden i)" by (simp add: aux_ej12)<br />
also have "… = extremo_izquierda i" using HI by simp<br />
finally show "?p (N t i d)" by simp<br />
qed<br />
<br />
(* migtermor *)<br />
theorem "hd (inOrden a) = extremo_izquierda a" (is "?P a")<br />
proof (induct a)<br />
fix h<br />
show "?P (H h)" by simp<br />
next<br />
fix n d<br />
fix i assume HId: "?P i"<br />
have AUX: "¬ (inOrden i = [])" (is "?Q i")<br />
proof (induct i)<br />
fix hi<br />
show "?Q (H hi)" by simp<br />
next<br />
fix ni<br />
fix ii assume HIid: "?Q ii"<br />
fix di assume HIdd: "?Q di"<br />
show "?Q (N ni ii di)" using HIid HIdd by simp<br />
qed<br />
have "hd (inOrden (N n i d)) = hd (inOrden i @[n]@inOrden d)" by simp<br />
also have "… = hd (inOrden i)" using AUX by simp<br />
also have "… = extremo_izquierda i" using HId by simp<br />
finally show "?P (N n i d)" by simp<br />
qed<br />
<br />
(* ivamenjim marpoldia1 wilmorort *)<br />
<br />
theorem "hd (inOrden a) = extremo_izquierda a" (is "?P a")<br />
proof (induct a)<br />
fix x <br />
show "?P (H x)" by simp <br />
next<br />
fix x <br />
fix i assume h1: "?P i"<br />
fix d assume h2: "?P d"<br />
have "hd (inOrden (N x i d)) = hd ((inOrden i) @ [x] @ (inOrden d))" by simp<br />
also have "... = hd (inOrden i)" by (simp add: aux_ej12_1)<br />
also have "... = extremo_izquierda i" using h1 by simp <br />
finally show "hd (inOrden (N x i d)) = extremo_izquierda (N x i d)" by simp<br />
qed<br />
<br />
text {* <br />
--------------------------------------------------------------------- <br />
Ejercicio 14. Demostrar o refutar<br />
hd (preOrden a) = last (postOrden a)<br />
--------------------------------------------------------------------- <br />
*}<br />
<br />
(* danrodcha *)<br />
theorem "hd (preOrden a) = last (postOrden a)" (is "?P a")<br />
proof (induct a)<br />
fix x<br />
show "?P (H x)" by simp<br />
fix x i d<br />
show "?P (N x i d)" by simp<br />
qed<br />
<br />
(* danrodcha *)<br />
theorem "hd (preOrden a) = last (postOrden a)" (is "?P a")<br />
proof (induct a)<br />
fix x<br />
show "?P (H x)" by simp<br />
fix x i d<br />
assume HIi: "?P i"<br />
assume HId: "?P d"<br />
have "hd (preOrden (N x i d)) = hd (x#preOrden i @ preOrden d)"<br />
by (simp only: preOrden.simps(2))<br />
also have "… = x" by simp<br />
also have "… = last (postOrden i @ postOrden d @ [x])" by simp<br />
also have "… = last (postOrden (N x i d))" <br />
by (simp only: postOrden.simps(2))<br />
finally show "?P (N x i d)" by simp<br />
qed<br />
<br />
(* pablucoto crigomgom bowma marpoldia1 wilmorort*) (*Similar al anterior*)<br />
theorem "hd (preOrden a) = last (postOrden a)" (is "?P a")<br />
proof (induct a)<br />
fix x<br />
show "?P (H x)" by simp<br />
next <br />
fix x i d<br />
assume HI1: "?P i"<br />
assume HI2: "?P d"<br />
have " hd (preOrden (N x i d)) = hd ([x] @ preOrden i @ preOrden d)" by simp<br />
also have "... = x" by simp<br />
also have "... = last ( postOrden i @ postOrden d @ [x]) " by simp<br />
also have "... = last ( postOrden (N x i d) )" by simp <br />
finally show "?P (N x i d)" by simp<br />
qed<br />
<br />
(* migtermor *)<br />
theorem "hd (preOrden a) = last (postOrden a)" (is "?P a")<br />
proof (induct a)<br />
fix h<br />
show "?P (H h)" by simp<br />
next<br />
fix n i d<br />
have "hd (preOrden (N n (i :: 'a arbol) (d :: 'a arbol))) = hd ([n]@preOrden i@preOrden d)" <br />
by simp<br />
(* Si no especifico que i y d son árboles, salta un error de tipo. Supongo que será por<br />
no haber asumido hipótesis sobre ellos *)<br />
also have "… = last (postOrden (N n i d))" by simp<br />
show "?P (N n i d)" by simp<br />
qed<br />
<br />
(* ivamenjim *)<br />
<br />
theorem "hd (preOrden a) = last (postOrden a)" (is "?P a")<br />
proof (induct a)<br />
fix x <br />
show "?P (H x)" by simp <br />
next<br />
fix x <br />
fix i assume h1: "?P i"<br />
fix d assume h2: "?P d"<br />
have "hd (preOrden (N x i d)) = hd ([x] @ (preOrden i) @ (preOrden d))" by simp<br />
also have "... = hd ([x])" by simp<br />
finally show "hd (preOrden (N x i d)) = last (postOrden (N x i d))" by simp<br />
qed<br />
<br />
text {* <br />
--------------------------------------------------------------------- <br />
Ejercicio 15. Demostrar o refutar<br />
hd (preOrden a) = raiz a<br />
--------------------------------------------------------------------- <br />
*}<br />
<br />
(* danrodcha *)<br />
theorem "hd (preOrden a) = raiz a" (is "?P a")<br />
proof (induct a)<br />
fix x<br />
show "?P (H x)" by simp<br />
fix x i d<br />
assume HIi: "?P i"<br />
assume HId: "?P d"<br />
have "hd (preOrden (N x i d)) = hd (x#preOrden i @ preOrden d)"<br />
by (simp only: preOrden.simps(2))<br />
also have "… = x" by simp<br />
also have "… = raiz (N x i d)" by simp<br />
finally show "?P (N x i d)" by simp<br />
qed<br />
<br />
(* danrodcha *)<br />
theorem "hd (preOrden a) = raiz a" (is "?P a")<br />
proof (induct a)<br />
fix x<br />
show "?P (H x)" by simp<br />
fix x i d<br />
show "?P (N x i d)" by simp<br />
qed<br />
<br />
(* pablucoto crigomgom ivamenjim marpoldia1 wilmorort *)<br />
theorem "hd (preOrden a) = raiz a" (is "?P a")<br />
proof (induct a) <br />
fix x<br />
show "?P (H x)" by simp<br />
next<br />
fix x i d<br />
assume HI1: " ?P i"<br />
assume HI2: " ?P d"<br />
have " hd (preOrden (N x i d)) = hd ([x] @ preOrden i @ preOrden d) " by simp<br />
also have "... = x" by simp<br />
also have "... = raiz (N x i d)" by simp<br />
finally show " ?P (N x i d)" by simp<br />
qed<br />
<br />
(* bowma *)<br />
(* similar al anterior pero sin suponer "?p d" *)<br />
theorem "hd (preOrden a) = last (postOrden a)" (is "?p a")<br />
proof (induct a)<br />
fix t<br />
show "?p (H t)" by simp<br />
next<br />
fix t i d<br />
assume HI: "?p i"<br />
have "hd (preOrden (N t i d)) = hd ([t] @ preOrden i @ preOrden d)" by simp<br />
also have "… = t" by simp<br />
also have "… = last (postOrden i @ postOrden d @ [t])" by simp<br />
also have "… = last (postOrden (N t i d))" by simp<br />
finally show "?p (N t i d)" by simp<br />
qed<br />
<br />
(* migtermor *)<br />
theorem "hd (preOrden a) = raiz a" (is "?P a")<br />
proof (induct a)<br />
fix h<br />
show "?P (H h)" by simp<br />
next<br />
fix n i d<br />
have "hd (preOrden (N n (i :: 'a arbol) (d :: 'a arbol))) = hd ([n]@preOrden i@preOrden d)"<br />
by simp<br />
also have "… = raiz (N n i d)" by simp<br />
finally show "?P (N n i d)" by simp<br />
qed<br />
<br />
(* ivamenjim: sin usar patrones *)<br />
<br />
theorem "hd (preOrden a) = raiz a" <br />
proof (induct a)<br />
fix x ::"'a"<br />
show "hd (preOrden (H x)) = raiz (H x)" by simp <br />
next<br />
fix x ::"'a"<br />
fix i ::"'a arbol" assume h1: "hd (preOrden i) = raiz i"<br />
fix d ::"'a arbol" assume h2: "hd (preOrden d) = raiz d"<br />
have "hd (preOrden (N x i d)) = hd ([x] @ (preOrden i) @ (preOrden d))" by simp<br />
also have "... = hd ([x])" by simp<br />
finally show "hd (preOrden (N x i d)) = raiz (N x i d)" by simp<br />
qed<br />
<br />
text {* <br />
--------------------------------------------------------------------- <br />
Ejercicio 16. Demostrar o refutar<br />
hd (inOrden a) = raiz a<br />
--------------------------------------------------------------------- <br />
*}<br />
<br />
(*crigomgom pablucoto bowma migtermor ivamenjim wilmorort *)<br />
theorem "hd (inOrden a) = raiz a"<br />
quickcheck<br />
oops<br />
<br />
(* danrodcha:<br />
Auto Quickcheck found a counterexample:<br />
a = N a⇩1 (H a⇩2) (H a⇩1)<br />
Evaluated terms:<br />
hd (inOrden a) = a⇩2<br />
raiz a = a⇩1 *)<br />
<br />
(* ivamenjim marpoldia1 *)<br />
<br />
theorem "hd (inOrden a) = raiz a" (is "?P a")<br />
proof (induct a)<br />
fix x <br />
show "?P (H x)" by simp <br />
next<br />
fix x <br />
fix i assume h1: "?P i"<br />
fix d assume h2: "?P d"<br />
have "hd (inOrden (N x i d)) = hd ((inOrden i) @ [x] @ (inOrden d))" by simp<br />
also have "... = hd (inOrden i)" by (simp add: aux_ej12_1) <br />
(* Perdemos la x, luego se refuta el enunciado del teorema *)<br />
qed<br />
<br />
text {* <br />
--------------------------------------------------------------------- <br />
Ejercicio 17. Demostrar o refutar<br />
last (postOrden a) = raiz a<br />
--------------------------------------------------------------------- <br />
*}<br />
<br />
(* danrodcha *)<br />
theorem "last (postOrden a) = raiz a" (is "?P a")<br />
proof (induct a)<br />
fix x<br />
show "?P (H x)" by simp<br />
fix x i d<br />
show "?P (N x i d)" by simp<br />
qed<br />
<br />
(* danrodcha *)<br />
theorem "last (postOrden a) = raiz a" (is "?P a")<br />
proof (induct a)<br />
fix x<br />
show "?P (H x)" by simp<br />
fix x i d<br />
assume HIi: "?P i"<br />
assume HId: "?P d"<br />
have "last (postOrden (N x i d)) = last (postOrden i @ postOrden d @ [x])"<br />
by (simp only: postOrden.simps(2))<br />
also have "… = x" by simp<br />
also have "… = raiz (N x i d)" by (simp only: raiz.simps(2))<br />
finally show "?P (N x i d)" by simp<br />
qed<br />
<br />
(* pablucoto crigomgom ivamenjim marpoldia1 wilmorort*) (*Similar al anterior*)<br />
<br />
theorem "last (postOrden a) = raiz a" (is "?P a")<br />
proof (induct a )<br />
fix x<br />
show "?P (H x)" by simp<br />
next <br />
fix x i d<br />
assume HI1: "?P i"<br />
assume HI2: "?P d"<br />
have "last (postOrden (N x i d)) = last ( postOrden i @ postOrden d @ [x])" by simp<br />
also have "... = x" by simp<br />
also have "... = raiz (N x i d) " by simp<br />
finally show " ?P (N x i d)" by simp<br />
qed<br />
<br />
(* bowma *)<br />
(* También sin usar el supuesto "?p d" *)<br />
theorem "last (postOrden a) = raiz a" (is "?p a")<br />
proof (induct a)<br />
fix t<br />
show "?p (H t)" by simp<br />
next<br />
fix t i d<br />
assume "?p i"<br />
(* si quito este supuesto, hay error pero no sé dónde se lo está usando *)<br />
have "last (postOrden (N t i d)) = last (postOrden i @ postOrden d @ [t])" by simp<br />
also have "... = t" by simp<br />
also have "... = raiz (N t i d)" by simp<br />
finally show "?p (N t i d)" by simp<br />
qed<br />
<br />
(* migtermor *)<br />
theorem "last (postOrden a) = raiz a" (is "?P a")<br />
proof (induct a)<br />
fix h<br />
show "?P (H h)" by simp<br />
next<br />
fix n i d<br />
have "last (postOrden (N n (i :: 'a arbol) (d :: 'a arbol))) = <br />
last (postOrden i@postOrden d@[n])" by simp<br />
also have "… = raiz (N n i d)" by simp<br />
finally show "?P (N n i d)" by simp<br />
qed<br />
<br />
(* ivamenjim: sin usar patrones *)<br />
<br />
theorem "last (postOrden a) = raiz a" <br />
proof (induct a)<br />
fix x::"'a" <br />
show "last (postOrden (H x)) = raiz (H x)" by simp <br />
next<br />
fix x::"'a" <br />
fix i::"'a arbol" assume h1: "last (postOrden i) = raiz i"<br />
fix d::"'a arbol" assume h2: "last (postOrden d) = raiz d"<br />
have "last (postOrden (N x i d)) = last ((postOrden i) @ (postOrden d) @ [x])" by simp<br />
also have "... = last ([x])" by simp <br />
finally show "last (postOrden (N x i d)) = raiz (N x i d)" by simp<br />
qed<br />
<br />
end<br />
</source></div>Lucnovdoshttps://www.glc.us.es/~jalonso/RA2016/index.php?title=Relaci%C3%B3n_6&diff=968Relación 62016-12-06T03:06:28Z<p>Lucnovdos: </p>
<hr />
<div><source lang="isar"><br />
chapter {* R6: Recorridos de árboles *}<br />
<br />
theory R6_Recorridos_de_arboles<br />
imports Main <br />
begin <br />
<br />
text {* <br />
--------------------------------------------------------------------- <br />
Ejercicio 1. Definir el tipo de datos arbol para representar los<br />
árboles binarios que tiene información en los nodos y en las hojas. <br />
Por ejemplo, el árbol<br />
e<br />
/ \<br />
/ \<br />
c g<br />
/ \ / \<br />
a d f h <br />
se representa por "N e (N c (H a) (H d)) (N g (H f) (H h))".<br />
--------------------------------------------------------------------- <br />
*}<br />
<br />
(* ivamenjim marpoldia1 manmorjim1 bowma migtermor wilmorort *)<br />
<br />
datatype 'a arbol = H "'a" | N "'a" "'a arbol" "'a arbol"<br />
<br />
value "N e (N c (H a) (H d)) (N g (H f) (H h))" <br />
<br />
text {* <br />
--------------------------------------------------------------------- <br />
Ejercicio 2. Definir la función <br />
preOrden :: "'a arbol ⇒ 'a list"<br />
tal que (preOrden a) es el recorrido pre orden del árbol a. Por<br />
ejemplo, <br />
preOrden (N e (N c (H a) (H d)) (N g (H f) (H h)))<br />
= [e,c,a,d,g,f,h] <br />
--------------------------------------------------------------------- <br />
*}<br />
<br />
(* ivamenjim marpoldia1 pablucoto bowma fraortmoy migtermor wilmorort lucnovdos *)<br />
<br />
fun preOrden :: "'a arbol ⇒ 'a list" where<br />
"preOrden (H t) = [t]"<br />
| "preOrden (N t i d) = [t] @ (preOrden i) @ (preOrden d)"<br />
<br />
(* danrodcha crigomgom manmorjim1 bowma *)<br />
fun preOrden1 :: "'a arbol ⇒ 'a list" where<br />
"preOrden1 (H x) = [x]"<br />
| "preOrden1 (N x i d) = x#preOrden1 i @ preOrden1 d"<br />
<br />
value "preOrden (N e (N c (H a) (H d)) (N g (H f) (H h))) <br />
= [e,c,a,d,g,f,h]" <br />
value "preOrden1 (N e (N c (H a) (H d)) (N g (H f) (H h))) <br />
= [e,c,a,d,g,f,h]" <br />
<br />
(* danrodcha *)<br />
lemma "preOrden a = preOrden1 a"<br />
by (induct a) simp_all<br />
<br />
text {* <br />
--------------------------------------------------------------------- <br />
Ejercicio 3. Definir la función <br />
postOrden :: "'a arbol ⇒ 'a list"<br />
tal que (postOrden a) es el recorrido post orden del árbol a. Por<br />
ejemplo, <br />
postOrden (N e (N c (H a) (H d)) (N g (H f) (H h)))<br />
= [e,c,a,d,g,f,h] <br />
--------------------------------------------------------------------- <br />
*}<br />
<br />
(* ivamenjim danrodcha crigomgom marpoldia1 manmorjim1 pablucoto bowma fraortmoy migtermor wilmorort lucnovdos *)<br />
<br />
fun postOrden :: "'a arbol ⇒ 'a list" where<br />
"postOrden (H t) = [t]"<br />
| "postOrden (N t i d) = (postOrden i) @ (postOrden d) @ [t]"<br />
<br />
value "postOrden (N e (N c (H a) (H d)) (N g (H f) (H h)))<br />
= [a,d,c,f,h,g,e]"<br />
<br />
text {* <br />
--------------------------------------------------------------------- <br />
Ejercicio 4. Definir la función <br />
inOrden :: "'a arbol ⇒ 'a list"<br />
tal que (inOrden a) es el recorrido in orden del árbol a. Por<br />
ejemplo, <br />
inOrden (N e (N c (H a) (H d)) (N g (H f) (H h)))<br />
= [a,c,d,e,f,g,h]<br />
--------------------------------------------------------------------- <br />
*}<br />
<br />
(* ivamenjim crigomgom marpoldia1 pablucoto bowma fraortmoy migtermor wilmorort lucnovdos*)<br />
<br />
fun inOrden :: "'a arbol ⇒ 'a list" where<br />
"inOrden (H t) = [t]"<br />
| "inOrden (N t i d) = (inOrden i) @ [t] @ (inOrden d)"<br />
<br />
<br />
(* danrodcha manmorjim1 *)<br />
fun inOrden1 :: "'a arbol ⇒ 'a list" where<br />
"inOrden1 (H t) = [t]"<br />
| "inOrden1 (N t i d) = inOrden1 i @ t#inOrden1 d"<br />
<br />
value "inOrden (N e (N c (H a) (H d)) (N g (H f) (H h)))<br />
= [a,c,d,e,f,g,h]"<br />
value "inOrden1 (N e (N c (H a) (H d)) (N g (H f) (H h)))<br />
= [a,c,d,e,f,g,h]"<br />
<br />
(* manmorjim1 *)<br />
lemma "inOrden t = inOrden1 t"<br />
apply (induct t)<br />
apply auto<br />
done<br />
<br />
text {* <br />
--------------------------------------------------------------------- <br />
Ejercicio 5. Definir la función <br />
espejo :: "'a arbol ⇒ 'a arbol"<br />
tal que (espejo a) es la imagen especular del árbol a. Por ejemplo, <br />
espejo (N e (N c (H a) (H d)) (N g (H f) (H h)))<br />
= N e (N g (H h) (H f)) (N c (H d) (H a))<br />
--------------------------------------------------------------------- <br />
*}<br />
<br />
(* ivamenjim danrodcha crigomgom marpoldia1 manmorjim1 pablucoto bowma fraortmoy migtermor wilmorort lucnovdos *)<br />
<br />
fun espejo :: "'a arbol ⇒ 'a arbol" where<br />
"espejo (H t) = H t"<br />
| "espejo (N t i d) = N t (espejo d) (espejo i)"<br />
<br />
value "espejo (N e (N c (H a) (H d)) (N g (H f) (H h))) <br />
= N e (N g (H h) (H f)) (N c (H d) (H a))"<br />
<br />
<br />
text {* <br />
--------------------------------------------------------------------- <br />
Ejercicio 6. Demostrar que<br />
preOrden (espejo a) = rev (postOrden a)<br />
--------------------------------------------------------------------- <br />
*}<br />
<br />
(* ivamenjim migtermor wilmorort *)<br />
<br />
lemma "preOrden (espejo a) = rev (postOrden a)" (is "?P a")<br />
proof (induct a)<br />
fix x <br />
show "?P (H x)" by simp <br />
next<br />
fix x <br />
fix i assume h1: "?P i"<br />
fix d assume h2: "?P d"<br />
have "preOrden (espejo (N x i d)) = preOrden (N x (espejo d) (espejo i))" by simp<br />
also have "... = [x] @ (preOrden (espejo d)) @ (preOrden (espejo i))" by simp<br />
also have "... = [x] @ rev (postOrden d) @ rev (postOrden i)" using h1 h2 by simp <br />
finally show "preOrden (espejo (N x i d)) = rev (postOrden (N x i d))" by simp<br />
qed<br />
<br />
(* danrodcha *)<br />
lemma "preOrden (espejo a) = rev (postOrden a)"<br />
by (induct a) simp_all<br />
<br />
(* danrodcha crigomgom*)<br />
lemma "preOrden (espejo a) = rev (postOrden a)" (is "?P a")<br />
proof (induct a)<br />
fix x<br />
show "?P (H x)" by simp<br />
next<br />
fix x i d <br />
assume HIi: "?P i"<br />
assume HId: "?P d"<br />
have "preOrden (espejo (N x i d)) = preOrden (N x (espejo d) (espejo i))"<br />
by (simp only: espejo.simps(2))<br />
also have "… = x#preOrden (espejo d) @ preOrden (espejo i)"<br />
by (simp only: preOrden.simps(2))<br />
also have"… = x#rev (postOrden d) @ rev (postOrden i)" <br />
using HIi HId by simp<br />
also have "… = rev (postOrden (N x i d))" by simp<br />
finally show "?P (N x i d)" by simp<br />
qed<br />
<br />
(* danrodcha fraortmoy *)<br />
lemma "preOrden (espejo a) = rev (postOrden a)" (is "?P a")<br />
proof (induct a)<br />
fix x<br />
show "?P (H x)" by simp<br />
next<br />
fix x i d <br />
assume HIi: "?P i"<br />
assume HId: "?P d"<br />
show "?P (N x i d)" using HIi HId by simp<br />
qed<br />
<br />
(* bowma *)<br />
lemma "preOrden (espejo a) = rev (postOrden a)"<br />
apply (induct a)<br />
apply simp_all<br />
done<br />
<br />
(* pablucoto marpoldia1*)<br />
<br />
lemma "preOrden (espejo a) = rev (postOrden a)" (is "?P a")<br />
proof (induct a)<br />
fix x <br />
show "?P (H x)" by simp <br />
next<br />
fix x i d <br />
assume h1: "?P i"<br />
assume h2: "?P d"<br />
have "preOrden (espejo (N x i d)) = preOrden (N x (espejo d) (espejo i))" by simp<br />
also have "... = [x] @ (preOrden (espejo d)) @ (preOrden (espejo i))" by simp<br />
also have "... = [x] @ rev (postOrden d) @ rev (postOrden i)" using h1 h2 by simp<br />
also have "... = [x] @ rev (postOrden i @ postOrden d)" by simp <br />
also have "... = rev ( postOrden i @ postOrden d @ [x] ) " by simp<br />
also have "... = rev (postOrden (N x i d)) " by simp<br />
finally show "?P (N x i d)" by simp <br />
qed<br />
<br />
(* bowma *)<br />
lemma "preOrden (espejo a) = rev (postOrden a)" (is "?p a")<br />
proof (induct a)<br />
fix t<br />
show "?p (H t)" by simp<br />
(* Aquí si le diga "preOrden (espejo (H t)) = rev (postOrden (H t))",isabelle dice: <br />
proof (prove)<br />
goal (1 subgoal):<br />
1. preOrden (espejo (H t)) = rev (postOrden (H t)) <br />
Introduced fixed type variable(s): 'b in "t__" <br />
No entiendo porqué *)<br />
next <br />
fix t i d<br />
assume H1: "?p i"<br />
assume H2: "?p d"<br />
have "preOrden (espejo (N t i d)) = preOrden (N t (espejo d) (espejo i))" by simp<br />
also have "... = [t] @ (preOrden (espejo d)) @ (preOrden (espejo i))" by simp<br />
also have "... = [t] @ rev (postOrden d) @ rev (postOrden i)" using H1 H2 by simp<br />
finally show "?p (N t i d)" by simp<br />
qed <br />
<br />
(* fraortmoy lucnovdos *)<br />
<br />
lemma "preOrden (espejo a) = rev (postOrden a)"<br />
by (induct a) auto<br />
<br />
text {* <br />
--------------------------------------------------------------------- <br />
Ejercicio 7. Demostrar que<br />
postOrden (espejo a) = rev (preOrden a)<br />
--------------------------------------------------------------------- <br />
*}<br />
<br />
(* ivamenjim crigomgom bowma migtermor wilmorort*)<br />
<br />
lemma "postOrden (espejo a) = rev (preOrden a)" (is "?P a")<br />
proof (induct a)<br />
fix x <br />
show "?P (H x)" by simp <br />
next<br />
fix x <br />
fix i assume h1: "?P i"<br />
fix d assume h2: "?P d"<br />
have "postOrden (espejo (N x i d)) = postOrden (N x (espejo d) (espejo i))" by simp<br />
also have "... = (postOrden (espejo d)) @ (postOrden (espejo i)) @ [x]" by simp<br />
also have "... = rev (preOrden d) @ rev (preOrden i) @ [x]" using h1 h2 by simp <br />
finally show "postOrden (espejo (N x i d)) = rev (preOrden (N x i d))" by simp<br />
(* "?p (N x i d)" más corto *)<br />
qed<br />
<br />
(* danrodcha fraortmoy *)<br />
lemma "postOrden (espejo a) = rev (preOrden a)" (is "?P a")<br />
proof (induct a)<br />
fix x<br />
show "?P (H x)" by simp<br />
next<br />
fix x i d <br />
assume HIi: "?P i"<br />
assume HId: "?P d"<br />
show "?P (N x i d)" using HIi HId by simp<br />
qed<br />
<br />
(* pablucoto marpoldia1*)<br />
<br />
lemma "postOrden (espejo a) = rev (preOrden a)" (is "?P a")<br />
proof (induct a)<br />
fix x<br />
show "?P (H x)" by simp <br />
next <br />
fix x i d<br />
assume H1: "?P i"<br />
assume H2: "?P d"<br />
have " postOrden (espejo (N x i d)) = postOrden ( N x (espejo d) (espejo i)) " by simp<br />
also have "... = postOrden (espejo d) @ postOrden (espejo i) @ [x] " by simp<br />
also have "... = rev (preOrden d) @ rev (preOrden i) @ [x] " using H1 H2 by simp<br />
also have "... = rev (preOrden d) @ rev (x # preOrden i)" by simp<br />
also have "... = rev (x # preOrden i @ preOrden d)" by simp<br />
also have "... = rev (preOrden (N x i d)) " by simp<br />
finally show "?P (N x i d)" by simp<br />
qed<br />
<br />
(* fraortmoy lucnovdos *)<br />
lemma "postOrden (espejo a) = rev (preOrden a)"<br />
by (induct a) auto<br />
<br />
text {* <br />
--------------------------------------------------------------------- <br />
Ejercicio 8. Demostrar que<br />
inOrden (espejo a) = rev (inOrden a)<br />
--------------------------------------------------------------------- <br />
*}<br />
<br />
(* ivamenjim crigomgom bowma migtermor wilmorort*)<br />
<br />
theorem "inOrden (espejo a) = rev (inOrden a)" (is "?P a")<br />
proof (induct a)<br />
fix x <br />
show "?P (H x)" by simp <br />
next<br />
fix x <br />
fix i assume h1: "?P i"<br />
fix d assume h2: "?P d"<br />
have "inOrden (espejo (N x i d)) = inOrden (N x (espejo d) (espejo i))" by simp<br />
also have "... = (inOrden (espejo d)) @ [x] @ (inOrden (espejo i))" by simp<br />
also have "... = rev (inOrden d) @ [x] @ rev (inOrden i)" using h1 h2 by simp <br />
finally show "inOrden (espejo (N x i d)) = rev (inOrden (N x i d))" by simp<br />
qed<br />
<br />
(* danrodcha *)<br />
theorem "inOrden (espejo a) = rev (inOrden a)"<br />
by (induct a) simp_all<br />
<br />
(* danrodcha *)<br />
theorem "inOrden (espejo a) = rev (inOrden a)" (is "?P a")<br />
proof (induct a)<br />
fix x<br />
show "?P (H x)" by simp<br />
next<br />
fix x i d <br />
assume HIi: "?P i"<br />
assume HId: "?P d"<br />
show "?P (N x i d)" using HIi HId by simp<br />
qed<br />
<br />
(* pablucoto marpoldia1 *)<br />
theorem "inOrden (espejo a) = rev (inOrden a)" (is "?P a")<br />
proof (induct a)<br />
fix x<br />
show "?P (H x) " by simp<br />
next <br />
fix x i d <br />
assume HI1: "?P i"<br />
assume HI2: "?P d"<br />
have " inOrden (espejo (N x i d)) = inOrden ( N x (espejo d) (espejo i) )" by simp<br />
also have "... = inOrden (espejo d) @ [x] @ inOrden (espejo i) " by simp<br />
also have "... = rev (inOrden d) @ [x] @ rev (inOrden i)" using HI1 HI2 by simp<br />
also have "... = rev (x # inOrden d ) @ rev (inOrden i)" by simp<br />
also have "... = rev ( inOrden i @ x # inOrden d) " by simp<br />
also have "... = rev (inOrden (N x i d))" by simp<br />
finally show "?P (N x i d)" by simp<br />
qed<br />
<br />
(* lucnovdos *)<br />
theorem "inOrden (espejo a) = rev (inOrden a)"<br />
by (induct a) auto<br />
<br />
text {* <br />
--------------------------------------------------------------------- <br />
Ejercicio 9. Definir la función <br />
raiz :: "'a arbol ⇒ 'a"<br />
tal que (raiz a) es la raiz del árbol a. Por ejemplo, <br />
raiz (N e (N c (H a) (H d)) (N g (H f) (H h))) = e<br />
--------------------------------------------------------------------- <br />
*}<br />
<br />
(* danrodcha crigomgom manmorjim1 ivamenjim bowma pablucoto migtermor marpoldia1 wilmorort lucnovdos *)<br />
fun raiz :: "'a arbol ⇒ 'a" where<br />
"raiz (H x) = x"<br />
| "raiz (N x i d) = x"<br />
<br />
value "raiz (N e (N c (H a) (H d)) (N g (H f) (H h))) = e"<br />
<br />
text {* <br />
--------------------------------------------------------------------- <br />
Ejercicio 10. Definir la función <br />
extremo_izquierda :: "'a arbol ⇒ 'a"<br />
tal que (extremo_izquierda a) es el nodo más a la izquierda del árbol<br />
a. Por ejemplo, <br />
extremo_izquierda (N e (N c (H a) (H d)) (N g (H f) (H h))) = a<br />
--------------------------------------------------------------------- <br />
*}<br />
<br />
(* danrodcha crigomgom manmorjim1 ivamenjim bowma pablucoto migtermor marpoldia1 wilmorort lucnovdos *)<br />
fun extremo_izquierda :: "'a arbol ⇒ 'a" where<br />
"extremo_izquierda (H x) = x"<br />
| "extremo_izquierda (N x i d) = extremo_izquierda i"<br />
<br />
value "extremo_izquierda (N e (N c (H a) (H d)) (N g (H f) (H h))) = a"<br />
<br />
(* bowma *)<br />
fun extremo_izquierda_1 :: "'a arbol ⇒ 'a" where<br />
"extremo_izquierda_1 (H t) = t"<br />
| "extremo_izquierda_1 (N t i d) = hd (inOrden (N t i d))"<br />
<br />
(* ivamenjim *)<br />
(* Metaejercicio de demostración. Llamando teorema_13 al teorema del ejercicio 13 *)<br />
<br />
lemma "extremo_izquierda a = extremo_izquierda_1 a"<br />
by (induct a, simp_all add: aux_ej12_1 teorema_13)<br />
<br />
text {* <br />
--------------------------------------------------------------------- <br />
Ejercicio 11. Definir la función <br />
extremo_derecha :: "'a arbol ⇒ 'a"<br />
tal que (extremo_derecha a) es el nodo más a la derecha del árbol<br />
a. Por ejemplo, <br />
extremo_derecha (N e (N c (H a) (H d)) (N g (H f) (H h))) = h<br />
--------------------------------------------------------------------- <br />
*}<br />
<br />
(* danrodcha crigomgom manmorjim1 ivamenjim bowma pablucoto migtermor marpoldia1 wilmorort lucnovdos *)<br />
fun extremo_derecha :: "'a arbol ⇒ 'a" where<br />
"extremo_derecha (H x) = x"<br />
| "extremo_derecha (N x i d) = extremo_derecha d"<br />
<br />
value "extremo_derecha (N e (N c (H a) (H d)) (N g (H f) (H h))) = h"<br />
<br />
(* bowma *)<br />
fun extremo_derecha_1 :: "'a arbol ⇒ 'a" where<br />
"extremo_derecha_1 (H t) = t"<br />
| "extremo_derecha_1 (N t i d) = last (inOrden (N t i d))"<br />
<br />
(* ivamenjim *)<br />
(* Metaejercicio de demostración. Llamando teorema_12 al teorema del ejercicio 12 *)<br />
<br />
lemma "extremo_derecha a = extremo_derecha_1 a"<br />
by (induct a, simp_all add: aux_ej12_1 teorema_12)<br />
<br />
text {* <br />
--------------------------------------------------------------------- <br />
Ejercicio 12. Demostrar o refutar<br />
last (inOrden a) = extremo_derecha a<br />
--------------------------------------------------------------------- <br />
*}<br />
<br />
(* danrodcha *)<br />
lemma aux_ej12: "inOrden a ≠ []"<br />
apply (induct a) <br />
apply simp<br />
apply simp<br />
done<br />
<br />
(* danrodcha pablucoto crigomgom wilmorort*)<br />
theorem "last (inOrden a) = extremo_derecha a" (is "?P a")<br />
proof (induct a)<br />
fix x<br />
show "?P (H x)" by simp<br />
next<br />
fix x i d <br />
assume HIi: "?P i"<br />
assume HId: "?P d"<br />
have "last (inOrden (N x i d)) = last (inOrden i @ [x] @ inOrden d)" <br />
by (simp only: inOrden.simps(2))<br />
also have "… = last (inOrden d)" by (simp add: aux_ej12)<br />
also have "… = extremo_derecha d" using HId by simp<br />
also have "… = extremo_derecha (N x i d)" by simp<br />
finally show "?P (N x i d)" by simp<br />
qed<br />
<br />
(* ivamenjim *)<br />
<br />
lemma aux_ej12_1: "inOrden a ≠ []"<br />
by (induct a) simp_all <br />
<br />
(* ivamenjim marpoldia1 *)<br />
(* Igual que la anterior, pero poniendo solo by simp en el primer have *)<br />
<br />
theorem "last (inOrden a) = extremo_derecha a" (is "?P a")<br />
proof (induct a)<br />
fix x <br />
show "?P (H x)" by simp <br />
next<br />
fix x <br />
fix i assume h1: "?P i"<br />
fix d assume h2: "?P d"<br />
have "last (inOrden (N x i d)) = last ((inOrden i) @ [x] @ (inOrden d))" by simp <br />
also have "... = last (inOrden d)" by (simp add: aux_ej12_1)<br />
also have "... = extremo_derecha d" using h2 by simp <br />
finally show "last (inOrden (N x i d)) = extremo_derecha (N x i d)" by simp<br />
qed<br />
<br />
(* bowma *)<br />
(* Casi lo mismo que el anterior,pero no hace falta suponer "?p i" *)<br />
theorem "last (inOrden a) = extremo_derecha a" (is "?p a")<br />
proof (induct a)<br />
fix t<br />
show "?p (H t)" by simp<br />
next <br />
fix t i d<br />
assume HI: "?p d"<br />
have "last (inOrden (N t i d)) = last (inOrden i @ [t] @ inOrden d)" by simp<br />
also have "... = last (inOrden d)" by (simp add:aux_ej12)<br />
also have "... = extremo_derecha d" using HI by simp<br />
finally show "?p (N t i d)" by simp<br />
qed<br />
<br />
(* lucnovdos*)<br />
(* El mismo que el anterior,pero sin usar padrones *)<br />
theorem "last (inOrden a) = extremo_derecha a"<br />
proof (induct a)<br />
fix x ::"'a"<br />
show "last (inOrden (H x)) = extremo_derecha (H x)" by simp<br />
next<br />
fix x ::"'a"<br />
fix i d ::"'a arbol"<br />
assume HI: "last (inOrden d) = extremo_derecha d"<br />
have "last (inOrden (N x i d)) = last ((inOrden i) @ [x] @ (inOrden d))" by simp <br />
also have "… = last (inOrden d)" by (simp add: aux_ej12)<br />
also have "… = extremo_derecha d" using HI by simp<br />
also have "… = extremo_derecha (N x i d)" by simp<br />
finally show "last (inOrden (N x i d)) = extremo_derecha (N x i d)" by simp<br />
qed<br />
<br />
(* migtermor *)<br />
theorem "last (inOrden a) = extremo_derecha a" (is "?P a")<br />
proof (induct a)<br />
fix h<br />
show "?P (H h)" by simp<br />
next<br />
fix n i<br />
fix d assume HId: "?P d"<br />
have AUX: "¬ (inOrden d = [])" (is "?Q d")<br />
proof (induct d)<br />
fix hd<br />
show "?Q (H hd)" by simp<br />
next<br />
fix nd<br />
fix id assume HIid: "?Q id"<br />
fix dd assume HIdd: "?Q dd"<br />
show "?Q (N nd id dd)" using HIid HIdd by simp<br />
qed<br />
have "last (inOrden (N n i d)) = last (inOrden i @[n]@inOrden d)" by simp<br />
also have "… = last (inOrden d)" using AUX by simp<br />
also have "… = extremo_derecha d" using HId by simp<br />
finally show "?P (N n i d)" by simp<br />
qed<br />
<br />
text {* <br />
--------------------------------------------------------------------- <br />
Ejercicio 13. Demostrar o refutar<br />
hd (inOrden a) = extremo_izquierda a<br />
--------------------------------------------------------------------- <br />
*}<br />
<br />
(* danrodcha pablucoto crigomgom *)<br />
theorem "hd (inOrden a) = extremo_izquierda a" (is "?P a")<br />
proof (induct a)<br />
fix x<br />
show "?P (H x)" by simp<br />
next<br />
fix x i d <br />
assume HIi: "?P i"<br />
assume HId: "?P d"<br />
have "hd (inOrden (N x i d)) = hd (inOrden i @ [x] @ inOrden d)" <br />
by (simp only: inOrden.simps(2))<br />
also have "… = hd (inOrden i)" by (simp add: aux_ej12)<br />
also have "… = extremo_izquierda i" using HIi by simp<br />
also have "… = extremo_izquierda (N x i d)" by simp<br />
finally show "?P (N x i d)" by simp<br />
qed<br />
<br />
(* bowma *)<br />
theorem "hd (inOrden a) = extremo_izquierda a" (is "?p a")<br />
proof (induct a)<br />
fix t<br />
show "?p (H t)" by simp<br />
next<br />
fix t i d <br />
assume HI: "?p i"<br />
have "hd (inOrden (N t i d)) = hd (inOrden i @ [t] @ inOrden d)" by simp<br />
also have "… = hd (inOrden i)" by (simp add: aux_ej12)<br />
also have "… = extremo_izquierda i" using HI by simp<br />
finally show "?p (N t i d)" by simp<br />
qed<br />
<br />
(* migtermor *)<br />
theorem "hd (inOrden a) = extremo_izquierda a" (is "?P a")<br />
proof (induct a)<br />
fix h<br />
show "?P (H h)" by simp<br />
next<br />
fix n d<br />
fix i assume HId: "?P i"<br />
have AUX: "¬ (inOrden i = [])" (is "?Q i")<br />
proof (induct i)<br />
fix hi<br />
show "?Q (H hi)" by simp<br />
next<br />
fix ni<br />
fix ii assume HIid: "?Q ii"<br />
fix di assume HIdd: "?Q di"<br />
show "?Q (N ni ii di)" using HIid HIdd by simp<br />
qed<br />
have "hd (inOrden (N n i d)) = hd (inOrden i @[n]@inOrden d)" by simp<br />
also have "… = hd (inOrden i)" using AUX by simp<br />
also have "… = extremo_izquierda i" using HId by simp<br />
finally show "?P (N n i d)" by simp<br />
qed<br />
<br />
(* ivamenjim marpoldia1 wilmorort *)<br />
<br />
theorem "hd (inOrden a) = extremo_izquierda a" (is "?P a")<br />
proof (induct a)<br />
fix x <br />
show "?P (H x)" by simp <br />
next<br />
fix x <br />
fix i assume h1: "?P i"<br />
fix d assume h2: "?P d"<br />
have "hd (inOrden (N x i d)) = hd ((inOrden i) @ [x] @ (inOrden d))" by simp<br />
also have "... = hd (inOrden i)" by (simp add: aux_ej12_1)<br />
also have "... = extremo_izquierda i" using h1 by simp <br />
finally show "hd (inOrden (N x i d)) = extremo_izquierda (N x i d)" by simp<br />
qed<br />
<br />
text {* <br />
--------------------------------------------------------------------- <br />
Ejercicio 14. Demostrar o refutar<br />
hd (preOrden a) = last (postOrden a)<br />
--------------------------------------------------------------------- <br />
*}<br />
<br />
(* danrodcha *)<br />
theorem "hd (preOrden a) = last (postOrden a)" (is "?P a")<br />
proof (induct a)<br />
fix x<br />
show "?P (H x)" by simp<br />
fix x i d<br />
show "?P (N x i d)" by simp<br />
qed<br />
<br />
(* danrodcha *)<br />
theorem "hd (preOrden a) = last (postOrden a)" (is "?P a")<br />
proof (induct a)<br />
fix x<br />
show "?P (H x)" by simp<br />
fix x i d<br />
assume HIi: "?P i"<br />
assume HId: "?P d"<br />
have "hd (preOrden (N x i d)) = hd (x#preOrden i @ preOrden d)"<br />
by (simp only: preOrden.simps(2))<br />
also have "… = x" by simp<br />
also have "… = last (postOrden i @ postOrden d @ [x])" by simp<br />
also have "… = last (postOrden (N x i d))" <br />
by (simp only: postOrden.simps(2))<br />
finally show "?P (N x i d)" by simp<br />
qed<br />
<br />
(* pablucoto crigomgom bowma marpoldia1 wilmorort*) (*Similar al anterior*)<br />
theorem "hd (preOrden a) = last (postOrden a)" (is "?P a")<br />
proof (induct a)<br />
fix x<br />
show "?P (H x)" by simp<br />
next <br />
fix x i d<br />
assume HI1: "?P i"<br />
assume HI2: "?P d"<br />
have " hd (preOrden (N x i d)) = hd ([x] @ preOrden i @ preOrden d)" by simp<br />
also have "... = x" by simp<br />
also have "... = last ( postOrden i @ postOrden d @ [x]) " by simp<br />
also have "... = last ( postOrden (N x i d) )" by simp <br />
finally show "?P (N x i d)" by simp<br />
qed<br />
<br />
(* migtermor *)<br />
theorem "hd (preOrden a) = last (postOrden a)" (is "?P a")<br />
proof (induct a)<br />
fix h<br />
show "?P (H h)" by simp<br />
next<br />
fix n i d<br />
have "hd (preOrden (N n (i :: 'a arbol) (d :: 'a arbol))) = hd ([n]@preOrden i@preOrden d)" <br />
by simp<br />
(* Si no especifico que i y d son árboles, salta un error de tipo. Supongo que será por<br />
no haber asumido hipótesis sobre ellos *)<br />
also have "… = last (postOrden (N n i d))" by simp<br />
show "?P (N n i d)" by simp<br />
qed<br />
<br />
(* ivamenjim *)<br />
<br />
theorem "hd (preOrden a) = last (postOrden a)" (is "?P a")<br />
proof (induct a)<br />
fix x <br />
show "?P (H x)" by simp <br />
next<br />
fix x <br />
fix i assume h1: "?P i"<br />
fix d assume h2: "?P d"<br />
have "hd (preOrden (N x i d)) = hd ([x] @ (preOrden i) @ (preOrden d))" by simp<br />
also have "... = hd ([x])" by simp<br />
finally show "hd (preOrden (N x i d)) = last (postOrden (N x i d))" by simp<br />
qed<br />
<br />
text {* <br />
--------------------------------------------------------------------- <br />
Ejercicio 15. Demostrar o refutar<br />
hd (preOrden a) = raiz a<br />
--------------------------------------------------------------------- <br />
*}<br />
<br />
(* danrodcha *)<br />
theorem "hd (preOrden a) = raiz a" (is "?P a")<br />
proof (induct a)<br />
fix x<br />
show "?P (H x)" by simp<br />
fix x i d<br />
assume HIi: "?P i"<br />
assume HId: "?P d"<br />
have "hd (preOrden (N x i d)) = hd (x#preOrden i @ preOrden d)"<br />
by (simp only: preOrden.simps(2))<br />
also have "… = x" by simp<br />
also have "… = raiz (N x i d)" by simp<br />
finally show "?P (N x i d)" by simp<br />
qed<br />
<br />
(* danrodcha *)<br />
theorem "hd (preOrden a) = raiz a" (is "?P a")<br />
proof (induct a)<br />
fix x<br />
show "?P (H x)" by simp<br />
fix x i d<br />
show "?P (N x i d)" by simp<br />
qed<br />
<br />
(* pablucoto crigomgom ivamenjim marpoldia1 wilmorort *)<br />
theorem "hd (preOrden a) = raiz a" (is "?P a")<br />
proof (induct a) <br />
fix x<br />
show "?P (H x)" by simp<br />
next<br />
fix x i d<br />
assume HI1: " ?P i"<br />
assume HI2: " ?P d"<br />
have " hd (preOrden (N x i d)) = hd ([x] @ preOrden i @ preOrden d) " by simp<br />
also have "... = x" by simp<br />
also have "... = raiz (N x i d)" by simp<br />
finally show " ?P (N x i d)" by simp<br />
qed<br />
<br />
(* bowma *)<br />
(* similar al anterior pero sin suponer "?p d" *)<br />
theorem "hd (preOrden a) = last (postOrden a)" (is "?p a")<br />
proof (induct a)<br />
fix t<br />
show "?p (H t)" by simp<br />
next<br />
fix t i d<br />
assume HI: "?p i"<br />
have "hd (preOrden (N t i d)) = hd ([t] @ preOrden i @ preOrden d)" by simp<br />
also have "… = t" by simp<br />
also have "… = last (postOrden i @ postOrden d @ [t])" by simp<br />
also have "… = last (postOrden (N t i d))" by simp<br />
finally show "?p (N t i d)" by simp<br />
qed<br />
<br />
(* migtermor *)<br />
theorem "hd (preOrden a) = raiz a" (is "?P a")<br />
proof (induct a)<br />
fix h<br />
show "?P (H h)" by simp<br />
next<br />
fix n i d<br />
have "hd (preOrden (N n (i :: 'a arbol) (d :: 'a arbol))) = hd ([n]@preOrden i@preOrden d)"<br />
by simp<br />
also have "… = raiz (N n i d)" by simp<br />
finally show "?P (N n i d)" by simp<br />
qed<br />
<br />
(* ivamenjim: sin usar patrones *)<br />
<br />
theorem "hd (preOrden a) = raiz a" <br />
proof (induct a)<br />
fix x ::"'a"<br />
show "hd (preOrden (H x)) = raiz (H x)" by simp <br />
next<br />
fix x ::"'a"<br />
fix i ::"'a arbol" assume h1: "hd (preOrden i) = raiz i"<br />
fix d ::"'a arbol" assume h2: "hd (preOrden d) = raiz d"<br />
have "hd (preOrden (N x i d)) = hd ([x] @ (preOrden i) @ (preOrden d))" by simp<br />
also have "... = hd ([x])" by simp<br />
finally show "hd (preOrden (N x i d)) = raiz (N x i d)" by simp<br />
qed<br />
<br />
text {* <br />
--------------------------------------------------------------------- <br />
Ejercicio 16. Demostrar o refutar<br />
hd (inOrden a) = raiz a<br />
--------------------------------------------------------------------- <br />
*}<br />
<br />
(*crigomgom pablucoto bowma migtermor ivamenjim wilmorort *)<br />
theorem "hd (inOrden a) = raiz a"<br />
quickcheck<br />
oops<br />
<br />
(* danrodcha:<br />
Auto Quickcheck found a counterexample:<br />
a = N a⇩1 (H a⇩2) (H a⇩1)<br />
Evaluated terms:<br />
hd (inOrden a) = a⇩2<br />
raiz a = a⇩1 *)<br />
<br />
(* ivamenjim marpoldia1 *)<br />
<br />
theorem "hd (inOrden a) = raiz a" (is "?P a")<br />
proof (induct a)<br />
fix x <br />
show "?P (H x)" by simp <br />
next<br />
fix x <br />
fix i assume h1: "?P i"<br />
fix d assume h2: "?P d"<br />
have "hd (inOrden (N x i d)) = hd ((inOrden i) @ [x] @ (inOrden d))" by simp<br />
also have "... = hd (inOrden i)" by (simp add: aux_ej12_1) <br />
(* Perdemos la x, luego se refuta el enunciado del teorema *)<br />
qed<br />
<br />
text {* <br />
--------------------------------------------------------------------- <br />
Ejercicio 17. Demostrar o refutar<br />
last (postOrden a) = raiz a<br />
--------------------------------------------------------------------- <br />
*}<br />
<br />
(* danrodcha *)<br />
theorem "last (postOrden a) = raiz a" (is "?P a")<br />
proof (induct a)<br />
fix x<br />
show "?P (H x)" by simp<br />
fix x i d<br />
show "?P (N x i d)" by simp<br />
qed<br />
<br />
(* danrodcha *)<br />
theorem "last (postOrden a) = raiz a" (is "?P a")<br />
proof (induct a)<br />
fix x<br />
show "?P (H x)" by simp<br />
fix x i d<br />
assume HIi: "?P i"<br />
assume HId: "?P d"<br />
have "last (postOrden (N x i d)) = last (postOrden i @ postOrden d @ [x])"<br />
by (simp only: postOrden.simps(2))<br />
also have "… = x" by simp<br />
also have "… = raiz (N x i d)" by (simp only: raiz.simps(2))<br />
finally show "?P (N x i d)" by simp<br />
qed<br />
<br />
(* pablucoto crigomgom ivamenjim marpoldia1 wilmorort*) (*Similar al anterior*)<br />
<br />
theorem "last (postOrden a) = raiz a" (is "?P a")<br />
proof (induct a )<br />
fix x<br />
show "?P (H x)" by simp<br />
next <br />
fix x i d<br />
assume HI1: "?P i"<br />
assume HI2: "?P d"<br />
have "last (postOrden (N x i d)) = last ( postOrden i @ postOrden d @ [x])" by simp<br />
also have "... = x" by simp<br />
also have "... = raiz (N x i d) " by simp<br />
finally show " ?P (N x i d)" by simp<br />
qed<br />
<br />
(* bowma *)<br />
(* También sin usar el supuesto "?p d" *)<br />
theorem "last (postOrden a) = raiz a" (is "?p a")<br />
proof (induct a)<br />
fix t<br />
show "?p (H t)" by simp<br />
next<br />
fix t i d<br />
assume "?p i"<br />
(* si quito este supuesto, hay error pero no sé dónde se lo está usando *)<br />
have "last (postOrden (N t i d)) = last (postOrden i @ postOrden d @ [t])" by simp<br />
also have "... = t" by simp<br />
also have "... = raiz (N t i d)" by simp<br />
finally show "?p (N t i d)" by simp<br />
qed<br />
<br />
(* migtermor *)<br />
theorem "last (postOrden a) = raiz a" (is "?P a")<br />
proof (induct a)<br />
fix h<br />
show "?P (H h)" by simp<br />
next<br />
fix n i d<br />
have "last (postOrden (N n (i :: 'a arbol) (d :: 'a arbol))) = <br />
last (postOrden i@postOrden d@[n])" by simp<br />
also have "… = raiz (N n i d)" by simp<br />
finally show "?P (N n i d)" by simp<br />
qed<br />
<br />
(* ivamenjim: sin usar patrones *)<br />
<br />
theorem "last (postOrden a) = raiz a" <br />
proof (induct a)<br />
fix x::"'a" <br />
show "last (postOrden (H x)) = raiz (H x)" by simp <br />
next<br />
fix x::"'a" <br />
fix i::"'a arbol" assume h1: "last (postOrden i) = raiz i"<br />
fix d::"'a arbol" assume h2: "last (postOrden d) = raiz d"<br />
have "last (postOrden (N x i d)) = last ((postOrden i) @ (postOrden d) @ [x])" by simp<br />
also have "... = last ([x])" by simp <br />
finally show "last (postOrden (N x i d)) = raiz (N x i d)" by simp<br />
qed<br />
<br />
end<br />
</source></div>Lucnovdoshttps://www.glc.us.es/~jalonso/RA2016/index.php?title=Relaci%C3%B3n_6&diff=967Relación 62016-12-06T01:30:04Z<p>Lucnovdos: </p>
<hr />
<div><source lang="isar"><br />
chapter {* R6: Recorridos de árboles *}<br />
<br />
theory R6_Recorridos_de_arboles<br />
imports Main <br />
begin <br />
<br />
text {* <br />
--------------------------------------------------------------------- <br />
Ejercicio 1. Definir el tipo de datos arbol para representar los<br />
árboles binarios que tiene información en los nodos y en las hojas. <br />
Por ejemplo, el árbol<br />
e<br />
/ \<br />
/ \<br />
c g<br />
/ \ / \<br />
a d f h <br />
se representa por "N e (N c (H a) (H d)) (N g (H f) (H h))".<br />
--------------------------------------------------------------------- <br />
*}<br />
<br />
(* ivamenjim marpoldia1 manmorjim1 bowma migtermor wilmorort *)<br />
<br />
datatype 'a arbol = H "'a" | N "'a" "'a arbol" "'a arbol"<br />
<br />
value "N e (N c (H a) (H d)) (N g (H f) (H h))" <br />
<br />
text {* <br />
--------------------------------------------------------------------- <br />
Ejercicio 2. Definir la función <br />
preOrden :: "'a arbol ⇒ 'a list"<br />
tal que (preOrden a) es el recorrido pre orden del árbol a. Por<br />
ejemplo, <br />
preOrden (N e (N c (H a) (H d)) (N g (H f) (H h)))<br />
= [e,c,a,d,g,f,h] <br />
--------------------------------------------------------------------- <br />
*}<br />
<br />
(* ivamenjim marpoldia1 pablucoto bowma fraortmoy migtermor wilmorort lucnovdos *)<br />
<br />
fun preOrden :: "'a arbol ⇒ 'a list" where<br />
"preOrden (H t) = [t]"<br />
| "preOrden (N t i d) = [t] @ (preOrden i) @ (preOrden d)"<br />
<br />
(* danrodcha crigomgom manmorjim1 bowma *)<br />
fun preOrden1 :: "'a arbol ⇒ 'a list" where<br />
"preOrden1 (H x) = [x]"<br />
| "preOrden1 (N x i d) = x#preOrden1 i @ preOrden1 d"<br />
<br />
value "preOrden (N e (N c (H a) (H d)) (N g (H f) (H h))) <br />
= [e,c,a,d,g,f,h]" <br />
value "preOrden1 (N e (N c (H a) (H d)) (N g (H f) (H h))) <br />
= [e,c,a,d,g,f,h]" <br />
<br />
(* danrodcha *)<br />
lemma "preOrden a = preOrden1 a"<br />
by (induct a) simp_all<br />
<br />
text {* <br />
--------------------------------------------------------------------- <br />
Ejercicio 3. Definir la función <br />
postOrden :: "'a arbol ⇒ 'a list"<br />
tal que (postOrden a) es el recorrido post orden del árbol a. Por<br />
ejemplo, <br />
postOrden (N e (N c (H a) (H d)) (N g (H f) (H h)))<br />
= [e,c,a,d,g,f,h] <br />
--------------------------------------------------------------------- <br />
*}<br />
<br />
(* ivamenjim danrodcha crigomgom marpoldia1 manmorjim1 pablucoto bowma fraortmoy migtermor wilmorort lucnovdos *)<br />
<br />
fun postOrden :: "'a arbol ⇒ 'a list" where<br />
"postOrden (H t) = [t]"<br />
| "postOrden (N t i d) = (postOrden i) @ (postOrden d) @ [t]"<br />
<br />
value "postOrden (N e (N c (H a) (H d)) (N g (H f) (H h)))<br />
= [a,d,c,f,h,g,e]"<br />
<br />
text {* <br />
--------------------------------------------------------------------- <br />
Ejercicio 4. Definir la función <br />
inOrden :: "'a arbol ⇒ 'a list"<br />
tal que (inOrden a) es el recorrido in orden del árbol a. Por<br />
ejemplo, <br />
inOrden (N e (N c (H a) (H d)) (N g (H f) (H h)))<br />
= [a,c,d,e,f,g,h]<br />
--------------------------------------------------------------------- <br />
*}<br />
<br />
(* ivamenjim crigomgom marpoldia1 pablucoto bowma fraortmoy migtermor wilmorort lucnovdos*)<br />
<br />
fun inOrden :: "'a arbol ⇒ 'a list" where<br />
"inOrden (H t) = [t]"<br />
| "inOrden (N t i d) = (inOrden i) @ [t] @ (inOrden d)"<br />
<br />
<br />
(* danrodcha manmorjim1 *)<br />
fun inOrden1 :: "'a arbol ⇒ 'a list" where<br />
"inOrden1 (H t) = [t]"<br />
| "inOrden1 (N t i d) = inOrden1 i @ t#inOrden1 d"<br />
<br />
value "inOrden (N e (N c (H a) (H d)) (N g (H f) (H h)))<br />
= [a,c,d,e,f,g,h]"<br />
value "inOrden1 (N e (N c (H a) (H d)) (N g (H f) (H h)))<br />
= [a,c,d,e,f,g,h]"<br />
<br />
(* manmorjim1 *)<br />
lemma "inOrden t = inOrden1 t"<br />
apply (induct t)<br />
apply auto<br />
done<br />
<br />
text {* <br />
--------------------------------------------------------------------- <br />
Ejercicio 5. Definir la función <br />
espejo :: "'a arbol ⇒ 'a arbol"<br />
tal que (espejo a) es la imagen especular del árbol a. Por ejemplo, <br />
espejo (N e (N c (H a) (H d)) (N g (H f) (H h)))<br />
= N e (N g (H h) (H f)) (N c (H d) (H a))<br />
--------------------------------------------------------------------- <br />
*}<br />
<br />
(* ivamenjim danrodcha crigomgom marpoldia1 manmorjim1 pablucoto bowma fraortmoy migtermor wilmorort lucnovdos *)<br />
<br />
fun espejo :: "'a arbol ⇒ 'a arbol" where<br />
"espejo (H t) = H t"<br />
| "espejo (N t i d) = N t (espejo d) (espejo i)"<br />
<br />
value "espejo (N e (N c (H a) (H d)) (N g (H f) (H h))) <br />
= N e (N g (H h) (H f)) (N c (H d) (H a))"<br />
<br />
<br />
text {* <br />
--------------------------------------------------------------------- <br />
Ejercicio 6. Demostrar que<br />
preOrden (espejo a) = rev (postOrden a)<br />
--------------------------------------------------------------------- <br />
*}<br />
<br />
(* ivamenjim migtermor wilmorort *)<br />
<br />
lemma "preOrden (espejo a) = rev (postOrden a)" (is "?P a")<br />
proof (induct a)<br />
fix x <br />
show "?P (H x)" by simp <br />
next<br />
fix x <br />
fix i assume h1: "?P i"<br />
fix d assume h2: "?P d"<br />
have "preOrden (espejo (N x i d)) = preOrden (N x (espejo d) (espejo i))" by simp<br />
also have "... = [x] @ (preOrden (espejo d)) @ (preOrden (espejo i))" by simp<br />
also have "... = [x] @ rev (postOrden d) @ rev (postOrden i)" using h1 h2 by simp <br />
finally show "preOrden (espejo (N x i d)) = rev (postOrden (N x i d))" by simp<br />
qed<br />
<br />
(* danrodcha *)<br />
lemma "preOrden (espejo a) = rev (postOrden a)"<br />
by (induct a) simp_all<br />
<br />
(* danrodcha crigomgom*)<br />
lemma "preOrden (espejo a) = rev (postOrden a)" (is "?P a")<br />
proof (induct a)<br />
fix x<br />
show "?P (H x)" by simp<br />
next<br />
fix x i d <br />
assume HIi: "?P i"<br />
assume HId: "?P d"<br />
have "preOrden (espejo (N x i d)) = preOrden (N x (espejo d) (espejo i))"<br />
by (simp only: espejo.simps(2))<br />
also have "… = x#preOrden (espejo d) @ preOrden (espejo i)"<br />
by (simp only: preOrden.simps(2))<br />
also have"… = x#rev (postOrden d) @ rev (postOrden i)" <br />
using HIi HId by simp<br />
also have "… = rev (postOrden (N x i d))" by simp<br />
finally show "?P (N x i d)" by simp<br />
qed<br />
<br />
(* danrodcha fraortmoy *)<br />
lemma "preOrden (espejo a) = rev (postOrden a)" (is "?P a")<br />
proof (induct a)<br />
fix x<br />
show "?P (H x)" by simp<br />
next<br />
fix x i d <br />
assume HIi: "?P i"<br />
assume HId: "?P d"<br />
show "?P (N x i d)" using HIi HId by simp<br />
qed<br />
<br />
(* bowma *)<br />
lemma "preOrden (espejo a) = rev (postOrden a)"<br />
apply (induct a)<br />
apply simp_all<br />
done<br />
<br />
(* pablucoto marpoldia1*)<br />
<br />
lemma "preOrden (espejo a) = rev (postOrden a)" (is "?P a")<br />
proof (induct a)<br />
fix x <br />
show "?P (H x)" by simp <br />
next<br />
fix x i d <br />
assume h1: "?P i"<br />
assume h2: "?P d"<br />
have "preOrden (espejo (N x i d)) = preOrden (N x (espejo d) (espejo i))" by simp<br />
also have "... = [x] @ (preOrden (espejo d)) @ (preOrden (espejo i))" by simp<br />
also have "... = [x] @ rev (postOrden d) @ rev (postOrden i)" using h1 h2 by simp<br />
also have "... = [x] @ rev (postOrden i @ postOrden d)" by simp <br />
also have "... = rev ( postOrden i @ postOrden d @ [x] ) " by simp<br />
also have "... = rev (postOrden (N x i d)) " by simp<br />
finally show "?P (N x i d)" by simp <br />
qed<br />
<br />
(* bowma *)<br />
lemma "preOrden (espejo a) = rev (postOrden a)" (is "?p a")<br />
proof (induct a)<br />
fix t<br />
show "?p (H t)" by simp<br />
(* Aquí si le diga "preOrden (espejo (H t)) = rev (postOrden (H t))",isabelle dice: <br />
proof (prove)<br />
goal (1 subgoal):<br />
1. preOrden (espejo (H t)) = rev (postOrden (H t)) <br />
Introduced fixed type variable(s): 'b in "t__" <br />
No entiendo porqué *)<br />
next <br />
fix t i d<br />
assume H1: "?p i"<br />
assume H2: "?p d"<br />
have "preOrden (espejo (N t i d)) = preOrden (N t (espejo d) (espejo i))" by simp<br />
also have "... = [t] @ (preOrden (espejo d)) @ (preOrden (espejo i))" by simp<br />
also have "... = [t] @ rev (postOrden d) @ rev (postOrden i)" using H1 H2 by simp<br />
finally show "?p (N t i d)" by simp<br />
qed <br />
<br />
(* fraortmoy lucnovdos *)<br />
<br />
lemma "preOrden (espejo a) = rev (postOrden a)"<br />
by (induct a) auto<br />
<br />
text {* <br />
--------------------------------------------------------------------- <br />
Ejercicio 7. Demostrar que<br />
postOrden (espejo a) = rev (preOrden a)<br />
--------------------------------------------------------------------- <br />
*}<br />
<br />
(* ivamenjim crigomgom bowma migtermor wilmorort*)<br />
<br />
lemma "postOrden (espejo a) = rev (preOrden a)" (is "?P a")<br />
proof (induct a)<br />
fix x <br />
show "?P (H x)" by simp <br />
next<br />
fix x <br />
fix i assume h1: "?P i"<br />
fix d assume h2: "?P d"<br />
have "postOrden (espejo (N x i d)) = postOrden (N x (espejo d) (espejo i))" by simp<br />
also have "... = (postOrden (espejo d)) @ (postOrden (espejo i)) @ [x]" by simp<br />
also have "... = rev (preOrden d) @ rev (preOrden i) @ [x]" using h1 h2 by simp <br />
finally show "postOrden (espejo (N x i d)) = rev (preOrden (N x i d))" by simp<br />
(* "?p (N x i d)" más corto *)<br />
qed<br />
<br />
(* danrodcha fraortmoy *)<br />
lemma "postOrden (espejo a) = rev (preOrden a)" (is "?P a")<br />
proof (induct a)<br />
fix x<br />
show "?P (H x)" by simp<br />
next<br />
fix x i d <br />
assume HIi: "?P i"<br />
assume HId: "?P d"<br />
show "?P (N x i d)" using HIi HId by simp<br />
qed<br />
<br />
(* pablucoto marpoldia1*)<br />
<br />
lemma "postOrden (espejo a) = rev (preOrden a)" (is "?P a")<br />
proof (induct a)<br />
fix x<br />
show "?P (H x)" by simp <br />
next <br />
fix x i d<br />
assume H1: "?P i"<br />
assume H2: "?P d"<br />
have " postOrden (espejo (N x i d)) = postOrden ( N x (espejo d) (espejo i)) " by simp<br />
also have "... = postOrden (espejo d) @ postOrden (espejo i) @ [x] " by simp<br />
also have "... = rev (preOrden d) @ rev (preOrden i) @ [x] " using H1 H2 by simp<br />
also have "... = rev (preOrden d) @ rev (x # preOrden i)" by simp<br />
also have "... = rev (x # preOrden i @ preOrden d)" by simp<br />
also have "... = rev (preOrden (N x i d)) " by simp<br />
finally show "?P (N x i d)" by simp<br />
qed<br />
<br />
(* fraortmoy lucnovdos *)<br />
lemma "postOrden (espejo a) = rev (preOrden a)"<br />
by (induct a) auto<br />
<br />
text {* <br />
--------------------------------------------------------------------- <br />
Ejercicio 8. Demostrar que<br />
inOrden (espejo a) = rev (inOrden a)<br />
--------------------------------------------------------------------- <br />
*}<br />
<br />
(* ivamenjim crigomgom bowma migtermor wilmorort*)<br />
<br />
theorem "inOrden (espejo a) = rev (inOrden a)" (is "?P a")<br />
proof (induct a)<br />
fix x <br />
show "?P (H x)" by simp <br />
next<br />
fix x <br />
fix i assume h1: "?P i"<br />
fix d assume h2: "?P d"<br />
have "inOrden (espejo (N x i d)) = inOrden (N x (espejo d) (espejo i))" by simp<br />
also have "... = (inOrden (espejo d)) @ [x] @ (inOrden (espejo i))" by simp<br />
also have "... = rev (inOrden d) @ [x] @ rev (inOrden i)" using h1 h2 by simp <br />
finally show "inOrden (espejo (N x i d)) = rev (inOrden (N x i d))" by simp<br />
qed<br />
<br />
(* danrodcha *)<br />
theorem "inOrden (espejo a) = rev (inOrden a)"<br />
by (induct a) simp_all<br />
<br />
(* danrodcha *)<br />
theorem "inOrden (espejo a) = rev (inOrden a)" (is "?P a")<br />
proof (induct a)<br />
fix x<br />
show "?P (H x)" by simp<br />
next<br />
fix x i d <br />
assume HIi: "?P i"<br />
assume HId: "?P d"<br />
show "?P (N x i d)" using HIi HId by simp<br />
qed<br />
<br />
(* pablucoto marpoldia1 *)<br />
theorem "inOrden (espejo a) = rev (inOrden a)" (is "?P a")<br />
proof (induct a)<br />
fix x<br />
show "?P (H x) " by simp<br />
next <br />
fix x i d <br />
assume HI1: "?P i"<br />
assume HI2: "?P d"<br />
have " inOrden (espejo (N x i d)) = inOrden ( N x (espejo d) (espejo i) )" by simp<br />
also have "... = inOrden (espejo d) @ [x] @ inOrden (espejo i) " by simp<br />
also have "... = rev (inOrden d) @ [x] @ rev (inOrden i)" using HI1 HI2 by simp<br />
also have "... = rev (x # inOrden d ) @ rev (inOrden i)" by simp<br />
also have "... = rev ( inOrden i @ x # inOrden d) " by simp<br />
also have "... = rev (inOrden (N x i d))" by simp<br />
finally show "?P (N x i d)" by simp<br />
qed<br />
<br />
(* lucnovdos *)<br />
theorem "inOrden (espejo a) = rev (inOrden a)"<br />
by (induct a) auto<br />
<br />
text {* <br />
--------------------------------------------------------------------- <br />
Ejercicio 9. Definir la función <br />
raiz :: "'a arbol ⇒ 'a"<br />
tal que (raiz a) es la raiz del árbol a. Por ejemplo, <br />
raiz (N e (N c (H a) (H d)) (N g (H f) (H h))) = e<br />
--------------------------------------------------------------------- <br />
*}<br />
<br />
(* danrodcha crigomgom manmorjim1 ivamenjim bowma pablucoto migtermor marpoldia1 wilmorort lucnovdos *)<br />
fun raiz :: "'a arbol ⇒ 'a" where<br />
"raiz (H x) = x"<br />
| "raiz (N x i d) = x"<br />
<br />
value "raiz (N e (N c (H a) (H d)) (N g (H f) (H h))) = e"<br />
<br />
text {* <br />
--------------------------------------------------------------------- <br />
Ejercicio 10. Definir la función <br />
extremo_izquierda :: "'a arbol ⇒ 'a"<br />
tal que (extremo_izquierda a) es el nodo más a la izquierda del árbol<br />
a. Por ejemplo, <br />
extremo_izquierda (N e (N c (H a) (H d)) (N g (H f) (H h))) = a<br />
--------------------------------------------------------------------- <br />
*}<br />
<br />
(* danrodcha crigomgom manmorjim1 ivamenjim bowma pablucoto migtermor marpoldia1 wilmorort lucnovdos *)<br />
fun extremo_izquierda :: "'a arbol ⇒ 'a" where<br />
"extremo_izquierda (H x) = x"<br />
| "extremo_izquierda (N x i d) = extremo_izquierda i"<br />
<br />
value "extremo_izquierda (N e (N c (H a) (H d)) (N g (H f) (H h))) = a"<br />
<br />
(* bowma *)<br />
fun extremo_izquierda_1 :: "'a arbol ⇒ 'a" where<br />
"extremo_izquierda_1 (H t) = t"<br />
| "extremo_izquierda_1 (N t i d) = hd (inOrden (N t i d))"<br />
<br />
(* ivamenjim *)<br />
(* Metaejercicio de demostración. Llamando teorema_13 al teorema del ejercicio 13 *)<br />
<br />
lemma "extremo_izquierda a = extremo_izquierda_1 a"<br />
by (induct a, simp_all add: aux_ej12_1 teorema_13)<br />
<br />
text {* <br />
--------------------------------------------------------------------- <br />
Ejercicio 11. Definir la función <br />
extremo_derecha :: "'a arbol ⇒ 'a"<br />
tal que (extremo_derecha a) es el nodo más a la derecha del árbol<br />
a. Por ejemplo, <br />
extremo_derecha (N e (N c (H a) (H d)) (N g (H f) (H h))) = h<br />
--------------------------------------------------------------------- <br />
*}<br />
<br />
(* danrodcha crigomgom manmorjim1 ivamenjim bowma pablucoto migtermor marpoldia1 wilmorort lucnovdos *)<br />
fun extremo_derecha :: "'a arbol ⇒ 'a" where<br />
"extremo_derecha (H x) = x"<br />
| "extremo_derecha (N x i d) = extremo_derecha d"<br />
<br />
value "extremo_derecha (N e (N c (H a) (H d)) (N g (H f) (H h))) = h"<br />
<br />
(* bowma *)<br />
fun extremo_derecha_1 :: "'a arbol ⇒ 'a" where<br />
"extremo_derecha_1 (H t) = t"<br />
| "extremo_derecha_1 (N t i d) = last (inOrden (N t i d))"<br />
<br />
(* ivamenjim *)<br />
(* Metaejercicio de demostración. Llamando teorema_12 al teorema del ejercicio 12 *)<br />
<br />
lemma "extremo_derecha a = extremo_derecha_1 a"<br />
by (induct a, simp_all add: aux_ej12_1 teorema_12)<br />
<br />
text {* <br />
--------------------------------------------------------------------- <br />
Ejercicio 12. Demostrar o refutar<br />
last (inOrden a) = extremo_derecha a<br />
--------------------------------------------------------------------- <br />
*}<br />
<br />
(* danrodcha *)<br />
lemma aux_ej12: "inOrden a ≠ []"<br />
apply (induct a) <br />
apply simp<br />
apply simp<br />
done<br />
<br />
(* danrodcha pablucoto crigomgom wilmorort*)<br />
theorem "last (inOrden a) = extremo_derecha a" (is "?P a")<br />
proof (induct a)<br />
fix x<br />
show "?P (H x)" by simp<br />
next<br />
fix x i d <br />
assume HIi: "?P i"<br />
assume HId: "?P d"<br />
have "last (inOrden (N x i d)) = last (inOrden i @ [x] @ inOrden d)" <br />
by (simp only: inOrden.simps(2))<br />
also have "… = last (inOrden d)" by (simp add: aux_ej12)<br />
also have "… = extremo_derecha d" using HId by simp<br />
also have "… = extremo_derecha (N x i d)" by simp<br />
finally show "?P (N x i d)" by simp<br />
qed<br />
<br />
(* ivamenjim *)<br />
<br />
lemma aux_ej12_1: "inOrden a ≠ []"<br />
by (induct a) simp_all <br />
<br />
(* ivamenjim marpoldia1 *)<br />
(* Igual que la anterior, pero poniendo solo by simp en el primer have *)<br />
<br />
theorem "last (inOrden a) = extremo_derecha a" (is "?P a")<br />
proof (induct a)<br />
fix x <br />
show "?P (H x)" by simp <br />
next<br />
fix x <br />
fix i assume h1: "?P i"<br />
fix d assume h2: "?P d"<br />
have "last (inOrden (N x i d)) = last ((inOrden i) @ [x] @ (inOrden d))" by simp <br />
also have "... = last (inOrden d)" by (simp add: aux_ej12_1)<br />
also have "... = extremo_derecha d" using h2 by simp <br />
finally show "last (inOrden (N x i d)) = extremo_derecha (N x i d)" by simp<br />
qed<br />
<br />
(* bowma *)<br />
(* Casi lo mismo que el anterior,pero no hace falta suponer "?p i" *)<br />
theorem "last (inOrden a) = extremo_derecha a" (is "?p a")<br />
proof (induct a)<br />
fix t<br />
show "?p (H t)" by simp<br />
next <br />
fix t i d<br />
assume HI: "?p d"<br />
have "last (inOrden (N t i d)) = last (inOrden i @ [t] @ inOrden d)" by simp<br />
also have "... = last (inOrden d)" by (simp add:aux_ej12)<br />
also have "... = extremo_derecha d" using HI by simp<br />
finally show "?p (N t i d)" by simp<br />
qed<br />
<br />
(* migtermor *)<br />
theorem "last (inOrden a) = extremo_derecha a" (is "?P a")<br />
proof (induct a)<br />
fix h<br />
show "?P (H h)" by simp<br />
next<br />
fix n i<br />
fix d assume HId: "?P d"<br />
have AUX: "¬ (inOrden d = [])" (is "?Q d")<br />
proof (induct d)<br />
fix hd<br />
show "?Q (H hd)" by simp<br />
next<br />
fix nd<br />
fix id assume HIid: "?Q id"<br />
fix dd assume HIdd: "?Q dd"<br />
show "?Q (N nd id dd)" using HIid HIdd by simp<br />
qed<br />
have "last (inOrden (N n i d)) = last (inOrden i @[n]@inOrden d)" by simp<br />
also have "… = last (inOrden d)" using AUX by simp<br />
also have "… = extremo_derecha d" using HId by simp<br />
finally show "?P (N n i d)" by simp<br />
qed<br />
<br />
text {* <br />
--------------------------------------------------------------------- <br />
Ejercicio 13. Demostrar o refutar<br />
hd (inOrden a) = extremo_izquierda a<br />
--------------------------------------------------------------------- <br />
*}<br />
<br />
(* danrodcha pablucoto crigomgom *)<br />
theorem "hd (inOrden a) = extremo_izquierda a" (is "?P a")<br />
proof (induct a)<br />
fix x<br />
show "?P (H x)" by simp<br />
next<br />
fix x i d <br />
assume HIi: "?P i"<br />
assume HId: "?P d"<br />
have "hd (inOrden (N x i d)) = hd (inOrden i @ [x] @ inOrden d)" <br />
by (simp only: inOrden.simps(2))<br />
also have "… = hd (inOrden i)" by (simp add: aux_ej12)<br />
also have "… = extremo_izquierda i" using HIi by simp<br />
also have "… = extremo_izquierda (N x i d)" by simp<br />
finally show "?P (N x i d)" by simp<br />
qed<br />
<br />
(* bowma *)<br />
theorem "hd (inOrden a) = extremo_izquierda a" (is "?p a")<br />
proof (induct a)<br />
fix t<br />
show "?p (H t)" by simp<br />
next<br />
fix t i d <br />
assume HI: "?p i"<br />
have "hd (inOrden (N t i d)) = hd (inOrden i @ [t] @ inOrden d)" by simp<br />
also have "… = hd (inOrden i)" by (simp add: aux_ej12)<br />
also have "… = extremo_izquierda i" using HI by simp<br />
finally show "?p (N t i d)" by simp<br />
qed<br />
<br />
(* migtermor *)<br />
theorem "hd (inOrden a) = extremo_izquierda a" (is "?P a")<br />
proof (induct a)<br />
fix h<br />
show "?P (H h)" by simp<br />
next<br />
fix n d<br />
fix i assume HId: "?P i"<br />
have AUX: "¬ (inOrden i = [])" (is "?Q i")<br />
proof (induct i)<br />
fix hi<br />
show "?Q (H hi)" by simp<br />
next<br />
fix ni<br />
fix ii assume HIid: "?Q ii"<br />
fix di assume HIdd: "?Q di"<br />
show "?Q (N ni ii di)" using HIid HIdd by simp<br />
qed<br />
have "hd (inOrden (N n i d)) = hd (inOrden i @[n]@inOrden d)" by simp<br />
also have "… = hd (inOrden i)" using AUX by simp<br />
also have "… = extremo_izquierda i" using HId by simp<br />
finally show "?P (N n i d)" by simp<br />
qed<br />
<br />
(* ivamenjim marpoldia1 wilmorort *)<br />
<br />
theorem "hd (inOrden a) = extremo_izquierda a" (is "?P a")<br />
proof (induct a)<br />
fix x <br />
show "?P (H x)" by simp <br />
next<br />
fix x <br />
fix i assume h1: "?P i"<br />
fix d assume h2: "?P d"<br />
have "hd (inOrden (N x i d)) = hd ((inOrden i) @ [x] @ (inOrden d))" by simp<br />
also have "... = hd (inOrden i)" by (simp add: aux_ej12_1)<br />
also have "... = extremo_izquierda i" using h1 by simp <br />
finally show "hd (inOrden (N x i d)) = extremo_izquierda (N x i d)" by simp<br />
qed<br />
<br />
text {* <br />
--------------------------------------------------------------------- <br />
Ejercicio 14. Demostrar o refutar<br />
hd (preOrden a) = last (postOrden a)<br />
--------------------------------------------------------------------- <br />
*}<br />
<br />
(* danrodcha *)<br />
theorem "hd (preOrden a) = last (postOrden a)" (is "?P a")<br />
proof (induct a)<br />
fix x<br />
show "?P (H x)" by simp<br />
fix x i d<br />
show "?P (N x i d)" by simp<br />
qed<br />
<br />
(* danrodcha *)<br />
theorem "hd (preOrden a) = last (postOrden a)" (is "?P a")<br />
proof (induct a)<br />
fix x<br />
show "?P (H x)" by simp<br />
fix x i d<br />
assume HIi: "?P i"<br />
assume HId: "?P d"<br />
have "hd (preOrden (N x i d)) = hd (x#preOrden i @ preOrden d)"<br />
by (simp only: preOrden.simps(2))<br />
also have "… = x" by simp<br />
also have "… = last (postOrden i @ postOrden d @ [x])" by simp<br />
also have "… = last (postOrden (N x i d))" <br />
by (simp only: postOrden.simps(2))<br />
finally show "?P (N x i d)" by simp<br />
qed<br />
<br />
(* pablucoto crigomgom bowma marpoldia1 wilmorort*) (*Similar al anterior*)<br />
theorem "hd (preOrden a) = last (postOrden a)" (is "?P a")<br />
proof (induct a)<br />
fix x<br />
show "?P (H x)" by simp<br />
next <br />
fix x i d<br />
assume HI1: "?P i"<br />
assume HI2: "?P d"<br />
have " hd (preOrden (N x i d)) = hd ([x] @ preOrden i @ preOrden d)" by simp<br />
also have "... = x" by simp<br />
also have "... = last ( postOrden i @ postOrden d @ [x]) " by simp<br />
also have "... = last ( postOrden (N x i d) )" by simp <br />
finally show "?P (N x i d)" by simp<br />
qed<br />
<br />
(* migtermor *)<br />
theorem "hd (preOrden a) = last (postOrden a)" (is "?P a")<br />
proof (induct a)<br />
fix h<br />
show "?P (H h)" by simp<br />
next<br />
fix n i d<br />
have "hd (preOrden (N n (i :: 'a arbol) (d :: 'a arbol))) = hd ([n]@preOrden i@preOrden d)" <br />
by simp<br />
(* Si no especifico que i y d son árboles, salta un error de tipo. Supongo que será por<br />
no haber asumido hipótesis sobre ellos *)<br />
also have "… = last (postOrden (N n i d))" by simp<br />
show "?P (N n i d)" by simp<br />
qed<br />
<br />
(* ivamenjim *)<br />
<br />
theorem "hd (preOrden a) = last (postOrden a)" (is "?P a")<br />
proof (induct a)<br />
fix x <br />
show "?P (H x)" by simp <br />
next<br />
fix x <br />
fix i assume h1: "?P i"<br />
fix d assume h2: "?P d"<br />
have "hd (preOrden (N x i d)) = hd ([x] @ (preOrden i) @ (preOrden d))" by simp<br />
also have "... = hd ([x])" by simp<br />
finally show "hd (preOrden (N x i d)) = last (postOrden (N x i d))" by simp<br />
qed<br />
<br />
text {* <br />
--------------------------------------------------------------------- <br />
Ejercicio 15. Demostrar o refutar<br />
hd (preOrden a) = raiz a<br />
--------------------------------------------------------------------- <br />
*}<br />
<br />
(* danrodcha *)<br />
theorem "hd (preOrden a) = raiz a" (is "?P a")<br />
proof (induct a)<br />
fix x<br />
show "?P (H x)" by simp<br />
fix x i d<br />
assume HIi: "?P i"<br />
assume HId: "?P d"<br />
have "hd (preOrden (N x i d)) = hd (x#preOrden i @ preOrden d)"<br />
by (simp only: preOrden.simps(2))<br />
also have "… = x" by simp<br />
also have "… = raiz (N x i d)" by simp<br />
finally show "?P (N x i d)" by simp<br />
qed<br />
<br />
(* danrodcha *)<br />
theorem "hd (preOrden a) = raiz a" (is "?P a")<br />
proof (induct a)<br />
fix x<br />
show "?P (H x)" by simp<br />
fix x i d<br />
show "?P (N x i d)" by simp<br />
qed<br />
<br />
(* pablucoto crigomgom ivamenjim marpoldia1 wilmorort *)<br />
theorem "hd (preOrden a) = raiz a" (is "?P a")<br />
proof (induct a) <br />
fix x<br />
show "?P (H x)" by simp<br />
next<br />
fix x i d<br />
assume HI1: " ?P i"<br />
assume HI2: " ?P d"<br />
have " hd (preOrden (N x i d)) = hd ([x] @ preOrden i @ preOrden d) " by simp<br />
also have "... = x" by simp<br />
also have "... = raiz (N x i d)" by simp<br />
finally show " ?P (N x i d)" by simp<br />
qed<br />
<br />
(* bowma *)<br />
(* similar al anterior pero sin suponer "?p d" *)<br />
theorem "hd (preOrden a) = last (postOrden a)" (is "?p a")<br />
proof (induct a)<br />
fix t<br />
show "?p (H t)" by simp<br />
next<br />
fix t i d<br />
assume HI: "?p i"<br />
have "hd (preOrden (N t i d)) = hd ([t] @ preOrden i @ preOrden d)" by simp<br />
also have "… = t" by simp<br />
also have "… = last (postOrden i @ postOrden d @ [t])" by simp<br />
also have "… = last (postOrden (N t i d))" by simp<br />
finally show "?p (N t i d)" by simp<br />
qed<br />
<br />
(* migtermor *)<br />
theorem "hd (preOrden a) = raiz a" (is "?P a")<br />
proof (induct a)<br />
fix h<br />
show "?P (H h)" by simp<br />
next<br />
fix n i d<br />
have "hd (preOrden (N n (i :: 'a arbol) (d :: 'a arbol))) = hd ([n]@preOrden i@preOrden d)"<br />
by simp<br />
also have "… = raiz (N n i d)" by simp<br />
finally show "?P (N n i d)" by simp<br />
qed<br />
<br />
(* ivamenjim: sin usar patrones *)<br />
<br />
theorem "hd (preOrden a) = raiz a" <br />
proof (induct a)<br />
fix x ::"'a"<br />
show "hd (preOrden (H x)) = raiz (H x)" by simp <br />
next<br />
fix x ::"'a"<br />
fix i ::"'a arbol" assume h1: "hd (preOrden i) = raiz i"<br />
fix d ::"'a arbol" assume h2: "hd (preOrden d) = raiz d"<br />
have "hd (preOrden (N x i d)) = hd ([x] @ (preOrden i) @ (preOrden d))" by simp<br />
also have "... = hd ([x])" by simp<br />
finally show "hd (preOrden (N x i d)) = raiz (N x i d)" by simp<br />
qed<br />
<br />
text {* <br />
--------------------------------------------------------------------- <br />
Ejercicio 16. Demostrar o refutar<br />
hd (inOrden a) = raiz a<br />
--------------------------------------------------------------------- <br />
*}<br />
<br />
(*crigomgom pablucoto bowma migtermor ivamenjim wilmorort *)<br />
theorem "hd (inOrden a) = raiz a"<br />
quickcheck<br />
oops<br />
<br />
(* danrodcha:<br />
Auto Quickcheck found a counterexample:<br />
a = N a⇩1 (H a⇩2) (H a⇩1)<br />
Evaluated terms:<br />
hd (inOrden a) = a⇩2<br />
raiz a = a⇩1 *)<br />
<br />
(* ivamenjim marpoldia1 *)<br />
<br />
theorem "hd (inOrden a) = raiz a" (is "?P a")<br />
proof (induct a)<br />
fix x <br />
show "?P (H x)" by simp <br />
next<br />
fix x <br />
fix i assume h1: "?P i"<br />
fix d assume h2: "?P d"<br />
have "hd (inOrden (N x i d)) = hd ((inOrden i) @ [x] @ (inOrden d))" by simp<br />
also have "... = hd (inOrden i)" by (simp add: aux_ej12_1) <br />
(* Perdemos la x, luego se refuta el enunciado del teorema *)<br />
qed<br />
<br />
text {* <br />
--------------------------------------------------------------------- <br />
Ejercicio 17. Demostrar o refutar<br />
last (postOrden a) = raiz a<br />
--------------------------------------------------------------------- <br />
*}<br />
<br />
(* danrodcha *)<br />
theorem "last (postOrden a) = raiz a" (is "?P a")<br />
proof (induct a)<br />
fix x<br />
show "?P (H x)" by simp<br />
fix x i d<br />
show "?P (N x i d)" by simp<br />
qed<br />
<br />
(* danrodcha *)<br />
theorem "last (postOrden a) = raiz a" (is "?P a")<br />
proof (induct a)<br />
fix x<br />
show "?P (H x)" by simp<br />
fix x i d<br />
assume HIi: "?P i"<br />
assume HId: "?P d"<br />
have "last (postOrden (N x i d)) = last (postOrden i @ postOrden d @ [x])"<br />
by (simp only: postOrden.simps(2))<br />
also have "… = x" by simp<br />
also have "… = raiz (N x i d)" by (simp only: raiz.simps(2))<br />
finally show "?P (N x i d)" by simp<br />
qed<br />
<br />
(* pablucoto crigomgom ivamenjim marpoldia1 wilmorort*) (*Similar al anterior*)<br />
<br />
theorem "last (postOrden a) = raiz a" (is "?P a")<br />
proof (induct a )<br />
fix x<br />
show "?P (H x)" by simp<br />
next <br />
fix x i d<br />
assume HI1: "?P i"<br />
assume HI2: "?P d"<br />
have "last (postOrden (N x i d)) = last ( postOrden i @ postOrden d @ [x])" by simp<br />
also have "... = x" by simp<br />
also have "... = raiz (N x i d) " by simp<br />
finally show " ?P (N x i d)" by simp<br />
qed<br />
<br />
(* bowma *)<br />
(* También sin usar el supuesto "?p d" *)<br />
theorem "last (postOrden a) = raiz a" (is "?p a")<br />
proof (induct a)<br />
fix t<br />
show "?p (H t)" by simp<br />
next<br />
fix t i d<br />
assume "?p i"<br />
(* si quito este supuesto, hay error pero no sé dónde se lo está usando *)<br />
have "last (postOrden (N t i d)) = last (postOrden i @ postOrden d @ [t])" by simp<br />
also have "... = t" by simp<br />
also have "... = raiz (N t i d)" by simp<br />
finally show "?p (N t i d)" by simp<br />
qed<br />
<br />
(* migtermor *)<br />
theorem "last (postOrden a) = raiz a" (is "?P a")<br />
proof (induct a)<br />
fix h<br />
show "?P (H h)" by simp<br />
next<br />
fix n i d<br />
have "last (postOrden (N n (i :: 'a arbol) (d :: 'a arbol))) = <br />
last (postOrden i@postOrden d@[n])" by simp<br />
also have "… = raiz (N n i d)" by simp<br />
finally show "?P (N n i d)" by simp<br />
qed<br />
<br />
(* ivamenjim: sin usar patrones *)<br />
<br />
theorem "last (postOrden a) = raiz a" <br />
proof (induct a)<br />
fix x::"'a" <br />
show "last (postOrden (H x)) = raiz (H x)" by simp <br />
next<br />
fix x::"'a" <br />
fix i::"'a arbol" assume h1: "last (postOrden i) = raiz i"<br />
fix d::"'a arbol" assume h2: "last (postOrden d) = raiz d"<br />
have "last (postOrden (N x i d)) = last ((postOrden i) @ (postOrden d) @ [x])" by simp<br />
also have "... = last ([x])" by simp <br />
finally show "last (postOrden (N x i d)) = raiz (N x i d)" by simp<br />
qed<br />
<br />
end<br />
</source></div>Lucnovdoshttps://www.glc.us.es/~jalonso/RA2016/index.php?title=Relaci%C3%B3n_6&diff=966Relación 62016-12-06T01:26:43Z<p>Lucnovdos: </p>
<hr />
<div><source lang="isar"><br />
chapter {* R6: Recorridos de árboles *}<br />
<br />
theory R6_Recorridos_de_arboles<br />
imports Main <br />
begin <br />
<br />
text {* <br />
--------------------------------------------------------------------- <br />
Ejercicio 1. Definir el tipo de datos arbol para representar los<br />
árboles binarios que tiene información en los nodos y en las hojas. <br />
Por ejemplo, el árbol<br />
e<br />
/ \<br />
/ \<br />
c g<br />
/ \ / \<br />
a d f h <br />
se representa por "N e (N c (H a) (H d)) (N g (H f) (H h))".<br />
--------------------------------------------------------------------- <br />
*}<br />
<br />
(* ivamenjim marpoldia1 manmorjim1 bowma migtermor wilmorort *)<br />
<br />
datatype 'a arbol = H "'a" | N "'a" "'a arbol" "'a arbol"<br />
<br />
value "N e (N c (H a) (H d)) (N g (H f) (H h))" <br />
<br />
text {* <br />
--------------------------------------------------------------------- <br />
Ejercicio 2. Definir la función <br />
preOrden :: "'a arbol ⇒ 'a list"<br />
tal que (preOrden a) es el recorrido pre orden del árbol a. Por<br />
ejemplo, <br />
preOrden (N e (N c (H a) (H d)) (N g (H f) (H h)))<br />
= [e,c,a,d,g,f,h] <br />
--------------------------------------------------------------------- <br />
*}<br />
<br />
(* ivamenjim marpoldia1 pablucoto bowma fraortmoy migtermor wilmorort lucnovdos *)<br />
<br />
fun preOrden :: "'a arbol ⇒ 'a list" where<br />
"preOrden (H t) = [t]"<br />
| "preOrden (N t i d) = [t] @ (preOrden i) @ (preOrden d)"<br />
<br />
(* danrodcha crigomgom manmorjim1 bowma *)<br />
fun preOrden1 :: "'a arbol ⇒ 'a list" where<br />
"preOrden1 (H x) = [x]"<br />
| "preOrden1 (N x i d) = x#preOrden1 i @ preOrden1 d"<br />
<br />
value "preOrden (N e (N c (H a) (H d)) (N g (H f) (H h))) <br />
= [e,c,a,d,g,f,h]" <br />
value "preOrden1 (N e (N c (H a) (H d)) (N g (H f) (H h))) <br />
= [e,c,a,d,g,f,h]" <br />
<br />
(* danrodcha *)<br />
lemma "preOrden a = preOrden1 a"<br />
by (induct a) simp_all<br />
<br />
text {* <br />
--------------------------------------------------------------------- <br />
Ejercicio 3. Definir la función <br />
postOrden :: "'a arbol ⇒ 'a list"<br />
tal que (postOrden a) es el recorrido post orden del árbol a. Por<br />
ejemplo, <br />
postOrden (N e (N c (H a) (H d)) (N g (H f) (H h)))<br />
= [e,c,a,d,g,f,h] <br />
--------------------------------------------------------------------- <br />
*}<br />
<br />
(* ivamenjim danrodcha crigomgom marpoldia1 manmorjim1 pablucoto bowma fraortmoy migtermor wilmorort lucnovdos *)<br />
<br />
fun postOrden :: "'a arbol ⇒ 'a list" where<br />
"postOrden (H t) = [t]"<br />
| "postOrden (N t i d) = (postOrden i) @ (postOrden d) @ [t]"<br />
<br />
value "postOrden (N e (N c (H a) (H d)) (N g (H f) (H h)))<br />
= [a,d,c,f,h,g,e]"<br />
<br />
text {* <br />
--------------------------------------------------------------------- <br />
Ejercicio 4. Definir la función <br />
inOrden :: "'a arbol ⇒ 'a list"<br />
tal que (inOrden a) es el recorrido in orden del árbol a. Por<br />
ejemplo, <br />
inOrden (N e (N c (H a) (H d)) (N g (H f) (H h)))<br />
= [a,c,d,e,f,g,h]<br />
--------------------------------------------------------------------- <br />
*}<br />
<br />
(* ivamenjim crigomgom marpoldia1 pablucoto bowma fraortmoy migtermor wilmorort lucnovdos*)<br />
<br />
fun inOrden :: "'a arbol ⇒ 'a list" where<br />
"inOrden (H t) = [t]"<br />
| "inOrden (N t i d) = (inOrden i) @ [t] @ (inOrden d)"<br />
<br />
<br />
(* danrodcha manmorjim1 *)<br />
fun inOrden1 :: "'a arbol ⇒ 'a list" where<br />
"inOrden1 (H t) = [t]"<br />
| "inOrden1 (N t i d) = inOrden1 i @ t#inOrden1 d"<br />
<br />
value "inOrden (N e (N c (H a) (H d)) (N g (H f) (H h)))<br />
= [a,c,d,e,f,g,h]"<br />
value "inOrden1 (N e (N c (H a) (H d)) (N g (H f) (H h)))<br />
= [a,c,d,e,f,g,h]"<br />
<br />
(* manmorjim1 *)<br />
lemma "inOrden t = inOrden1 t"<br />
apply (induct t)<br />
apply auto<br />
done<br />
<br />
text {* <br />
--------------------------------------------------------------------- <br />
Ejercicio 5. Definir la función <br />
espejo :: "'a arbol ⇒ 'a arbol"<br />
tal que (espejo a) es la imagen especular del árbol a. Por ejemplo, <br />
espejo (N e (N c (H a) (H d)) (N g (H f) (H h)))<br />
= N e (N g (H h) (H f)) (N c (H d) (H a))<br />
--------------------------------------------------------------------- <br />
*}<br />
<br />
(* ivamenjim danrodcha crigomgom marpoldia1 manmorjim1 pablucoto bowma fraortmoy migtermor wilmorort lucnovdos *)<br />
<br />
fun espejo :: "'a arbol ⇒ 'a arbol" where<br />
"espejo (H t) = H t"<br />
| "espejo (N t i d) = N t (espejo d) (espejo i)"<br />
<br />
value "espejo (N e (N c (H a) (H d)) (N g (H f) (H h))) <br />
= N e (N g (H h) (H f)) (N c (H d) (H a))"<br />
<br />
<br />
text {* <br />
--------------------------------------------------------------------- <br />
Ejercicio 6. Demostrar que<br />
preOrden (espejo a) = rev (postOrden a)<br />
--------------------------------------------------------------------- <br />
*}<br />
<br />
(* ivamenjim migtermor wilmorort *)<br />
<br />
lemma "preOrden (espejo a) = rev (postOrden a)" (is "?P a")<br />
proof (induct a)<br />
fix x <br />
show "?P (H x)" by simp <br />
next<br />
fix x <br />
fix i assume h1: "?P i"<br />
fix d assume h2: "?P d"<br />
have "preOrden (espejo (N x i d)) = preOrden (N x (espejo d) (espejo i))" by simp<br />
also have "... = [x] @ (preOrden (espejo d)) @ (preOrden (espejo i))" by simp<br />
also have "... = [x] @ rev (postOrden d) @ rev (postOrden i)" using h1 h2 by simp <br />
finally show "preOrden (espejo (N x i d)) = rev (postOrden (N x i d))" by simp<br />
qed<br />
<br />
(* danrodcha *)<br />
lemma "preOrden (espejo a) = rev (postOrden a)"<br />
by (induct a) simp_all<br />
<br />
(* danrodcha crigomgom*)<br />
lemma "preOrden (espejo a) = rev (postOrden a)" (is "?P a")<br />
proof (induct a)<br />
fix x<br />
show "?P (H x)" by simp<br />
next<br />
fix x i d <br />
assume HIi: "?P i"<br />
assume HId: "?P d"<br />
have "preOrden (espejo (N x i d)) = preOrden (N x (espejo d) (espejo i))"<br />
by (simp only: espejo.simps(2))<br />
also have "… = x#preOrden (espejo d) @ preOrden (espejo i)"<br />
by (simp only: preOrden.simps(2))<br />
also have"… = x#rev (postOrden d) @ rev (postOrden i)" <br />
using HIi HId by simp<br />
also have "… = rev (postOrden (N x i d))" by simp<br />
finally show "?P (N x i d)" by simp<br />
qed<br />
<br />
(* danrodcha fraortmoy *)<br />
lemma "preOrden (espejo a) = rev (postOrden a)" (is "?P a")<br />
proof (induct a)<br />
fix x<br />
show "?P (H x)" by simp<br />
next<br />
fix x i d <br />
assume HIi: "?P i"<br />
assume HId: "?P d"<br />
show "?P (N x i d)" using HIi HId by simp<br />
qed<br />
<br />
(* bowma *)<br />
lemma "preOrden (espejo a) = rev (postOrden a)"<br />
apply (induct a)<br />
apply simp_all<br />
done<br />
<br />
(* pablucoto marpoldia1*)<br />
<br />
lemma "preOrden (espejo a) = rev (postOrden a)" (is "?P a")<br />
proof (induct a)<br />
fix x <br />
show "?P (H x)" by simp <br />
next<br />
fix x i d <br />
assume h1: "?P i"<br />
assume h2: "?P d"<br />
have "preOrden (espejo (N x i d)) = preOrden (N x (espejo d) (espejo i))" by simp<br />
also have "... = [x] @ (preOrden (espejo d)) @ (preOrden (espejo i))" by simp<br />
also have "... = [x] @ rev (postOrden d) @ rev (postOrden i)" using h1 h2 by simp<br />
also have "... = [x] @ rev (postOrden i @ postOrden d)" by simp <br />
also have "... = rev ( postOrden i @ postOrden d @ [x] ) " by simp<br />
also have "... = rev (postOrden (N x i d)) " by simp<br />
finally show "?P (N x i d)" by simp <br />
qed<br />
<br />
(* bowma *)<br />
lemma "preOrden (espejo a) = rev (postOrden a)" (is "?p a")<br />
proof (induct a)<br />
fix t<br />
show "?p (H t)" by simp<br />
(* Aquí si le diga "preOrden (espejo (H t)) = rev (postOrden (H t))",isabelle dice: <br />
proof (prove)<br />
goal (1 subgoal):<br />
1. preOrden (espejo (H t)) = rev (postOrden (H t)) <br />
Introduced fixed type variable(s): 'b in "t__" <br />
No entiendo porqué *)<br />
next <br />
fix t i d<br />
assume H1: "?p i"<br />
assume H2: "?p d"<br />
have "preOrden (espejo (N t i d)) = preOrden (N t (espejo d) (espejo i))" by simp<br />
also have "... = [t] @ (preOrden (espejo d)) @ (preOrden (espejo i))" by simp<br />
also have "... = [t] @ rev (postOrden d) @ rev (postOrden i)" using H1 H2 by simp<br />
finally show "?p (N t i d)" by simp<br />
qed <br />
<br />
(* fraortmoy lucnovdos *)<br />
<br />
lemma "preOrden (espejo a) = rev (postOrden a)"<br />
by (induct a) auto<br />
<br />
text {* <br />
--------------------------------------------------------------------- <br />
Ejercicio 7. Demostrar que<br />
postOrden (espejo a) = rev (preOrden a)<br />
--------------------------------------------------------------------- <br />
*}<br />
<br />
(* ivamenjim crigomgom bowma migtermor wilmorort*)<br />
<br />
lemma "postOrden (espejo a) = rev (preOrden a)" (is "?P a")<br />
proof (induct a)<br />
fix x <br />
show "?P (H x)" by simp <br />
next<br />
fix x <br />
fix i assume h1: "?P i"<br />
fix d assume h2: "?P d"<br />
have "postOrden (espejo (N x i d)) = postOrden (N x (espejo d) (espejo i))" by simp<br />
also have "... = (postOrden (espejo d)) @ (postOrden (espejo i)) @ [x]" by simp<br />
also have "... = rev (preOrden d) @ rev (preOrden i) @ [x]" using h1 h2 by simp <br />
finally show "postOrden (espejo (N x i d)) = rev (preOrden (N x i d))" by simp<br />
(* "?p (N x i d)" más corto *)<br />
qed<br />
<br />
(* danrodcha fraortmoy *)<br />
lemma "postOrden (espejo a) = rev (preOrden a)" (is "?P a")<br />
proof (induct a)<br />
fix x<br />
show "?P (H x)" by simp<br />
next<br />
fix x i d <br />
assume HIi: "?P i"<br />
assume HId: "?P d"<br />
show "?P (N x i d)" using HIi HId by simp<br />
qed<br />
<br />
(* pablucoto marpoldia1*)<br />
<br />
lemma "postOrden (espejo a) = rev (preOrden a)" (is "?P a")<br />
proof (induct a)<br />
fix x<br />
show "?P (H x)" by simp <br />
next <br />
fix x i d<br />
assume H1: "?P i"<br />
assume H2: "?P d"<br />
have " postOrden (espejo (N x i d)) = postOrden ( N x (espejo d) (espejo i)) " by simp<br />
also have "... = postOrden (espejo d) @ postOrden (espejo i) @ [x] " by simp<br />
also have "... = rev (preOrden d) @ rev (preOrden i) @ [x] " using H1 H2 by simp<br />
also have "... = rev (preOrden d) @ rev (x # preOrden i)" by simp<br />
also have "... = rev (x # preOrden i @ preOrden d)" by simp<br />
also have "... = rev (preOrden (N x i d)) " by simp<br />
finally show "?P (N x i d)" by simp<br />
qed<br />
<br />
(* fraortmoy lucnovdos *)<br />
lemma "postOrden (espejo a) = rev (preOrden a)"<br />
by (induct a) auto<br />
<br />
text {* <br />
--------------------------------------------------------------------- <br />
Ejercicio 8. Demostrar que<br />
inOrden (espejo a) = rev (inOrden a)<br />
--------------------------------------------------------------------- <br />
*}<br />
<br />
(* ivamenjim crigomgom bowma migtermor wilmorort*)<br />
<br />
theorem "inOrden (espejo a) = rev (inOrden a)" (is "?P a")<br />
proof (induct a)<br />
fix x <br />
show "?P (H x)" by simp <br />
next<br />
fix x <br />
fix i assume h1: "?P i"<br />
fix d assume h2: "?P d"<br />
have "inOrden (espejo (N x i d)) = inOrden (N x (espejo d) (espejo i))" by simp<br />
also have "... = (inOrden (espejo d)) @ [x] @ (inOrden (espejo i))" by simp<br />
also have "... = rev (inOrden d) @ [x] @ rev (inOrden i)" using h1 h2 by simp <br />
finally show "inOrden (espejo (N x i d)) = rev (inOrden (N x i d))" by simp<br />
qed<br />
<br />
(* danrodcha *)<br />
theorem "inOrden (espejo a) = rev (inOrden a)"<br />
by (induct a) simp_all<br />
<br />
(* danrodcha *)<br />
theorem "inOrden (espejo a) = rev (inOrden a)" (is "?P a")<br />
proof (induct a)<br />
fix x<br />
show "?P (H x)" by simp<br />
next<br />
fix x i d <br />
assume HIi: "?P i"<br />
assume HId: "?P d"<br />
show "?P (N x i d)" using HIi HId by simp<br />
qed<br />
<br />
(* pablucoto marpoldia1 *)<br />
theorem "inOrden (espejo a) = rev (inOrden a)" (is "?P a")<br />
proof (induct a)<br />
fix x<br />
show "?P (H x) " by simp<br />
next <br />
fix x i d <br />
assume HI1: "?P i"<br />
assume HI2: "?P d"<br />
have " inOrden (espejo (N x i d)) = inOrden ( N x (espejo d) (espejo i) )" by simp<br />
also have "... = inOrden (espejo d) @ [x] @ inOrden (espejo i) " by simp<br />
also have "... = rev (inOrden d) @ [x] @ rev (inOrden i)" using HI1 HI2 by simp<br />
also have "... = rev (x # inOrden d ) @ rev (inOrden i)" by simp<br />
also have "... = rev ( inOrden i @ x # inOrden d) " by simp<br />
also have "... = rev (inOrden (N x i d))" by simp<br />
finally show "?P (N x i d)" by simp<br />
qed<br />
<br />
(* lucnovdos *)<br />
theorem "inOrden (espejo a) = rev (inOrden a)"<br />
by (induct a) auto<br />
<br />
text {* <br />
--------------------------------------------------------------------- <br />
Ejercicio 9. Definir la función <br />
raiz :: "'a arbol ⇒ 'a"<br />
tal que (raiz a) es la raiz del árbol a. Por ejemplo, <br />
raiz (N e (N c (H a) (H d)) (N g (H f) (H h))) = e<br />
--------------------------------------------------------------------- <br />
*}<br />
<br />
(* danrodcha crigomgom manmorjim1 ivamenjim bowma pablucoto migtermor marpoldia1 wilmorort lucnovdos *)<br />
fun raiz :: "'a arbol ⇒ 'a" where<br />
"raiz (H x) = x"<br />
| "raiz (N x i d) = x"<br />
<br />
value "raiz (N e (N c (H a) (H d)) (N g (H f) (H h))) = e"<br />
<br />
text {* <br />
--------------------------------------------------------------------- <br />
Ejercicio 10. Definir la función <br />
extremo_izquierda :: "'a arbol ⇒ 'a"<br />
tal que (extremo_izquierda a) es el nodo más a la izquierda del árbol<br />
a. Por ejemplo, <br />
extremo_izquierda (N e (N c (H a) (H d)) (N g (H f) (H h))) = a<br />
--------------------------------------------------------------------- <br />
*}<br />
<br />
(* danrodcha crigomgom manmorjim1 ivamenjim bowma pablucoto migtermor marpoldia1 wilmorort *)<br />
fun extremo_izquierda :: "'a arbol ⇒ 'a" where<br />
"extremo_izquierda (H x) = x"<br />
| "extremo_izquierda (N x i d) = extremo_izquierda i"<br />
<br />
value "extremo_izquierda (N e (N c (H a) (H d)) (N g (H f) (H h))) = a"<br />
<br />
(* bowma *)<br />
fun extremo_izquierda_1 :: "'a arbol ⇒ 'a" where<br />
"extremo_izquierda_1 (H t) = t"<br />
| "extremo_izquierda_1 (N t i d) = hd (inOrden (N t i d))"<br />
<br />
(* ivamenjim *)<br />
(* Metaejercicio de demostración. Llamando teorema_13 al teorema del ejercicio 13 *)<br />
<br />
lemma "extremo_izquierda a = extremo_izquierda_1 a"<br />
by (induct a, simp_all add: aux_ej12_1 teorema_13)<br />
<br />
text {* <br />
--------------------------------------------------------------------- <br />
Ejercicio 11. Definir la función <br />
extremo_derecha :: "'a arbol ⇒ 'a"<br />
tal que (extremo_derecha a) es el nodo más a la derecha del árbol<br />
a. Por ejemplo, <br />
extremo_derecha (N e (N c (H a) (H d)) (N g (H f) (H h))) = h<br />
--------------------------------------------------------------------- <br />
*}<br />
<br />
(* danrodcha crigomgom manmorjim1 ivamenjim bowma pablucoto migtermor marpoldia1 wilmorort *)<br />
fun extremo_derecha :: "'a arbol ⇒ 'a" where<br />
"extremo_derecha (H x) = x"<br />
| "extremo_derecha (N x i d) = extremo_derecha d"<br />
<br />
value "extremo_derecha (N e (N c (H a) (H d)) (N g (H f) (H h))) = h"<br />
<br />
(* bowma *)<br />
fun extremo_derecha_1 :: "'a arbol ⇒ 'a" where<br />
"extremo_derecha_1 (H t) = t"<br />
| "extremo_derecha_1 (N t i d) = last (inOrden (N t i d))"<br />
<br />
(* ivamenjim *)<br />
(* Metaejercicio de demostración. Llamando teorema_12 al teorema del ejercicio 12 *)<br />
<br />
lemma "extremo_derecha a = extremo_derecha_1 a"<br />
by (induct a, simp_all add: aux_ej12_1 teorema_12)<br />
<br />
text {* <br />
--------------------------------------------------------------------- <br />
Ejercicio 12. Demostrar o refutar<br />
last (inOrden a) = extremo_derecha a<br />
--------------------------------------------------------------------- <br />
*}<br />
<br />
(* danrodcha *)<br />
lemma aux_ej12: "inOrden a ≠ []"<br />
apply (induct a) <br />
apply simp<br />
apply simp<br />
done<br />
<br />
(* danrodcha pablucoto crigomgom wilmorort*)<br />
theorem "last (inOrden a) = extremo_derecha a" (is "?P a")<br />
proof (induct a)<br />
fix x<br />
show "?P (H x)" by simp<br />
next<br />
fix x i d <br />
assume HIi: "?P i"<br />
assume HId: "?P d"<br />
have "last (inOrden (N x i d)) = last (inOrden i @ [x] @ inOrden d)" <br />
by (simp only: inOrden.simps(2))<br />
also have "… = last (inOrden d)" by (simp add: aux_ej12)<br />
also have "… = extremo_derecha d" using HId by simp<br />
also have "… = extremo_derecha (N x i d)" by simp<br />
finally show "?P (N x i d)" by simp<br />
qed<br />
<br />
(* ivamenjim *)<br />
<br />
lemma aux_ej12_1: "inOrden a ≠ []"<br />
by (induct a) simp_all <br />
<br />
(* ivamenjim marpoldia1 *)<br />
(* Igual que la anterior, pero poniendo solo by simp en el primer have *)<br />
<br />
theorem "last (inOrden a) = extremo_derecha a" (is "?P a")<br />
proof (induct a)<br />
fix x <br />
show "?P (H x)" by simp <br />
next<br />
fix x <br />
fix i assume h1: "?P i"<br />
fix d assume h2: "?P d"<br />
have "last (inOrden (N x i d)) = last ((inOrden i) @ [x] @ (inOrden d))" by simp <br />
also have "... = last (inOrden d)" by (simp add: aux_ej12_1)<br />
also have "... = extremo_derecha d" using h2 by simp <br />
finally show "last (inOrden (N x i d)) = extremo_derecha (N x i d)" by simp<br />
qed<br />
<br />
(* bowma *)<br />
(* Casi lo mismo que el anterior,pero no hace falta suponer "?p i" *)<br />
theorem "last (inOrden a) = extremo_derecha a" (is "?p a")<br />
proof (induct a)<br />
fix t<br />
show "?p (H t)" by simp<br />
next <br />
fix t i d<br />
assume HI: "?p d"<br />
have "last (inOrden (N t i d)) = last (inOrden i @ [t] @ inOrden d)" by simp<br />
also have "... = last (inOrden d)" by (simp add:aux_ej12)<br />
also have "... = extremo_derecha d" using HI by simp<br />
finally show "?p (N t i d)" by simp<br />
qed<br />
<br />
(* migtermor *)<br />
theorem "last (inOrden a) = extremo_derecha a" (is "?P a")<br />
proof (induct a)<br />
fix h<br />
show "?P (H h)" by simp<br />
next<br />
fix n i<br />
fix d assume HId: "?P d"<br />
have AUX: "¬ (inOrden d = [])" (is "?Q d")<br />
proof (induct d)<br />
fix hd<br />
show "?Q (H hd)" by simp<br />
next<br />
fix nd<br />
fix id assume HIid: "?Q id"<br />
fix dd assume HIdd: "?Q dd"<br />
show "?Q (N nd id dd)" using HIid HIdd by simp<br />
qed<br />
have "last (inOrden (N n i d)) = last (inOrden i @[n]@inOrden d)" by simp<br />
also have "… = last (inOrden d)" using AUX by simp<br />
also have "… = extremo_derecha d" using HId by simp<br />
finally show "?P (N n i d)" by simp<br />
qed<br />
<br />
text {* <br />
--------------------------------------------------------------------- <br />
Ejercicio 13. Demostrar o refutar<br />
hd (inOrden a) = extremo_izquierda a<br />
--------------------------------------------------------------------- <br />
*}<br />
<br />
(* danrodcha pablucoto crigomgom *)<br />
theorem "hd (inOrden a) = extremo_izquierda a" (is "?P a")<br />
proof (induct a)<br />
fix x<br />
show "?P (H x)" by simp<br />
next<br />
fix x i d <br />
assume HIi: "?P i"<br />
assume HId: "?P d"<br />
have "hd (inOrden (N x i d)) = hd (inOrden i @ [x] @ inOrden d)" <br />
by (simp only: inOrden.simps(2))<br />
also have "… = hd (inOrden i)" by (simp add: aux_ej12)<br />
also have "… = extremo_izquierda i" using HIi by simp<br />
also have "… = extremo_izquierda (N x i d)" by simp<br />
finally show "?P (N x i d)" by simp<br />
qed<br />
<br />
(* bowma *)<br />
theorem "hd (inOrden a) = extremo_izquierda a" (is "?p a")<br />
proof (induct a)<br />
fix t<br />
show "?p (H t)" by simp<br />
next<br />
fix t i d <br />
assume HI: "?p i"<br />
have "hd (inOrden (N t i d)) = hd (inOrden i @ [t] @ inOrden d)" by simp<br />
also have "… = hd (inOrden i)" by (simp add: aux_ej12)<br />
also have "… = extremo_izquierda i" using HI by simp<br />
finally show "?p (N t i d)" by simp<br />
qed<br />
<br />
(* migtermor *)<br />
theorem "hd (inOrden a) = extremo_izquierda a" (is "?P a")<br />
proof (induct a)<br />
fix h<br />
show "?P (H h)" by simp<br />
next<br />
fix n d<br />
fix i assume HId: "?P i"<br />
have AUX: "¬ (inOrden i = [])" (is "?Q i")<br />
proof (induct i)<br />
fix hi<br />
show "?Q (H hi)" by simp<br />
next<br />
fix ni<br />
fix ii assume HIid: "?Q ii"<br />
fix di assume HIdd: "?Q di"<br />
show "?Q (N ni ii di)" using HIid HIdd by simp<br />
qed<br />
have "hd (inOrden (N n i d)) = hd (inOrden i @[n]@inOrden d)" by simp<br />
also have "… = hd (inOrden i)" using AUX by simp<br />
also have "… = extremo_izquierda i" using HId by simp<br />
finally show "?P (N n i d)" by simp<br />
qed<br />
<br />
(* ivamenjim marpoldia1 wilmorort *)<br />
<br />
theorem "hd (inOrden a) = extremo_izquierda a" (is "?P a")<br />
proof (induct a)<br />
fix x <br />
show "?P (H x)" by simp <br />
next<br />
fix x <br />
fix i assume h1: "?P i"<br />
fix d assume h2: "?P d"<br />
have "hd (inOrden (N x i d)) = hd ((inOrden i) @ [x] @ (inOrden d))" by simp<br />
also have "... = hd (inOrden i)" by (simp add: aux_ej12_1)<br />
also have "... = extremo_izquierda i" using h1 by simp <br />
finally show "hd (inOrden (N x i d)) = extremo_izquierda (N x i d)" by simp<br />
qed<br />
<br />
text {* <br />
--------------------------------------------------------------------- <br />
Ejercicio 14. Demostrar o refutar<br />
hd (preOrden a) = last (postOrden a)<br />
--------------------------------------------------------------------- <br />
*}<br />
<br />
(* danrodcha *)<br />
theorem "hd (preOrden a) = last (postOrden a)" (is "?P a")<br />
proof (induct a)<br />
fix x<br />
show "?P (H x)" by simp<br />
fix x i d<br />
show "?P (N x i d)" by simp<br />
qed<br />
<br />
(* danrodcha *)<br />
theorem "hd (preOrden a) = last (postOrden a)" (is "?P a")<br />
proof (induct a)<br />
fix x<br />
show "?P (H x)" by simp<br />
fix x i d<br />
assume HIi: "?P i"<br />
assume HId: "?P d"<br />
have "hd (preOrden (N x i d)) = hd (x#preOrden i @ preOrden d)"<br />
by (simp only: preOrden.simps(2))<br />
also have "… = x" by simp<br />
also have "… = last (postOrden i @ postOrden d @ [x])" by simp<br />
also have "… = last (postOrden (N x i d))" <br />
by (simp only: postOrden.simps(2))<br />
finally show "?P (N x i d)" by simp<br />
qed<br />
<br />
(* pablucoto crigomgom bowma marpoldia1 wilmorort*) (*Similar al anterior*)<br />
theorem "hd (preOrden a) = last (postOrden a)" (is "?P a")<br />
proof (induct a)<br />
fix x<br />
show "?P (H x)" by simp<br />
next <br />
fix x i d<br />
assume HI1: "?P i"<br />
assume HI2: "?P d"<br />
have " hd (preOrden (N x i d)) = hd ([x] @ preOrden i @ preOrden d)" by simp<br />
also have "... = x" by simp<br />
also have "... = last ( postOrden i @ postOrden d @ [x]) " by simp<br />
also have "... = last ( postOrden (N x i d) )" by simp <br />
finally show "?P (N x i d)" by simp<br />
qed<br />
<br />
(* migtermor *)<br />
theorem "hd (preOrden a) = last (postOrden a)" (is "?P a")<br />
proof (induct a)<br />
fix h<br />
show "?P (H h)" by simp<br />
next<br />
fix n i d<br />
have "hd (preOrden (N n (i :: 'a arbol) (d :: 'a arbol))) = hd ([n]@preOrden i@preOrden d)" <br />
by simp<br />
(* Si no especifico que i y d son árboles, salta un error de tipo. Supongo que será por<br />
no haber asumido hipótesis sobre ellos *)<br />
also have "… = last (postOrden (N n i d))" by simp<br />
show "?P (N n i d)" by simp<br />
qed<br />
<br />
(* ivamenjim *)<br />
<br />
theorem "hd (preOrden a) = last (postOrden a)" (is "?P a")<br />
proof (induct a)<br />
fix x <br />
show "?P (H x)" by simp <br />
next<br />
fix x <br />
fix i assume h1: "?P i"<br />
fix d assume h2: "?P d"<br />
have "hd (preOrden (N x i d)) = hd ([x] @ (preOrden i) @ (preOrden d))" by simp<br />
also have "... = hd ([x])" by simp<br />
finally show "hd (preOrden (N x i d)) = last (postOrden (N x i d))" by simp<br />
qed<br />
<br />
text {* <br />
--------------------------------------------------------------------- <br />
Ejercicio 15. Demostrar o refutar<br />
hd (preOrden a) = raiz a<br />
--------------------------------------------------------------------- <br />
*}<br />
<br />
(* danrodcha *)<br />
theorem "hd (preOrden a) = raiz a" (is "?P a")<br />
proof (induct a)<br />
fix x<br />
show "?P (H x)" by simp<br />
fix x i d<br />
assume HIi: "?P i"<br />
assume HId: "?P d"<br />
have "hd (preOrden (N x i d)) = hd (x#preOrden i @ preOrden d)"<br />
by (simp only: preOrden.simps(2))<br />
also have "… = x" by simp<br />
also have "… = raiz (N x i d)" by simp<br />
finally show "?P (N x i d)" by simp<br />
qed<br />
<br />
(* danrodcha *)<br />
theorem "hd (preOrden a) = raiz a" (is "?P a")<br />
proof (induct a)<br />
fix x<br />
show "?P (H x)" by simp<br />
fix x i d<br />
show "?P (N x i d)" by simp<br />
qed<br />
<br />
(* pablucoto crigomgom ivamenjim marpoldia1 wilmorort *)<br />
theorem "hd (preOrden a) = raiz a" (is "?P a")<br />
proof (induct a) <br />
fix x<br />
show "?P (H x)" by simp<br />
next<br />
fix x i d<br />
assume HI1: " ?P i"<br />
assume HI2: " ?P d"<br />
have " hd (preOrden (N x i d)) = hd ([x] @ preOrden i @ preOrden d) " by simp<br />
also have "... = x" by simp<br />
also have "... = raiz (N x i d)" by simp<br />
finally show " ?P (N x i d)" by simp<br />
qed<br />
<br />
(* bowma *)<br />
(* similar al anterior pero sin suponer "?p d" *)<br />
theorem "hd (preOrden a) = last (postOrden a)" (is "?p a")<br />
proof (induct a)<br />
fix t<br />
show "?p (H t)" by simp<br />
next<br />
fix t i d<br />
assume HI: "?p i"<br />
have "hd (preOrden (N t i d)) = hd ([t] @ preOrden i @ preOrden d)" by simp<br />
also have "… = t" by simp<br />
also have "… = last (postOrden i @ postOrden d @ [t])" by simp<br />
also have "… = last (postOrden (N t i d))" by simp<br />
finally show "?p (N t i d)" by simp<br />
qed<br />
<br />
(* migtermor *)<br />
theorem "hd (preOrden a) = raiz a" (is "?P a")<br />
proof (induct a)<br />
fix h<br />
show "?P (H h)" by simp<br />
next<br />
fix n i d<br />
have "hd (preOrden (N n (i :: 'a arbol) (d :: 'a arbol))) = hd ([n]@preOrden i@preOrden d)"<br />
by simp<br />
also have "… = raiz (N n i d)" by simp<br />
finally show "?P (N n i d)" by simp<br />
qed<br />
<br />
(* ivamenjim: sin usar patrones *)<br />
<br />
theorem "hd (preOrden a) = raiz a" <br />
proof (induct a)<br />
fix x ::"'a"<br />
show "hd (preOrden (H x)) = raiz (H x)" by simp <br />
next<br />
fix x ::"'a"<br />
fix i ::"'a arbol" assume h1: "hd (preOrden i) = raiz i"<br />
fix d ::"'a arbol" assume h2: "hd (preOrden d) = raiz d"<br />
have "hd (preOrden (N x i d)) = hd ([x] @ (preOrden i) @ (preOrden d))" by simp<br />
also have "... = hd ([x])" by simp<br />
finally show "hd (preOrden (N x i d)) = raiz (N x i d)" by simp<br />
qed<br />
<br />
text {* <br />
--------------------------------------------------------------------- <br />
Ejercicio 16. Demostrar o refutar<br />
hd (inOrden a) = raiz a<br />
--------------------------------------------------------------------- <br />
*}<br />
<br />
(*crigomgom pablucoto bowma migtermor ivamenjim wilmorort *)<br />
theorem "hd (inOrden a) = raiz a"<br />
quickcheck<br />
oops<br />
<br />
(* danrodcha:<br />
Auto Quickcheck found a counterexample:<br />
a = N a⇩1 (H a⇩2) (H a⇩1)<br />
Evaluated terms:<br />
hd (inOrden a) = a⇩2<br />
raiz a = a⇩1 *)<br />
<br />
(* ivamenjim marpoldia1 *)<br />
<br />
theorem "hd (inOrden a) = raiz a" (is "?P a")<br />
proof (induct a)<br />
fix x <br />
show "?P (H x)" by simp <br />
next<br />
fix x <br />
fix i assume h1: "?P i"<br />
fix d assume h2: "?P d"<br />
have "hd (inOrden (N x i d)) = hd ((inOrden i) @ [x] @ (inOrden d))" by simp<br />
also have "... = hd (inOrden i)" by (simp add: aux_ej12_1) <br />
(* Perdemos la x, luego se refuta el enunciado del teorema *)<br />
qed<br />
<br />
text {* <br />
--------------------------------------------------------------------- <br />
Ejercicio 17. Demostrar o refutar<br />
last (postOrden a) = raiz a<br />
--------------------------------------------------------------------- <br />
*}<br />
<br />
(* danrodcha *)<br />
theorem "last (postOrden a) = raiz a" (is "?P a")<br />
proof (induct a)<br />
fix x<br />
show "?P (H x)" by simp<br />
fix x i d<br />
show "?P (N x i d)" by simp<br />
qed<br />
<br />
(* danrodcha *)<br />
theorem "last (postOrden a) = raiz a" (is "?P a")<br />
proof (induct a)<br />
fix x<br />
show "?P (H x)" by simp<br />
fix x i d<br />
assume HIi: "?P i"<br />
assume HId: "?P d"<br />
have "last (postOrden (N x i d)) = last (postOrden i @ postOrden d @ [x])"<br />
by (simp only: postOrden.simps(2))<br />
also have "… = x" by simp<br />
also have "… = raiz (N x i d)" by (simp only: raiz.simps(2))<br />
finally show "?P (N x i d)" by simp<br />
qed<br />
<br />
(* pablucoto crigomgom ivamenjim marpoldia1 wilmorort*) (*Similar al anterior*)<br />
<br />
theorem "last (postOrden a) = raiz a" (is "?P a")<br />
proof (induct a )<br />
fix x<br />
show "?P (H x)" by simp<br />
next <br />
fix x i d<br />
assume HI1: "?P i"<br />
assume HI2: "?P d"<br />
have "last (postOrden (N x i d)) = last ( postOrden i @ postOrden d @ [x])" by simp<br />
also have "... = x" by simp<br />
also have "... = raiz (N x i d) " by simp<br />
finally show " ?P (N x i d)" by simp<br />
qed<br />
<br />
(* bowma *)<br />
(* También sin usar el supuesto "?p d" *)<br />
theorem "last (postOrden a) = raiz a" (is "?p a")<br />
proof (induct a)<br />
fix t<br />
show "?p (H t)" by simp<br />
next<br />
fix t i d<br />
assume "?p i"<br />
(* si quito este supuesto, hay error pero no sé dónde se lo está usando *)<br />
have "last (postOrden (N t i d)) = last (postOrden i @ postOrden d @ [t])" by simp<br />
also have "... = t" by simp<br />
also have "... = raiz (N t i d)" by simp<br />
finally show "?p (N t i d)" by simp<br />
qed<br />
<br />
(* migtermor *)<br />
theorem "last (postOrden a) = raiz a" (is "?P a")<br />
proof (induct a)<br />
fix h<br />
show "?P (H h)" by simp<br />
next<br />
fix n i d<br />
have "last (postOrden (N n (i :: 'a arbol) (d :: 'a arbol))) = <br />
last (postOrden i@postOrden d@[n])" by simp<br />
also have "… = raiz (N n i d)" by simp<br />
finally show "?P (N n i d)" by simp<br />
qed<br />
<br />
(* ivamenjim: sin usar patrones *)<br />
<br />
theorem "last (postOrden a) = raiz a" <br />
proof (induct a)<br />
fix x::"'a" <br />
show "last (postOrden (H x)) = raiz (H x)" by simp <br />
next<br />
fix x::"'a" <br />
fix i::"'a arbol" assume h1: "last (postOrden i) = raiz i"<br />
fix d::"'a arbol" assume h2: "last (postOrden d) = raiz d"<br />
have "last (postOrden (N x i d)) = last ((postOrden i) @ (postOrden d) @ [x])" by simp<br />
also have "... = last ([x])" by simp <br />
finally show "last (postOrden (N x i d)) = raiz (N x i d)" by simp<br />
qed<br />
<br />
end<br />
</source></div>Lucnovdoshttps://www.glc.us.es/~jalonso/RA2016/index.php?title=Relaci%C3%B3n_6&diff=965Relación 62016-12-06T01:23:03Z<p>Lucnovdos: </p>
<hr />
<div><source lang="isar"><br />
chapter {* R6: Recorridos de árboles *}<br />
<br />
theory R6_Recorridos_de_arboles<br />
imports Main <br />
begin <br />
<br />
text {* <br />
--------------------------------------------------------------------- <br />
Ejercicio 1. Definir el tipo de datos arbol para representar los<br />
árboles binarios que tiene información en los nodos y en las hojas. <br />
Por ejemplo, el árbol<br />
e<br />
/ \<br />
/ \<br />
c g<br />
/ \ / \<br />
a d f h <br />
se representa por "N e (N c (H a) (H d)) (N g (H f) (H h))".<br />
--------------------------------------------------------------------- <br />
*}<br />
<br />
(* ivamenjim marpoldia1 manmorjim1 bowma migtermor wilmorort *)<br />
<br />
datatype 'a arbol = H "'a" | N "'a" "'a arbol" "'a arbol"<br />
<br />
value "N e (N c (H a) (H d)) (N g (H f) (H h))" <br />
<br />
text {* <br />
--------------------------------------------------------------------- <br />
Ejercicio 2. Definir la función <br />
preOrden :: "'a arbol ⇒ 'a list"<br />
tal que (preOrden a) es el recorrido pre orden del árbol a. Por<br />
ejemplo, <br />
preOrden (N e (N c (H a) (H d)) (N g (H f) (H h)))<br />
= [e,c,a,d,g,f,h] <br />
--------------------------------------------------------------------- <br />
*}<br />
<br />
(* ivamenjim marpoldia1 pablucoto bowma fraortmoy migtermor wilmorort lucnovdos *)<br />
<br />
fun preOrden :: "'a arbol ⇒ 'a list" where<br />
"preOrden (H t) = [t]"<br />
| "preOrden (N t i d) = [t] @ (preOrden i) @ (preOrden d)"<br />
<br />
(* danrodcha crigomgom manmorjim1 bowma *)<br />
fun preOrden1 :: "'a arbol ⇒ 'a list" where<br />
"preOrden1 (H x) = [x]"<br />
| "preOrden1 (N x i d) = x#preOrden1 i @ preOrden1 d"<br />
<br />
value "preOrden (N e (N c (H a) (H d)) (N g (H f) (H h))) <br />
= [e,c,a,d,g,f,h]" <br />
value "preOrden1 (N e (N c (H a) (H d)) (N g (H f) (H h))) <br />
= [e,c,a,d,g,f,h]" <br />
<br />
(* danrodcha *)<br />
lemma "preOrden a = preOrden1 a"<br />
by (induct a) simp_all<br />
<br />
text {* <br />
--------------------------------------------------------------------- <br />
Ejercicio 3. Definir la función <br />
postOrden :: "'a arbol ⇒ 'a list"<br />
tal que (postOrden a) es el recorrido post orden del árbol a. Por<br />
ejemplo, <br />
postOrden (N e (N c (H a) (H d)) (N g (H f) (H h)))<br />
= [e,c,a,d,g,f,h] <br />
--------------------------------------------------------------------- <br />
*}<br />
<br />
(* ivamenjim danrodcha crigomgom marpoldia1 manmorjim1 pablucoto bowma fraortmoy migtermor wilmorort lucnovdos *)<br />
<br />
fun postOrden :: "'a arbol ⇒ 'a list" where<br />
"postOrden (H t) = [t]"<br />
| "postOrden (N t i d) = (postOrden i) @ (postOrden d) @ [t]"<br />
<br />
value "postOrden (N e (N c (H a) (H d)) (N g (H f) (H h)))<br />
= [a,d,c,f,h,g,e]"<br />
<br />
text {* <br />
--------------------------------------------------------------------- <br />
Ejercicio 4. Definir la función <br />
inOrden :: "'a arbol ⇒ 'a list"<br />
tal que (inOrden a) es el recorrido in orden del árbol a. Por<br />
ejemplo, <br />
inOrden (N e (N c (H a) (H d)) (N g (H f) (H h)))<br />
= [a,c,d,e,f,g,h]<br />
--------------------------------------------------------------------- <br />
*}<br />
<br />
(* ivamenjim crigomgom marpoldia1 pablucoto bowma fraortmoy migtermor wilmorort lucnovdos*)<br />
<br />
fun inOrden :: "'a arbol ⇒ 'a list" where<br />
"inOrden (H t) = [t]"<br />
| "inOrden (N t i d) = (inOrden i) @ [t] @ (inOrden d)"<br />
<br />
<br />
(* danrodcha manmorjim1 *)<br />
fun inOrden1 :: "'a arbol ⇒ 'a list" where<br />
"inOrden1 (H t) = [t]"<br />
| "inOrden1 (N t i d) = inOrden1 i @ t#inOrden1 d"<br />
<br />
value "inOrden (N e (N c (H a) (H d)) (N g (H f) (H h)))<br />
= [a,c,d,e,f,g,h]"<br />
value "inOrden1 (N e (N c (H a) (H d)) (N g (H f) (H h)))<br />
= [a,c,d,e,f,g,h]"<br />
<br />
(* manmorjim1 *)<br />
lemma "inOrden t = inOrden1 t"<br />
apply (induct t)<br />
apply auto<br />
done<br />
<br />
text {* <br />
--------------------------------------------------------------------- <br />
Ejercicio 5. Definir la función <br />
espejo :: "'a arbol ⇒ 'a arbol"<br />
tal que (espejo a) es la imagen especular del árbol a. Por ejemplo, <br />
espejo (N e (N c (H a) (H d)) (N g (H f) (H h)))<br />
= N e (N g (H h) (H f)) (N c (H d) (H a))<br />
--------------------------------------------------------------------- <br />
*}<br />
<br />
(* ivamenjim danrodcha crigomgom marpoldia1 manmorjim1 pablucoto bowma fraortmoy migtermor wilmorort lucnovdos *)<br />
<br />
fun espejo :: "'a arbol ⇒ 'a arbol" where<br />
"espejo (H t) = H t"<br />
| "espejo (N t i d) = N t (espejo d) (espejo i)"<br />
<br />
value "espejo (N e (N c (H a) (H d)) (N g (H f) (H h))) <br />
= N e (N g (H h) (H f)) (N c (H d) (H a))"<br />
<br />
<br />
text {* <br />
--------------------------------------------------------------------- <br />
Ejercicio 6. Demostrar que<br />
preOrden (espejo a) = rev (postOrden a)<br />
--------------------------------------------------------------------- <br />
*}<br />
<br />
(* ivamenjim migtermor wilmorort *)<br />
<br />
lemma "preOrden (espejo a) = rev (postOrden a)" (is "?P a")<br />
proof (induct a)<br />
fix x <br />
show "?P (H x)" by simp <br />
next<br />
fix x <br />
fix i assume h1: "?P i"<br />
fix d assume h2: "?P d"<br />
have "preOrden (espejo (N x i d)) = preOrden (N x (espejo d) (espejo i))" by simp<br />
also have "... = [x] @ (preOrden (espejo d)) @ (preOrden (espejo i))" by simp<br />
also have "... = [x] @ rev (postOrden d) @ rev (postOrden i)" using h1 h2 by simp <br />
finally show "preOrden (espejo (N x i d)) = rev (postOrden (N x i d))" by simp<br />
qed<br />
<br />
(* danrodcha *)<br />
lemma "preOrden (espejo a) = rev (postOrden a)"<br />
by (induct a) simp_all<br />
<br />
(* danrodcha crigomgom*)<br />
lemma "preOrden (espejo a) = rev (postOrden a)" (is "?P a")<br />
proof (induct a)<br />
fix x<br />
show "?P (H x)" by simp<br />
next<br />
fix x i d <br />
assume HIi: "?P i"<br />
assume HId: "?P d"<br />
have "preOrden (espejo (N x i d)) = preOrden (N x (espejo d) (espejo i))"<br />
by (simp only: espejo.simps(2))<br />
also have "… = x#preOrden (espejo d) @ preOrden (espejo i)"<br />
by (simp only: preOrden.simps(2))<br />
also have"… = x#rev (postOrden d) @ rev (postOrden i)" <br />
using HIi HId by simp<br />
also have "… = rev (postOrden (N x i d))" by simp<br />
finally show "?P (N x i d)" by simp<br />
qed<br />
<br />
(* danrodcha fraortmoy *)<br />
lemma "preOrden (espejo a) = rev (postOrden a)" (is "?P a")<br />
proof (induct a)<br />
fix x<br />
show "?P (H x)" by simp<br />
next<br />
fix x i d <br />
assume HIi: "?P i"<br />
assume HId: "?P d"<br />
show "?P (N x i d)" using HIi HId by simp<br />
qed<br />
<br />
(* bowma *)<br />
lemma "preOrden (espejo a) = rev (postOrden a)"<br />
apply (induct a)<br />
apply simp_all<br />
done<br />
<br />
(* pablucoto marpoldia1*)<br />
<br />
lemma "preOrden (espejo a) = rev (postOrden a)" (is "?P a")<br />
proof (induct a)<br />
fix x <br />
show "?P (H x)" by simp <br />
next<br />
fix x i d <br />
assume h1: "?P i"<br />
assume h2: "?P d"<br />
have "preOrden (espejo (N x i d)) = preOrden (N x (espejo d) (espejo i))" by simp<br />
also have "... = [x] @ (preOrden (espejo d)) @ (preOrden (espejo i))" by simp<br />
also have "... = [x] @ rev (postOrden d) @ rev (postOrden i)" using h1 h2 by simp<br />
also have "... = [x] @ rev (postOrden i @ postOrden d)" by simp <br />
also have "... = rev ( postOrden i @ postOrden d @ [x] ) " by simp<br />
also have "... = rev (postOrden (N x i d)) " by simp<br />
finally show "?P (N x i d)" by simp <br />
qed<br />
<br />
(* bowma *)<br />
lemma "preOrden (espejo a) = rev (postOrden a)" (is "?p a")<br />
proof (induct a)<br />
fix t<br />
show "?p (H t)" by simp<br />
(* Aquí si le diga "preOrden (espejo (H t)) = rev (postOrden (H t))",isabelle dice: <br />
proof (prove)<br />
goal (1 subgoal):<br />
1. preOrden (espejo (H t)) = rev (postOrden (H t)) <br />
Introduced fixed type variable(s): 'b in "t__" <br />
No entiendo porqué *)<br />
next <br />
fix t i d<br />
assume H1: "?p i"<br />
assume H2: "?p d"<br />
have "preOrden (espejo (N t i d)) = preOrden (N t (espejo d) (espejo i))" by simp<br />
also have "... = [t] @ (preOrden (espejo d)) @ (preOrden (espejo i))" by simp<br />
also have "... = [t] @ rev (postOrden d) @ rev (postOrden i)" using H1 H2 by simp<br />
finally show "?p (N t i d)" by simp<br />
qed <br />
<br />
(* fraortmoy lucnovdos *)<br />
<br />
lemma "preOrden (espejo a) = rev (postOrden a)"<br />
by (induct a) auto<br />
<br />
text {* <br />
--------------------------------------------------------------------- <br />
Ejercicio 7. Demostrar que<br />
postOrden (espejo a) = rev (preOrden a)<br />
--------------------------------------------------------------------- <br />
*}<br />
<br />
(* ivamenjim crigomgom bowma migtermor wilmorort*)<br />
<br />
lemma "postOrden (espejo a) = rev (preOrden a)" (is "?P a")<br />
proof (induct a)<br />
fix x <br />
show "?P (H x)" by simp <br />
next<br />
fix x <br />
fix i assume h1: "?P i"<br />
fix d assume h2: "?P d"<br />
have "postOrden (espejo (N x i d)) = postOrden (N x (espejo d) (espejo i))" by simp<br />
also have "... = (postOrden (espejo d)) @ (postOrden (espejo i)) @ [x]" by simp<br />
also have "... = rev (preOrden d) @ rev (preOrden i) @ [x]" using h1 h2 by simp <br />
finally show "postOrden (espejo (N x i d)) = rev (preOrden (N x i d))" by simp<br />
(* "?p (N x i d)" más corto *)<br />
qed<br />
<br />
(* danrodcha fraortmoy *)<br />
lemma "postOrden (espejo a) = rev (preOrden a)" (is "?P a")<br />
proof (induct a)<br />
fix x<br />
show "?P (H x)" by simp<br />
next<br />
fix x i d <br />
assume HIi: "?P i"<br />
assume HId: "?P d"<br />
show "?P (N x i d)" using HIi HId by simp<br />
qed<br />
<br />
(* pablucoto marpoldia1*)<br />
<br />
lemma "postOrden (espejo a) = rev (preOrden a)" (is "?P a")<br />
proof (induct a)<br />
fix x<br />
show "?P (H x)" by simp <br />
next <br />
fix x i d<br />
assume H1: "?P i"<br />
assume H2: "?P d"<br />
have " postOrden (espejo (N x i d)) = postOrden ( N x (espejo d) (espejo i)) " by simp<br />
also have "... = postOrden (espejo d) @ postOrden (espejo i) @ [x] " by simp<br />
also have "... = rev (preOrden d) @ rev (preOrden i) @ [x] " using H1 H2 by simp<br />
also have "... = rev (preOrden d) @ rev (x # preOrden i)" by simp<br />
also have "... = rev (x # preOrden i @ preOrden d)" by simp<br />
also have "... = rev (preOrden (N x i d)) " by simp<br />
finally show "?P (N x i d)" by simp<br />
qed<br />
<br />
(* fraortmoy lucnovdos *)<br />
lemma "postOrden (espejo a) = rev (preOrden a)"<br />
by (induct a) auto<br />
<br />
text {* <br />
--------------------------------------------------------------------- <br />
Ejercicio 8. Demostrar que<br />
inOrden (espejo a) = rev (inOrden a)<br />
--------------------------------------------------------------------- <br />
*}<br />
<br />
(* ivamenjim crigomgom bowma migtermor wilmorort*)<br />
<br />
theorem "inOrden (espejo a) = rev (inOrden a)" (is "?P a")<br />
proof (induct a)<br />
fix x <br />
show "?P (H x)" by simp <br />
next<br />
fix x <br />
fix i assume h1: "?P i"<br />
fix d assume h2: "?P d"<br />
have "inOrden (espejo (N x i d)) = inOrden (N x (espejo d) (espejo i))" by simp<br />
also have "... = (inOrden (espejo d)) @ [x] @ (inOrden (espejo i))" by simp<br />
also have "... = rev (inOrden d) @ [x] @ rev (inOrden i)" using h1 h2 by simp <br />
finally show "inOrden (espejo (N x i d)) = rev (inOrden (N x i d))" by simp<br />
qed<br />
<br />
(* danrodcha *)<br />
theorem "inOrden (espejo a) = rev (inOrden a)"<br />
by (induct a) simp_all<br />
<br />
(* danrodcha *)<br />
theorem "inOrden (espejo a) = rev (inOrden a)" (is "?P a")<br />
proof (induct a)<br />
fix x<br />
show "?P (H x)" by simp<br />
next<br />
fix x i d <br />
assume HIi: "?P i"<br />
assume HId: "?P d"<br />
show "?P (N x i d)" using HIi HId by simp<br />
qed<br />
<br />
(* pablucoto marpoldia1 *)<br />
theorem "inOrden (espejo a) = rev (inOrden a)" (is "?P a")<br />
proof (induct a)<br />
fix x<br />
show "?P (H x) " by simp<br />
next <br />
fix x i d <br />
assume HI1: "?P i"<br />
assume HI2: "?P d"<br />
have " inOrden (espejo (N x i d)) = inOrden ( N x (espejo d) (espejo i) )" by simp<br />
also have "... = inOrden (espejo d) @ [x] @ inOrden (espejo i) " by simp<br />
also have "... = rev (inOrden d) @ [x] @ rev (inOrden i)" using HI1 HI2 by simp<br />
also have "... = rev (x # inOrden d ) @ rev (inOrden i)" by simp<br />
also have "... = rev ( inOrden i @ x # inOrden d) " by simp<br />
also have "... = rev (inOrden (N x i d))" by simp<br />
finally show "?P (N x i d)" by simp<br />
qed<br />
<br />
text {* <br />
--------------------------------------------------------------------- <br />
Ejercicio 9. Definir la función <br />
raiz :: "'a arbol ⇒ 'a"<br />
tal que (raiz a) es la raiz del árbol a. Por ejemplo, <br />
raiz (N e (N c (H a) (H d)) (N g (H f) (H h))) = e<br />
--------------------------------------------------------------------- <br />
*}<br />
<br />
(* danrodcha crigomgom manmorjim1 ivamenjim bowma pablucoto migtermor marpoldia1 wilmorort *)<br />
fun raiz :: "'a arbol ⇒ 'a" where<br />
"raiz (H x) = x"<br />
| "raiz (N x i d) = x"<br />
<br />
value "raiz (N e (N c (H a) (H d)) (N g (H f) (H h))) = e"<br />
<br />
text {* <br />
--------------------------------------------------------------------- <br />
Ejercicio 10. Definir la función <br />
extremo_izquierda :: "'a arbol ⇒ 'a"<br />
tal que (extremo_izquierda a) es el nodo más a la izquierda del árbol<br />
a. Por ejemplo, <br />
extremo_izquierda (N e (N c (H a) (H d)) (N g (H f) (H h))) = a<br />
--------------------------------------------------------------------- <br />
*}<br />
<br />
(* danrodcha crigomgom manmorjim1 ivamenjim bowma pablucoto migtermor marpoldia1 wilmorort *)<br />
fun extremo_izquierda :: "'a arbol ⇒ 'a" where<br />
"extremo_izquierda (H x) = x"<br />
| "extremo_izquierda (N x i d) = extremo_izquierda i"<br />
<br />
value "extremo_izquierda (N e (N c (H a) (H d)) (N g (H f) (H h))) = a"<br />
<br />
(* bowma *)<br />
fun extremo_izquierda_1 :: "'a arbol ⇒ 'a" where<br />
"extremo_izquierda_1 (H t) = t"<br />
| "extremo_izquierda_1 (N t i d) = hd (inOrden (N t i d))"<br />
<br />
(* ivamenjim *)<br />
(* Metaejercicio de demostración. Llamando teorema_13 al teorema del ejercicio 13 *)<br />
<br />
lemma "extremo_izquierda a = extremo_izquierda_1 a"<br />
by (induct a, simp_all add: aux_ej12_1 teorema_13)<br />
<br />
text {* <br />
--------------------------------------------------------------------- <br />
Ejercicio 11. Definir la función <br />
extremo_derecha :: "'a arbol ⇒ 'a"<br />
tal que (extremo_derecha a) es el nodo más a la derecha del árbol<br />
a. Por ejemplo, <br />
extremo_derecha (N e (N c (H a) (H d)) (N g (H f) (H h))) = h<br />
--------------------------------------------------------------------- <br />
*}<br />
<br />
(* danrodcha crigomgom manmorjim1 ivamenjim bowma pablucoto migtermor marpoldia1 wilmorort *)<br />
fun extremo_derecha :: "'a arbol ⇒ 'a" where<br />
"extremo_derecha (H x) = x"<br />
| "extremo_derecha (N x i d) = extremo_derecha d"<br />
<br />
value "extremo_derecha (N e (N c (H a) (H d)) (N g (H f) (H h))) = h"<br />
<br />
(* bowma *)<br />
fun extremo_derecha_1 :: "'a arbol ⇒ 'a" where<br />
"extremo_derecha_1 (H t) = t"<br />
| "extremo_derecha_1 (N t i d) = last (inOrden (N t i d))"<br />
<br />
(* ivamenjim *)<br />
(* Metaejercicio de demostración. Llamando teorema_12 al teorema del ejercicio 12 *)<br />
<br />
lemma "extremo_derecha a = extremo_derecha_1 a"<br />
by (induct a, simp_all add: aux_ej12_1 teorema_12)<br />
<br />
text {* <br />
--------------------------------------------------------------------- <br />
Ejercicio 12. Demostrar o refutar<br />
last (inOrden a) = extremo_derecha a<br />
--------------------------------------------------------------------- <br />
*}<br />
<br />
(* danrodcha *)<br />
lemma aux_ej12: "inOrden a ≠ []"<br />
apply (induct a) <br />
apply simp<br />
apply simp<br />
done<br />
<br />
(* danrodcha pablucoto crigomgom wilmorort*)<br />
theorem "last (inOrden a) = extremo_derecha a" (is "?P a")<br />
proof (induct a)<br />
fix x<br />
show "?P (H x)" by simp<br />
next<br />
fix x i d <br />
assume HIi: "?P i"<br />
assume HId: "?P d"<br />
have "last (inOrden (N x i d)) = last (inOrden i @ [x] @ inOrden d)" <br />
by (simp only: inOrden.simps(2))<br />
also have "… = last (inOrden d)" by (simp add: aux_ej12)<br />
also have "… = extremo_derecha d" using HId by simp<br />
also have "… = extremo_derecha (N x i d)" by simp<br />
finally show "?P (N x i d)" by simp<br />
qed<br />
<br />
(* ivamenjim *)<br />
<br />
lemma aux_ej12_1: "inOrden a ≠ []"<br />
by (induct a) simp_all <br />
<br />
(* ivamenjim marpoldia1 *)<br />
(* Igual que la anterior, pero poniendo solo by simp en el primer have *)<br />
<br />
theorem "last (inOrden a) = extremo_derecha a" (is "?P a")<br />
proof (induct a)<br />
fix x <br />
show "?P (H x)" by simp <br />
next<br />
fix x <br />
fix i assume h1: "?P i"<br />
fix d assume h2: "?P d"<br />
have "last (inOrden (N x i d)) = last ((inOrden i) @ [x] @ (inOrden d))" by simp <br />
also have "... = last (inOrden d)" by (simp add: aux_ej12_1)<br />
also have "... = extremo_derecha d" using h2 by simp <br />
finally show "last (inOrden (N x i d)) = extremo_derecha (N x i d)" by simp<br />
qed<br />
<br />
(* bowma *)<br />
(* Casi lo mismo que el anterior,pero no hace falta suponer "?p i" *)<br />
theorem "last (inOrden a) = extremo_derecha a" (is "?p a")<br />
proof (induct a)<br />
fix t<br />
show "?p (H t)" by simp<br />
next <br />
fix t i d<br />
assume HI: "?p d"<br />
have "last (inOrden (N t i d)) = last (inOrden i @ [t] @ inOrden d)" by simp<br />
also have "... = last (inOrden d)" by (simp add:aux_ej12)<br />
also have "... = extremo_derecha d" using HI by simp<br />
finally show "?p (N t i d)" by simp<br />
qed<br />
<br />
(* migtermor *)<br />
theorem "last (inOrden a) = extremo_derecha a" (is "?P a")<br />
proof (induct a)<br />
fix h<br />
show "?P (H h)" by simp<br />
next<br />
fix n i<br />
fix d assume HId: "?P d"<br />
have AUX: "¬ (inOrden d = [])" (is "?Q d")<br />
proof (induct d)<br />
fix hd<br />
show "?Q (H hd)" by simp<br />
next<br />
fix nd<br />
fix id assume HIid: "?Q id"<br />
fix dd assume HIdd: "?Q dd"<br />
show "?Q (N nd id dd)" using HIid HIdd by simp<br />
qed<br />
have "last (inOrden (N n i d)) = last (inOrden i @[n]@inOrden d)" by simp<br />
also have "… = last (inOrden d)" using AUX by simp<br />
also have "… = extremo_derecha d" using HId by simp<br />
finally show "?P (N n i d)" by simp<br />
qed<br />
<br />
text {* <br />
--------------------------------------------------------------------- <br />
Ejercicio 13. Demostrar o refutar<br />
hd (inOrden a) = extremo_izquierda a<br />
--------------------------------------------------------------------- <br />
*}<br />
<br />
(* danrodcha pablucoto crigomgom *)<br />
theorem "hd (inOrden a) = extremo_izquierda a" (is "?P a")<br />
proof (induct a)<br />
fix x<br />
show "?P (H x)" by simp<br />
next<br />
fix x i d <br />
assume HIi: "?P i"<br />
assume HId: "?P d"<br />
have "hd (inOrden (N x i d)) = hd (inOrden i @ [x] @ inOrden d)" <br />
by (simp only: inOrden.simps(2))<br />
also have "… = hd (inOrden i)" by (simp add: aux_ej12)<br />
also have "… = extremo_izquierda i" using HIi by simp<br />
also have "… = extremo_izquierda (N x i d)" by simp<br />
finally show "?P (N x i d)" by simp<br />
qed<br />
<br />
(* bowma *)<br />
theorem "hd (inOrden a) = extremo_izquierda a" (is "?p a")<br />
proof (induct a)<br />
fix t<br />
show "?p (H t)" by simp<br />
next<br />
fix t i d <br />
assume HI: "?p i"<br />
have "hd (inOrden (N t i d)) = hd (inOrden i @ [t] @ inOrden d)" by simp<br />
also have "… = hd (inOrden i)" by (simp add: aux_ej12)<br />
also have "… = extremo_izquierda i" using HI by simp<br />
finally show "?p (N t i d)" by simp<br />
qed<br />
<br />
(* migtermor *)<br />
theorem "hd (inOrden a) = extremo_izquierda a" (is "?P a")<br />
proof (induct a)<br />
fix h<br />
show "?P (H h)" by simp<br />
next<br />
fix n d<br />
fix i assume HId: "?P i"<br />
have AUX: "¬ (inOrden i = [])" (is "?Q i")<br />
proof (induct i)<br />
fix hi<br />
show "?Q (H hi)" by simp<br />
next<br />
fix ni<br />
fix ii assume HIid: "?Q ii"<br />
fix di assume HIdd: "?Q di"<br />
show "?Q (N ni ii di)" using HIid HIdd by simp<br />
qed<br />
have "hd (inOrden (N n i d)) = hd (inOrden i @[n]@inOrden d)" by simp<br />
also have "… = hd (inOrden i)" using AUX by simp<br />
also have "… = extremo_izquierda i" using HId by simp<br />
finally show "?P (N n i d)" by simp<br />
qed<br />
<br />
(* ivamenjim marpoldia1 wilmorort *)<br />
<br />
theorem "hd (inOrden a) = extremo_izquierda a" (is "?P a")<br />
proof (induct a)<br />
fix x <br />
show "?P (H x)" by simp <br />
next<br />
fix x <br />
fix i assume h1: "?P i"<br />
fix d assume h2: "?P d"<br />
have "hd (inOrden (N x i d)) = hd ((inOrden i) @ [x] @ (inOrden d))" by simp<br />
also have "... = hd (inOrden i)" by (simp add: aux_ej12_1)<br />
also have "... = extremo_izquierda i" using h1 by simp <br />
finally show "hd (inOrden (N x i d)) = extremo_izquierda (N x i d)" by simp<br />
qed<br />
<br />
text {* <br />
--------------------------------------------------------------------- <br />
Ejercicio 14. Demostrar o refutar<br />
hd (preOrden a) = last (postOrden a)<br />
--------------------------------------------------------------------- <br />
*}<br />
<br />
(* danrodcha *)<br />
theorem "hd (preOrden a) = last (postOrden a)" (is "?P a")<br />
proof (induct a)<br />
fix x<br />
show "?P (H x)" by simp<br />
fix x i d<br />
show "?P (N x i d)" by simp<br />
qed<br />
<br />
(* danrodcha *)<br />
theorem "hd (preOrden a) = last (postOrden a)" (is "?P a")<br />
proof (induct a)<br />
fix x<br />
show "?P (H x)" by simp<br />
fix x i d<br />
assume HIi: "?P i"<br />
assume HId: "?P d"<br />
have "hd (preOrden (N x i d)) = hd (x#preOrden i @ preOrden d)"<br />
by (simp only: preOrden.simps(2))<br />
also have "… = x" by simp<br />
also have "… = last (postOrden i @ postOrden d @ [x])" by simp<br />
also have "… = last (postOrden (N x i d))" <br />
by (simp only: postOrden.simps(2))<br />
finally show "?P (N x i d)" by simp<br />
qed<br />
<br />
(* pablucoto crigomgom bowma marpoldia1 wilmorort*) (*Similar al anterior*)<br />
theorem "hd (preOrden a) = last (postOrden a)" (is "?P a")<br />
proof (induct a)<br />
fix x<br />
show "?P (H x)" by simp<br />
next <br />
fix x i d<br />
assume HI1: "?P i"<br />
assume HI2: "?P d"<br />
have " hd (preOrden (N x i d)) = hd ([x] @ preOrden i @ preOrden d)" by simp<br />
also have "... = x" by simp<br />
also have "... = last ( postOrden i @ postOrden d @ [x]) " by simp<br />
also have "... = last ( postOrden (N x i d) )" by simp <br />
finally show "?P (N x i d)" by simp<br />
qed<br />
<br />
(* migtermor *)<br />
theorem "hd (preOrden a) = last (postOrden a)" (is "?P a")<br />
proof (induct a)<br />
fix h<br />
show "?P (H h)" by simp<br />
next<br />
fix n i d<br />
have "hd (preOrden (N n (i :: 'a arbol) (d :: 'a arbol))) = hd ([n]@preOrden i@preOrden d)" <br />
by simp<br />
(* Si no especifico que i y d son árboles, salta un error de tipo. Supongo que será por<br />
no haber asumido hipótesis sobre ellos *)<br />
also have "… = last (postOrden (N n i d))" by simp<br />
show "?P (N n i d)" by simp<br />
qed<br />
<br />
(* ivamenjim *)<br />
<br />
theorem "hd (preOrden a) = last (postOrden a)" (is "?P a")<br />
proof (induct a)<br />
fix x <br />
show "?P (H x)" by simp <br />
next<br />
fix x <br />
fix i assume h1: "?P i"<br />
fix d assume h2: "?P d"<br />
have "hd (preOrden (N x i d)) = hd ([x] @ (preOrden i) @ (preOrden d))" by simp<br />
also have "... = hd ([x])" by simp<br />
finally show "hd (preOrden (N x i d)) = last (postOrden (N x i d))" by simp<br />
qed<br />
<br />
text {* <br />
--------------------------------------------------------------------- <br />
Ejercicio 15. Demostrar o refutar<br />
hd (preOrden a) = raiz a<br />
--------------------------------------------------------------------- <br />
*}<br />
<br />
(* danrodcha *)<br />
theorem "hd (preOrden a) = raiz a" (is "?P a")<br />
proof (induct a)<br />
fix x<br />
show "?P (H x)" by simp<br />
fix x i d<br />
assume HIi: "?P i"<br />
assume HId: "?P d"<br />
have "hd (preOrden (N x i d)) = hd (x#preOrden i @ preOrden d)"<br />
by (simp only: preOrden.simps(2))<br />
also have "… = x" by simp<br />
also have "… = raiz (N x i d)" by simp<br />
finally show "?P (N x i d)" by simp<br />
qed<br />
<br />
(* danrodcha *)<br />
theorem "hd (preOrden a) = raiz a" (is "?P a")<br />
proof (induct a)<br />
fix x<br />
show "?P (H x)" by simp<br />
fix x i d<br />
show "?P (N x i d)" by simp<br />
qed<br />
<br />
(* pablucoto crigomgom ivamenjim marpoldia1 wilmorort *)<br />
theorem "hd (preOrden a) = raiz a" (is "?P a")<br />
proof (induct a) <br />
fix x<br />
show "?P (H x)" by simp<br />
next<br />
fix x i d<br />
assume HI1: " ?P i"<br />
assume HI2: " ?P d"<br />
have " hd (preOrden (N x i d)) = hd ([x] @ preOrden i @ preOrden d) " by simp<br />
also have "... = x" by simp<br />
also have "... = raiz (N x i d)" by simp<br />
finally show " ?P (N x i d)" by simp<br />
qed<br />
<br />
(* bowma *)<br />
(* similar al anterior pero sin suponer "?p d" *)<br />
theorem "hd (preOrden a) = last (postOrden a)" (is "?p a")<br />
proof (induct a)<br />
fix t<br />
show "?p (H t)" by simp<br />
next<br />
fix t i d<br />
assume HI: "?p i"<br />
have "hd (preOrden (N t i d)) = hd ([t] @ preOrden i @ preOrden d)" by simp<br />
also have "… = t" by simp<br />
also have "… = last (postOrden i @ postOrden d @ [t])" by simp<br />
also have "… = last (postOrden (N t i d))" by simp<br />
finally show "?p (N t i d)" by simp<br />
qed<br />
<br />
(* migtermor *)<br />
theorem "hd (preOrden a) = raiz a" (is "?P a")<br />
proof (induct a)<br />
fix h<br />
show "?P (H h)" by simp<br />
next<br />
fix n i d<br />
have "hd (preOrden (N n (i :: 'a arbol) (d :: 'a arbol))) = hd ([n]@preOrden i@preOrden d)"<br />
by simp<br />
also have "… = raiz (N n i d)" by simp<br />
finally show "?P (N n i d)" by simp<br />
qed<br />
<br />
(* ivamenjim: sin usar patrones *)<br />
<br />
theorem "hd (preOrden a) = raiz a" <br />
proof (induct a)<br />
fix x ::"'a"<br />
show "hd (preOrden (H x)) = raiz (H x)" by simp <br />
next<br />
fix x ::"'a"<br />
fix i ::"'a arbol" assume h1: "hd (preOrden i) = raiz i"<br />
fix d ::"'a arbol" assume h2: "hd (preOrden d) = raiz d"<br />
have "hd (preOrden (N x i d)) = hd ([x] @ (preOrden i) @ (preOrden d))" by simp<br />
also have "... = hd ([x])" by simp<br />
finally show "hd (preOrden (N x i d)) = raiz (N x i d)" by simp<br />
qed<br />
<br />
text {* <br />
--------------------------------------------------------------------- <br />
Ejercicio 16. Demostrar o refutar<br />
hd (inOrden a) = raiz a<br />
--------------------------------------------------------------------- <br />
*}<br />
<br />
(*crigomgom pablucoto bowma migtermor ivamenjim wilmorort *)<br />
theorem "hd (inOrden a) = raiz a"<br />
quickcheck<br />
oops<br />
<br />
(* danrodcha:<br />
Auto Quickcheck found a counterexample:<br />
a = N a⇩1 (H a⇩2) (H a⇩1)<br />
Evaluated terms:<br />
hd (inOrden a) = a⇩2<br />
raiz a = a⇩1 *)<br />
<br />
(* ivamenjim marpoldia1 *)<br />
<br />
theorem "hd (inOrden a) = raiz a" (is "?P a")<br />
proof (induct a)<br />
fix x <br />
show "?P (H x)" by simp <br />
next<br />
fix x <br />
fix i assume h1: "?P i"<br />
fix d assume h2: "?P d"<br />
have "hd (inOrden (N x i d)) = hd ((inOrden i) @ [x] @ (inOrden d))" by simp<br />
also have "... = hd (inOrden i)" by (simp add: aux_ej12_1) <br />
(* Perdemos la x, luego se refuta el enunciado del teorema *)<br />
qed<br />
<br />
text {* <br />
--------------------------------------------------------------------- <br />
Ejercicio 17. Demostrar o refutar<br />
last (postOrden a) = raiz a<br />
--------------------------------------------------------------------- <br />
*}<br />
<br />
(* danrodcha *)<br />
theorem "last (postOrden a) = raiz a" (is "?P a")<br />
proof (induct a)<br />
fix x<br />
show "?P (H x)" by simp<br />
fix x i d<br />
show "?P (N x i d)" by simp<br />
qed<br />
<br />
(* danrodcha *)<br />
theorem "last (postOrden a) = raiz a" (is "?P a")<br />
proof (induct a)<br />
fix x<br />
show "?P (H x)" by simp<br />
fix x i d<br />
assume HIi: "?P i"<br />
assume HId: "?P d"<br />
have "last (postOrden (N x i d)) = last (postOrden i @ postOrden d @ [x])"<br />
by (simp only: postOrden.simps(2))<br />
also have "… = x" by simp<br />
also have "… = raiz (N x i d)" by (simp only: raiz.simps(2))<br />
finally show "?P (N x i d)" by simp<br />
qed<br />
<br />
(* pablucoto crigomgom ivamenjim marpoldia1 wilmorort*) (*Similar al anterior*)<br />
<br />
theorem "last (postOrden a) = raiz a" (is "?P a")<br />
proof (induct a )<br />
fix x<br />
show "?P (H x)" by simp<br />
next <br />
fix x i d<br />
assume HI1: "?P i"<br />
assume HI2: "?P d"<br />
have "last (postOrden (N x i d)) = last ( postOrden i @ postOrden d @ [x])" by simp<br />
also have "... = x" by simp<br />
also have "... = raiz (N x i d) " by simp<br />
finally show " ?P (N x i d)" by simp<br />
qed<br />
<br />
(* bowma *)<br />
(* También sin usar el supuesto "?p d" *)<br />
theorem "last (postOrden a) = raiz a" (is "?p a")<br />
proof (induct a)<br />
fix t<br />
show "?p (H t)" by simp<br />
next<br />
fix t i d<br />
assume "?p i"<br />
(* si quito este supuesto, hay error pero no sé dónde se lo está usando *)<br />
have "last (postOrden (N t i d)) = last (postOrden i @ postOrden d @ [t])" by simp<br />
also have "... = t" by simp<br />
also have "... = raiz (N t i d)" by simp<br />
finally show "?p (N t i d)" by simp<br />
qed<br />
<br />
(* migtermor *)<br />
theorem "last (postOrden a) = raiz a" (is "?P a")<br />
proof (induct a)<br />
fix h<br />
show "?P (H h)" by simp<br />
next<br />
fix n i d<br />
have "last (postOrden (N n (i :: 'a arbol) (d :: 'a arbol))) = <br />
last (postOrden i@postOrden d@[n])" by simp<br />
also have "… = raiz (N n i d)" by simp<br />
finally show "?P (N n i d)" by simp<br />
qed<br />
<br />
(* ivamenjim: sin usar patrones *)<br />
<br />
theorem "last (postOrden a) = raiz a" <br />
proof (induct a)<br />
fix x::"'a" <br />
show "last (postOrden (H x)) = raiz (H x)" by simp <br />
next<br />
fix x::"'a" <br />
fix i::"'a arbol" assume h1: "last (postOrden i) = raiz i"<br />
fix d::"'a arbol" assume h2: "last (postOrden d) = raiz d"<br />
have "last (postOrden (N x i d)) = last ((postOrden i) @ (postOrden d) @ [x])" by simp<br />
also have "... = last ([x])" by simp <br />
finally show "last (postOrden (N x i d)) = raiz (N x i d)" by simp<br />
qed<br />
<br />
end<br />
</source></div>Lucnovdoshttps://www.glc.us.es/~jalonso/RA2016/index.php?title=Relaci%C3%B3n_6&diff=964Relación 62016-12-06T01:16:49Z<p>Lucnovdos: </p>
<hr />
<div><source lang="isar"><br />
chapter {* R6: Recorridos de árboles *}<br />
<br />
theory R6_Recorridos_de_arboles<br />
imports Main <br />
begin <br />
<br />
text {* <br />
--------------------------------------------------------------------- <br />
Ejercicio 1. Definir el tipo de datos arbol para representar los<br />
árboles binarios que tiene información en los nodos y en las hojas. <br />
Por ejemplo, el árbol<br />
e<br />
/ \<br />
/ \<br />
c g<br />
/ \ / \<br />
a d f h <br />
se representa por "N e (N c (H a) (H d)) (N g (H f) (H h))".<br />
--------------------------------------------------------------------- <br />
*}<br />
<br />
(* ivamenjim marpoldia1 manmorjim1 bowma migtermor wilmorort *)<br />
<br />
datatype 'a arbol = H "'a" | N "'a" "'a arbol" "'a arbol"<br />
<br />
value "N e (N c (H a) (H d)) (N g (H f) (H h))" <br />
<br />
text {* <br />
--------------------------------------------------------------------- <br />
Ejercicio 2. Definir la función <br />
preOrden :: "'a arbol ⇒ 'a list"<br />
tal que (preOrden a) es el recorrido pre orden del árbol a. Por<br />
ejemplo, <br />
preOrden (N e (N c (H a) (H d)) (N g (H f) (H h)))<br />
= [e,c,a,d,g,f,h] <br />
--------------------------------------------------------------------- <br />
*}<br />
<br />
(* ivamenjim marpoldia1 pablucoto bowma fraortmoy migtermor wilmorort lucnovdos *)<br />
<br />
fun preOrden :: "'a arbol ⇒ 'a list" where<br />
"preOrden (H t) = [t]"<br />
| "preOrden (N t i d) = [t] @ (preOrden i) @ (preOrden d)"<br />
<br />
(* danrodcha crigomgom manmorjim1 bowma *)<br />
fun preOrden1 :: "'a arbol ⇒ 'a list" where<br />
"preOrden1 (H x) = [x]"<br />
| "preOrden1 (N x i d) = x#preOrden1 i @ preOrden1 d"<br />
<br />
value "preOrden (N e (N c (H a) (H d)) (N g (H f) (H h))) <br />
= [e,c,a,d,g,f,h]" <br />
value "preOrden1 (N e (N c (H a) (H d)) (N g (H f) (H h))) <br />
= [e,c,a,d,g,f,h]" <br />
<br />
(* danrodcha *)<br />
lemma "preOrden a = preOrden1 a"<br />
by (induct a) simp_all<br />
<br />
text {* <br />
--------------------------------------------------------------------- <br />
Ejercicio 3. Definir la función <br />
postOrden :: "'a arbol ⇒ 'a list"<br />
tal que (postOrden a) es el recorrido post orden del árbol a. Por<br />
ejemplo, <br />
postOrden (N e (N c (H a) (H d)) (N g (H f) (H h)))<br />
= [e,c,a,d,g,f,h] <br />
--------------------------------------------------------------------- <br />
*}<br />
<br />
(* ivamenjim danrodcha crigomgom marpoldia1 manmorjim1 pablucoto bowma fraortmoy migtermor wilmorort lucnovdos *)<br />
<br />
fun postOrden :: "'a arbol ⇒ 'a list" where<br />
"postOrden (H t) = [t]"<br />
| "postOrden (N t i d) = (postOrden i) @ (postOrden d) @ [t]"<br />
<br />
value "postOrden (N e (N c (H a) (H d)) (N g (H f) (H h)))<br />
= [a,d,c,f,h,g,e]"<br />
<br />
text {* <br />
--------------------------------------------------------------------- <br />
Ejercicio 4. Definir la función <br />
inOrden :: "'a arbol ⇒ 'a list"<br />
tal que (inOrden a) es el recorrido in orden del árbol a. Por<br />
ejemplo, <br />
inOrden (N e (N c (H a) (H d)) (N g (H f) (H h)))<br />
= [a,c,d,e,f,g,h]<br />
--------------------------------------------------------------------- <br />
*}<br />
<br />
(* ivamenjim crigomgom marpoldia1 pablucoto bowma fraortmoy migtermor wilmorort lucnovdos*)<br />
<br />
fun inOrden :: "'a arbol ⇒ 'a list" where<br />
"inOrden (H t) = [t]"<br />
| "inOrden (N t i d) = (inOrden i) @ [t] @ (inOrden d)"<br />
<br />
<br />
(* danrodcha manmorjim1 *)<br />
fun inOrden1 :: "'a arbol ⇒ 'a list" where<br />
"inOrden1 (H t) = [t]"<br />
| "inOrden1 (N t i d) = inOrden1 i @ t#inOrden1 d"<br />
<br />
value "inOrden (N e (N c (H a) (H d)) (N g (H f) (H h)))<br />
= [a,c,d,e,f,g,h]"<br />
value "inOrden1 (N e (N c (H a) (H d)) (N g (H f) (H h)))<br />
= [a,c,d,e,f,g,h]"<br />
<br />
(* manmorjim1 *)<br />
lemma "inOrden t = inOrden1 t"<br />
apply (induct t)<br />
apply auto<br />
done<br />
<br />
text {* <br />
--------------------------------------------------------------------- <br />
Ejercicio 5. Definir la función <br />
espejo :: "'a arbol ⇒ 'a arbol"<br />
tal que (espejo a) es la imagen especular del árbol a. Por ejemplo, <br />
espejo (N e (N c (H a) (H d)) (N g (H f) (H h)))<br />
= N e (N g (H h) (H f)) (N c (H d) (H a))<br />
--------------------------------------------------------------------- <br />
*}<br />
<br />
(* ivamenjim danrodcha crigomgom marpoldia1 manmorjim1 pablucoto bowma fraortmoy migtermor wilmorort*)<br />
<br />
fun espejo :: "'a arbol ⇒ 'a arbol" where<br />
"espejo (H t) = H t"<br />
| "espejo (N t i d) = N t (espejo d) (espejo i)"<br />
<br />
value "espejo (N e (N c (H a) (H d)) (N g (H f) (H h))) <br />
= N e (N g (H h) (H f)) (N c (H d) (H a))"<br />
<br />
<br />
text {* <br />
--------------------------------------------------------------------- <br />
Ejercicio 6. Demostrar que<br />
preOrden (espejo a) = rev (postOrden a)<br />
--------------------------------------------------------------------- <br />
*}<br />
<br />
(* ivamenjim migtermor wilmorort *)<br />
<br />
lemma "preOrden (espejo a) = rev (postOrden a)" (is "?P a")<br />
proof (induct a)<br />
fix x <br />
show "?P (H x)" by simp <br />
next<br />
fix x <br />
fix i assume h1: "?P i"<br />
fix d assume h2: "?P d"<br />
have "preOrden (espejo (N x i d)) = preOrden (N x (espejo d) (espejo i))" by simp<br />
also have "... = [x] @ (preOrden (espejo d)) @ (preOrden (espejo i))" by simp<br />
also have "... = [x] @ rev (postOrden d) @ rev (postOrden i)" using h1 h2 by simp <br />
finally show "preOrden (espejo (N x i d)) = rev (postOrden (N x i d))" by simp<br />
qed<br />
<br />
(* danrodcha *)<br />
lemma "preOrden (espejo a) = rev (postOrden a)"<br />
by (induct a) simp_all<br />
<br />
(* danrodcha crigomgom*)<br />
lemma "preOrden (espejo a) = rev (postOrden a)" (is "?P a")<br />
proof (induct a)<br />
fix x<br />
show "?P (H x)" by simp<br />
next<br />
fix x i d <br />
assume HIi: "?P i"<br />
assume HId: "?P d"<br />
have "preOrden (espejo (N x i d)) = preOrden (N x (espejo d) (espejo i))"<br />
by (simp only: espejo.simps(2))<br />
also have "… = x#preOrden (espejo d) @ preOrden (espejo i)"<br />
by (simp only: preOrden.simps(2))<br />
also have"… = x#rev (postOrden d) @ rev (postOrden i)" <br />
using HIi HId by simp<br />
also have "… = rev (postOrden (N x i d))" by simp<br />
finally show "?P (N x i d)" by simp<br />
qed<br />
<br />
(* danrodcha fraortmoy *)<br />
lemma "preOrden (espejo a) = rev (postOrden a)" (is "?P a")<br />
proof (induct a)<br />
fix x<br />
show "?P (H x)" by simp<br />
next<br />
fix x i d <br />
assume HIi: "?P i"<br />
assume HId: "?P d"<br />
show "?P (N x i d)" using HIi HId by simp<br />
qed<br />
<br />
(* bowma *)<br />
lemma "preOrden (espejo a) = rev (postOrden a)"<br />
apply (induct a)<br />
apply simp_all<br />
done<br />
<br />
(* pablucoto marpoldia1*)<br />
<br />
lemma "preOrden (espejo a) = rev (postOrden a)" (is "?P a")<br />
proof (induct a)<br />
fix x <br />
show "?P (H x)" by simp <br />
next<br />
fix x i d <br />
assume h1: "?P i"<br />
assume h2: "?P d"<br />
have "preOrden (espejo (N x i d)) = preOrden (N x (espejo d) (espejo i))" by simp<br />
also have "... = [x] @ (preOrden (espejo d)) @ (preOrden (espejo i))" by simp<br />
also have "... = [x] @ rev (postOrden d) @ rev (postOrden i)" using h1 h2 by simp<br />
also have "... = [x] @ rev (postOrden i @ postOrden d)" by simp <br />
also have "... = rev ( postOrden i @ postOrden d @ [x] ) " by simp<br />
also have "... = rev (postOrden (N x i d)) " by simp<br />
finally show "?P (N x i d)" by simp <br />
qed<br />
<br />
(* bowma *)<br />
lemma "preOrden (espejo a) = rev (postOrden a)" (is "?p a")<br />
proof (induct a)<br />
fix t<br />
show "?p (H t)" by simp<br />
(* Aquí si le diga "preOrden (espejo (H t)) = rev (postOrden (H t))",isabelle dice: <br />
proof (prove)<br />
goal (1 subgoal):<br />
1. preOrden (espejo (H t)) = rev (postOrden (H t)) <br />
Introduced fixed type variable(s): 'b in "t__" <br />
No entiendo porqué *)<br />
next <br />
fix t i d<br />
assume H1: "?p i"<br />
assume H2: "?p d"<br />
have "preOrden (espejo (N t i d)) = preOrden (N t (espejo d) (espejo i))" by simp<br />
also have "... = [t] @ (preOrden (espejo d)) @ (preOrden (espejo i))" by simp<br />
also have "... = [t] @ rev (postOrden d) @ rev (postOrden i)" using H1 H2 by simp<br />
finally show "?p (N t i d)" by simp<br />
qed <br />
<br />
(* fraortmoy *)<br />
<br />
lemma "preOrden (espejo a) = rev (postOrden a)"<br />
by (induct a) auto<br />
<br />
text {* <br />
--------------------------------------------------------------------- <br />
Ejercicio 7. Demostrar que<br />
postOrden (espejo a) = rev (preOrden a)<br />
--------------------------------------------------------------------- <br />
*}<br />
<br />
(* ivamenjim crigomgom bowma migtermor wilmorort*)<br />
<br />
lemma "postOrden (espejo a) = rev (preOrden a)" (is "?P a")<br />
proof (induct a)<br />
fix x <br />
show "?P (H x)" by simp <br />
next<br />
fix x <br />
fix i assume h1: "?P i"<br />
fix d assume h2: "?P d"<br />
have "postOrden (espejo (N x i d)) = postOrden (N x (espejo d) (espejo i))" by simp<br />
also have "... = (postOrden (espejo d)) @ (postOrden (espejo i)) @ [x]" by simp<br />
also have "... = rev (preOrden d) @ rev (preOrden i) @ [x]" using h1 h2 by simp <br />
finally show "postOrden (espejo (N x i d)) = rev (preOrden (N x i d))" by simp<br />
(* "?p (N x i d)" más corto *)<br />
qed<br />
<br />
(* danrodcha fraortmoy *)<br />
lemma "postOrden (espejo a) = rev (preOrden a)" (is "?P a")<br />
proof (induct a)<br />
fix x<br />
show "?P (H x)" by simp<br />
next<br />
fix x i d <br />
assume HIi: "?P i"<br />
assume HId: "?P d"<br />
show "?P (N x i d)" using HIi HId by simp<br />
qed<br />
<br />
(* pablucoto marpoldia1*)<br />
<br />
lemma "postOrden (espejo a) = rev (preOrden a)" (is "?P a")<br />
proof (induct a)<br />
fix x<br />
show "?P (H x)" by simp <br />
next <br />
fix x i d<br />
assume H1: "?P i"<br />
assume H2: "?P d"<br />
have " postOrden (espejo (N x i d)) = postOrden ( N x (espejo d) (espejo i)) " by simp<br />
also have "... = postOrden (espejo d) @ postOrden (espejo i) @ [x] " by simp<br />
also have "... = rev (preOrden d) @ rev (preOrden i) @ [x] " using H1 H2 by simp<br />
also have "... = rev (preOrden d) @ rev (x # preOrden i)" by simp<br />
also have "... = rev (x # preOrden i @ preOrden d)" by simp<br />
also have "... = rev (preOrden (N x i d)) " by simp<br />
finally show "?P (N x i d)" by simp<br />
qed<br />
<br />
(* fraortmoy *)<br />
lemma "postOrden (espejo a) = rev (preOrden a)"<br />
by (induct a) auto<br />
<br />
text {* <br />
--------------------------------------------------------------------- <br />
Ejercicio 8. Demostrar que<br />
inOrden (espejo a) = rev (inOrden a)<br />
--------------------------------------------------------------------- <br />
*}<br />
<br />
(* ivamenjim crigomgom bowma migtermor wilmorort*)<br />
<br />
theorem "inOrden (espejo a) = rev (inOrden a)" (is "?P a")<br />
proof (induct a)<br />
fix x <br />
show "?P (H x)" by simp <br />
next<br />
fix x <br />
fix i assume h1: "?P i"<br />
fix d assume h2: "?P d"<br />
have "inOrden (espejo (N x i d)) = inOrden (N x (espejo d) (espejo i))" by simp<br />
also have "... = (inOrden (espejo d)) @ [x] @ (inOrden (espejo i))" by simp<br />
also have "... = rev (inOrden d) @ [x] @ rev (inOrden i)" using h1 h2 by simp <br />
finally show "inOrden (espejo (N x i d)) = rev (inOrden (N x i d))" by simp<br />
qed<br />
<br />
(* danrodcha *)<br />
theorem "inOrden (espejo a) = rev (inOrden a)"<br />
by (induct a) simp_all<br />
<br />
(* danrodcha *)<br />
theorem "inOrden (espejo a) = rev (inOrden a)" (is "?P a")<br />
proof (induct a)<br />
fix x<br />
show "?P (H x)" by simp<br />
next<br />
fix x i d <br />
assume HIi: "?P i"<br />
assume HId: "?P d"<br />
show "?P (N x i d)" using HIi HId by simp<br />
qed<br />
<br />
(* pablucoto marpoldia1 *)<br />
theorem "inOrden (espejo a) = rev (inOrden a)" (is "?P a")<br />
proof (induct a)<br />
fix x<br />
show "?P (H x) " by simp<br />
next <br />
fix x i d <br />
assume HI1: "?P i"<br />
assume HI2: "?P d"<br />
have " inOrden (espejo (N x i d)) = inOrden ( N x (espejo d) (espejo i) )" by simp<br />
also have "... = inOrden (espejo d) @ [x] @ inOrden (espejo i) " by simp<br />
also have "... = rev (inOrden d) @ [x] @ rev (inOrden i)" using HI1 HI2 by simp<br />
also have "... = rev (x # inOrden d ) @ rev (inOrden i)" by simp<br />
also have "... = rev ( inOrden i @ x # inOrden d) " by simp<br />
also have "... = rev (inOrden (N x i d))" by simp<br />
finally show "?P (N x i d)" by simp<br />
qed<br />
<br />
text {* <br />
--------------------------------------------------------------------- <br />
Ejercicio 9. Definir la función <br />
raiz :: "'a arbol ⇒ 'a"<br />
tal que (raiz a) es la raiz del árbol a. Por ejemplo, <br />
raiz (N e (N c (H a) (H d)) (N g (H f) (H h))) = e<br />
--------------------------------------------------------------------- <br />
*}<br />
<br />
(* danrodcha crigomgom manmorjim1 ivamenjim bowma pablucoto migtermor marpoldia1 wilmorort *)<br />
fun raiz :: "'a arbol ⇒ 'a" where<br />
"raiz (H x) = x"<br />
| "raiz (N x i d) = x"<br />
<br />
value "raiz (N e (N c (H a) (H d)) (N g (H f) (H h))) = e"<br />
<br />
text {* <br />
--------------------------------------------------------------------- <br />
Ejercicio 10. Definir la función <br />
extremo_izquierda :: "'a arbol ⇒ 'a"<br />
tal que (extremo_izquierda a) es el nodo más a la izquierda del árbol<br />
a. Por ejemplo, <br />
extremo_izquierda (N e (N c (H a) (H d)) (N g (H f) (H h))) = a<br />
--------------------------------------------------------------------- <br />
*}<br />
<br />
(* danrodcha crigomgom manmorjim1 ivamenjim bowma pablucoto migtermor marpoldia1 wilmorort *)<br />
fun extremo_izquierda :: "'a arbol ⇒ 'a" where<br />
"extremo_izquierda (H x) = x"<br />
| "extremo_izquierda (N x i d) = extremo_izquierda i"<br />
<br />
value "extremo_izquierda (N e (N c (H a) (H d)) (N g (H f) (H h))) = a"<br />
<br />
(* bowma *)<br />
fun extremo_izquierda_1 :: "'a arbol ⇒ 'a" where<br />
"extremo_izquierda_1 (H t) = t"<br />
| "extremo_izquierda_1 (N t i d) = hd (inOrden (N t i d))"<br />
<br />
(* ivamenjim *)<br />
(* Metaejercicio de demostración. Llamando teorema_13 al teorema del ejercicio 13 *)<br />
<br />
lemma "extremo_izquierda a = extremo_izquierda_1 a"<br />
by (induct a, simp_all add: aux_ej12_1 teorema_13)<br />
<br />
text {* <br />
--------------------------------------------------------------------- <br />
Ejercicio 11. Definir la función <br />
extremo_derecha :: "'a arbol ⇒ 'a"<br />
tal que (extremo_derecha a) es el nodo más a la derecha del árbol<br />
a. Por ejemplo, <br />
extremo_derecha (N e (N c (H a) (H d)) (N g (H f) (H h))) = h<br />
--------------------------------------------------------------------- <br />
*}<br />
<br />
(* danrodcha crigomgom manmorjim1 ivamenjim bowma pablucoto migtermor marpoldia1 wilmorort *)<br />
fun extremo_derecha :: "'a arbol ⇒ 'a" where<br />
"extremo_derecha (H x) = x"<br />
| "extremo_derecha (N x i d) = extremo_derecha d"<br />
<br />
value "extremo_derecha (N e (N c (H a) (H d)) (N g (H f) (H h))) = h"<br />
<br />
(* bowma *)<br />
fun extremo_derecha_1 :: "'a arbol ⇒ 'a" where<br />
"extremo_derecha_1 (H t) = t"<br />
| "extremo_derecha_1 (N t i d) = last (inOrden (N t i d))"<br />
<br />
(* ivamenjim *)<br />
(* Metaejercicio de demostración. Llamando teorema_12 al teorema del ejercicio 12 *)<br />
<br />
lemma "extremo_derecha a = extremo_derecha_1 a"<br />
by (induct a, simp_all add: aux_ej12_1 teorema_12)<br />
<br />
text {* <br />
--------------------------------------------------------------------- <br />
Ejercicio 12. Demostrar o refutar<br />
last (inOrden a) = extremo_derecha a<br />
--------------------------------------------------------------------- <br />
*}<br />
<br />
(* danrodcha *)<br />
lemma aux_ej12: "inOrden a ≠ []"<br />
apply (induct a) <br />
apply simp<br />
apply simp<br />
done<br />
<br />
(* danrodcha pablucoto crigomgom wilmorort*)<br />
theorem "last (inOrden a) = extremo_derecha a" (is "?P a")<br />
proof (induct a)<br />
fix x<br />
show "?P (H x)" by simp<br />
next<br />
fix x i d <br />
assume HIi: "?P i"<br />
assume HId: "?P d"<br />
have "last (inOrden (N x i d)) = last (inOrden i @ [x] @ inOrden d)" <br />
by (simp only: inOrden.simps(2))<br />
also have "… = last (inOrden d)" by (simp add: aux_ej12)<br />
also have "… = extremo_derecha d" using HId by simp<br />
also have "… = extremo_derecha (N x i d)" by simp<br />
finally show "?P (N x i d)" by simp<br />
qed<br />
<br />
(* ivamenjim *)<br />
<br />
lemma aux_ej12_1: "inOrden a ≠ []"<br />
by (induct a) simp_all <br />
<br />
(* ivamenjim marpoldia1 *)<br />
(* Igual que la anterior, pero poniendo solo by simp en el primer have *)<br />
<br />
theorem "last (inOrden a) = extremo_derecha a" (is "?P a")<br />
proof (induct a)<br />
fix x <br />
show "?P (H x)" by simp <br />
next<br />
fix x <br />
fix i assume h1: "?P i"<br />
fix d assume h2: "?P d"<br />
have "last (inOrden (N x i d)) = last ((inOrden i) @ [x] @ (inOrden d))" by simp <br />
also have "... = last (inOrden d)" by (simp add: aux_ej12_1)<br />
also have "... = extremo_derecha d" using h2 by simp <br />
finally show "last (inOrden (N x i d)) = extremo_derecha (N x i d)" by simp<br />
qed<br />
<br />
(* bowma *)<br />
(* Casi lo mismo que el anterior,pero no hace falta suponer "?p i" *)<br />
theorem "last (inOrden a) = extremo_derecha a" (is "?p a")<br />
proof (induct a)<br />
fix t<br />
show "?p (H t)" by simp<br />
next <br />
fix t i d<br />
assume HI: "?p d"<br />
have "last (inOrden (N t i d)) = last (inOrden i @ [t] @ inOrden d)" by simp<br />
also have "... = last (inOrden d)" by (simp add:aux_ej12)<br />
also have "... = extremo_derecha d" using HI by simp<br />
finally show "?p (N t i d)" by simp<br />
qed<br />
<br />
(* migtermor *)<br />
theorem "last (inOrden a) = extremo_derecha a" (is "?P a")<br />
proof (induct a)<br />
fix h<br />
show "?P (H h)" by simp<br />
next<br />
fix n i<br />
fix d assume HId: "?P d"<br />
have AUX: "¬ (inOrden d = [])" (is "?Q d")<br />
proof (induct d)<br />
fix hd<br />
show "?Q (H hd)" by simp<br />
next<br />
fix nd<br />
fix id assume HIid: "?Q id"<br />
fix dd assume HIdd: "?Q dd"<br />
show "?Q (N nd id dd)" using HIid HIdd by simp<br />
qed<br />
have "last (inOrden (N n i d)) = last (inOrden i @[n]@inOrden d)" by simp<br />
also have "… = last (inOrden d)" using AUX by simp<br />
also have "… = extremo_derecha d" using HId by simp<br />
finally show "?P (N n i d)" by simp<br />
qed<br />
<br />
text {* <br />
--------------------------------------------------------------------- <br />
Ejercicio 13. Demostrar o refutar<br />
hd (inOrden a) = extremo_izquierda a<br />
--------------------------------------------------------------------- <br />
*}<br />
<br />
(* danrodcha pablucoto crigomgom *)<br />
theorem "hd (inOrden a) = extremo_izquierda a" (is "?P a")<br />
proof (induct a)<br />
fix x<br />
show "?P (H x)" by simp<br />
next<br />
fix x i d <br />
assume HIi: "?P i"<br />
assume HId: "?P d"<br />
have "hd (inOrden (N x i d)) = hd (inOrden i @ [x] @ inOrden d)" <br />
by (simp only: inOrden.simps(2))<br />
also have "… = hd (inOrden i)" by (simp add: aux_ej12)<br />
also have "… = extremo_izquierda i" using HIi by simp<br />
also have "… = extremo_izquierda (N x i d)" by simp<br />
finally show "?P (N x i d)" by simp<br />
qed<br />
<br />
(* bowma *)<br />
theorem "hd (inOrden a) = extremo_izquierda a" (is "?p a")<br />
proof (induct a)<br />
fix t<br />
show "?p (H t)" by simp<br />
next<br />
fix t i d <br />
assume HI: "?p i"<br />
have "hd (inOrden (N t i d)) = hd (inOrden i @ [t] @ inOrden d)" by simp<br />
also have "… = hd (inOrden i)" by (simp add: aux_ej12)<br />
also have "… = extremo_izquierda i" using HI by simp<br />
finally show "?p (N t i d)" by simp<br />
qed<br />
<br />
(* migtermor *)<br />
theorem "hd (inOrden a) = extremo_izquierda a" (is "?P a")<br />
proof (induct a)<br />
fix h<br />
show "?P (H h)" by simp<br />
next<br />
fix n d<br />
fix i assume HId: "?P i"<br />
have AUX: "¬ (inOrden i = [])" (is "?Q i")<br />
proof (induct i)<br />
fix hi<br />
show "?Q (H hi)" by simp<br />
next<br />
fix ni<br />
fix ii assume HIid: "?Q ii"<br />
fix di assume HIdd: "?Q di"<br />
show "?Q (N ni ii di)" using HIid HIdd by simp<br />
qed<br />
have "hd (inOrden (N n i d)) = hd (inOrden i @[n]@inOrden d)" by simp<br />
also have "… = hd (inOrden i)" using AUX by simp<br />
also have "… = extremo_izquierda i" using HId by simp<br />
finally show "?P (N n i d)" by simp<br />
qed<br />
<br />
(* ivamenjim marpoldia1 wilmorort *)<br />
<br />
theorem "hd (inOrden a) = extremo_izquierda a" (is "?P a")<br />
proof (induct a)<br />
fix x <br />
show "?P (H x)" by simp <br />
next<br />
fix x <br />
fix i assume h1: "?P i"<br />
fix d assume h2: "?P d"<br />
have "hd (inOrden (N x i d)) = hd ((inOrden i) @ [x] @ (inOrden d))" by simp<br />
also have "... = hd (inOrden i)" by (simp add: aux_ej12_1)<br />
also have "... = extremo_izquierda i" using h1 by simp <br />
finally show "hd (inOrden (N x i d)) = extremo_izquierda (N x i d)" by simp<br />
qed<br />
<br />
text {* <br />
--------------------------------------------------------------------- <br />
Ejercicio 14. Demostrar o refutar<br />
hd (preOrden a) = last (postOrden a)<br />
--------------------------------------------------------------------- <br />
*}<br />
<br />
(* danrodcha *)<br />
theorem "hd (preOrden a) = last (postOrden a)" (is "?P a")<br />
proof (induct a)<br />
fix x<br />
show "?P (H x)" by simp<br />
fix x i d<br />
show "?P (N x i d)" by simp<br />
qed<br />
<br />
(* danrodcha *)<br />
theorem "hd (preOrden a) = last (postOrden a)" (is "?P a")<br />
proof (induct a)<br />
fix x<br />
show "?P (H x)" by simp<br />
fix x i d<br />
assume HIi: "?P i"<br />
assume HId: "?P d"<br />
have "hd (preOrden (N x i d)) = hd (x#preOrden i @ preOrden d)"<br />
by (simp only: preOrden.simps(2))<br />
also have "… = x" by simp<br />
also have "… = last (postOrden i @ postOrden d @ [x])" by simp<br />
also have "… = last (postOrden (N x i d))" <br />
by (simp only: postOrden.simps(2))<br />
finally show "?P (N x i d)" by simp<br />
qed<br />
<br />
(* pablucoto crigomgom bowma marpoldia1 wilmorort*) (*Similar al anterior*)<br />
theorem "hd (preOrden a) = last (postOrden a)" (is "?P a")<br />
proof (induct a)<br />
fix x<br />
show "?P (H x)" by simp<br />
next <br />
fix x i d<br />
assume HI1: "?P i"<br />
assume HI2: "?P d"<br />
have " hd (preOrden (N x i d)) = hd ([x] @ preOrden i @ preOrden d)" by simp<br />
also have "... = x" by simp<br />
also have "... = last ( postOrden i @ postOrden d @ [x]) " by simp<br />
also have "... = last ( postOrden (N x i d) )" by simp <br />
finally show "?P (N x i d)" by simp<br />
qed<br />
<br />
(* migtermor *)<br />
theorem "hd (preOrden a) = last (postOrden a)" (is "?P a")<br />
proof (induct a)<br />
fix h<br />
show "?P (H h)" by simp<br />
next<br />
fix n i d<br />
have "hd (preOrden (N n (i :: 'a arbol) (d :: 'a arbol))) = hd ([n]@preOrden i@preOrden d)" <br />
by simp<br />
(* Si no especifico que i y d son árboles, salta un error de tipo. Supongo que será por<br />
no haber asumido hipótesis sobre ellos *)<br />
also have "… = last (postOrden (N n i d))" by simp<br />
show "?P (N n i d)" by simp<br />
qed<br />
<br />
(* ivamenjim *)<br />
<br />
theorem "hd (preOrden a) = last (postOrden a)" (is "?P a")<br />
proof (induct a)<br />
fix x <br />
show "?P (H x)" by simp <br />
next<br />
fix x <br />
fix i assume h1: "?P i"<br />
fix d assume h2: "?P d"<br />
have "hd (preOrden (N x i d)) = hd ([x] @ (preOrden i) @ (preOrden d))" by simp<br />
also have "... = hd ([x])" by simp<br />
finally show "hd (preOrden (N x i d)) = last (postOrden (N x i d))" by simp<br />
qed<br />
<br />
text {* <br />
--------------------------------------------------------------------- <br />
Ejercicio 15. Demostrar o refutar<br />
hd (preOrden a) = raiz a<br />
--------------------------------------------------------------------- <br />
*}<br />
<br />
(* danrodcha *)<br />
theorem "hd (preOrden a) = raiz a" (is "?P a")<br />
proof (induct a)<br />
fix x<br />
show "?P (H x)" by simp<br />
fix x i d<br />
assume HIi: "?P i"<br />
assume HId: "?P d"<br />
have "hd (preOrden (N x i d)) = hd (x#preOrden i @ preOrden d)"<br />
by (simp only: preOrden.simps(2))<br />
also have "… = x" by simp<br />
also have "… = raiz (N x i d)" by simp<br />
finally show "?P (N x i d)" by simp<br />
qed<br />
<br />
(* danrodcha *)<br />
theorem "hd (preOrden a) = raiz a" (is "?P a")<br />
proof (induct a)<br />
fix x<br />
show "?P (H x)" by simp<br />
fix x i d<br />
show "?P (N x i d)" by simp<br />
qed<br />
<br />
(* pablucoto crigomgom ivamenjim marpoldia1 wilmorort *)<br />
theorem "hd (preOrden a) = raiz a" (is "?P a")<br />
proof (induct a) <br />
fix x<br />
show "?P (H x)" by simp<br />
next<br />
fix x i d<br />
assume HI1: " ?P i"<br />
assume HI2: " ?P d"<br />
have " hd (preOrden (N x i d)) = hd ([x] @ preOrden i @ preOrden d) " by simp<br />
also have "... = x" by simp<br />
also have "... = raiz (N x i d)" by simp<br />
finally show " ?P (N x i d)" by simp<br />
qed<br />
<br />
(* bowma *)<br />
(* similar al anterior pero sin suponer "?p d" *)<br />
theorem "hd (preOrden a) = last (postOrden a)" (is "?p a")<br />
proof (induct a)<br />
fix t<br />
show "?p (H t)" by simp<br />
next<br />
fix t i d<br />
assume HI: "?p i"<br />
have "hd (preOrden (N t i d)) = hd ([t] @ preOrden i @ preOrden d)" by simp<br />
also have "… = t" by simp<br />
also have "… = last (postOrden i @ postOrden d @ [t])" by simp<br />
also have "… = last (postOrden (N t i d))" by simp<br />
finally show "?p (N t i d)" by simp<br />
qed<br />
<br />
(* migtermor *)<br />
theorem "hd (preOrden a) = raiz a" (is "?P a")<br />
proof (induct a)<br />
fix h<br />
show "?P (H h)" by simp<br />
next<br />
fix n i d<br />
have "hd (preOrden (N n (i :: 'a arbol) (d :: 'a arbol))) = hd ([n]@preOrden i@preOrden d)"<br />
by simp<br />
also have "… = raiz (N n i d)" by simp<br />
finally show "?P (N n i d)" by simp<br />
qed<br />
<br />
(* ivamenjim: sin usar patrones *)<br />
<br />
theorem "hd (preOrden a) = raiz a" <br />
proof (induct a)<br />
fix x ::"'a"<br />
show "hd (preOrden (H x)) = raiz (H x)" by simp <br />
next<br />
fix x ::"'a"<br />
fix i ::"'a arbol" assume h1: "hd (preOrden i) = raiz i"<br />
fix d ::"'a arbol" assume h2: "hd (preOrden d) = raiz d"<br />
have "hd (preOrden (N x i d)) = hd ([x] @ (preOrden i) @ (preOrden d))" by simp<br />
also have "... = hd ([x])" by simp<br />
finally show "hd (preOrden (N x i d)) = raiz (N x i d)" by simp<br />
qed<br />
<br />
text {* <br />
--------------------------------------------------------------------- <br />
Ejercicio 16. Demostrar o refutar<br />
hd (inOrden a) = raiz a<br />
--------------------------------------------------------------------- <br />
*}<br />
<br />
(*crigomgom pablucoto bowma migtermor ivamenjim wilmorort *)<br />
theorem "hd (inOrden a) = raiz a"<br />
quickcheck<br />
oops<br />
<br />
(* danrodcha:<br />
Auto Quickcheck found a counterexample:<br />
a = N a⇩1 (H a⇩2) (H a⇩1)<br />
Evaluated terms:<br />
hd (inOrden a) = a⇩2<br />
raiz a = a⇩1 *)<br />
<br />
(* ivamenjim marpoldia1 *)<br />
<br />
theorem "hd (inOrden a) = raiz a" (is "?P a")<br />
proof (induct a)<br />
fix x <br />
show "?P (H x)" by simp <br />
next<br />
fix x <br />
fix i assume h1: "?P i"<br />
fix d assume h2: "?P d"<br />
have "hd (inOrden (N x i d)) = hd ((inOrden i) @ [x] @ (inOrden d))" by simp<br />
also have "... = hd (inOrden i)" by (simp add: aux_ej12_1) <br />
(* Perdemos la x, luego se refuta el enunciado del teorema *)<br />
qed<br />
<br />
text {* <br />
--------------------------------------------------------------------- <br />
Ejercicio 17. Demostrar o refutar<br />
last (postOrden a) = raiz a<br />
--------------------------------------------------------------------- <br />
*}<br />
<br />
(* danrodcha *)<br />
theorem "last (postOrden a) = raiz a" (is "?P a")<br />
proof (induct a)<br />
fix x<br />
show "?P (H x)" by simp<br />
fix x i d<br />
show "?P (N x i d)" by simp<br />
qed<br />
<br />
(* danrodcha *)<br />
theorem "last (postOrden a) = raiz a" (is "?P a")<br />
proof (induct a)<br />
fix x<br />
show "?P (H x)" by simp<br />
fix x i d<br />
assume HIi: "?P i"<br />
assume HId: "?P d"<br />
have "last (postOrden (N x i d)) = last (postOrden i @ postOrden d @ [x])"<br />
by (simp only: postOrden.simps(2))<br />
also have "… = x" by simp<br />
also have "… = raiz (N x i d)" by (simp only: raiz.simps(2))<br />
finally show "?P (N x i d)" by simp<br />
qed<br />
<br />
(* pablucoto crigomgom ivamenjim marpoldia1 wilmorort*) (*Similar al anterior*)<br />
<br />
theorem "last (postOrden a) = raiz a" (is "?P a")<br />
proof (induct a )<br />
fix x<br />
show "?P (H x)" by simp<br />
next <br />
fix x i d<br />
assume HI1: "?P i"<br />
assume HI2: "?P d"<br />
have "last (postOrden (N x i d)) = last ( postOrden i @ postOrden d @ [x])" by simp<br />
also have "... = x" by simp<br />
also have "... = raiz (N x i d) " by simp<br />
finally show " ?P (N x i d)" by simp<br />
qed<br />
<br />
(* bowma *)<br />
(* También sin usar el supuesto "?p d" *)<br />
theorem "last (postOrden a) = raiz a" (is "?p a")<br />
proof (induct a)<br />
fix t<br />
show "?p (H t)" by simp<br />
next<br />
fix t i d<br />
assume "?p i"<br />
(* si quito este supuesto, hay error pero no sé dónde se lo está usando *)<br />
have "last (postOrden (N t i d)) = last (postOrden i @ postOrden d @ [t])" by simp<br />
also have "... = t" by simp<br />
also have "... = raiz (N t i d)" by simp<br />
finally show "?p (N t i d)" by simp<br />
qed<br />
<br />
(* migtermor *)<br />
theorem "last (postOrden a) = raiz a" (is "?P a")<br />
proof (induct a)<br />
fix h<br />
show "?P (H h)" by simp<br />
next<br />
fix n i d<br />
have "last (postOrden (N n (i :: 'a arbol) (d :: 'a arbol))) = <br />
last (postOrden i@postOrden d@[n])" by simp<br />
also have "… = raiz (N n i d)" by simp<br />
finally show "?P (N n i d)" by simp<br />
qed<br />
<br />
(* ivamenjim: sin usar patrones *)<br />
<br />
theorem "last (postOrden a) = raiz a" <br />
proof (induct a)<br />
fix x::"'a" <br />
show "last (postOrden (H x)) = raiz (H x)" by simp <br />
next<br />
fix x::"'a" <br />
fix i::"'a arbol" assume h1: "last (postOrden i) = raiz i"<br />
fix d::"'a arbol" assume h2: "last (postOrden d) = raiz d"<br />
have "last (postOrden (N x i d)) = last ((postOrden i) @ (postOrden d) @ [x])" by simp<br />
also have "... = last ([x])" by simp <br />
finally show "last (postOrden (N x i d)) = raiz (N x i d)" by simp<br />
qed<br />
<br />
end<br />
</source></div>Lucnovdoshttps://www.glc.us.es/~jalonso/RA2016/index.php?title=Relaci%C3%B3n_6&diff=963Relación 62016-12-06T01:06:00Z<p>Lucnovdos: </p>
<hr />
<div><source lang="isar"><br />
chapter {* R6: Recorridos de árboles *}<br />
<br />
theory R6_Recorridos_de_arboles<br />
imports Main <br />
begin <br />
<br />
text {* <br />
--------------------------------------------------------------------- <br />
Ejercicio 1. Definir el tipo de datos arbol para representar los<br />
árboles binarios que tiene información en los nodos y en las hojas. <br />
Por ejemplo, el árbol<br />
e<br />
/ \<br />
/ \<br />
c g<br />
/ \ / \<br />
a d f h <br />
se representa por "N e (N c (H a) (H d)) (N g (H f) (H h))".<br />
--------------------------------------------------------------------- <br />
*}<br />
<br />
(* ivamenjim marpoldia1 manmorjim1 bowma migtermor wilmorort lucnovdos *)<br />
<br />
datatype 'a arbol = H "'a" | N "'a" "'a arbol" "'a arbol"<br />
<br />
value "N e (N c (H a) (H d)) (N g (H f) (H h))" <br />
<br />
text {* <br />
--------------------------------------------------------------------- <br />
Ejercicio 2. Definir la función <br />
preOrden :: "'a arbol ⇒ 'a list"<br />
tal que (preOrden a) es el recorrido pre orden del árbol a. Por<br />
ejemplo, <br />
preOrden (N e (N c (H a) (H d)) (N g (H f) (H h)))<br />
= [e,c,a,d,g,f,h] <br />
--------------------------------------------------------------------- <br />
*}<br />
<br />
(* ivamenjim marpoldia1 pablucoto bowma fraortmoy migtermor wilmorort lucnovdos *)<br />
<br />
fun preOrden :: "'a arbol ⇒ 'a list" where<br />
"preOrden (H t) = [t]"<br />
| "preOrden (N t i d) = [t] @ (preOrden i) @ (preOrden d)"<br />
<br />
(* danrodcha crigomgom manmorjim1 bowma *)<br />
fun preOrden1 :: "'a arbol ⇒ 'a list" where<br />
"preOrden1 (H x) = [x]"<br />
| "preOrden1 (N x i d) = x#preOrden1 i @ preOrden1 d"<br />
<br />
value "preOrden (N e (N c (H a) (H d)) (N g (H f) (H h))) <br />
= [e,c,a,d,g,f,h]" <br />
value "preOrden1 (N e (N c (H a) (H d)) (N g (H f) (H h))) <br />
= [e,c,a,d,g,f,h]" <br />
<br />
(* danrodcha *)<br />
lemma "preOrden a = preOrden1 a"<br />
by (induct a) simp_all<br />
<br />
text {* <br />
--------------------------------------------------------------------- <br />
Ejercicio 3. Definir la función <br />
postOrden :: "'a arbol ⇒ 'a list"<br />
tal que (postOrden a) es el recorrido post orden del árbol a. Por<br />
ejemplo, <br />
postOrden (N e (N c (H a) (H d)) (N g (H f) (H h)))<br />
= [e,c,a,d,g,f,h] <br />
--------------------------------------------------------------------- <br />
*}<br />
<br />
(* ivamenjim danrodcha crigomgom marpoldia1 manmorjim1 pablucoto bowma fraortmoy migtermor wilmorort lucnovdos *)<br />
<br />
fun postOrden :: "'a arbol ⇒ 'a list" where<br />
"postOrden (H t) = [t]"<br />
| "postOrden (N t i d) = (postOrden i) @ (postOrden d) @ [t]"<br />
<br />
value "postOrden (N e (N c (H a) (H d)) (N g (H f) (H h)))<br />
= [a,d,c,f,h,g,e]"<br />
<br />
text {* <br />
--------------------------------------------------------------------- <br />
Ejercicio 4. Definir la función <br />
inOrden :: "'a arbol ⇒ 'a list"<br />
tal que (inOrden a) es el recorrido in orden del árbol a. Por<br />
ejemplo, <br />
inOrden (N e (N c (H a) (H d)) (N g (H f) (H h)))<br />
= [a,c,d,e,f,g,h]<br />
--------------------------------------------------------------------- <br />
*}<br />
<br />
(* ivamenjim crigomgom marpoldia1 pablucoto bowma fraortmoy migtermor wilmorort lucnovdos*)<br />
<br />
fun inOrden :: "'a arbol ⇒ 'a list" where<br />
"inOrden (H t) = [t]"<br />
| "inOrden (N t i d) = (inOrden i) @ [t] @ (inOrden d)"<br />
<br />
<br />
(* danrodcha manmorjim1 *)<br />
fun inOrden1 :: "'a arbol ⇒ 'a list" where<br />
"inOrden1 (H t) = [t]"<br />
| "inOrden1 (N t i d) = inOrden1 i @ t#inOrden1 d"<br />
<br />
value "inOrden (N e (N c (H a) (H d)) (N g (H f) (H h)))<br />
= [a,c,d,e,f,g,h]"<br />
value "inOrden1 (N e (N c (H a) (H d)) (N g (H f) (H h)))<br />
= [a,c,d,e,f,g,h]"<br />
<br />
(* manmorjim1 *)<br />
lemma "inOrden t = inOrden1 t"<br />
apply (induct t)<br />
apply auto<br />
done<br />
<br />
text {* <br />
--------------------------------------------------------------------- <br />
Ejercicio 5. Definir la función <br />
espejo :: "'a arbol ⇒ 'a arbol"<br />
tal que (espejo a) es la imagen especular del árbol a. Por ejemplo, <br />
espejo (N e (N c (H a) (H d)) (N g (H f) (H h)))<br />
= N e (N g (H h) (H f)) (N c (H d) (H a))<br />
--------------------------------------------------------------------- <br />
*}<br />
<br />
(* ivamenjim danrodcha crigomgom marpoldia1 manmorjim1 pablucoto bowma fraortmoy migtermor wilmorort*)<br />
<br />
fun espejo :: "'a arbol ⇒ 'a arbol" where<br />
"espejo (H t) = H t"<br />
| "espejo (N t i d) = N t (espejo d) (espejo i)"<br />
<br />
value "espejo (N e (N c (H a) (H d)) (N g (H f) (H h))) <br />
= N e (N g (H h) (H f)) (N c (H d) (H a))"<br />
<br />
<br />
text {* <br />
--------------------------------------------------------------------- <br />
Ejercicio 6. Demostrar que<br />
preOrden (espejo a) = rev (postOrden a)<br />
--------------------------------------------------------------------- <br />
*}<br />
<br />
(* ivamenjim migtermor wilmorort *)<br />
<br />
lemma "preOrden (espejo a) = rev (postOrden a)" (is "?P a")<br />
proof (induct a)<br />
fix x <br />
show "?P (H x)" by simp <br />
next<br />
fix x <br />
fix i assume h1: "?P i"<br />
fix d assume h2: "?P d"<br />
have "preOrden (espejo (N x i d)) = preOrden (N x (espejo d) (espejo i))" by simp<br />
also have "... = [x] @ (preOrden (espejo d)) @ (preOrden (espejo i))" by simp<br />
also have "... = [x] @ rev (postOrden d) @ rev (postOrden i)" using h1 h2 by simp <br />
finally show "preOrden (espejo (N x i d)) = rev (postOrden (N x i d))" by simp<br />
qed<br />
<br />
(* danrodcha *)<br />
lemma "preOrden (espejo a) = rev (postOrden a)"<br />
by (induct a) simp_all<br />
<br />
(* danrodcha crigomgom*)<br />
lemma "preOrden (espejo a) = rev (postOrden a)" (is "?P a")<br />
proof (induct a)<br />
fix x<br />
show "?P (H x)" by simp<br />
next<br />
fix x i d <br />
assume HIi: "?P i"<br />
assume HId: "?P d"<br />
have "preOrden (espejo (N x i d)) = preOrden (N x (espejo d) (espejo i))"<br />
by (simp only: espejo.simps(2))<br />
also have "… = x#preOrden (espejo d) @ preOrden (espejo i)"<br />
by (simp only: preOrden.simps(2))<br />
also have"… = x#rev (postOrden d) @ rev (postOrden i)" <br />
using HIi HId by simp<br />
also have "… = rev (postOrden (N x i d))" by simp<br />
finally show "?P (N x i d)" by simp<br />
qed<br />
<br />
(* danrodcha fraortmoy *)<br />
lemma "preOrden (espejo a) = rev (postOrden a)" (is "?P a")<br />
proof (induct a)<br />
fix x<br />
show "?P (H x)" by simp<br />
next<br />
fix x i d <br />
assume HIi: "?P i"<br />
assume HId: "?P d"<br />
show "?P (N x i d)" using HIi HId by simp<br />
qed<br />
<br />
(* bowma *)<br />
lemma "preOrden (espejo a) = rev (postOrden a)"<br />
apply (induct a)<br />
apply simp_all<br />
done<br />
<br />
(* pablucoto marpoldia1*)<br />
<br />
lemma "preOrden (espejo a) = rev (postOrden a)" (is "?P a")<br />
proof (induct a)<br />
fix x <br />
show "?P (H x)" by simp <br />
next<br />
fix x i d <br />
assume h1: "?P i"<br />
assume h2: "?P d"<br />
have "preOrden (espejo (N x i d)) = preOrden (N x (espejo d) (espejo i))" by simp<br />
also have "... = [x] @ (preOrden (espejo d)) @ (preOrden (espejo i))" by simp<br />
also have "... = [x] @ rev (postOrden d) @ rev (postOrden i)" using h1 h2 by simp<br />
also have "... = [x] @ rev (postOrden i @ postOrden d)" by simp <br />
also have "... = rev ( postOrden i @ postOrden d @ [x] ) " by simp<br />
also have "... = rev (postOrden (N x i d)) " by simp<br />
finally show "?P (N x i d)" by simp <br />
qed<br />
<br />
(* bowma *)<br />
lemma "preOrden (espejo a) = rev (postOrden a)" (is "?p a")<br />
proof (induct a)<br />
fix t<br />
show "?p (H t)" by simp<br />
(* Aquí si le diga "preOrden (espejo (H t)) = rev (postOrden (H t))",isabelle dice: <br />
proof (prove)<br />
goal (1 subgoal):<br />
1. preOrden (espejo (H t)) = rev (postOrden (H t)) <br />
Introduced fixed type variable(s): 'b in "t__" <br />
No entiendo porqué *)<br />
next <br />
fix t i d<br />
assume H1: "?p i"<br />
assume H2: "?p d"<br />
have "preOrden (espejo (N t i d)) = preOrden (N t (espejo d) (espejo i))" by simp<br />
also have "... = [t] @ (preOrden (espejo d)) @ (preOrden (espejo i))" by simp<br />
also have "... = [t] @ rev (postOrden d) @ rev (postOrden i)" using H1 H2 by simp<br />
finally show "?p (N t i d)" by simp<br />
qed <br />
<br />
(* fraortmoy *)<br />
<br />
lemma "preOrden (espejo a) = rev (postOrden a)"<br />
by (induct a) auto<br />
<br />
text {* <br />
--------------------------------------------------------------------- <br />
Ejercicio 7. Demostrar que<br />
postOrden (espejo a) = rev (preOrden a)<br />
--------------------------------------------------------------------- <br />
*}<br />
<br />
(* ivamenjim crigomgom bowma migtermor wilmorort*)<br />
<br />
lemma "postOrden (espejo a) = rev (preOrden a)" (is "?P a")<br />
proof (induct a)<br />
fix x <br />
show "?P (H x)" by simp <br />
next<br />
fix x <br />
fix i assume h1: "?P i"<br />
fix d assume h2: "?P d"<br />
have "postOrden (espejo (N x i d)) = postOrden (N x (espejo d) (espejo i))" by simp<br />
also have "... = (postOrden (espejo d)) @ (postOrden (espejo i)) @ [x]" by simp<br />
also have "... = rev (preOrden d) @ rev (preOrden i) @ [x]" using h1 h2 by simp <br />
finally show "postOrden (espejo (N x i d)) = rev (preOrden (N x i d))" by simp<br />
(* "?p (N x i d)" más corto *)<br />
qed<br />
<br />
(* danrodcha fraortmoy *)<br />
lemma "postOrden (espejo a) = rev (preOrden a)" (is "?P a")<br />
proof (induct a)<br />
fix x<br />
show "?P (H x)" by simp<br />
next<br />
fix x i d <br />
assume HIi: "?P i"<br />
assume HId: "?P d"<br />
show "?P (N x i d)" using HIi HId by simp<br />
qed<br />
<br />
(* pablucoto marpoldia1*)<br />
<br />
lemma "postOrden (espejo a) = rev (preOrden a)" (is "?P a")<br />
proof (induct a)<br />
fix x<br />
show "?P (H x)" by simp <br />
next <br />
fix x i d<br />
assume H1: "?P i"<br />
assume H2: "?P d"<br />
have " postOrden (espejo (N x i d)) = postOrden ( N x (espejo d) (espejo i)) " by simp<br />
also have "... = postOrden (espejo d) @ postOrden (espejo i) @ [x] " by simp<br />
also have "... = rev (preOrden d) @ rev (preOrden i) @ [x] " using H1 H2 by simp<br />
also have "... = rev (preOrden d) @ rev (x # preOrden i)" by simp<br />
also have "... = rev (x # preOrden i @ preOrden d)" by simp<br />
also have "... = rev (preOrden (N x i d)) " by simp<br />
finally show "?P (N x i d)" by simp<br />
qed<br />
<br />
(* fraortmoy *)<br />
lemma "postOrden (espejo a) = rev (preOrden a)"<br />
by (induct a) auto<br />
<br />
text {* <br />
--------------------------------------------------------------------- <br />
Ejercicio 8. Demostrar que<br />
inOrden (espejo a) = rev (inOrden a)<br />
--------------------------------------------------------------------- <br />
*}<br />
<br />
(* ivamenjim crigomgom bowma migtermor wilmorort*)<br />
<br />
theorem "inOrden (espejo a) = rev (inOrden a)" (is "?P a")<br />
proof (induct a)<br />
fix x <br />
show "?P (H x)" by simp <br />
next<br />
fix x <br />
fix i assume h1: "?P i"<br />
fix d assume h2: "?P d"<br />
have "inOrden (espejo (N x i d)) = inOrden (N x (espejo d) (espejo i))" by simp<br />
also have "... = (inOrden (espejo d)) @ [x] @ (inOrden (espejo i))" by simp<br />
also have "... = rev (inOrden d) @ [x] @ rev (inOrden i)" using h1 h2 by simp <br />
finally show "inOrden (espejo (N x i d)) = rev (inOrden (N x i d))" by simp<br />
qed<br />
<br />
(* danrodcha *)<br />
theorem "inOrden (espejo a) = rev (inOrden a)"<br />
by (induct a) simp_all<br />
<br />
(* danrodcha *)<br />
theorem "inOrden (espejo a) = rev (inOrden a)" (is "?P a")<br />
proof (induct a)<br />
fix x<br />
show "?P (H x)" by simp<br />
next<br />
fix x i d <br />
assume HIi: "?P i"<br />
assume HId: "?P d"<br />
show "?P (N x i d)" using HIi HId by simp<br />
qed<br />
<br />
(* pablucoto marpoldia1 *)<br />
theorem "inOrden (espejo a) = rev (inOrden a)" (is "?P a")<br />
proof (induct a)<br />
fix x<br />
show "?P (H x) " by simp<br />
next <br />
fix x i d <br />
assume HI1: "?P i"<br />
assume HI2: "?P d"<br />
have " inOrden (espejo (N x i d)) = inOrden ( N x (espejo d) (espejo i) )" by simp<br />
also have "... = inOrden (espejo d) @ [x] @ inOrden (espejo i) " by simp<br />
also have "... = rev (inOrden d) @ [x] @ rev (inOrden i)" using HI1 HI2 by simp<br />
also have "... = rev (x # inOrden d ) @ rev (inOrden i)" by simp<br />
also have "... = rev ( inOrden i @ x # inOrden d) " by simp<br />
also have "... = rev (inOrden (N x i d))" by simp<br />
finally show "?P (N x i d)" by simp<br />
qed<br />
<br />
text {* <br />
--------------------------------------------------------------------- <br />
Ejercicio 9. Definir la función <br />
raiz :: "'a arbol ⇒ 'a"<br />
tal que (raiz a) es la raiz del árbol a. Por ejemplo, <br />
raiz (N e (N c (H a) (H d)) (N g (H f) (H h))) = e<br />
--------------------------------------------------------------------- <br />
*}<br />
<br />
(* danrodcha crigomgom manmorjim1 ivamenjim bowma pablucoto migtermor marpoldia1 wilmorort *)<br />
fun raiz :: "'a arbol ⇒ 'a" where<br />
"raiz (H x) = x"<br />
| "raiz (N x i d) = x"<br />
<br />
value "raiz (N e (N c (H a) (H d)) (N g (H f) (H h))) = e"<br />
<br />
text {* <br />
--------------------------------------------------------------------- <br />
Ejercicio 10. Definir la función <br />
extremo_izquierda :: "'a arbol ⇒ 'a"<br />
tal que (extremo_izquierda a) es el nodo más a la izquierda del árbol<br />
a. Por ejemplo, <br />
extremo_izquierda (N e (N c (H a) (H d)) (N g (H f) (H h))) = a<br />
--------------------------------------------------------------------- <br />
*}<br />
<br />
(* danrodcha crigomgom manmorjim1 ivamenjim bowma pablucoto migtermor marpoldia1 wilmorort *)<br />
fun extremo_izquierda :: "'a arbol ⇒ 'a" where<br />
"extremo_izquierda (H x) = x"<br />
| "extremo_izquierda (N x i d) = extremo_izquierda i"<br />
<br />
value "extremo_izquierda (N e (N c (H a) (H d)) (N g (H f) (H h))) = a"<br />
<br />
(* bowma *)<br />
fun extremo_izquierda_1 :: "'a arbol ⇒ 'a" where<br />
"extremo_izquierda_1 (H t) = t"<br />
| "extremo_izquierda_1 (N t i d) = hd (inOrden (N t i d))"<br />
<br />
(* ivamenjim *)<br />
(* Metaejercicio de demostración. Llamando teorema_13 al teorema del ejercicio 13 *)<br />
<br />
lemma "extremo_izquierda a = extremo_izquierda_1 a"<br />
by (induct a, simp_all add: aux_ej12_1 teorema_13)<br />
<br />
text {* <br />
--------------------------------------------------------------------- <br />
Ejercicio 11. Definir la función <br />
extremo_derecha :: "'a arbol ⇒ 'a"<br />
tal que (extremo_derecha a) es el nodo más a la derecha del árbol<br />
a. Por ejemplo, <br />
extremo_derecha (N e (N c (H a) (H d)) (N g (H f) (H h))) = h<br />
--------------------------------------------------------------------- <br />
*}<br />
<br />
(* danrodcha crigomgom manmorjim1 ivamenjim bowma pablucoto migtermor marpoldia1 wilmorort *)<br />
fun extremo_derecha :: "'a arbol ⇒ 'a" where<br />
"extremo_derecha (H x) = x"<br />
| "extremo_derecha (N x i d) = extremo_derecha d"<br />
<br />
value "extremo_derecha (N e (N c (H a) (H d)) (N g (H f) (H h))) = h"<br />
<br />
(* bowma *)<br />
fun extremo_derecha_1 :: "'a arbol ⇒ 'a" where<br />
"extremo_derecha_1 (H t) = t"<br />
| "extremo_derecha_1 (N t i d) = last (inOrden (N t i d))"<br />
<br />
(* ivamenjim *)<br />
(* Metaejercicio de demostración. Llamando teorema_12 al teorema del ejercicio 12 *)<br />
<br />
lemma "extremo_derecha a = extremo_derecha_1 a"<br />
by (induct a, simp_all add: aux_ej12_1 teorema_12)<br />
<br />
text {* <br />
--------------------------------------------------------------------- <br />
Ejercicio 12. Demostrar o refutar<br />
last (inOrden a) = extremo_derecha a<br />
--------------------------------------------------------------------- <br />
*}<br />
<br />
(* danrodcha *)<br />
lemma aux_ej12: "inOrden a ≠ []"<br />
apply (induct a) <br />
apply simp<br />
apply simp<br />
done<br />
<br />
(* danrodcha pablucoto crigomgom wilmorort*)<br />
theorem "last (inOrden a) = extremo_derecha a" (is "?P a")<br />
proof (induct a)<br />
fix x<br />
show "?P (H x)" by simp<br />
next<br />
fix x i d <br />
assume HIi: "?P i"<br />
assume HId: "?P d"<br />
have "last (inOrden (N x i d)) = last (inOrden i @ [x] @ inOrden d)" <br />
by (simp only: inOrden.simps(2))<br />
also have "… = last (inOrden d)" by (simp add: aux_ej12)<br />
also have "… = extremo_derecha d" using HId by simp<br />
also have "… = extremo_derecha (N x i d)" by simp<br />
finally show "?P (N x i d)" by simp<br />
qed<br />
<br />
(* ivamenjim *)<br />
<br />
lemma aux_ej12_1: "inOrden a ≠ []"<br />
by (induct a) simp_all <br />
<br />
(* ivamenjim marpoldia1 *)<br />
(* Igual que la anterior, pero poniendo solo by simp en el primer have *)<br />
<br />
theorem "last (inOrden a) = extremo_derecha a" (is "?P a")<br />
proof (induct a)<br />
fix x <br />
show "?P (H x)" by simp <br />
next<br />
fix x <br />
fix i assume h1: "?P i"<br />
fix d assume h2: "?P d"<br />
have "last (inOrden (N x i d)) = last ((inOrden i) @ [x] @ (inOrden d))" by simp <br />
also have "... = last (inOrden d)" by (simp add: aux_ej12_1)<br />
also have "... = extremo_derecha d" using h2 by simp <br />
finally show "last (inOrden (N x i d)) = extremo_derecha (N x i d)" by simp<br />
qed<br />
<br />
(* bowma *)<br />
(* Casi lo mismo que el anterior,pero no hace falta suponer "?p i" *)<br />
theorem "last (inOrden a) = extremo_derecha a" (is "?p a")<br />
proof (induct a)<br />
fix t<br />
show "?p (H t)" by simp<br />
next <br />
fix t i d<br />
assume HI: "?p d"<br />
have "last (inOrden (N t i d)) = last (inOrden i @ [t] @ inOrden d)" by simp<br />
also have "... = last (inOrden d)" by (simp add:aux_ej12)<br />
also have "... = extremo_derecha d" using HI by simp<br />
finally show "?p (N t i d)" by simp<br />
qed<br />
<br />
(* migtermor *)<br />
theorem "last (inOrden a) = extremo_derecha a" (is "?P a")<br />
proof (induct a)<br />
fix h<br />
show "?P (H h)" by simp<br />
next<br />
fix n i<br />
fix d assume HId: "?P d"<br />
have AUX: "¬ (inOrden d = [])" (is "?Q d")<br />
proof (induct d)<br />
fix hd<br />
show "?Q (H hd)" by simp<br />
next<br />
fix nd<br />
fix id assume HIid: "?Q id"<br />
fix dd assume HIdd: "?Q dd"<br />
show "?Q (N nd id dd)" using HIid HIdd by simp<br />
qed<br />
have "last (inOrden (N n i d)) = last (inOrden i @[n]@inOrden d)" by simp<br />
also have "… = last (inOrden d)" using AUX by simp<br />
also have "… = extremo_derecha d" using HId by simp<br />
finally show "?P (N n i d)" by simp<br />
qed<br />
<br />
text {* <br />
--------------------------------------------------------------------- <br />
Ejercicio 13. Demostrar o refutar<br />
hd (inOrden a) = extremo_izquierda a<br />
--------------------------------------------------------------------- <br />
*}<br />
<br />
(* danrodcha pablucoto crigomgom *)<br />
theorem "hd (inOrden a) = extremo_izquierda a" (is "?P a")<br />
proof (induct a)<br />
fix x<br />
show "?P (H x)" by simp<br />
next<br />
fix x i d <br />
assume HIi: "?P i"<br />
assume HId: "?P d"<br />
have "hd (inOrden (N x i d)) = hd (inOrden i @ [x] @ inOrden d)" <br />
by (simp only: inOrden.simps(2))<br />
also have "… = hd (inOrden i)" by (simp add: aux_ej12)<br />
also have "… = extremo_izquierda i" using HIi by simp<br />
also have "… = extremo_izquierda (N x i d)" by simp<br />
finally show "?P (N x i d)" by simp<br />
qed<br />
<br />
(* bowma *)<br />
theorem "hd (inOrden a) = extremo_izquierda a" (is "?p a")<br />
proof (induct a)<br />
fix t<br />
show "?p (H t)" by simp<br />
next<br />
fix t i d <br />
assume HI: "?p i"<br />
have "hd (inOrden (N t i d)) = hd (inOrden i @ [t] @ inOrden d)" by simp<br />
also have "… = hd (inOrden i)" by (simp add: aux_ej12)<br />
also have "… = extremo_izquierda i" using HI by simp<br />
finally show "?p (N t i d)" by simp<br />
qed<br />
<br />
(* migtermor *)<br />
theorem "hd (inOrden a) = extremo_izquierda a" (is "?P a")<br />
proof (induct a)<br />
fix h<br />
show "?P (H h)" by simp<br />
next<br />
fix n d<br />
fix i assume HId: "?P i"<br />
have AUX: "¬ (inOrden i = [])" (is "?Q i")<br />
proof (induct i)<br />
fix hi<br />
show "?Q (H hi)" by simp<br />
next<br />
fix ni<br />
fix ii assume HIid: "?Q ii"<br />
fix di assume HIdd: "?Q di"<br />
show "?Q (N ni ii di)" using HIid HIdd by simp<br />
qed<br />
have "hd (inOrden (N n i d)) = hd (inOrden i @[n]@inOrden d)" by simp<br />
also have "… = hd (inOrden i)" using AUX by simp<br />
also have "… = extremo_izquierda i" using HId by simp<br />
finally show "?P (N n i d)" by simp<br />
qed<br />
<br />
(* ivamenjim marpoldia1 wilmorort *)<br />
<br />
theorem "hd (inOrden a) = extremo_izquierda a" (is "?P a")<br />
proof (induct a)<br />
fix x <br />
show "?P (H x)" by simp <br />
next<br />
fix x <br />
fix i assume h1: "?P i"<br />
fix d assume h2: "?P d"<br />
have "hd (inOrden (N x i d)) = hd ((inOrden i) @ [x] @ (inOrden d))" by simp<br />
also have "... = hd (inOrden i)" by (simp add: aux_ej12_1)<br />
also have "... = extremo_izquierda i" using h1 by simp <br />
finally show "hd (inOrden (N x i d)) = extremo_izquierda (N x i d)" by simp<br />
qed<br />
<br />
text {* <br />
--------------------------------------------------------------------- <br />
Ejercicio 14. Demostrar o refutar<br />
hd (preOrden a) = last (postOrden a)<br />
--------------------------------------------------------------------- <br />
*}<br />
<br />
(* danrodcha *)<br />
theorem "hd (preOrden a) = last (postOrden a)" (is "?P a")<br />
proof (induct a)<br />
fix x<br />
show "?P (H x)" by simp<br />
fix x i d<br />
show "?P (N x i d)" by simp<br />
qed<br />
<br />
(* danrodcha *)<br />
theorem "hd (preOrden a) = last (postOrden a)" (is "?P a")<br />
proof (induct a)<br />
fix x<br />
show "?P (H x)" by simp<br />
fix x i d<br />
assume HIi: "?P i"<br />
assume HId: "?P d"<br />
have "hd (preOrden (N x i d)) = hd (x#preOrden i @ preOrden d)"<br />
by (simp only: preOrden.simps(2))<br />
also have "… = x" by simp<br />
also have "… = last (postOrden i @ postOrden d @ [x])" by simp<br />
also have "… = last (postOrden (N x i d))" <br />
by (simp only: postOrden.simps(2))<br />
finally show "?P (N x i d)" by simp<br />
qed<br />
<br />
(* pablucoto crigomgom bowma marpoldia1 wilmorort*) (*Similar al anterior*)<br />
theorem "hd (preOrden a) = last (postOrden a)" (is "?P a")<br />
proof (induct a)<br />
fix x<br />
show "?P (H x)" by simp<br />
next <br />
fix x i d<br />
assume HI1: "?P i"<br />
assume HI2: "?P d"<br />
have " hd (preOrden (N x i d)) = hd ([x] @ preOrden i @ preOrden d)" by simp<br />
also have "... = x" by simp<br />
also have "... = last ( postOrden i @ postOrden d @ [x]) " by simp<br />
also have "... = last ( postOrden (N x i d) )" by simp <br />
finally show "?P (N x i d)" by simp<br />
qed<br />
<br />
(* migtermor *)<br />
theorem "hd (preOrden a) = last (postOrden a)" (is "?P a")<br />
proof (induct a)<br />
fix h<br />
show "?P (H h)" by simp<br />
next<br />
fix n i d<br />
have "hd (preOrden (N n (i :: 'a arbol) (d :: 'a arbol))) = hd ([n]@preOrden i@preOrden d)" <br />
by simp<br />
(* Si no especifico que i y d son árboles, salta un error de tipo. Supongo que será por<br />
no haber asumido hipótesis sobre ellos *)<br />
also have "… = last (postOrden (N n i d))" by simp<br />
show "?P (N n i d)" by simp<br />
qed<br />
<br />
(* ivamenjim *)<br />
<br />
theorem "hd (preOrden a) = last (postOrden a)" (is "?P a")<br />
proof (induct a)<br />
fix x <br />
show "?P (H x)" by simp <br />
next<br />
fix x <br />
fix i assume h1: "?P i"<br />
fix d assume h2: "?P d"<br />
have "hd (preOrden (N x i d)) = hd ([x] @ (preOrden i) @ (preOrden d))" by simp<br />
also have "... = hd ([x])" by simp<br />
finally show "hd (preOrden (N x i d)) = last (postOrden (N x i d))" by simp<br />
qed<br />
<br />
text {* <br />
--------------------------------------------------------------------- <br />
Ejercicio 15. Demostrar o refutar<br />
hd (preOrden a) = raiz a<br />
--------------------------------------------------------------------- <br />
*}<br />
<br />
(* danrodcha *)<br />
theorem "hd (preOrden a) = raiz a" (is "?P a")<br />
proof (induct a)<br />
fix x<br />
show "?P (H x)" by simp<br />
fix x i d<br />
assume HIi: "?P i"<br />
assume HId: "?P d"<br />
have "hd (preOrden (N x i d)) = hd (x#preOrden i @ preOrden d)"<br />
by (simp only: preOrden.simps(2))<br />
also have "… = x" by simp<br />
also have "… = raiz (N x i d)" by simp<br />
finally show "?P (N x i d)" by simp<br />
qed<br />
<br />
(* danrodcha *)<br />
theorem "hd (preOrden a) = raiz a" (is "?P a")<br />
proof (induct a)<br />
fix x<br />
show "?P (H x)" by simp<br />
fix x i d<br />
show "?P (N x i d)" by simp<br />
qed<br />
<br />
(* pablucoto crigomgom ivamenjim marpoldia1 wilmorort *)<br />
theorem "hd (preOrden a) = raiz a" (is "?P a")<br />
proof (induct a) <br />
fix x<br />
show "?P (H x)" by simp<br />
next<br />
fix x i d<br />
assume HI1: " ?P i"<br />
assume HI2: " ?P d"<br />
have " hd (preOrden (N x i d)) = hd ([x] @ preOrden i @ preOrden d) " by simp<br />
also have "... = x" by simp<br />
also have "... = raiz (N x i d)" by simp<br />
finally show " ?P (N x i d)" by simp<br />
qed<br />
<br />
(* bowma *)<br />
(* similar al anterior pero sin suponer "?p d" *)<br />
theorem "hd (preOrden a) = last (postOrden a)" (is "?p a")<br />
proof (induct a)<br />
fix t<br />
show "?p (H t)" by simp<br />
next<br />
fix t i d<br />
assume HI: "?p i"<br />
have "hd (preOrden (N t i d)) = hd ([t] @ preOrden i @ preOrden d)" by simp<br />
also have "… = t" by simp<br />
also have "… = last (postOrden i @ postOrden d @ [t])" by simp<br />
also have "… = last (postOrden (N t i d))" by simp<br />
finally show "?p (N t i d)" by simp<br />
qed<br />
<br />
(* migtermor *)<br />
theorem "hd (preOrden a) = raiz a" (is "?P a")<br />
proof (induct a)<br />
fix h<br />
show "?P (H h)" by simp<br />
next<br />
fix n i d<br />
have "hd (preOrden (N n (i :: 'a arbol) (d :: 'a arbol))) = hd ([n]@preOrden i@preOrden d)"<br />
by simp<br />
also have "… = raiz (N n i d)" by simp<br />
finally show "?P (N n i d)" by simp<br />
qed<br />
<br />
(* ivamenjim: sin usar patrones *)<br />
<br />
theorem "hd (preOrden a) = raiz a" <br />
proof (induct a)<br />
fix x ::"'a"<br />
show "hd (preOrden (H x)) = raiz (H x)" by simp <br />
next<br />
fix x ::"'a"<br />
fix i ::"'a arbol" assume h1: "hd (preOrden i) = raiz i"<br />
fix d ::"'a arbol" assume h2: "hd (preOrden d) = raiz d"<br />
have "hd (preOrden (N x i d)) = hd ([x] @ (preOrden i) @ (preOrden d))" by simp<br />
also have "... = hd ([x])" by simp<br />
finally show "hd (preOrden (N x i d)) = raiz (N x i d)" by simp<br />
qed<br />
<br />
text {* <br />
--------------------------------------------------------------------- <br />
Ejercicio 16. Demostrar o refutar<br />
hd (inOrden a) = raiz a<br />
--------------------------------------------------------------------- <br />
*}<br />
<br />
(*crigomgom pablucoto bowma migtermor ivamenjim wilmorort *)<br />
theorem "hd (inOrden a) = raiz a"<br />
quickcheck<br />
oops<br />
<br />
(* danrodcha:<br />
Auto Quickcheck found a counterexample:<br />
a = N a⇩1 (H a⇩2) (H a⇩1)<br />
Evaluated terms:<br />
hd (inOrden a) = a⇩2<br />
raiz a = a⇩1 *)<br />
<br />
(* ivamenjim marpoldia1 *)<br />
<br />
theorem "hd (inOrden a) = raiz a" (is "?P a")<br />
proof (induct a)<br />
fix x <br />
show "?P (H x)" by simp <br />
next<br />
fix x <br />
fix i assume h1: "?P i"<br />
fix d assume h2: "?P d"<br />
have "hd (inOrden (N x i d)) = hd ((inOrden i) @ [x] @ (inOrden d))" by simp<br />
also have "... = hd (inOrden i)" by (simp add: aux_ej12_1) <br />
(* Perdemos la x, luego se refuta el enunciado del teorema *)<br />
qed<br />
<br />
text {* <br />
--------------------------------------------------------------------- <br />
Ejercicio 17. Demostrar o refutar<br />
last (postOrden a) = raiz a<br />
--------------------------------------------------------------------- <br />
*}<br />
<br />
(* danrodcha *)<br />
theorem "last (postOrden a) = raiz a" (is "?P a")<br />
proof (induct a)<br />
fix x<br />
show "?P (H x)" by simp<br />
fix x i d<br />
show "?P (N x i d)" by simp<br />
qed<br />
<br />
(* danrodcha *)<br />
theorem "last (postOrden a) = raiz a" (is "?P a")<br />
proof (induct a)<br />
fix x<br />
show "?P (H x)" by simp<br />
fix x i d<br />
assume HIi: "?P i"<br />
assume HId: "?P d"<br />
have "last (postOrden (N x i d)) = last (postOrden i @ postOrden d @ [x])"<br />
by (simp only: postOrden.simps(2))<br />
also have "… = x" by simp<br />
also have "… = raiz (N x i d)" by (simp only: raiz.simps(2))<br />
finally show "?P (N x i d)" by simp<br />
qed<br />
<br />
(* pablucoto crigomgom ivamenjim marpoldia1 wilmorort*) (*Similar al anterior*)<br />
<br />
theorem "last (postOrden a) = raiz a" (is "?P a")<br />
proof (induct a )<br />
fix x<br />
show "?P (H x)" by simp<br />
next <br />
fix x i d<br />
assume HI1: "?P i"<br />
assume HI2: "?P d"<br />
have "last (postOrden (N x i d)) = last ( postOrden i @ postOrden d @ [x])" by simp<br />
also have "... = x" by simp<br />
also have "... = raiz (N x i d) " by simp<br />
finally show " ?P (N x i d)" by simp<br />
qed<br />
<br />
(* bowma *)<br />
(* También sin usar el supuesto "?p d" *)<br />
theorem "last (postOrden a) = raiz a" (is "?p a")<br />
proof (induct a)<br />
fix t<br />
show "?p (H t)" by simp<br />
next<br />
fix t i d<br />
assume "?p i"<br />
(* si quito este supuesto, hay error pero no sé dónde se lo está usando *)<br />
have "last (postOrden (N t i d)) = last (postOrden i @ postOrden d @ [t])" by simp<br />
also have "... = t" by simp<br />
also have "... = raiz (N t i d)" by simp<br />
finally show "?p (N t i d)" by simp<br />
qed<br />
<br />
(* migtermor *)<br />
theorem "last (postOrden a) = raiz a" (is "?P a")<br />
proof (induct a)<br />
fix h<br />
show "?P (H h)" by simp<br />
next<br />
fix n i d<br />
have "last (postOrden (N n (i :: 'a arbol) (d :: 'a arbol))) = <br />
last (postOrden i@postOrden d@[n])" by simp<br />
also have "… = raiz (N n i d)" by simp<br />
finally show "?P (N n i d)" by simp<br />
qed<br />
<br />
(* ivamenjim: sin usar patrones *)<br />
<br />
theorem "last (postOrden a) = raiz a" <br />
proof (induct a)<br />
fix x::"'a" <br />
show "last (postOrden (H x)) = raiz (H x)" by simp <br />
next<br />
fix x::"'a" <br />
fix i::"'a arbol" assume h1: "last (postOrden i) = raiz i"<br />
fix d::"'a arbol" assume h2: "last (postOrden d) = raiz d"<br />
have "last (postOrden (N x i d)) = last ((postOrden i) @ (postOrden d) @ [x])" by simp<br />
also have "... = last ([x])" by simp <br />
finally show "last (postOrden (N x i d)) = raiz (N x i d)" by simp<br />
qed<br />
<br />
end<br />
</source></div>Lucnovdoshttps://www.glc.us.es/~jalonso/RA2016/index.php?title=Relaci%C3%B3n_5&diff=837Relación 52016-11-30T20:27:47Z<p>Lucnovdos: </p>
<hr />
<div><source lang="isar"><br />
chapter {* R5: Eliminación de duplicados *}<br />
<br />
theory R5_Eliminacion_de_duplicados<br />
imports Main <br />
begin<br />
<br />
text {*<br />
--------------------------------------------------------------------- <br />
Ejercicio 1. Definir la funcion primitiva recursiva <br />
estaEn :: 'a ⇒ 'a list ⇒ bool<br />
tal que (estaEn x xs) se verifica si el elemento x está en la lista<br />
xs. Por ejemplo, <br />
estaEn (2::nat) [3,2,4] = True<br />
estaEn (1::nat) [3,2,4] = False<br />
--------------------------------------------------------------------- <br />
*}<br />
<br />
(* crigomgom rubgonmar bowma wilmorort pablucoto serrodcal <br />
anaprarod migtermor paupeddeg fraortmoy marpoldia1<br />
danrodcha manmorjim1 jeamacpov *)<br />
fun estaEn :: "'a ⇒ 'a list ⇒ bool" where<br />
"estaEn _ [] = False"<br />
| "estaEn x (a#xs) = ((a = x) ∨ (estaEn x xs))"<br />
<br />
value "estaEn (2::nat) [3,2,4] = True"<br />
value "estaEn (1::nat) [3,2,4] = False"<br />
<br />
(* ivamenjim ferrenseg josgarsan juacabsou dancorgar pabrodmac lucnovdos<br />
fracorjim1 *) <br />
(* Igual que la anterior pero con x en lugar de _ en el caso base *)<br />
fun estaEn1 :: "'a ⇒ 'a list ⇒ bool" where<br />
"estaEn1 x [] = False" <br />
| "estaEn1 x (a#xs) = ((x=a) ∨ estaEn1 x xs)"<br />
<br />
value "estaEn1 (2::nat) [3,2,4] = True"<br />
value "estaEn1 (1::nat) [3,2,4] = False"<br />
<br />
(* wilmorort *)<br />
(* reutilizando la funcion "algunos" de R4.thy*)<br />
fun algunos :: "('a ⇒ bool) ⇒ 'a list ⇒ bool" where<br />
"algunos p [] = False"<br />
| "algunos p (x#xs) = (p x ∨ algunos p xs)"<br />
<br />
fun estaEn2 :: "'a ⇒ 'a list ⇒ bool" where<br />
"estaEn2 a xs = algunos (λx. x = a) xs"<br />
<br />
value "estaEn2 (2::nat) [3,2,4] = True"<br />
value "estaEn2 (1::nat) [3,2,4] = False"<br />
<br />
<br />
text {* <br />
--------------------------------------------------------------------- <br />
Ejercicio 2. Definir la función primitiva recursiva <br />
sinDuplicados :: 'a list ⇒ bool<br />
tal que (sinDuplicados xs) se verifica si la lista xs no contiene<br />
duplicados. Por ejemplo, <br />
sinDuplicados [1::nat,4,2] = True<br />
sinDuplicados [1::nat,4,2,4] = False<br />
--------------------------------------------------------------------- <br />
*}<br />
<br />
(* crigomgom rubgonmar ivamenjim wilmorort bowma pablucoto <br />
serrodcal anaprarod migtermor paupeddeg fraortmoy marpoldia1 <br />
ferrenseg josgarsan danrodcha manmorjim1 juacabsou dancorgar<br />
pabrodmac lucnovdos jeamacpov *) <br />
fun sinDuplicados :: "'a list ⇒ bool" where<br />
"sinDuplicados [] = True"<br />
| "sinDuplicados (x#xs) = (¬ estaEn x xs ∧ sinDuplicados xs)"<br />
<br />
value "sinDuplicados [1::nat,4,2] = True"<br />
value "sinDuplicados [1::nat,4,2,4] = False"<br />
<br />
(* fracorjim1 - La versión anterior no prueba el segundo enunciado. La<br />
que propongo demuestra ambos. *) <br />
fun sinDuplicados1 :: "'a list ⇒ bool" where<br />
"sinDuplicados1 [] = True"<br />
| "sinDuplicados1 (x#xs) = (¬(estaEn x xs ∧ sinDuplicados xs))"<br />
<br />
value "sinDuplicados1 [1::nat,4,2] = True"<br />
value "sinDuplicados1 [1::nat,4,2,4] = False"<br />
<br />
(* Comentario: La definición sinDuplicados1 no cumple el segundo<br />
ejemplo. *) <br />
<br />
(* paupeddeg *)<br />
(* Utilizando la función ∉ de Isabelle *)<br />
fun sinDuplicados2 :: "'a list ⇒ bool" where<br />
"sinDuplicados2 [] = True"<br />
| "sinDuplicados2 (a#xs) = ((a ∉ set xs) ∧ sinDuplicados2 xs ) "<br />
<br />
(* Comentario: Uso de ∉ y set *)<br />
<br />
text {* <br />
--------------------------------------------------------------------- <br />
Ejercicio 3. Definir la función primitiva recursiva <br />
borraDuplicados :: 'a list ⇒ bool<br />
tal que (borraDuplicados xs) es la lista obtenida eliminando los<br />
elementos duplicados de la lista xs. Por ejemplo, <br />
borraDuplicados [1::nat,2,4,2,3] = [1,4,2,3]<br />
<br />
Nota: La función borraDuplicados es equivalente a la predefinida<br />
remdups. <br />
--------------------------------------------------------------------- <br />
*}<br />
<br />
(* crigomgom rubgonmar wilmorort bowma pablucoto serrodcal <br />
anaprarod migtermor paupeddeg fraortmoy marpoldia1 ferrenseg <br />
josgarsan danrodcha juacabsou dancorgar manmorjim1 pabrodmac<br />
lucnovdos jeamacpov *) <br />
fun borraDuplicados0 :: "'a list ⇒ 'a list" where<br />
"borraDuplicados0 [] = []"<br />
| "borraDuplicados0 (x#xs) = (if estaEn x xs <br />
then borraDuplicados0 xs <br />
else x#borraDuplicados0 xs)"<br />
<br />
value "borraDuplicados0 [1::nat,2,4,2,3] = [1,4,2,3]"<br />
<br />
(* rubgonmar *)<br />
(* Otra forma Sin usar if <br />
Utilizando case aunque se le sacaría más partido con más de 2 casos *)<br />
fun borraDuplicados1 :: "'a list ⇒ 'a list" where<br />
"borraDuplicados1 [] = []" <br />
| "borraDuplicados1 (x#xs) = (case estaEn x xs of <br />
False => x#borraDuplicados1 xs <br />
| True => borraDuplicados1 xs )"<br />
<br />
value "borraDuplicados1 [1::nat,2,4,2,3] = [1,4,2,3]"<br />
<br />
(* rubgonmar *)<br />
(* Otra forma utilizando let *)<br />
fun borraDuplicados2 :: "'a list ⇒ 'a list" where<br />
"borraDuplicados2 [] = []" <br />
| "borraDuplicados2 (x#xs) = <br />
(let condicion = estaEn x xs::bool in <br />
if condicion <br />
then borraDuplicados2 xs <br />
else x # borraDuplicados2 xs)"<br />
<br />
value "borraDuplicados2 [1::nat,2,4,2,3] = [1,4,2,3]"<br />
<br />
(* ivamenjim *)<br />
(* Utilizando la negación primero *)<br />
fun borraDuplicados3 :: "'a list ⇒ 'a list" where<br />
"borraDuplicados3 [] = []"<br />
| "borraDuplicados3 (x#xs) = (if ¬(estaEn x xs) <br />
then (x#(borraDuplicados3 xs)) <br />
else borraDuplicados3 xs)"<br />
<br />
value "borraDuplicados3 [1::nat,2,4,2,3] = [1,4,2,3]"<br />
<br />
fun borraDuplicados :: "'a list ⇒ 'a list" where<br />
"borraDuplicados xs = borraDuplicados0 xs"<br />
<br />
<br />
(* fracorjim1. Utilizo un acumulador para optimizar la eficiencia *)<br />
<br />
fun borraDuplicadosAc :: "'a list ⇒ 'a list ⇒ 'a list" where<br />
"borraDuplicadosAc [] ys = ys"<br />
| "borraDuplicadosAc (x#xs) ys = (if (estaEn x ys) then (borraDuplicadosAc xs ys)<br />
else (borraDuplicadosAc xs (x#ys)))"<br />
<br />
(* fracorjim1. Uso un caso base *)<br />
fun borraDuplicados4 :: "'a list ⇒ 'a list" where<br />
"borraDuplicados4 xs = (if (sinDuplicados xs) then xs else borraDuplicadosAc xs [])"<br />
<br />
<br />
<br />
text {*<br />
--------------------------------------------------------------------- <br />
Ejercicio 4.1. Demostrar o refutar automáticamente<br />
length (borraDuplicados xs) ≤ length xs<br />
--------------------------------------------------------------------- <br />
*}<br />
<br />
(* crigomgom anaprarod ferrenseg *)<br />
lemma length_borraDuplicados:<br />
"length (borraDuplicados xs) ≤ length xs"<br />
by (induct xs, simp_all)<br />
<br />
(* Comentario: Falla para borraDuplicados1. *)<br />
<br />
(* rubgonmar wilmorort pablucoto serrodcal migtermor paupeddeg <br />
fraortmoy marpoldia1 danrodcha juacabsou dancorgar josgarsan<br />
pabrodmac lucnovdos *) <br />
lemma length_borraDuplicados2:<br />
"length ( borraDuplicados xs ) ≤ length xs"<br />
by (induct xs) auto<br />
<br />
(* manmorjim1 *)<br />
(* soy de ponerlo mejor por pasos *)<br />
lemma length_borraDuplicados3:<br />
"length ( borraDuplicados xs ) ≤ length xs"<br />
apply (induct xs)<br />
apply auto<br />
done<br />
<br />
(* ivamenjim *)<br />
(* Demostrando objetivo a objetivo *)<br />
lemma length_borraDuplicados4:<br />
"length (borraDuplicados xs) ≤ length xs"<br />
apply (induct xs)<br />
apply simp <br />
apply auto<br />
done<br />
<br />
(* bowma anaprarod *)<br />
lemma length_borraDuplicados5:<br />
"length (borraDuplicados xs) ≤ length xs"<br />
apply (induct xs)<br />
apply (simp, simp) (* creo que es mejor poner aquí simp_all *)<br />
done<br />
<br />
(* ferrenseg *)<br />
lemma length_borraDuplicados6:<br />
"length (borraDuplicados xs) ≤ length xs"<br />
by (induct xs) simp_all (* Creo que se puede poner simp_all fuera de<br />
paréntesis *) <br />
<br />
text {*<br />
--------------------------------------------------------------------- <br />
Ejercicio 4.2. Demostrar o refutar detalladamente<br />
length (borraDuplicados xs) ≤ length xs<br />
--------------------------------------------------------------------- <br />
*}<br />
<br />
(* crigomgom *)<br />
lemma length_borraDuplicados_2: <br />
"length (borraDuplicados xs) ≤ length xs"<br />
proof (induct xs)<br />
show "length (borraDuplicados []) ≤ length []" by simp<br />
next<br />
fix x xs<br />
assume HI: "length (borraDuplicados xs) ≤ length xs"<br />
show "length (borraDuplicados (x#xs)) ≤ length (x#xs)"<br />
proof (cases)<br />
assume "estaEn x xs"<br />
then have "length (borraDuplicados (x#xs)) = <br />
length (borraDuplicados xs)" by simp<br />
also have "... ≤ length xs" using HI by simp<br />
also have "... ≤ length (x#xs)" by simp<br />
finally show "length (borraDuplicados (x#xs)) ≤ length (x#xs)" <br />
by simp<br />
next<br />
assume "(¬ estaEn x xs)"<br />
then have "length (borraDuplicados (x#xs)) = <br />
length (x#borraDuplicados xs)" by simp<br />
also have "... = 1 + length (borraDuplicados xs)" by simp<br />
also have "... ≤ 1 + length xs" using HI by simp<br />
also have "... = length (x#xs)" by simp<br />
finally show "length (borraDuplicados (x#xs)) ≤ length (x#xs)" <br />
by simp<br />
qed<br />
qed<br />
<br />
(* ivamenjim wilmorort ferrenseg rubgonmar juacabsou dancorgar<br />
josgarsan lucnovdos *) <br />
lemma length_borraDuplicados_2b: <br />
"length (borraDuplicados xs) ≤ length xs"<br />
proof (induct xs)<br />
show "length (borraDuplicados []) ≤ length []" by simp<br />
next <br />
fix a xs<br />
assume HI: "length (borraDuplicados xs) ≤ length xs"<br />
have "length (borraDuplicados (a # xs)) ≤ <br />
1+length (borraDuplicados xs)" by simp<br />
also have "... ≤ 1+length xs" using HI by simp<br />
finally show "length (borraDuplicados (a # xs)) ≤ length (a # xs)" <br />
by simp<br />
qed<br />
<br />
(* serrodcal anaprarod danrodcha *)<br />
lemma length_borraDuplicados_2c: <br />
"length (borraDuplicados xs) ≤ length xs" (is "?P xs")<br />
proof (induct xs)<br />
show "?P []" by simp<br />
next <br />
fix a xs<br />
assume HI: "?P xs"<br />
have "length (borraDuplicados (a # xs)) ≤ <br />
1+length (borraDuplicados xs)" by simp<br />
also have "... ≤ 1+length xs" using HI by simp<br />
finally show "?P (a # xs)" by simp<br />
qed<br />
<br />
(* pablucoto jeamacpov *)<br />
lemma length_borraDuplicados_2d: <br />
"length (borraDuplicados xs) ≤ length xs" <br />
proof(induct xs)<br />
show "length (borraDuplicados []) ≤ length [] " by simp<br />
next <br />
fix a xs<br />
assume HI: " length (borraDuplicados xs) ≤ length xs "<br />
have "length (borraDuplicados (a # xs)) ≤ <br />
1 + length(borraDuplicados xs)" by simp<br />
also have "... ≤ 1 + length xs" using HI by simp <br />
also have "... ≤ length (a#xs)" by simp<br />
finally show "length (borraDuplicados (a # xs)) ≤ length (a # xs) " <br />
by simp<br />
qed<br />
<br />
(* bowma *)<br />
lemma length_borraDuplicados_2f: <br />
"length (borraDuplicados xs) ≤ length xs" (is "?p xs")<br />
proof (induct xs)<br />
show "length (borraDuplicados []) ≤ length []" by simp<br />
next<br />
fix a xs<br />
assume HI: "?p xs"<br />
have c1: "1+length xs = length (a#xs)" by simp<br />
also have "length(borraDuplicados (a#xs)) ≤ <br />
1 + length(borraDuplicados xs)" by simp<br />
also have "... ≤ 1+length xs" using HI by simp<br />
then show "length(borraDuplicados (a#xs)) ≤ length (a#xs)" <br />
using c1 by simp<br />
(* ¿Aquí porqué no puedo usar "finally show "?p (a # xs)" using c1 by simp? <br />
Y porque no puedo añadir "finally show "?p (a # xs)" by simp al final? *)<br />
qed<br />
<br />
(* Comentario: Responder la pregunta. *)<br />
<br />
(* migtermor *)<br />
lemma length_borraDuplicados_2g: <br />
"length (borraDuplicados xs) ≤ length xs" (is "?P xs")<br />
proof (induct xs)<br />
show "?P []" by simp<br />
next<br />
fix a xs<br />
assume HI: "?P xs"<br />
have "length (borraDuplicados (a#xs)) ≤ (length (a#xs))"<br />
proof (cases)<br />
assume "(estaEn a xs)"<br />
then have Aux: "length (borraDuplicados (a#xs)) = <br />
length (borraDuplicados xs)" by simp<br />
also have "… ≤ length (a#xs)" using HI by simp<br />
then show "length (borraDuplicados (a#xs)) ≤ (length (a#xs))" <br />
using Aux by simp<br />
next<br />
assume "¬ (estaEn a xs)"<br />
then have Aux: "length (borraDuplicados (a#xs)) = <br />
1+ length (borraDuplicados xs)" by simp<br />
also have "… ≤ length (a#xs)" using HI by simp<br />
then show "length (borraDuplicados (a#xs)) ≤ (length (a#xs))" <br />
using Aux by simp<br />
qed<br />
then show "?P (a#xs)" by simp<br />
qed<br />
<br />
(* paupeddeg marpoldia1 pabrodmac*)<br />
lemma length_borraDuplicados_2h:<br />
"length (borraDuplicados xs) ≤ length xs"<br />
proof (induct xs)<br />
show "length (borraDuplicados []) ≤ length []" by simp<br />
next<br />
fix a xs <br />
assume HI: "length (borraDuplicados xs) ≤ length xs"<br />
have "length (borraDuplicados (a # xs)) ≤ <br />
length [a] + length (borraDuplicados xs) " by simp<br />
also have "... ≤ 1 + length (borraDuplicados xs)" by simp<br />
also have "... ≤ 1 + length xs" using HI by simp<br />
finally show "length (borraDuplicados (a # xs)) ≤ <br />
length (a # xs)" by simp<br />
qed<br />
<br />
(* fraortmoy *)<br />
(* muy parecida a alguna anterior, pero yo dí más pasos *)<br />
lemma length_borraDuplicados_2i: <br />
"length (borraDuplicados xs) ≤ length xs"<br />
proof (induct xs)<br />
show "length (borraDuplicados []) ≤ length []" by simp<br />
next<br />
fix a xs<br />
assume H1: "length (borraDuplicados xs) ≤ length xs"<br />
have "length (borraDuplicados (a # xs)) ≤ <br />
length(borraDuplicados [a])+length (borraDuplicados xs)" <br />
by simp<br />
also have "… ≤ 1 + length (borraDuplicados xs)" by simp<br />
also have "… ≤ 1 + length xs" using H1 by simp<br />
finally show "length (borraDuplicados (a # xs)) ≤ length (a # xs)" <br />
by simp<br />
qed<br />
<br />
text {*<br />
--------------------------------------------------------------------- <br />
Ejercicio 5.1. Demostrar o refutar automáticamente<br />
estaEn a (borraDuplicados xs) = estaEn a xs<br />
--------------------------------------------------------------------- <br />
*}<br />
<br />
(* crigomgom rubgonmar wilmorort pablucoto serrodcal bowma <br />
migtermor fraortmoy marpoldia1 ferrenseg danrodcha juacabsou paupeddeg pabrodmac lucnovdos dancorgar jeamacpov *)<br />
lemma estaEn_borraDuplicados: <br />
"estaEn a (borraDuplicados xs) = estaEn a xs"<br />
by (induct xs) auto<br />
<br />
(* ivamenjim manmorjim1 *)<br />
lemma estaEn_borraDuplicados2: <br />
"estaEn a (borraDuplicados xs) = estaEn a xs"<br />
apply (induct xs) <br />
apply auto<br />
done<br />
<br />
(* anaprarod *)<br />
lemma estaEn_borraDuplicados3: <br />
"estaEn a (borraDuplicados xs) = estaEn a xs"<br />
by (induct xs, simp_all, blast)<br />
<br />
(* anaprarod *)<br />
lemma estaEn_borraDuplicados4: <br />
"estaEn a (borraDuplicados xs) = estaEn a xs"<br />
apply (induct xs)<br />
apply (cases "estaEn x xs")<br />
apply (simp_all)<br />
apply blast<br />
done<br />
<br />
(* bowma *)<br />
lemma estaEn_borraDuplicados5: <br />
"estaEn a (borraDuplicados xs) = estaEn a xs"<br />
apply (induct xs)<br />
apply simp<br />
apply (simp, blast)<br />
done<br />
<br />
text {*<br />
--------------------------------------------------------------------- <br />
Ejercicio 5.2. Demostrar o refutar detalladamente<br />
estaEn a (borraDuplicados xs) = estaEn a xs<br />
Nota: Para la demostración de la equivalencia se puede usar<br />
proof (rule iffI)<br />
La regla iffI es<br />
⟦P ⟹ Q ; Q ⟹ P⟧ ⟹ P = Q<br />
--------------------------------------------------------------------- <br />
*}<br />
<br />
(* wilmorort *)<br />
lemma estaEn_borraDuplicados_2: <br />
"estaEn a (borraDuplicados xs) = estaEn a xs"<br />
proof (induct xs)<br />
show "estaEn a (borraDuplicados []) = estaEn a []" by simp<br />
next<br />
fix b xs<br />
assume HI: "estaEn a (borraDuplicados xs) = estaEn a xs"<br />
show "estaEn a (borraDuplicados (b#xs)) = estaEn a (b#xs)"<br />
proof (rule iffI)<br />
assume H1: "estaEn a (borraDuplicados (b#xs))"<br />
show "estaEn a (b#xs)"<br />
proof (cases)<br />
assume "estaEn b xs"<br />
then have "estaEn a (borraDuplicados xs)" using H1 by simp<br />
then have "estaEn a xs" using HI by simp<br />
then show "estaEn a (b#xs)" by simp<br />
next<br />
assume "¬ estaEn b xs"<br />
then have "estaEn a (b#(borraDuplicados xs))" using H1 by simp<br />
then have "a=b ∨ (estaEn a (borraDuplicados xs))" by simp<br />
then have " a=b ∨ (estaEn a xs)" using HI by simp<br />
then show "estaEn a (b#xs)" by simp<br />
qed<br />
next<br />
assume H2: "estaEn a (b#xs)"<br />
show "estaEn a (borraDuplicados (b#xs))"<br />
proof (cases)<br />
assume "a=b"<br />
then have "estaEn b (borraDuplicados xs) = estaEn b xs" <br />
using HI by simp<br />
then have "(estaEn b xs ⟶ estaEn b (borraDuplicados xs)) ∧<br />
(¬ estaEn b xs ⟶ estaEn b (b # borraDuplicados xs))" <br />
by simp <br />
then have "estaEn b (borraDuplicados (b#xs))" by simp<br />
then show "estaEn a (borraDuplicados (b#xs))" using `a=b` by simp<br />
next<br />
assume "a≠b"<br />
then have "estaEn a (b#xs)" using H2 by simp<br />
then have "a = b ∨ estaEn a xs" by simp<br />
then have "False ∨ estaEn a xs " using `a≠b` by simp<br />
then have "estaEn a xs" by simp<br />
then have "estaEn a (borraDuplicados xs)" using HI by simp<br />
then show "estaEn a (borraDuplicados (b#xs))" using `a≠b` by simp<br />
qed<br />
qed<br />
qed<br />
<br />
(* anaprarod marpoldia1 ferrenseg juacabsou *)<br />
lemma estaEn_borraDuplicados_2b:<br />
"estaEn a (borraDuplicados xs) = estaEn a xs" (is "?P xs")<br />
proof (induct xs)<br />
show "?P []" by simp<br />
next<br />
fix x xs<br />
assume HI: "?P xs"<br />
show "?P (x#xs)"<br />
proof (cases)<br />
assume "estaEn x xs"<br />
then have "estaEn a (borraDuplicados (x#xs)) = <br />
estaEn a (borraDuplicados xs)" by simp<br />
also have "...= estaEn a xs" using HI by simp<br />
finally show "?P (x#xs)" by auto<br />
next<br />
assume "¬estaEn x xs"<br />
then have "estaEn a (borraDuplicados (x#xs)) = <br />
estaEn a (x#borraDuplicados xs)" by simp<br />
also have "...= (x = a ∨ estaEn a (borraDuplicados xs))" by simp<br />
finally show "?P (x#xs)" using HI by simp<br />
qed<br />
qed<br />
<br />
(* migtermor *)<br />
lemma estaEn_borraDuplicados_2c: <br />
"estaEn a (borraDuplicados xs) = estaEn a xs" (is "?P xs")<br />
proof (induct xs)<br />
show "?P []" by simp<br />
next <br />
fix aa xs<br />
assume HI: "?P xs"<br />
have P1: "estaEn a (borraDuplicados (aa#xs)) = estaEn a (aa#xs)"<br />
proof (cases)<br />
assume C1: "(estaEn aa xs)"<br />
have "estaEn a (borraDuplicados (aa#xs)) = <br />
estaEn a (borraDuplicados xs)" <br />
using C1 by simp<br />
also have P3: "… = estaEn a xs" using HI by simp<br />
also have "… = estaEn a (aa#xs)" <br />
proof (cases)<br />
assume "(a=aa)"<br />
then show "estaEn a xs = estaEn a (aa#xs)" using C1 by simp<br />
next<br />
assume "¬(a=aa)"<br />
then show "estaEn a xs = estaEn a (aa#xs)" by simp<br />
qed<br />
then show "estaEn a (borraDuplicados (aa#xs)) = estaEn a (aa#xs)" <br />
using P3 by simp<br />
next<br />
assume C2: "¬(estaEn aa xs)"<br />
then show "estaEn a (borraDuplicados (aa#xs)) = <br />
estaEn a (aa#xs)" using HI by simp<br />
qed<br />
also have Conc: "estaEn a (borraDuplicados (aa#xs)) = estaEn a (aa#xs)" <br />
using P1 by simp<br />
finally show "?P (aa#xs)" using Conc by simp<br />
qed<br />
<br />
(* crigomgom paupeddeg pabrodmac*)<br />
lemma estaEn_borraDuplicados_2d: <br />
"estaEn a (borraDuplicados xs) = estaEn a xs"<br />
proof (induct xs)<br />
show "estaEn a (borraDuplicados []) = estaEn a []" by simp<br />
next<br />
fix x xs<br />
assume HI: "estaEn a (borraDuplicados xs) = estaEn a xs"<br />
show "estaEn a (borraDuplicados (x#xs)) = estaEn a (x#xs)"<br />
proof (rule iffI)<br />
assume a1: "estaEn a (borraDuplicados (x#xs))"<br />
show "estaEn a (x#xs)"<br />
proof (cases)<br />
assume "estaEn x xs"<br />
then have "estaEn a (borraDuplicados xs)" using a1 by simp<br />
then have "estaEn a xs" using HI by simp<br />
then show "estaEn a (x#xs)" by simp<br />
next<br />
assume "¬ estaEn x xs"<br />
then have "estaEn a (x#(borraDuplicados xs))" using a1 by simp<br />
then have " x=a ∨ (estaEn a (borraDuplicados xs))" by simp<br />
then have " x=a ∨ (estaEn a xs)" using HI by simp<br />
then show "estaEn a (x#xs)" by simp<br />
qed<br />
next<br />
assume a2: "estaEn a (x#xs)"<br />
show "estaEn a (borraDuplicados (x#xs))"<br />
proof (cases)<br />
assume "a=x"<br />
then show "estaEn a (borraDuplicados (x#xs))" using HI by simp<br />
next<br />
assume b1: "a≠x"<br />
then have "estaEn a (x#xs)" using a2 by simp<br />
then have "x = a ∨ estaEn a xs" by simp<br />
then have "estaEn a xs " using b1 by simp<br />
then have "estaEn a (borraDuplicados xs)" using HI by simp<br />
then show "estaEn a (borraDuplicados (x#xs))" using b1 by simp<br />
qed<br />
qed<br />
qed<br />
<br />
(* rubgonmar jeamacpov *)<br />
lemma estaEn_borraDuplicados_2e: <br />
"estaEn a ( borraDuplicados xs ) = estaEn a xs"<br />
proof (induct xs)<br />
show "estaEn a (borraDuplicados []) = estaEn a []" by simp<br />
next<br />
fix x<br />
fix xs<br />
assume HI: "estaEn a (borraDuplicados xs) = estaEn a xs"<br />
show "estaEn a (borraDuplicados (x#xs)) = estaEn a (x#xs)"<br />
proof (rule iffI) (* usamos proof de la regla dada iffI*)<br />
assume cprim: "estaEn a (borraDuplicados (x#xs))"<br />
show "estaEn a (x#xs)"<br />
proof (cases)<br />
assume "estaEn x xs"<br />
then show "estaEn a (x#xs)" using cprim HI by simp<br />
next<br />
assume "¬ estaEn x xs"<br />
then show "estaEn a (x#xs)" using cprim HI by simp<br />
qed<br />
next<br />
assume cseg: "estaEn a (x#xs)"<br />
show "estaEn a (borraDuplicados (x#xs))"<br />
proof (cases)<br />
assume "a=x"<br />
then show "estaEn a (borraDuplicados (x#xs))" using HI by auto<br />
next<br />
assume "a≠x"<br />
then show "estaEn a (borraDuplicados (x#xs))" using `a≠x` <br />
cseg HI by simp<br />
qed<br />
qed<br />
qed<br />
<br />
(* <br />
Aplico la regla iffI:<br />
⟦P ⟹ Q ; Q ⟹ P⟧ ⟹ P = Q<br />
Así:<br />
[estaEn a (borraDuplicados (x # xs)) <br />
⟹ estaEn a (x # xs); estaEn a (x # xs) <br />
⟹ estaEn a (borraDuplicados (x # xs))] <br />
⟹ estaEn a (borraDuplicados (x # xs)) = estaEn a (x # xs)<br />
*)<br />
<br />
(* bowma ivamenjim *)<br />
lemma estaEn_borraDuplicados_2f: <br />
"estaEn a (borraDuplicados xs) = estaEn a xs" (is "?p xs")<br />
proof (induct xs)<br />
show "?p []" by simp<br />
next <br />
fix x xs<br />
assume HI: "?p xs"<br />
show "?p (x#xs)"<br />
proof (cases)<br />
assume H1:"estaEn x xs"<br />
then have "estaEn a (borraDuplicados (x#xs)) = <br />
estaEn a (borraDuplicados xs)" by simp<br />
also have "... = estaEn a xs" using HI by simp<br />
also have "... = estaEn a (x#xs)" <br />
proof(cases)<br />
assume "x=a"<br />
then show "estaEn a xs = estaEn a (x#xs)" using H1 by simp<br />
next<br />
assume "x≠a"<br />
then show "estaEn a xs = estaEn a (x#xs)" by simp<br />
qed<br />
finally show "?p (x#xs)" by simp<br />
next<br />
assume H2:"¬estaEn x xs"<br />
then have "estaEn a (borraDuplicados (x#xs)) = <br />
estaEn a (x#borraDuplicados xs)" by simp<br />
also have "... = ((x=a) ∨ estaEn a (borraDuplicados xs))" by simp<br />
also have "... = ((x=a) ∨ estaEn a xs)" using HI by simp<br />
also have "... = estaEn a (x#xs)" by simp<br />
finally show "?p (x#xs)" by simp<br />
qed<br />
<br />
(* danrodcha *)<br />
(* es como la de ruben pero con diferencias de estilo *)<br />
lemma estaEn_borraDuplicados_2g: <br />
"estaEn a (borraDuplicados xs) = estaEn a xs" (is "?P xs")<br />
proof (induct xs)<br />
show "?P []" by simp<br />
next<br />
fix x xs assume HI: "?P xs"<br />
show "?P (x#xs)"<br />
proof (rule iffI)<br />
assume H1: "estaEn a (borraDuplicados (x # xs))"<br />
show "estaEn a (x#xs)"<br />
proof (cases "estaEn x xs")<br />
case True<br />
then show "estaEn a (x#xs)" using H1 HI by simp<br />
next<br />
case False<br />
then show "estaEn a (x#xs)" using H1 HI by simp<br />
qed<br />
next<br />
assume H2: "estaEn a (x#xs)"<br />
show "estaEn a (borraDuplicados (x # xs))"<br />
proof (cases "x=a")<br />
case True<br />
then show "estaEn a (borraDuplicados (x # xs))" using HI by simp<br />
next<br />
case False<br />
then show "estaEn a (borraDuplicados (x # xs))" using H2 HI <br />
by simp<br />
qed<br />
qed<br />
qed<br />
<br />
text {*<br />
--------------------------------------------------------------------- <br />
Ejercicio 6.1. Demostrar o refutar automáticamente<br />
sinDuplicados (borraDuplicados xs)<br />
--------------------------------------------------------------------- <br />
*}<br />
<br />
(* ivamenjim wilmorort serrodcal crigomgom anaprarod fraortmoy <br />
marpoldia1 ferrenseg danrodcha juacabsou paupeddeg josgarsan pabrodmac dancorgar jeamacpov *)<br />
lemma sinDuplicados_borraDuplicados:<br />
"sinDuplicados (borraDuplicados xs)"<br />
by (induct xs) (auto simp add: estaEn_borraDuplicados)<br />
<br />
(* migtermor bowma *)<br />
lemma sinDuplicados_borraDuplicados:<br />
"sinDuplicados (borraDuplicados xs)"<br />
by (induct xs, simp_all add: estaEn_borraDuplicados_2)<br />
<br />
(* manmorjim1 no caí en usar la demostración anterior y he realizado<br />
la demostración de que si un elemento no estaba en una lista seguirá sin estar<br />
después de eliminar los duplicados en esa lista... *)<br />
lemma noEsta_tras_borrarDuplicados:<br />
"(¬estaEn x xs) ⟶ (¬estaEn x (borraDuplicados xs))"<br />
apply (induct xs)<br />
apply auto<br />
done<br />
<br />
lemma sinDuplicados_borraDuplicados:<br />
"sinDuplicados (borraDuplicados xs)"<br />
apply (induct xs)<br />
apply simp<br />
apply (induct xs)<br />
apply auto<br />
apply (simp add: noEsta_tras_borrarDuplicados)<br />
done<br />
<br />
text {*<br />
--------------------------------------------------------------------- <br />
Ejercicio 6.2. Demostrar o refutar detalladamente<br />
sinDuplicados (borraDuplicados xs)<br />
--------------------------------------------------------------------- <br />
*}<br />
<br />
(*wilmorort*)<br />
<br />
-- "La demostración estructurada es"<br />
lemma sinDuplicados_borraDuplicados_2:<br />
"sinDuplicados (borraDuplicados xs)"<br />
proof (induct xs)<br />
show " sinDuplicados (borraDuplicados [])" by simp<br />
next<br />
fix a xs<br />
assume HI: "sinDuplicados (borraDuplicados xs)"<br />
show "sinDuplicados (borraDuplicados (a # xs))"<br />
proof (cases)<br />
assume "estaEn a xs"<br />
then show "sinDuplicados (borraDuplicados (a#xs))" using HI by simp<br />
next<br />
assume"¬ estaEn a xs"<br />
then have "¬ (estaEn a xs) ∧ sinDuplicados (borraDuplicados xs)" using HI by simp<br />
then have "¬ estaEn a (borraDuplicados xs) ∧ sinDuplicados (borraDuplicados xs)" <br />
by (simp add: estaEn_borraDuplicados)<br />
then have " sinDuplicados (a#borraDuplicados xs)" by simp<br />
then show " sinDuplicados (borraDuplicados(a #xs))" by simp<br />
qed<br />
qed<br />
<br />
(* ivamenjim migtermor crigomgom rubgonmar fraortmoy marpoldia1 ferrenseg bowma juacabsou serrodcal josgarsan pabrodmac dancorgar jeamacpov lucnovdos *)<br />
<br />
lemma sinDuplicados_borraDuplicados_2:<br />
"sinDuplicados (borraDuplicados xs)"<br />
proof (induct xs)<br />
show "sinDuplicados (borraDuplicados [])" by simp<br />
next<br />
fix a xs<br />
assume HI: "sinDuplicados (borraDuplicados xs)"<br />
show "sinDuplicados (borraDuplicados (a # xs))" <br />
proof (cases)<br />
assume "estaEn a xs" <br />
then show "sinDuplicados (borraDuplicados (a # xs))" using HI by simp<br />
next<br />
assume "¬(estaEn a xs)"<br />
then show "sinDuplicados (borraDuplicados (a # xs))" using HI by (simp add: estaEn_borraDuplicados)<br />
qed<br />
qed<br />
<br />
(* anaprarod paupeddeg*)<br />
lemma sinDuplicados_borraDuplicados_2:<br />
"sinDuplicados (borraDuplicados xs)" (is "?P xs")<br />
proof (induct xs)<br />
show "?P []" by simp<br />
next<br />
fix x xs<br />
assume HI: "?P xs"<br />
show "?P (x#xs)"<br />
proof (cases)<br />
assume c1: "estaEn x xs"<br />
then show "sinDuplicados (borraDuplicados (x#xs))" using HI by simp<br />
next<br />
assume c2: "¬ estaEn x xs"<br />
then have "sinDuplicados (borraDuplicados (x#xs)) =sinDuplicados (x#borraDuplicados xs)" by simp<br />
also have "…= (¬estaEn x (borraDuplicados xs) ∧ sinDuplicados (borraDuplicados xs))" by simp<br />
also have "… = (¬estaEn x (borraDuplicados xs))" using HI by simp<br />
also have "… = (¬(estaEn x xs))" by (simp add:estaEn_borraDuplicados)<br />
also have "… = True" using c2 by simp<br />
finally show "?P (x#xs)" by simp<br />
qed<br />
qed<br />
<br />
(* danrodcha *)<br />
lemma sinDuplicados_borraDuplicados_2:<br />
"sinDuplicados (borraDuplicados xs)" (is "?P xs")<br />
proof (induct xs)<br />
show "?P []" by simp<br />
next<br />
fix x xs assume HI: "?P xs"<br />
show "?P (x#xs)"<br />
proof (cases "estaEn x xs")<br />
case True<br />
then have 1: "sinDuplicados (borraDuplicados (x#xs)) <br />
= sinDuplicados (borraDuplicados xs)" by (simp add: estaEn_borraDuplicados_2)<br />
show "?P (x#xs)" using HI 1 by simp<br />
next<br />
case False<br />
then have "sinDuplicados (borraDuplicados (x#xs)) <br />
= sinDuplicados (x#borraDuplicados xs)" by simp<br />
also have "… = (¬ (estaEn x (borraDuplicados xs)) ∧ <br />
sinDuplicados (borraDuplicados xs))" by simp<br />
also have "… = True" using `¬ estaEn x xs` HI by (simp add:estaEn_borraDuplicados)<br />
finally show "?P (x#xs)" by simp<br />
qed<br />
qed<br />
<br />
text {*<br />
--------------------------------------------------------------------- <br />
Ejercicio 7. Demostrar o refutar:<br />
borraDuplicados (rev xs) = rev (borraDuplicados xs)<br />
--------------------------------------------------------------------- <br />
*}<br />
<br />
(*crigomgom rubgonmar ivamenjim wilmorort pablucoto migtermor <br />
anaprarod fraortmoy ferrenseg marpoldia1 bowma danrodcha juacabsou paupeddeg manmorjim1 serrodcal josgarsan pabrodmac lucnovdos dancorgar jeamacpov *)<br />
lemma "borraDuplicados (rev xs) = rev (borraDuplicados xs)"<br />
quickcheck<br />
oops<br />
<br />
(* ivamenjim: Quickcheck encuentra el siguiente contraejemplo: <br />
xs = [a1, a2, a1]<br />
Por lo que:<br />
· "borraDuplicados (rev xs) = [a2, a1]"<br />
· "rev (borraDuplicados xs) = [a1, a2]" *)<br />
<br />
end<br />
</source></div>Lucnovdoshttps://www.glc.us.es/~jalonso/RA2016/index.php?title=Relaci%C3%B3n_5&diff=826Relación 52016-11-30T12:26:53Z<p>Lucnovdos: </p>
<hr />
<div><source lang="isar"><br />
chapter {* R5: Eliminación de duplicados *}<br />
<br />
theory R5_Eliminacion_de_duplicados<br />
imports Main <br />
begin<br />
<br />
text {*<br />
--------------------------------------------------------------------- <br />
Ejercicio 1. Definir la funcion primitiva recursiva <br />
estaEn :: 'a ⇒ 'a list ⇒ bool<br />
tal que (estaEn x xs) se verifica si el elemento x está en la lista<br />
xs. Por ejemplo, <br />
estaEn (2::nat) [3,2,4] = True<br />
estaEn (1::nat) [3,2,4] = False<br />
--------------------------------------------------------------------- <br />
*}<br />
<br />
(* crigomgom rubgonmar bowma wilmorort pablucoto serrodcal <br />
anaprarod migtermor paupeddeg fraortmoy marpoldia1<br />
danrodcha manmorjim1 *)<br />
fun estaEn :: "'a ⇒ 'a list ⇒ bool" where<br />
"estaEn _ [] = False"<br />
| "estaEn x (a#xs) = ((a = x) ∨ (estaEn x xs))"<br />
<br />
value "estaEn (2::nat) [3,2,4] = True"<br />
value "estaEn (1::nat) [3,2,4] = False"<br />
<br />
(* ivamenjim, ferrenseg josgarsan juacabsou dancorgar pabrodmac lucnovdos *)<br />
(* Igual que la anterior pero con x en lugar de _ en el caso base *)<br />
<br />
fun estaEn1 :: "'a ⇒ 'a list ⇒ bool" where<br />
"estaEn1 x [] = False" <br />
| "estaEn1 x (a#xs) = ((x=a) ∨ estaEn1 x xs)"<br />
<br />
value "estaEn1 (2::nat) [3,2,4] = True"<br />
value "estaEn1 (1::nat) [3,2,4] = False"<br />
<br />
(* wilmorort *)<br />
(* reutilizando la funcion "algunos" de R4.thy*)<br />
fun estaEn2 :: "'a ⇒ 'a list ⇒ bool" where<br />
"estaEn2 a xs = algunos (λx. x = a) xs"<br />
<br />
value "estaEn2 (2::nat) [3,2,4] = True"<br />
value "estaEn2 (1::nat) [3,2,4] = False"<br />
<br />
<br />
text {* <br />
--------------------------------------------------------------------- <br />
Ejercicio 2. Definir la función primitiva recursiva <br />
sinDuplicados :: 'a list ⇒ bool<br />
tal que (sinDuplicados xs) se verifica si la lista xs no contiene<br />
duplicados. Por ejemplo, <br />
sinDuplicados [1::nat,4,2] = True<br />
sinDuplicados [1::nat,4,2,4] = False<br />
--------------------------------------------------------------------- <br />
*}<br />
<br />
(* crigomgom rubgonmar ivamenjim wilmorort bowma pablucoto <br />
serrodcal anaprarod migtermor paupeddeg fraortmoy marpoldia1 <br />
ferrenseg josgarsan danrodcha manmorjim1 juacabsou dancorgar pabrodmac lucnovdos*)<br />
fun sinDuplicados :: "'a list ⇒ bool" where<br />
"sinDuplicados [] = True"<br />
| "sinDuplicados (x#xs) = (¬ estaEn x xs ∧ sinDuplicados xs)"<br />
<br />
value "sinDuplicados [1::nat,4,2] = True"<br />
value "sinDuplicados [1::nat,4,2,4] = False"<br />
<br />
(* paupeddeg *)<br />
(* Utilizando la función ∉ de Isabelle *)<br />
fun sinDuplicados2 :: "'a list ⇒ bool" where<br />
"sinDuplicados2 [] = True"<br />
| "sinDuplicados2 (a#xs) = ((a ∉ set xs) ∧ sinDuplicados2 xs ) "<br />
<br />
<br />
<br />
text {* <br />
--------------------------------------------------------------------- <br />
Ejercicio 3. Definir la función primitiva recursiva <br />
borraDuplicados :: 'a list ⇒ bool<br />
tal que (borraDuplicados xs) es la lista obtenida eliminando los<br />
elementos duplicados de la lista xs. Por ejemplo, <br />
borraDuplicados [1::nat,2,4,2,3] = [1,4,2,3]<br />
<br />
Nota: La función borraDuplicados es equivalente a la predefinida<br />
remdups. <br />
--------------------------------------------------------------------- <br />
*}<br />
<br />
(* crigomgom rubgonmar wilmorort bowma pablucoto serrodcal <br />
anaprarod migtermor paupeddeg fraortmoy marpoldia1 ferrenseg <br />
josgarsan danrodcha juacabsou dancorgar manmorjim1 pabrodmac lucnovdos*)<br />
fun borraDuplicados :: "'a list ⇒ 'a list" where<br />
"borraDuplicados [] = []"<br />
| "borraDuplicados (x#xs) =( if estaEn x xs then borraDuplicados xs else x#borraDuplicados xs)"<br />
<br />
<br />
value "borraDuplicados [1::nat,2,4,2,3] = [1,4,2,3]"<br />
<br />
(* ivamenjim *)<br />
(* Utilizando la negación primero *)<br />
<br />
fun borraDuplicados :: "'a list ⇒ 'a list" where<br />
"borraDuplicados [] = []"<br />
| "borraDuplicados (x#xs) = (if ¬(estaEn x xs) then (x#(borraDuplicados xs)) else borraDuplicados xs)"<br />
<br />
<br />
(* rubgonmar *)<br />
(* Otra forma Sin usar if <br />
Utilizando case aunque se le sacaría más partido con más de 2 casos *)<br />
<br />
fun borraDuplicados1 :: "'a list ⇒ 'a list" where<br />
"borraDuplicados1 [] = []" |<br />
"borraDuplicados1 (x#xs) = ( case estaEn x xs of False => x#borraDuplicados1 xs | True => borraDuplicados1 xs )"<br />
<br />
<br />
(* rubgonmar *)<br />
(*Otra forma utilizando let*)<br />
fun borraDuplicados2 :: "'a list ⇒ 'a list" where<br />
"borraDuplicados2 [] = []" |<br />
"borraDuplicados2 (x#xs) = (let condicion = estaEn x xs::bool in <br />
if condicion then borraDuplicados2 xs else x # borraDuplicados2 xs)"<br />
<br />
value "borraDuplicados [1::nat,2,4,2,3] = [1,4,2,3]"<br />
<br />
text {*<br />
--------------------------------------------------------------------- <br />
Ejercicio 4.1. Demostrar o refutar automáticamente<br />
length (borraDuplicados xs) ≤ length xs<br />
--------------------------------------------------------------------- <br />
*}<br />
<br />
-- "La demostración automática es"<br />
(*crigomgom anaprarod ferrenseg*)<br />
lemma length_borraDuplicados:<br />
"length (borraDuplicados xs) ≤ length xs"<br />
by (induct xs, simp_all)<br />
<br />
(* rubgonmar wilmorort pablucoto serrodcal migtermor paupeddeg <br />
fraortmoy marpoldia1 danrodcha juacabsou dancorgar josgarsan pabrodmac lucnovdos *)<br />
lemma length_borraDuplicados:<br />
"length ( borraDuplicados xs ) ≤ length xs"<br />
by (induct xs) auto<br />
<br />
(* manmorjim1 *)<br />
(* soy de ponerlo mejor por pasos *)<br />
lemma length_borraDuplicados:<br />
"length ( borraDuplicados xs ) ≤ length xs"<br />
apply (induct xs)<br />
apply auto<br />
done<br />
<br />
(* ivamenjim *)<br />
(* Demostrando objetivo a objetivo *)<br />
lemma length_borraDuplicados:<br />
"length (borraDuplicados xs) ≤ length xs"<br />
apply (induct xs)<br />
apply simp <br />
apply auto<br />
done<br />
<br />
(* bowma anaprarod *)<br />
lemma length_borraDuplicados:<br />
"length (borraDuplicados xs) ≤ length xs"<br />
apply (induct xs)<br />
apply (simp, simp) (* creo que es mejor poner aquí simp_all *)<br />
done<br />
<br />
(* ferrenseg *)<br />
<br />
lemma length_borraDuplicados:<br />
"length (borraDuplicados xs) ≤ length xs"<br />
by (induct xs) simp_all (* Creo que se puede poner simp_all fuera de parentesis *)<br />
<br />
text {*<br />
--------------------------------------------------------------------- <br />
Ejercicio 4.2. Demostrar o refutar detalladamente<br />
length (borraDuplicados xs) ≤ length xs<br />
--------------------------------------------------------------------- <br />
*}<br />
<br />
-- "La demostración estructurada es"<br />
(* crigomgom *)<br />
lemma length_borraDuplicados_2: <br />
"length (borraDuplicados xs) ≤ length xs"<br />
proof (induct xs)<br />
show "length (borraDuplicados []) ≤ length []" by simp<br />
next<br />
fix x xs<br />
assume HI: "length (borraDuplicados xs) ≤ length xs"<br />
show "length (borraDuplicados (x#xs)) ≤ length (x#xs)"<br />
proof (cases)<br />
assume "estaEn x xs"<br />
then have "length (borraDuplicados (x#xs)) = length (borraDuplicados xs)" by simp<br />
also have "... ≤ length xs" using HI by simp<br />
also have "... ≤ length (x#xs)" by simp<br />
finally show "length (borraDuplicados (x#xs)) ≤ length (x#xs)" by simp<br />
next<br />
assume "(¬ estaEn x xs)"<br />
then have "length (borraDuplicados (x#xs)) = length (x#borraDuplicados xs)" by simp<br />
also have "... = 1 + length (borraDuplicados xs)" by simp<br />
also have "... ≤ 1 + length xs" using HI by simp<br />
also have "... = length (x#xs)" by simp<br />
finally show "length (borraDuplicados (x#xs)) ≤ length (x#xs)" by simp<br />
qed<br />
qed<br />
<br />
(* ivamenjim wilmorort ferrenseg rubgonmar juacabsou dancorgar josgarsan lucnovdos*)<br />
lemma length_borraDuplicados_2: <br />
"length (borraDuplicados xs) ≤ length xs"<br />
proof (induct xs)<br />
show "length (borraDuplicados []) ≤ length []" by simp<br />
next <br />
fix a xs<br />
assume HI: "length (borraDuplicados xs) ≤ length xs"<br />
have "length (borraDuplicados (a # xs)) ≤ 1+length (borraDuplicados xs)" by simp<br />
also have "... ≤ 1+length xs" using HI by simp<br />
finally show "length (borraDuplicados (a # xs)) ≤ length (a # xs)" by simp<br />
qed<br />
<br />
(* serrodcal anaprarod danrodcha *)<br />
lemma length_borraDuplicados_2: "length (borraDuplicados xs) ≤ length xs" (is "?P xs")<br />
proof (induct xs)<br />
show "?P []" by simp<br />
next <br />
fix a xs<br />
assume HI: "?P xs"<br />
have "length (borraDuplicados (a # xs)) ≤ 1+length (borraDuplicados xs)" by simp<br />
also have "... ≤ 1+length xs" using HI by simp<br />
finally show "?P (a # xs)" by simp<br />
qed<br />
<br />
(* pablucoto *)<br />
lemma length_borraDuplicados_2: <br />
"length (borraDuplicados xs) ≤ length xs" <br />
proof(induct xs)<br />
show "length (borraDuplicados []) ≤ length [] " by simp<br />
next <br />
fix a xs<br />
assume HI: " length (borraDuplicados xs) ≤ length xs "<br />
have "length (borraDuplicados (a # xs)) ≤ 1 + length(borraDuplicados xs)" by simp<br />
also have "... ≤ 1 + length xs" using HI by simp <br />
also have "... ≤ length (a#xs)" by simp<br />
finally show "length (borraDuplicados (a # xs)) ≤ length (a # xs) " by simp<br />
qed<br />
<br />
(* bowma *)<br />
lemma length_borraDuplicados_2: <br />
"length (borraDuplicados xs) ≤ length xs" (is "?p xs")<br />
proof (induct xs)<br />
show "length (borraDuplicados []) ≤ length []" by simp<br />
next<br />
fix a xs<br />
assume HI: "?p xs"<br />
have c1:"1+length xs = length (a#xs)" by simp<br />
also have "length(borraDuplicados (a#xs)) ≤ 1 + length(borraDuplicados xs)" by simp<br />
also have "... ≤ 1+length xs" using HI by simp<br />
then show "length(borraDuplicados (a#xs)) ≤ length (a#xs)" using c1 by simp<br />
(* ¿Aquí porqué no puedo usar "finally show "?p (a # xs)" using c1 by simp? <br />
Y porque no puedo añadir "finally show *) "?p (a # xs)" by simp al final? *)<br />
qed<br />
<br />
(* migtermor *)<br />
<br />
lemma length_borraDuplicados_2: <br />
"length (borraDuplicados xs) ≤ length xs" (is "?P xs")<br />
proof (induct xs)<br />
show "?P []" by simp<br />
next<br />
fix a xs<br />
assume HI: "?P xs"<br />
have "length (borraDuplicados (a#xs)) ≤ (length (a#xs))"<br />
proof (cases)<br />
assume "(estaEn a xs)"<br />
then have Aux: "length (borraDuplicados (a#xs)) = length (borraDuplicados xs)" by simp<br />
also have "… ≤ length (a#xs)" using HI by simp<br />
then show "length (borraDuplicados (a#xs)) ≤ (length (a#xs))" using Aux by simp<br />
next<br />
assume "¬ (estaEn a xs)"<br />
then have Aux: "length (borraDuplicados (a#xs)) = 1+ length (borraDuplicados xs)" by simp<br />
also have "… ≤ length (a#xs)" using HI by simp<br />
then show "length (borraDuplicados (a#xs)) ≤ (length (a#xs))" using Aux by simp<br />
qed<br />
then show "?P (a#xs)" by simp<br />
qed<br />
<br />
(* paupeddeg marpoldia1 pabrodmac*)<br />
lemma length_borraDuplicados_4:<br />
"length (borraDuplicados xs) ≤ length xs"<br />
proof (induct xs)<br />
show "length (borraDuplicados []) ≤ length []" by simp<br />
next<br />
fix a xs <br />
assume HI: "length (borraDuplicados xs) ≤ length xs"<br />
have "length (borraDuplicados (a # xs)) ≤ length [a] + length (borraDuplicados xs) " by simp<br />
also have "... ≤ 1 + length (borraDuplicados xs)" by simp<br />
also have "... ≤ 1 + length xs" using HI by simp<br />
finally show "length (borraDuplicados (a # xs)) ≤ length (a # xs)" by simp<br />
qed<br />
<br />
(* fraortmoy *)<br />
(* muy parecida a alguna anterior, pero yo dí mas pasos *)<br />
lemma length_borraDuplicados_2: <br />
"length (borraDuplicados xs) ≤ length xs"<br />
proof (induct xs)<br />
show "length (borraDuplicados []) ≤ length []" by simp<br />
next<br />
fix a xs<br />
assume H1: "length (borraDuplicados xs) ≤ length xs"<br />
have "length (borraDuplicados (a # xs)) ≤ length(borraDuplicados [a])+length (borraDuplicados xs)" by simp<br />
also have "… ≤ 1 + length (borraDuplicados xs)" by simp<br />
also have "… ≤ 1 + length xs" using H1 by simp<br />
finally show "length (borraDuplicados (a # xs)) ≤ length (a # xs)" by simp<br />
qed<br />
<br />
text {*<br />
--------------------------------------------------------------------- <br />
Ejercicio 5.1. Demostrar o refutar automáticamente<br />
estaEn a (borraDuplicados xs) = estaEn a xs<br />
--------------------------------------------------------------------- <br />
*}<br />
<br />
-- "La demostración automática es"<br />
(* crigomgom rubgonmar wilmorort pablucoto serrodcal bowma <br />
migtermor fraortmoy marpoldia1 ferrenseg danrodcha juacabsou paupeddeg pabrodmac lucnovdos*)<br />
lemma estaEn_borraDuplicados: <br />
"estaEn a (borraDuplicados xs) = estaEn a xs"<br />
by (induct xs) auto<br />
<br />
(* ivamenjim manmorjim1 *)<br />
lemma estaEn_borraDuplicados: <br />
"estaEn a (borraDuplicados xs) = estaEn a xs"<br />
apply (induct xs) <br />
apply auto<br />
done<br />
<br />
(* anaprarod *)<br />
lemma estaEn_borraDuplicados: <br />
"estaEn a (borraDuplicados xs) = estaEn a xs"<br />
by (induct xs, simp_all, blast)<br />
<br />
<br />
(* anaprarod *)<br />
lemma estaEn_borraDuplicados': <br />
"estaEn a (borraDuplicados xs) = estaEn a xs"<br />
apply (induct xs)<br />
apply (cases "estaEn x xs")<br />
apply (simp_all)<br />
apply blast<br />
done<br />
<br />
(* bowma *)<br />
lemma estaEn_borraDuplicados: <br />
"estaEn a (borraDuplicados xs) = estaEn a xs"<br />
apply (induct xs)<br />
apply simp<br />
apply (simp, blast)<br />
done<br />
<br />
text {*<br />
--------------------------------------------------------------------- <br />
Ejercicio 5.2. Demostrar o refutar detalladamente<br />
estaEn a (borraDuplicados xs) = estaEn a xs<br />
Nota: Para la demostración de la equivalencia se puede usar<br />
proof (rule iffI)<br />
La regla iffI es<br />
⟦P ⟹ Q ; Q ⟹ P⟧ ⟹ P = Q<br />
--------------------------------------------------------------------- <br />
*}<br />
<br />
-- "La demostración estructurada es"<br />
<br />
(* wilmorort *)<br />
<br />
lemma estaEn_borraDuplicados_2: <br />
"estaEn a (borraDuplicados xs) = estaEn a xs"<br />
proof (induct xs)<br />
show "estaEn a (borraDuplicados []) = estaEn a []" by simp<br />
next<br />
fix b xs<br />
assume HI: "estaEn a (borraDuplicados xs) = estaEn a xs"<br />
show "estaEn a (borraDuplicados (b#xs)) = estaEn a (b#xs)"<br />
proof (rule iffI)<br />
assume H1: "estaEn a (borraDuplicados (b#xs))"<br />
show "estaEn a (b#xs)"<br />
proof (cases)<br />
assume "estaEn b xs"<br />
then have "estaEn a (borraDuplicados xs)" using H1 by simp<br />
then have "estaEn a xs" using HI by simp<br />
then show "estaEn a (b#xs)" by simp<br />
next<br />
assume "¬ estaEn b xs"<br />
then have "estaEn a (b#(borraDuplicados xs))" using H1 by simp<br />
then have "a=b ∨ (estaEn a (borraDuplicados xs))" by simp<br />
then have " a=b ∨ (estaEn a xs)" using HI by simp<br />
then show "estaEn a (b#xs)" by simp<br />
qed<br />
next<br />
assume H2: "estaEn a (b#xs)"<br />
show "estaEn a (borraDuplicados (b#xs))"<br />
proof (cases)<br />
assume "a=b"<br />
then have "estaEn b (borraDuplicados xs) = estaEn b xs" using HI by simp<br />
then have "(estaEn b xs ⟶ estaEn b (borraDuplicados xs)) ∧<br />
(¬ estaEn b xs ⟶ estaEn b (b # borraDuplicados xs))" by simp <br />
then have "estaEn b (borraDuplicados (b#xs))" by simp<br />
then show "estaEn a (borraDuplicados (b#xs))" using `a=b` by simp<br />
next<br />
assume "a≠b"<br />
then have "estaEn a (b#xs)" using H2 by simp<br />
then have "a = b ∨ estaEn a xs" by simp<br />
then have "False ∨ estaEn a xs " using `a≠b` by simp<br />
then have "estaEn a xs" by simp<br />
then have "estaEn a (borraDuplicados xs)" using HI by simp<br />
then show "estaEn a (borraDuplicados (b#xs))" using `a≠b` by simp<br />
qed<br />
qed<br />
qed<br />
<br />
<br />
(* anaprarod marpoldia1 ferrenseg juacabsou *)<br />
lemma estaEn_borraDuplicados_2:<br />
"estaEn a (borraDuplicados xs) = estaEn a xs" (is "?P xs")<br />
proof (induct xs)<br />
show "?P []" by simp<br />
next<br />
fix x xs<br />
assume HI: "?P xs"<br />
show "?P (x#xs)"<br />
proof (cases)<br />
assume "estaEn x xs"<br />
then have "estaEn a (borraDuplicados (x#xs)) = estaEn a (borraDuplicados xs)" by simp<br />
also have "...= estaEn a xs" using HI by simp<br />
finally show "?P (x#xs)" by auto<br />
next<br />
assume "¬estaEn x xs"<br />
then have "estaEn a (borraDuplicados (x#xs)) = estaEn a (x#borraDuplicados xs)" by simp<br />
also have "...= (x = a ∨ estaEn a (borraDuplicados xs))" by simp<br />
finally show "?P (x#xs)" using HI by simp<br />
qed<br />
qed<br />
<br />
<br />
(* migtermor *)<br />
<br />
lemma estaEn_borraDuplicados_2: <br />
"estaEn a (borraDuplicados xs) = estaEn a xs" (is "?P xs")<br />
proof (induct xs)<br />
show "?P []" by simp<br />
next <br />
fix aa xs<br />
assume HI: "?P xs"<br />
have P1: "estaEn a (borraDuplicados (aa#xs)) = estaEn a (aa#xs)"<br />
proof (cases)<br />
assume C1: "(estaEn aa xs)"<br />
have "estaEn a (borraDuplicados (aa#xs)) = estaEn a (borraDuplicados xs)" <br />
using C1 by simp<br />
also have P3: "… = estaEn a xs" using HI by simp<br />
also have "… = estaEn a (aa#xs)" <br />
proof (cases)<br />
assume "(a=aa)"<br />
then show "estaEn a xs = estaEn a (aa#xs)" using C1 by simp<br />
next<br />
assume "¬(a=aa)"<br />
then show "estaEn a xs = estaEn a (aa#xs)" by simp<br />
qed<br />
then show "estaEn a (borraDuplicados (aa#xs)) = estaEn a (aa#xs)" using P3 by simp<br />
next<br />
assume C2: "¬(estaEn aa xs)"<br />
then show "estaEn a (borraDuplicados (aa#xs)) = estaEn a (aa#xs)" using HI by simp<br />
qed<br />
also have Conc: "estaEn a (borraDuplicados (aa#xs)) = estaEn a (aa#xs)" using P1 by simp<br />
finally show "?P (aa#xs)" using Conc by simp<br />
qed<br />
<br />
(* crigomgom paupeddeg pabrodmac*)<br />
-- "La demostración estructurada es"<br />
lemma estaEn_borraDuplicados_2: <br />
"estaEn a (borraDuplicados xs) = estaEn a xs"<br />
proof (induct xs)<br />
show "estaEn a (borraDuplicados []) = estaEn a []" by simp<br />
next<br />
fix x xs<br />
assume HI: "estaEn a (borraDuplicados xs) = estaEn a xs"<br />
show "estaEn a (borraDuplicados (x#xs)) = estaEn a (x#xs)"<br />
proof (rule iffI)<br />
assume a1: "estaEn a (borraDuplicados (x#xs))"<br />
show "estaEn a (x#xs)"<br />
proof (cases)<br />
assume "estaEn x xs"<br />
then have "estaEn a (borraDuplicados xs)" using a1 by simp<br />
then have "estaEn a xs" using HI by simp<br />
then show "estaEn a (x#xs)" by simp<br />
next<br />
assume "¬ estaEn x xs"<br />
then have "estaEn a (x#(borraDuplicados xs))" using a1 by simp<br />
then have " x=a ∨ (estaEn a (borraDuplicados xs))" by simp<br />
then have " x=a ∨ (estaEn a xs)" using HI by simp<br />
then show "estaEn a (x#xs)" by simp<br />
qed<br />
next<br />
assume a2: "estaEn a (x#xs)"<br />
show "estaEn a (borraDuplicados (x#xs))"<br />
proof (cases)<br />
assume "a=x"<br />
then show "estaEn a (borraDuplicados (x#xs))" using HI by simp<br />
next<br />
assume b1: "a≠x"<br />
then have "estaEn a (x#xs)" using a2 by simp<br />
then have "x = a ∨ estaEn a xs" by simp<br />
then have "estaEn a xs " using b1 by simp<br />
then have "estaEn a (borraDuplicados xs)" using HI by simp<br />
then show "estaEn a (borraDuplicados (x#xs))" using b1 by simp<br />
qed<br />
qed<br />
qed<br />
<br />
<br />
(* rubgonmar *)<br />
lemma estaEn_borraDuplicados_2: <br />
"estaEn a ( borraDuplicados xs ) = estaEn a xs"<br />
proof (induct xs)<br />
show "estaEn a (borraDuplicados []) = estaEn a []" by simp<br />
next<br />
fix x<br />
fix xs<br />
assume HI: "estaEn a (borraDuplicados xs) = estaEn a xs"<br />
show "estaEn a (borraDuplicados (x#xs)) = estaEn a (x#xs)"<br />
proof (rule iffI) (* usamos proof de la regla dada iffI*)<br />
assume cprim: "estaEn a (borraDuplicados (x#xs))"<br />
show "estaEn a (x#xs)"<br />
proof (cases)<br />
assume "estaEn x xs"<br />
then show "estaEn a (x#xs)" using cprim HI by simp<br />
next<br />
assume "¬ estaEn x xs"<br />
then show "estaEn a (x#xs)" using cprim HI by simp<br />
qed<br />
next<br />
assume cseg: "estaEn a (x#xs)"<br />
show "estaEn a (borraDuplicados (x#xs))"<br />
proof (cases)<br />
assume "a=x"<br />
then show "estaEn a (borraDuplicados (x#xs))" using HI by auto<br />
next<br />
assume "a≠x"<br />
then show "estaEn a (borraDuplicados (x#xs))" using `a≠x` cseg HI by simp<br />
qed<br />
qed<br />
qed<br />
<br />
(* <br />
Aplico la regla iffI:<br />
⟦P ⟹ Q ; Q ⟹ P⟧ ⟹ P = Q<br />
Así:<br />
[estaEn a (borraDuplicados (x # xs)) ⟹ estaEn a (x # xs); estaEn a (x # xs) ⟹ estaEn a (borraDuplicados (x # xs))] <br />
⟹ estaEn a (borraDuplicados (x # xs)) = estaEn a (x # xs)<br />
*)<br />
<br />
<br />
(* bowma ivamenjim *)<br />
lemma estaEn_borraDuplicados_2: <br />
"estaEn a (borraDuplicados xs) = estaEn a xs" (is "?p xs")<br />
proof (induct xs)<br />
show "?p []" by simp<br />
next <br />
fix x xs<br />
assume HI: "?p xs"<br />
show "?p (x#xs)"<br />
proof (cases)<br />
assume H1:"estaEn x xs"<br />
then have "estaEn a (borraDuplicados (x#xs)) = estaEn a (borraDuplicados xs)" by simp<br />
also have "... = estaEn a xs" using HI by simp<br />
also have "... = estaEn a (x#xs)" <br />
proof(cases)<br />
assume "x=a"<br />
then show "estaEn a xs = estaEn a (x#xs)" using H1 by simp<br />
next<br />
assume "x≠a"<br />
then show "estaEn a xs = estaEn a (x#xs)" by simp<br />
qed<br />
finally show "?p (x#xs)" by simp<br />
next<br />
assume H2:"¬estaEn x xs"<br />
then have "estaEn a (borraDuplicados (x#xs)) = estaEn a (x#borraDuplicados xs)" by simp<br />
also have "... = ((x=a) ∨ estaEn a (borraDuplicados xs))" by simp<br />
also have "... = ((x=a) ∨ estaEn a xs)" using HI by simp<br />
also have "... = estaEn a (x#xs)" by simp<br />
finally show "?p (x#xs)" by simp<br />
qed<br />
<br />
(*danrodcha *)<br />
(* es como la de ruben pero con diferencias de estilo *)<br />
lemma estaEn_borraDuplicados_2: <br />
"estaEn a (borraDuplicados xs) = estaEn a xs" (is "?P xs")<br />
proof (induct xs)<br />
show "?P []" by simp<br />
next<br />
fix x xs assume HI: "?P xs"<br />
show "?P (x#xs)"<br />
proof (rule iffI)<br />
assume H1: "estaEn a (borraDuplicados (x # xs))"<br />
show "estaEn a (x#xs)"<br />
proof (cases "estaEn x xs")<br />
case True<br />
then show "estaEn a (x#xs)" using H1 HI by simp<br />
next<br />
case False<br />
then show "estaEn a (x#xs)" using H1 HI by simp<br />
qed<br />
next<br />
assume H2: "estaEn a (x#xs)"<br />
show "estaEn a (borraDuplicados (x # xs))"<br />
proof (cases "x=a")<br />
case True<br />
then show "estaEn a (borraDuplicados (x # xs))" using HI by simp<br />
next<br />
case False<br />
then show "estaEn a (borraDuplicados (x # xs))" using H2 HI by simp<br />
qed<br />
qed<br />
qed<br />
<br />
text {*<br />
--------------------------------------------------------------------- <br />
Ejercicio 6.1. Demostrar o refutar automáticamente<br />
sinDuplicados (borraDuplicados xs)<br />
--------------------------------------------------------------------- <br />
*}<br />
<br />
-- "La demostración automática"<br />
<br />
(* ivamenjim wilmorort serrodcal crigomgom anaprarod fraortmoy <br />
marpoldia1 ferrenseg danrodcha juacabsou paupeddeg josgarsan pabrodmac *)<br />
lemma sinDuplicados_borraDuplicados:<br />
"sinDuplicados (borraDuplicados xs)"<br />
by (induct xs) (auto simp add: estaEn_borraDuplicados)<br />
<br />
(* migtermor bowma *)<br />
lemma sinDuplicados_borraDuplicados:<br />
"sinDuplicados (borraDuplicados xs)"<br />
by (induct xs, simp_all add: estaEn_borraDuplicados_2)<br />
<br />
(* manmorjim1 no caí en usar la demostración anterior y he realizado<br />
la demostración de que si un elemento no estaba en una lista seguirá sin estar<br />
después de eliminar los duplicados en esa lista... *)<br />
lemma noEsta_tras_borrarDuplicados:<br />
"(¬estaEn x xs) ⟶ (¬estaEn x (borraDuplicados xs))"<br />
apply (induct xs)<br />
apply auto<br />
done<br />
<br />
lemma sinDuplicados_borraDuplicados:<br />
"sinDuplicados (borraDuplicados xs)"<br />
apply (induct xs)<br />
apply simp<br />
apply (induct xs)<br />
apply auto<br />
apply (simp add: noEsta_tras_borrarDuplicados)<br />
done<br />
<br />
text {*<br />
--------------------------------------------------------------------- <br />
Ejercicio 6.2. Demostrar o refutar detalladamente<br />
sinDuplicados (borraDuplicados xs)<br />
--------------------------------------------------------------------- <br />
*}<br />
<br />
(*wilmorort*)<br />
<br />
-- "La demostración estructurada es"<br />
lemma sinDuplicados_borraDuplicados_2:<br />
"sinDuplicados (borraDuplicados xs)"<br />
proof (induct xs)<br />
show " sinDuplicados (borraDuplicados [])" by simp<br />
next<br />
fix a xs<br />
assume HI: "sinDuplicados (borraDuplicados xs)"<br />
show "sinDuplicados (borraDuplicados (a # xs))"<br />
proof (cases)<br />
assume "estaEn a xs"<br />
then show "sinDuplicados (borraDuplicados (a#xs))" using HI by simp<br />
next<br />
assume"¬ estaEn a xs"<br />
then have "¬ (estaEn a xs) ∧ sinDuplicados (borraDuplicados xs)" using HI by simp<br />
then have "¬ estaEn a (borraDuplicados xs) ∧ sinDuplicados (borraDuplicados xs)" <br />
by (simp add: estaEn_borraDuplicados)<br />
then have " sinDuplicados (a#borraDuplicados xs)" by simp<br />
then show " sinDuplicados (borraDuplicados(a #xs))" by simp<br />
qed<br />
qed<br />
<br />
(* ivamenjim migtermor crigomgom rubgonmar fraortmoy marpoldia1 ferrenseg bowma juacabsou serrodcal josgarsan pabrodmac*)<br />
<br />
lemma sinDuplicados_borraDuplicados_2:<br />
"sinDuplicados (borraDuplicados xs)"<br />
proof (induct xs)<br />
show "sinDuplicados (borraDuplicados [])" by simp<br />
next<br />
fix a xs<br />
assume HI: "sinDuplicados (borraDuplicados xs)"<br />
show "sinDuplicados (borraDuplicados (a # xs))" <br />
proof (cases)<br />
assume "estaEn a xs" <br />
then show "sinDuplicados (borraDuplicados (a # xs))" using HI by simp<br />
next<br />
assume "¬(estaEn a xs)"<br />
then show "sinDuplicados (borraDuplicados (a # xs))" using HI by (simp add: estaEn_borraDuplicados)<br />
qed<br />
qed<br />
<br />
(* anaprarod paupeddeg*)<br />
lemma sinDuplicados_borraDuplicados_2:<br />
"sinDuplicados (borraDuplicados xs)" (is "?P xs")<br />
proof (induct xs)<br />
show "?P []" by simp<br />
next<br />
fix x xs<br />
assume HI: "?P xs"<br />
show "?P (x#xs)"<br />
proof (cases)<br />
assume c1: "estaEn x xs"<br />
then show "sinDuplicados (borraDuplicados (x#xs))" using HI by simp<br />
next<br />
assume c2: "¬ estaEn x xs"<br />
then have "sinDuplicados (borraDuplicados (x#xs)) =sinDuplicados (x#borraDuplicados xs)" by simp<br />
also have "…= (¬estaEn x (borraDuplicados xs) ∧ sinDuplicados (borraDuplicados xs))" by simp<br />
also have "… = (¬estaEn x (borraDuplicados xs))" using HI by simp<br />
also have "… = (¬(estaEn x xs))" by (simp add:estaEn_borraDuplicados)<br />
also have "… = True" using c2 by simp<br />
finally show "?P (x#xs)" by simp<br />
qed<br />
qed<br />
<br />
(* danrodcha *)<br />
lemma sinDuplicados_borraDuplicados_2:<br />
"sinDuplicados (borraDuplicados xs)" (is "?P xs")<br />
proof (induct xs)<br />
show "?P []" by simp<br />
next<br />
fix x xs assume HI: "?P xs"<br />
show "?P (x#xs)"<br />
proof (cases "estaEn x xs")<br />
case True<br />
then have 1: "sinDuplicados (borraDuplicados (x#xs)) <br />
= sinDuplicados (borraDuplicados xs)" by (simp add: estaEn_borraDuplicados_2)<br />
show "?P (x#xs)" using HI 1 by simp<br />
next<br />
case False<br />
then have "sinDuplicados (borraDuplicados (x#xs)) <br />
= sinDuplicados (x#borraDuplicados xs)" by simp<br />
also have "… = (¬ (estaEn x (borraDuplicados xs)) ∧ <br />
sinDuplicados (borraDuplicados xs))" by simp<br />
also have "… = True" using `¬ estaEn x xs` HI by (simp add:estaEn_borraDuplicados)<br />
finally show "?P (x#xs)" by simp<br />
qed<br />
qed<br />
<br />
text {*<br />
--------------------------------------------------------------------- <br />
Ejercicio 7. Demostrar o refutar:<br />
borraDuplicados (rev xs) = rev (borraDuplicados xs)<br />
--------------------------------------------------------------------- <br />
*}<br />
<br />
(*crigomgom rubgonmar ivamenjim wilmorort pablucoto migtermor <br />
anaprarod fraortmoy ferrenseg marpoldia1 bowma danrodcha juacabsou paupeddeg manmorjim1 serrodcal josgarsan pabrodmac lucnovdos*)<br />
lemma "borraDuplicados (rev xs) = rev (borraDuplicados xs)"<br />
quickcheck<br />
oops<br />
<br />
(* ivamenjim: Quickcheck encuentra el siguiente contraejemplo: <br />
xs = [a1, a2, a1]<br />
Por lo que:<br />
· "borraDuplicados (rev xs) = [a2, a1]"<br />
· "rev (borraDuplicados xs) = [a1, a2]" *)<br />
<br />
end<br />
</source></div>Lucnovdoshttps://www.glc.us.es/~jalonso/RA2016/index.php?title=Relaci%C3%B3n_4&diff=654Relación 42016-11-23T20:14:53Z<p>Lucnovdos: </p>
<hr />
<div><source lang="isar"><br />
chapter {* R4: Cuantificadores sobre listas *}<br />
<br />
theory R4_Cuantificadores_sobre_listas<br />
imports Main <br />
begin<br />
<br />
text {* <br />
--------------------------------------------------------------------- <br />
Ejercicio 1. Definir la función <br />
todos :: ('a ⇒ bool) ⇒ 'a list ⇒ bool<br />
tal que (todos p xs) se verifica si todos los elementos de la lista <br />
xs cumplen la propiedad p. Por ejemplo, se verifica <br />
todos (λx. 1<length x) [[2,1,4],[1,3]]<br />
¬todos (λx. 1<length x) [[2,1,4],[3]]<br />
<br />
Nota: La función todos es equivalente a la predefinida list_all. <br />
--------------------------------------------------------------------- <br />
*}<br />
<br />
{*danrodcha*}<br />
fun todos :: "('a ⇒ bool) ⇒ 'a list ⇒ bool" where<br />
"todos p xs = "todos p [] = True"<br />
| "todos p (x#xs) = (p x ∧ todos p xs)""<br />
<br />
(* ivamenjim migtermor dancorgar wilmorort marpoldia1 ferrenseg paupeddeg pablucoto crigomgom anaprarod serrodcal juacabsou rubgonmar josgarsan fraortmoy lucnovdos pabrodmac fracorjim1*)<br />
fun todos :: "('a ⇒ bool) ⇒ 'a list ⇒ bool" where<br />
"todos p [] = True"<br />
| "todos p (x#xs) = (p x ∧ todos p xs)"<br />
<br />
text {* <br />
--------------------------------------------------------------------- <br />
Ejercicio 2. Definir la función <br />
algunos :: ('a ⇒ bool) ⇒ 'a list ⇒ bool<br />
tal que (algunos p xs) se verifica si algunos elementos de la lista <br />
xs cumplen la propiedad p. Por ejemplo, se verifica <br />
algunos (λx. 1<length x) [[2,1,4],[3]]<br />
¬algunos (λx. 1<length x) [[],[3]]"<br />
<br />
Nota: La función algunos es equivalente a la predefinida list_ex. <br />
--------------------------------------------------------------------- <br />
*}<br />
<br />
{*danrodcha ivamenjim migtermor dancorgar marpoldia1 ferrenseg wilmorort paupeddeg pablucoto crigomgom anaprarod serrodcal juacabsou rubgonmar josgarsan fraortmoy lucnovdos pabrodmac*}<br />
fun algunos :: "('a ⇒ bool) ⇒ 'a list ⇒ bool" where<br />
"algunos p [] = False"<br />
| "algunos p (x#xs) = (p x ∨ algunos p xs)"<br />
<br />
(* fracorjim1. En esta versión el procesamiento se detiene al encontrar una coincidencia *)<br />
<br />
fun algunos :: "('a ⇒ bool) ⇒ 'a list ⇒ bool" where<br />
"algunos p [] = False"<br />
| "algunos p (x#xs) = (if p x then True else algunos p xs)"<br />
<br />
<br />
text {*<br />
--------------------------------------------------------------------- <br />
Ejercicio 3.1. Demostrar o refutar automáticamente <br />
todos (λx. P x ∧ Q x) xs = (todos P xs ∧ todos Q xs)<br />
--------------------------------------------------------------------- <br />
*}<br />
<br />
{*danrodcha ivamenjim migtermor dancorgar marpoldia1 ferrenseg wilmorort paupeddeg pablucoto anaprarod serrodcal juacabsou rubgonmar josgarsan fraortmoy lucnovdos pabrodmac *}<br />
lemma "todos (λx. P x ∧ Q x) xs = (todos P xs ∧ todos Q xs)"<br />
by (induct xs) auto<br />
<br />
(* fracorjim1 *)<br />
<br />
lemma "todos (λx. P x ∧ Q x) xs = (todos P xs ∧ todos Q xs)"<br />
apply (induct xs)<br />
apply auto<br />
done<br />
<br />
text {*<br />
--------------------------------------------------------------------- <br />
Ejercicio 3.2. Demostrar o refutar detalladamente<br />
todos (λx. P x ∧ Q x) xs = (todos P xs ∧ todos Q xs)<br />
--------------------------------------------------------------------- <br />
*}<br />
<br />
{*danrodcha fracorjim1*}<br />
lemma "todos (λx. P x ∧ Q x) xs = (todos P xs ∧ todos Q xs)"<br />
proof (induct xs)<br />
show "?R []" by simp<br />
next<br />
fix a xs <br />
assume HI: "?R xs"<br />
have "todos (λx. P x ∧ Q x) (a#xs) = (P a ∧ Q a ∧ todos (λx. P x ∧ Q x) xs)" by simp<br />
also have "… = (P a ∧ Q a ∧ (todos P xs ∧ todos Q xs))" using HI by simp<br />
also have "… = ((P a ∧ todos P xs) ∧ (Q a ∧ todos Q xs))" by blast<br />
also have "… = (todos P (a#xs) ∧ todos Q (a#xs))" by simp<br />
finally show "?R (a#xs)" by simp<br />
qed<br />
<br />
(* ivamenjim wilmorort serrodcal josgarsan*)<br />
<br />
lemma "todos (λx. P x ∧ Q x) xs = (todos P xs ∧ todos Q xs)"<br />
proof (induct xs)<br />
show "todos (λx. P x ∧ Q x) [] = (todos P [] ∧ todos Q [])" by simp<br />
next <br />
fix a xs<br />
assume HI: "todos (λx. P x ∧ Q x) xs = (todos P xs ∧ todos Q xs)"<br />
have "todos (λx. P x ∧ Q x) (a # xs) = (P a ∧ Q a ∧ todos (λx. P x ∧ Q x) xs)" by simp<br />
also have "... = (P a ∧ Q a ∧ todos P xs ∧ todos Q xs)" using HI by simp<br />
finally show "todos (λx. P x ∧ Q x) (a # xs) = (todos P (a # xs) ∧ todos Q (a # xs))" by auto<br />
qed<br />
<br />
(* dancorgar paupeddeg *)<br />
lemma "todos (λx. P x ∧ Q x) xs = (todos P xs ∧ todos Q xs)"<br />
proof (induct xs)<br />
show "todos (λx. P x ∧ Q x) [] = (todos P [] ∧ todos Q [])" by simp<br />
next<br />
fix y xs<br />
assume HI: "todos (λx. P x ∧ Q x) xs = (todos P xs ∧ todos Q xs)"<br />
have "todos (λx. P x ∧ Q x) (y#xs) = ((P y ∧ Q y) ∧ (todos (λx. P x ∧ Q x) xs))" by simp<br />
also have "… = ((P y ∧ Q y) ∧ (todos P xs ∧ todos Q xs))" using HI by simp<br />
also have "… = ((P y ∧ todos P xs) ∧ (Q y ∧ todos Q xs))" by blast<br />
also have "… = ((todos P (y#xs)) ∧ (todos Q (y#xs)))" by simp<br />
finally show "todos (λx. P x ∧ Q x) (y#xs) = (todos P (y#xs) ∧ todos Q (y#xs))" by simp<br />
qed<br />
<br />
<br />
(* migtermor *)<br />
lemma "todos (λx. P x ∧ Q x) xs = (todos P xs ∧ todos Q xs)"<br />
proof (induct xs)<br />
show "todos (λx. P x ∧ Q x) [] = (todos P [] ∧ todos Q [])" by simp<br />
next <br />
fix x xs<br />
assume HI: "todos (λx. P x ∧ Q x) xs = (todos P xs ∧ todos Q xs)"<br />
have "(todos (λx. P x ∧ Q x) (x#xs)) = (( P x ∧ Q x) ∧ (todos (λx. P x ∧ Q x) xs))"<br />
by (simp only: todos.simps(2))<br />
also have "… = ((P x ∧ Q x) ∧ (todos P xs ∧ todos Q xs))" using HI by simp<br />
also have "… = (((P x)∧(todos P xs)) ∧ ((Q x) ∧ (todos Q xs)))" by arith<br />
also have "((P x ∧ Q x) ∧ (todos P xs ∧ todos Q xs)) = (((P x)∧(todos P xs)) ∧ ((Q x) ∧ (todos Q xs)))"<br />
by arith (* Este paso es exactamente el mismo que el anterior, pero sin cualquiera de los dos no funciona el "finally show" *)<br />
have "… = (((P x)∧(todos P xs))∧((Q x)∧(todos Q xs)))" by simp<br />
have "… = ((todos P (x#xs))∧(todos Q (x#xs)))" by simp<br />
finally show "(todos (λx. P x ∧ Q x) (x#xs)) = ((todos P (x#xs)) ∧ (todos Q (x#xs)))" by simp <br />
qed<br />
<br />
(* marpoldia1 *)<br />
lemma "todos (λx. P x ∧ Q x) xs = (todos P xs ∧ todos Q xs)"<br />
proof (induct xs)<br />
show "todos (λx. P x ∧ Q x) [] = (todos P [] ∧ todos Q [])" by simp<br />
next<br />
fix a xs<br />
assume HI: "todos (λx. P x ∧ Q x) xs = (todos P xs ∧ todos Q xs)" <br />
have "todos (λx. P x ∧ Q x) (a # xs) = ((P a ∧ Q a) ∧ (todos P xs ∧ todos Q xs))" using HI by simp<br />
also have "... = ((P a ∧ todos P xs) ∧ (Q a ∧ todos Q xs))" by blast<br />
also have "... = (todos P (a#xs) ∧ todos Q (a#xs)) " by simp<br />
finally show "todos (λx. P x ∧ Q x) (a#xs) = (todos P (a#xs) ∧ todos Q (a#xs))" by simp <br />
qed<br />
<br />
(* ferrenseg *)<br />
<br />
lemma "todos (λx. P x ∧ Q x) xs = (todos P xs ∧ todos Q xs)"<br />
proof (induct xs)<br />
show "todos (λx. P x ∧ Q x) [] = (todos P [] ∧ todos Q [])" by simp<br />
next <br />
fix n xs<br />
assume HI: "todos (λx. P x ∧ Q x) xs = (todos P xs ∧ todos Q xs)"<br />
have "todos (λx. P x ∧ Q x) (n # xs) = (P n ∧ Q n ∧ todos P xs ∧ todos Q xs)" using HI by simp<br />
also have "… = P n ∧ Q n ∧ todos P xs ∧ todos Q xs" by blast<br />
also have "⋯ = todos P (n # xs) ∧ todos Q (n # xs)" by simp <br />
finally show "todos (λx. P x ∧ Q x) (n # xs) = todos P (n # xs) ∧ todos Q (n # xs)" by simp<br />
qed<br />
<br />
(* lucnovdos *)<br />
<br />
lemma "todos (λx. P x ∧ Q x) xs = (todos P xs ∧ todos Q xs)"<br />
proof (induct xs)<br />
show "todos (λx. P x ∧ Q x) [] = (todos P [] ∧ todos Q [])" by simp<br />
next<br />
fix n xs<br />
assume HI: "todos (λx. P x ∧ Q x) xs = (todos P xs ∧ todos Q xs)" <br />
have "todos (λx. P x ∧ Q x) (n # xs) = ((P n ∧ Q n) ∧ (todos P xs ∧ todos Q xs))" using HI by simp<br />
also have "... = ((P n ∧ todos P xs) ∧ (Q n ∧ todos Q xs))" by arith<br />
also have "... = ((todos P(n#xs)) ∧ (todos Q(n#xs)))" by simp<br />
<br />
finally show "todos (λx. P x ∧ Q x) (n#xs) = (todos P (n#xs) ∧ todos Q (n#xs))" by simp <br />
qed<br />
<br />
{* wilmorort pablucoto crigomgom anaprarod juacabsou rubgonmar fraortmoy *}<br />
<br />
lemma "todos (λx. P x ∧ Q x) xs = (todos P xs ∧ todos Q xs)"<br />
proof (induct xs)<br />
show "todos (λx. P x ∧ Q x) [] = (todos P [] ∧ todos Q [])" by simp <br />
next<br />
fix a xs<br />
assume HI: "todos (λx. P x ∧ Q x) xs = (todos P xs ∧ todos Q xs)"<br />
have "todos (λx. P x ∧ Q x) (a # xs) = (P a ∧ Q a ∧ todos (λx. P x ∧ Q x) xs)" by simp<br />
also have "... = (P a ∧ Q a ∧ todos P xs ∧ todos Q xs)" using HI by simp<br />
also have "... = (P a ∧ todos P xs ∧ Q a ∧ todos Q xs)" by arith<br />
also have "... = (todos P (a # xs) ∧ todos Q (a # xs))" by simp (* Este paso se puede obviar*)<br />
finally show "todos (λx. P x ∧ Q x) (a # xs) = (todos P (a # xs) ∧ todos Q (a # xs))" by simp<br />
qed<br />
<br />
(* pabrodmac *)<br />
<br />
lemma "todos (λx. P x ∧ Q x) xs = (todos P xs ∧ todos Q xs)"<br />
proof (induct xs)<br />
show "todos (λx. P x ∧ Q x) [] = (todos P [] ∧ todos Q [])" by simp<br />
next <br />
fix x xs<br />
assume HI: "todos (λx. P x ∧ Q x) xs = (todos P xs ∧ todos Q xs)" <br />
have "todos (λx. P x ∧ Q x) (x#xs) = ( P x ∧ Q x ∧ todos (λx. P x ∧ Q x) xs)" by simp<br />
also have "... = (P x ∧ Q x ∧ (todos P xs ∧ todos Q xs))" using HI by simp<br />
also have "... = (P x ∧ todos P xs ∧ Q x ∧ todos Q xs)" by arith<br />
finally show "todos (λx. P x ∧ Q x) (x#xs) = (todos P (x#xs) ∧ todos Q (x#xs))" by simp<br />
qed<br />
<br />
text {*<br />
--------------------------------------------------------------------- <br />
Ejercicio 4.1. Demostrar o refutar automáticamente<br />
todos P (x @ y) = (todos P x ∧ todos P y)<br />
--------------------------------------------------------------------- <br />
*}<br />
{*danrodcha ivamenjim marpoldia1 migtermor ferrenseg wilmorort paupeddeg crigomgom anaprarod serrodcal juacabsou rubgonmar pablucoto fraortmoy josgarsan lucnovdos pabrodmac*}<br />
lemma "todos P (x @ y) = (todos P x ∧ todos P y)"<br />
by (induct x) auto<br />
<br />
{* anaprarod dancorgar fracorjim1*}<br />
lemma "todos P (x @ y) = (todos P x ∧ todos P y)"<br />
apply (induct x)<br />
apply auto<br />
done<br />
<br />
text {*<br />
--------------------------------------------------------------------- <br />
Ejercicio 4.2. Demostrar o refutar detalladamente<br />
todos P (x @ y) = (todos P x ∧ todos P y)<br />
--------------------------------------------------------------------- <br />
*}<br />
<br />
(* ivamenjim ferrenseg wilmorort dancorgar fraortmoy josgarsan lucnovdos*)<br />
<br />
lemma todos_append:<br />
"todos P (x @ y) = (todos P x ∧ todos P y)"<br />
proof (induct x)<br />
show "todos P ([] @ y) = (todos P [] ∧ todos P y)" by simp<br />
next <br />
fix a x<br />
assume HI: "todos P (x @ y) = (todos P x ∧ todos P y)"<br />
have "todos P ((a#x) @ y) = (P a ∧ (todos P (x @ y)))" by simp<br />
also have "... = (P a ∧ (todos P x ∧ todos P y))" using HI by simp<br />
finally show "todos P ((a # x) @ y) = (todos P (a # x) ∧ todos P y)" by simp<br />
qed<br />
<br />
(* migtermor serrodcal *)<br />
<br />
lemma todos_append1:<br />
"todos P (x @ y) = (todos P x ∧ todos P y)" (is "?P x")<br />
proof (induct x)<br />
show "?P []" by simp<br />
next <br />
fix a x<br />
assume HI: "?P x"<br />
have "todos P ((a#x) @ y) = (P a ∧ (todos P (x @ y)))" by simp<br />
also have "… = (P a ∧ (todos P x ∧ todos P y))" using HI by simp<br />
finally show "?P (a#x)" by simp<br />
qed <br />
<br />
(* marpoldia1 *)<br />
lemma todos_append:<br />
"todos P (x @ y) = (todos P x ∧ todos P y)"<br />
proof (induct x)<br />
show "todos P ([] @ y) = (todos P [] ∧ todos P y)" by simp<br />
next<br />
fix a x<br />
assume HI: "todos P (x @ y) = (todos P x ∧ todos P y)"<br />
have "todos P ((a # x) @ y) = todos P (a#(x@y))" by simp<br />
also have "... = (P a ∧ todos P (x @ y))" by simp<br />
also have "... = (P a ∧ (todos P x ∧ todos P y))" using HI by simp<br />
finally show "todos P ((a # x) @ y) = (todos P (a # x) ∧ todos P y)" by simp <br />
qed<br />
<br />
(* paupeddeg *)<br />
lemma todos_append3:<br />
"todos P (x @ y) = (todos P x ∧ todos P y)"<br />
proof (induct x)<br />
show "todos P ([] @ y) = (todos P [] ∧ todos P y)" by simp<br />
next<br />
fix a x<br />
assume HI: "todos P (x @ y) = (todos P x ∧ todos P y)"<br />
have "todos P ((a # x) @ y) = todos P (a # (x @ y))" by simp<br />
also have "... = (todos P [a] ∧ todos P (x @ y)) " by simp<br />
also have "... = (todos P [a] ∧ (todos P x ∧ todos P y))" using HI by simp<br />
finally show "todos P ((a # x) @ y) = (todos P (a # x) ∧ todos P y)" by simp<br />
qed<br />
<br />
(* crigomgom juacabsou anaprarod*)<br />
<br />
lemma todos_append4:<br />
"todos P (x @ y) = (todos P x ∧ todos P y)"<br />
proof (induct x)<br />
show "todos P ([] @ y) = (todos P [] ∧ todos P y)" by simp<br />
next<br />
fix x xs<br />
assume HI:"todos P (xs @ y) = (todos P xs ∧ todos P y)"<br />
have "todos P ((x # xs) @ y) = (P x ∧ todos P (xs @ y ))" by simp<br />
also have "... = (P x ∧ (todos P xs ∧ todos P y)) " using HI by simp<br />
also have "... = ((P x ∧ todos P xs) ∧ todos P y)" by simp<br />
finally show "todos P ((x # xs) @ y) = (todos P (x # xs) ∧ todos P y)" by simp<br />
qed<br />
<br />
(* pablucoto *)<br />
lemma todos_append5:<br />
"todos P (x @ y) = (todos P x ∧ todos P y)"<br />
proof (induct x )<br />
show "todos P ([] @ y) = (todos P [] ∧ todos P y)" by simp<br />
next<br />
fix a x<br />
assume HI:"todos P (x @ y) = (todos P x ∧ todos P y) " <br />
have "todos P ((a # x) @ y) = ( P a ∧ todos P (x @ y))" by simp<br />
also have " ... =( P a ∧ (todos P x ∧ todos P y))" using HI by simp<br />
also have "... = (todos P (a#x) ∧ todos P y)" by simp<br />
finally show "todos P ((a # x) @ y) = (todos P (a # x) ∧ todos P y)" by auto<br />
qed<br />
<br />
{*danrodcha*}<br />
lemma todos_append6:<br />
"todos P (x @ y) = (todos P x ∧ todos P y)" (is "?Q x")<br />
proof (induct x)<br />
show "?Q []" by simp<br />
next<br />
fix x a assume HI: "?Q x"<br />
have "todos P ((a # x) @ y) = todos P (a # x @ y)" by simp<br />
also have "… = ((P a) ∧ todos P (x @ y))" by simp<br />
also have "… = ((P a) ∧ (todos P x ∧ todos P y))" using HI by simp<br />
also have "… = (todos P (a # x) ∧ todos P y)" by simp<br />
finally show "?Q (a # x)" by simp<br />
qed<br />
<br />
{*pabrodmac*}<br />
lemma todos_append7:<br />
"todos P (x @ y) = (todos P x ∧ todos P y)"<br />
proof (induct x)<br />
show "todos P ([] @ y) = (todos P [] ∧ todos P y)" by simp<br />
next <br />
fix a x<br />
assume HI: "todos P (x @ y) = (todos P x ∧ todos P y)" <br />
have "todos P ((a#x) @ y) = (P a ∧ todos P (x @ y))" by simp<br />
also have "... = (P a ∧ (todos P x ∧ todos P y))" using HI by simp<br />
finally show "todos P ((a#x) @ y) = (todos P (a#x) ∧ todos P y)" by simp<br />
qed<br />
<br />
<br />
text {*<br />
--------------------------------------------------------------------- <br />
Ejercicio 5.1. Demostrar o refutar automáticamente <br />
todos P (rev xs) = todos P xs<br />
--------------------------------------------------------------------- <br />
*}<br />
<br />
lemma "todos P (rev xs) = todos P xs"<br />
oops<br />
<br />
(* migtermor ivamenjim marpoldia1 serrodcal anaprarod paupeddeg dancorgar *)<br />
lemma "todos P (rev xs) = todos P xs" <br />
apply (induct xs)<br />
apply simp<br />
apply (simp add: todos_append)<br />
apply auto<br />
done<br />
<br />
(* ferrenseg crigomgom rubgonmar fraortmoy josgarsan danrodcha lucnovdos pabrodmac*)<br />
<br />
lemma "todos P (rev xs) = todos P xs"<br />
by (induct xs) (auto simp add: todos_append)<br />
<br />
(* wilmorort *)<br />
lemma "todos P (rev xs) = todos P xs"<br />
by (induct xs,simp,simp add: todos_append,auto)<br />
<br />
(* juacabsou *)<br />
lemma "todos P (rev xs) = todos P xs"<br />
apply (induct xs,simp,simp add: todos_append,auto) done<br />
<br />
(* pablucoto * )<br />
lemma "todos P (rev xs) = todos P xs"<br />
by (induct xs) (auto, simp_all add: todos_append) <br />
lemma "todos P (rev xs) = todos P xs"<br />
by (induct xs) ( simp_all add: todos_append, auto)<br />
<br />
text {*<br />
--------------------------------------------------------------------- <br />
Ejercicio 5.2. Demostrar o refutar detalladamente<br />
todos P (rev xs) = todos P xs<br />
--------------------------------------------------------------------- <br />
*}<br />
<br />
lemma "todos P (rev xs) = todos P xs"<br />
oops<br />
<br />
(* migtermor *)<br />
<br />
lemma auxiliar:<br />
"rev (a#xs) = rev xs @ [a]"<br />
by auto<br />
<br />
lemma "todos P (rev xs) = todos P xs" (is "?Q xs")<br />
proof (induct xs)<br />
show "?Q []" by (simp only: rev.simps(1))<br />
next <br />
fix a xs<br />
assume HI: "?Q xs"<br />
have "todos P (rev (a#xs)) = (todos P (rev xs @ [a]))" by (simp add: auxiliar)<br />
have "… = ((todos P (rev xs)) ∧ (todos P [a]))" by (simp add: todos_append)<br />
have "… = (todos P (rev xs) ∧ P a)" by simp<br />
also have Aux: "… = (todos P xs ∧ P a)" using HI by simp<br />
have Aux1: "… = (P a ∧ todos P xs)" by arith<br />
have "(todos P (rev xs) ∧ P a) = (P a ∧ todos P xs)" using Aux Aux1 by simp<br />
finally show "todos P (rev (a#xs)) = todos P (a#xs)" by (simp add: todos_append)<br />
qed<br />
<br />
<br />
(* marpoldia1 ferrenseg crigomgom serrodcal juacabsou rubgonmar josgarsan pablucoto pabrodmac lucnovdos*)<br />
lemma "todos P (rev xs) = todos P xs"<br />
proof (induct xs)<br />
show "todos P (rev []) = todos P []" by simp<br />
next<br />
fix a xs<br />
assume HI: "todos P (rev xs) = todos P xs"<br />
have "todos P (rev (a # xs)) = (todos P ((rev xs)@[a]))" by simp<br />
also have "... = (todos P (rev xs) ∧ todos P [a])" by (simp add:todos_append)<br />
also have "... = (todos P xs ∧ P a)" using HI by simp<br />
also have "... = (P a ∧ todos P xs)" by arith<br />
also have "... = (todos P (a#xs))" by simp<br />
finally show "todos P (rev (a # xs)) = (todos P (a#xs))" by simp <br />
qed<br />
<br />
(* ivamenjim *)<br />
<br />
lemma "todos P (rev xs) = todos P xs" (is "?P xs")<br />
proof (induct xs)<br />
show "?P []" by simp<br />
next <br />
fix a xs<br />
assume HI: "?P xs"<br />
have "todos P (rev (a # xs)) = (todos P ((rev xs)@[a]))" by simp<br />
also have "... = ((todos P (rev xs)) ∧ todos P [a])" by (simp add: todos_append)<br />
also have "... = (todos P xs ∧ todos P [a])" using HI by simp<br />
also have "... = (todos P [a] ∧ todos P xs)" by arith<br />
finally show "todos P (rev (a # xs)) = todos P (a # xs)" by simp<br />
qed<br />
<br />
(* fraortmoy serrodcal *)<br />
<br />
lemma "todos P (rev xs) = todos P xs"<br />
proof (induct xs)<br />
show "todos P (rev []) = todos P []" by simp<br />
next<br />
fix a xs <br />
assume H1: "todos P (rev xs) = todos P xs"<br />
have " todos P (rev (a # xs)) = todos P (rev xs @ [a])" by simp<br />
also have "… = (todos P (rev xs) ∧ todos P [a])" by (simp add:todos_append)<br />
also have "… = (todos P xs ∧ todos P [a])" using H1 by simp<br />
also have "… = (todos P [a] ∧ todos P xs)" by arith<br />
finally show "todos P (rev (a # xs)) = todos P (a#xs)" by simp<br />
qed<br />
<br />
(* paupeddeg dancorgar *)<br />
lemma "todos P (rev xs) = todos P xs"<br />
proof (induct xs)<br />
show " todos P (rev []) = todos P []" by simp<br />
next<br />
fix a xs<br />
assume HI: "todos P (rev xs) = todos P xs "<br />
have "todos P (rev (a # xs)) = todos P (rev(xs) @ [a])" by simp<br />
also have "... = (todos P (rev xs) ∧ todos P [a]) " by (simp add: todos_append)<br />
also have "... = (todos P xs ∧ todos P [a])" using HI by simp<br />
also have "... = (todos P [a] ∧ todos P xs)" by arith<br />
also have "... = todos P ([a] @ xs)" by (simp add: todos_append)<br />
finally show "todos P (rev (a # xs)) = todos P (a # xs)" by simp<br />
qed <br />
<br />
(* wilmorort *)<br />
lemma "todos P (rev xs) = todos P xs" (is "?P xs")<br />
proof (induct xs) <br />
show "?P []" by simp <br />
next<br />
fix a xs<br />
assume HI: "?P xs"<br />
have "todos P (rev (a#xs)) = todos P ((rev xs) @ [a])" by simp<br />
also have "... = (todos P(rev xs) ∧ todos P [a])" by (simp add: todos_append)<br />
also have "... = (todos P xs ∧ todos P [a])" using HI by simp<br />
also have "... = (todos P [a] ∧ todos P xs)" by (simp add: HOL.conj_commute)<br />
also have "... = todos P([a]@(xs))" by (simp)<br />
finally show "todos P (rev (a#xs))= todos P (a # xs)" by simp<br />
qed<br />
<br />
(* anaprarod *) <br />
(* es igual que las anteriores pero con el final también con patrones *)<br />
<br />
lemma "todos P (rev xs) = todos P xs" (is "?P xs")<br />
proof (induct xs)<br />
show "?P []" by auto<br />
next<br />
fix a xs<br />
assume HI: "?P xs"<br />
have "todos P (rev (a#xs)) = todos P (rev xs @ [a])" by simp<br />
also have "… = (todos P (rev xs) ∧ todos P [a])" by (simp add: todos_append)<br />
also have "… = (todos P xs ∧ todos P [a])" using HI by simp<br />
also have "… = (todos P [a] ∧ todos P xs)" by arith<br />
also have "… = todos P (a#xs)" by simp<br />
finally show "?P (a # xs)" by simp<br />
qed<br />
<br />
{*danrodcha*}<br />
lemma "todos P (rev xs) = todos P xs" (is "?Q xs")<br />
proof (induct xs)<br />
show "?Q []" by simp<br />
next<br />
fix a xs assume HI: "?Q xs"<br />
have "todos P (rev (a#xs)) = todos P ((rev xs) @ [a])" by simp<br />
also have "… = (todos P (rev xs) ∧ todos P [a])" by (simp add:todos_append)<br />
also have "… = (todos P xs ∧ todos P [a])" using HI by simp<br />
also have "… = (todos P [a] ∧ todos P xs)" by (simp add: HOL.conj_commute)<br />
also have "… = todos P (a # xs)" by simp<br />
finally show "?Q (a # xs)" by simp<br />
qed<br />
<br />
text {*<br />
--------------------------------------------------------------------- <br />
Ejercicio 6. Demostrar o refutar:<br />
algunos (λx. P x ∧ Q x) xs = (algunos P xs ∧ algunos Q xs)<br />
--------------------------------------------------------------------- <br />
*}<br />
<br />
lemma "algunos (λx. P x ∧ Q x) xs = (algunos P xs ∧ algunos Q xs)"<br />
oops<br />
<br />
(* migtermor ivamenjim ferrenseg paupeddeg crigomgom serrodcal wilmorort juacabsou rubgonmar anaprarod marpoldia1 fraortmoy josgarsan danrodcha pablucoto lucnovdos*)<br />
lemma "algunos (λx. P x ∧ Q x) xs = (algunos P xs ∧ algunos Q xs)" <br />
quickcheck<br />
oops<br />
(* Quickcheck encuentra el siguiente contraejemplo: P={a1}, Q={a2}, xs={a1,a2}. En este ejemplo:<br />
· "algunos (λx. P x ∧ Q x) xs = False"<br />
· "(algunos P xs ∧ algunos Q xs) = True" *)<br />
<br />
(* dancorgar *)<br />
lemma "algunos (λx. P x ∧ Q x) xs = (algunos P xs ∧ algunos Q xs)"<br />
proof -<br />
assume H1: "xs = [a, b]"<br />
assume H2: "P a = True"<br />
assume H3: "Q a = False"<br />
assume H4: "P b = False"<br />
assume H5: "Q b = True" <br />
have F1: "(algunos P xs ∧ algunos Q xs) = True" using H1 H2 H3 H4 H5 by simp<br />
have F2: "algunos (λx. P x ∧ Q x) xs = False" using H1 H2 H3 H4 H5 by simp<br />
have "algunos (λx. P x ∧ Q x) xs ≠ (algunos P xs ∧ algunos Q xs)" using F1 F2 by simp<br />
oops<br />
<br />
text {*<br />
--------------------------------------------------------------------- <br />
Ejercicio 7.1. Demostrar o refutar automáticamente <br />
algunos P (map f xs) = algunos (P ∘ f) xs<br />
--------------------------------------------------------------------- <br />
*}<br />
<br />
(* ivamenjim migtermor marpoldia1 crigomgom rubgonmar wilmorort anaprarod <br />
fraortmoy juacabsou paupeddeg josgarsan danrodcha pablucoto lucnovdos *)<br />
<br />
lemma "algunos P (map f xs) = algunos (P o f) xs"<br />
by (induct xs) auto<br />
<br />
(* ferrenseg *)<br />
<br />
lemma "algunos P (map f xs) = algunos (P o f) xs"<br />
by (induct xs) simp_all<br />
<br />
(* anaprarod dancorgar serrodcal *)<br />
lemma "algunos P (map f xs) = algunos (P o f) xs"<br />
apply (induct xs)<br />
apply auto<br />
done<br />
<br />
text {*<br />
--------------------------------------------------------------------- <br />
Ejercicio 7.2. Demostrar o refutar detalladamente<br />
algunos P (map f xs) = algunos (P ∘ f) xs<br />
--------------------------------------------------------------------- <br />
*}<br />
<br />
lemma "algunos P (map f xs) = algunos (P o f) xs"<br />
oops<br />
<br />
(* migtermor *)<br />
<br />
lemma AUX: "algunos (λa. P (f a)) xs = algunos (P o f) xs"<br />
by (induct xs) auto<br />
<br />
lemma "algunos P (map f xs) = algunos (P o f) xs" (is "?Q xs")<br />
proof (induct xs)<br />
show "?Q []" by simp<br />
next<br />
fix x xs<br />
assume HI: "?Q xs"<br />
have "algunos P (map f (x#xs)) = (algunos P ((f x)#(map f xs)))" by simp<br />
also have "… = ((P (f x)) ∨ (algunos P (map f xs)))" by (simp only: algunos.simps(2))<br />
also have "… = ((P (f x)) ∨ (algunos (P o f) xs))" <br />
proof (cases)<br />
assume C1: "(P (f x))"<br />
have Aux: "((P (f x)) ∨ (algunos P (map f xs))) = True" using C1 by simp<br />
have Aux1: "… = ((P (f x)) ∨ (algunos (P o f) xs))" using C1 by simp<br />
then show "((P (f x)) ∨ (algunos P (map f xs))) = ((P (f x)) ∨ (algunos (P o f) xs))" <br />
using Aux Aux1 by simp<br />
next<br />
assume C2: "¬(P (f x))"<br />
have Aux2: "((P (f x)) ∨ (algunos P (map f xs))) = (algunos P (map f xs))" using C2 by simp<br />
have Aux3: "… = (algunos (P o f) xs)" using HI by (simp add: AUX)<br />
also have "… = ((P (f x)) ∨ (algunos (P o f) xs))" using C2 by simp<br />
then show "((P (f x)) ∨ (algunos P (map f xs))) = ((P (f x)) ∨ (algunos (P o f) xs))" <br />
using Aux2 Aux3 by simp<br />
qed<br />
also have "… = (((P o f) x) ∨ (algunos (P o f) xs))" by simp<br />
also have "… = (algunos (P o f) (x#xs))" by simp<br />
finally show "algunos P (map f (x#xs)) = (algunos (P o f) (x#xs))" by simp<br />
qed<br />
<br />
(* ferrenseg *)<br />
<br />
lemma "algunos P (map f xs) = algunos (P ∘ f) xs"<br />
proof (induct xs)<br />
show "algunos P (map f []) = algunos (P ∘ f) []" by simp<br />
next<br />
fix x xs<br />
assume HI: "algunos P (map f xs) = algunos (P ∘ f) xs"<br />
show "algunos P (map f (x # xs)) = algunos (P ∘ f) (x # xs)"<br />
proof -<br />
have "algunos P (map f (x # xs)) = algunos P ((f x) # (map f xs))" by simp<br />
also have "… = ((P (f x)) ∨ (algunos P (map f xs)))" by simp<br />
also have "… = (((P ∘ f) x) ∨ (algunos (P ∘ f) xs))" using HI by simp<br />
also have "… = algunos (P ∘ f) (x # xs)" by simp<br />
finally show ?thesis<br />
qed<br />
qed<br />
<br />
(* ivamenjim *)<br />
<br />
lemma "algunos P (map f xs) = algunos (P o f) xs"<br />
proof (induct xs)<br />
show "algunos P (map f []) = algunos (P ∘ f) []" by simp<br />
next <br />
fix a xs<br />
assume HI: "algunos P (map f xs) = algunos (P ∘ f) xs"<br />
have "algunos P (map f (a # xs)) = algunos P ((f a)#map f xs)" by simp<br />
also have "... = (algunos P (map f [a]) ∨ algunos P (map f xs))" by simp <br />
also have "... = (algunos P (map f [a]) ∨ algunos (P ∘ f) xs)" using HI by simp <br />
finally show "algunos P (map f (a # xs)) = algunos (P ∘ f) (a # xs)" by simp<br />
qed<br />
<br />
(* crigomgom rubgonmar anaprarod marpoldia1 juacabsou danrodcha paupeddeg josgarsan lucnovdos*)<br />
lemma "algunos P (map f xs) = algunos (P o f) xs"<br />
proof (induct xs)<br />
show "algunos P (map f []) = algunos (P ∘ f) []" by simp<br />
next<br />
fix x xs<br />
assume HI: "algunos P (map f xs) = algunos (P ∘ f) xs"<br />
have "algunos P (map f (x # xs)) = algunos P ((f x) # (map f xs))" by simp<br />
also have "... = (P (f x) ∨ algunos P (map f xs))" by simp<br />
also have "... = (P (f x) ∨ algunos (P ∘ f) xs)" using HI by simp<br />
also have "... = ((P ∘ f) x ∨ algunos (P ∘ f) xs)" by simp<br />
finally show "algunos P (map f (x # xs)) = algunos (P ∘ f) (x # xs)" by simp<br />
qed<br />
<br />
(* wilmorort pablucoto*)<br />
lemma "algunos P (map f xs) = algunos (P o f) xs" (is "?P xs" )<br />
proof (induct xs)<br />
show "?P []" by simp<br />
next<br />
fix a xs<br />
assume HI: "algunos P (map f xs) = algunos (P ∘ f) xs"<br />
have "algunos P (map f (a # xs)) = (P (f a) ∨ algunos P (map f xs))" by simp<br />
also have "... = (P (f a) ∨ algunos (P ∘ f) xs )" using HI by simp<br />
also have "... = ((P ∘ f) a ∨ algunos (P ∘ f)xs)" by simp<br />
finally show " algunos P (map f (a # xs)) = algunos (P ∘ f) (a # xs)" by simp<br />
qed<br />
<br />
(* fraortmoy serrodcal *)<br />
<br />
lemma "algunos P (map f xs) = algunos (P o f) xs"<br />
proof (induct xs)<br />
show "algunos P (map f []) = algunos (P ∘ f) []" by simp<br />
next<br />
fix a xs<br />
assume H1: "algunos P (map f xs) = algunos (P ∘ f) xs"<br />
have "algunos P (map f (a # xs)) = (algunos P (map f [a]) ∨ algunos P (map f xs))" by simp<br />
also have "… = (algunos (P ∘ f) [a] ∨ algunos (P ∘ f) xs )" using H1 by simp<br />
also have "… = algunos (P ∘ f) (a#xs)" by simp<br />
finally show "algunos P (map f (a # xs)) = algunos (P ∘ f) (a#xs)" by simp<br />
qed<br />
<br />
{*danrodcha*}<br />
lemma "algunos P (map f xs) = algunos (P o f) xs" (is "?Q xs")<br />
proof (induct xs)<br />
show "?Q []" by simp<br />
next<br />
fix a xs assume HI: "?Q xs"<br />
have " algunos P (map f (a # xs)) = algunos P ((f a)#(map f xs))" by simp<br />
also have "… = (P (f a) ∨ algunos P (map f xs))" by simp<br />
also have "… = (P (f a) ∨ algunos (P ∘ f) xs)" using HI by blast<br />
also have "… = ((P ∘ f) a ∨ algunos (P ∘ f) xs)" by simp<br />
also have "… = algunos (P ∘ f) (a # xs)" by simp<br />
finally show "?Q (a # xs)" by blast<br />
qed<br />
<br />
(* dancorgar *)<br />
lemma "algunos P (map f xs) = algunos (P o f) xs"<br />
proof (induct xs)<br />
show "algunos P (map f []) = algunos (P o f) []" by simp<br />
next<br />
fix x xs<br />
assume HI: "algunos P (map f xs) = algunos (P o f) xs"<br />
have "algunos P (map f (x#xs)) = ((P (f x)) ∨ (algunos P (map f xs)))" by simp<br />
also have "… = ((P (f x)) ∨ (algunos (P o f) xs))" using HI by simp<br />
finally show "algunos P (map f (x#xs)) = algunos (P o f) (x#xs)" by simp<br />
qed<br />
<br />
text {*<br />
--------------------------------------------------------------------- <br />
Ejercicio 8.1. Demostrar o refutar automáticamente <br />
algunos P (xs @ ys) = (algunos P xs ∨ algunos P ys)<br />
--------------------------------------------------------------------- <br />
*}<br />
<br />
(* ivamenjim marpoldia1 paupeddeg crigomgom rubgonmar wilmorort fraortmoy danrodcha pablucoto dancorgar josgarsan anaprarod juacabsou lucnovdos*)<br />
<br />
lemma "algunos P (xs @ ys) = (algunos P xs ∨ algunos P ys)"<br />
by (induct xs) auto<br />
<br />
(* ferrenseg *)<br />
lemma "algunos P (xs @ ys) = (algunos P xs ∨ algunos P ys)"<br />
by (induct xs) simp_all<br />
<br />
(* anaprarod *)<br />
lemma "algunos P (xs @ ys) = (algunos P xs ∨ algunos P ys)"<br />
apply (induct xs)<br />
apply auto<br />
done<br />
<br />
text {*<br />
--------------------------------------------------------------------- <br />
Ejercicio 8.2. Demostrar o refutar detalladamente<br />
algunos P (xs @ ys) = (algunos P xs ∨ algunos P ys)<br />
--------------------------------------------------------------------- <br />
*}<br />
<br />
(* ivamenjim ferrenseg marpoldia1 wilmorort dancorgar josgarsan juacabsou serrodcal lucnovdos*)<br />
<br />
lemma algunos_append:<br />
"algunos P (xs @ ys) = (algunos P xs ∨ algunos P ys)"<br />
proof (induct xs)<br />
show "algunos P ([] @ ys) = (algunos P [] ∨ algunos P ys)" by simp<br />
next <br />
fix a xs<br />
assume HI: "algunos P (xs @ ys) = (algunos P xs ∨ algunos P ys)"<br />
have "algunos P ((a # xs) @ ys) = (P a ∨ (algunos P (xs @ ys)))" by simp<br />
also have "... = (P a ∨ (algunos P xs ∨ algunos P ys))" using HI by simp<br />
finally show "algunos P ((a # xs) @ ys) = (algunos P (a # xs) ∨ algunos P ys)" by simp<br />
qed<br />
<br />
(* migtermor crigomgom *)<br />
<br />
lemma algunos_append2:<br />
"algunos P (xs @ ys) = (algunos P xs ∨ algunos P ys)" (is "?Q xs")<br />
proof (induct xs)<br />
show "?Q []" by simp<br />
next <br />
fix x xs<br />
assume HI: "?Q xs"<br />
have "algunos P ((x#xs) @ ys) = algunos P (x#(xs @ ys))" by simp<br />
also have "… = ((P x) ∨ (algunos P (xs @ ys)))" by simp<br />
also have "… = ((P x) ∨ (algunos P xs) ∨ (algunos P ys))" using HI by simp<br />
also have "… = ((algunos P (x#xs)) ∨ (algunos P ys))" by simp<br />
finally show "algunos P ((x#xs) @ ys) = (algunos P (x#xs) ∨ algunos P ys)" by simp<br />
qed<br />
<br />
(* paupeddeg*)<br />
lemma algunos_append3:<br />
"algunos P (xs @ ys) = (algunos P xs ∨ algunos P ys)"<br />
proof (induct xs)<br />
show "algunos P ([] @ ys) = (algunos P [] ∨ algunos P ys)" by simp<br />
next<br />
fix a xs<br />
assume HI: "algunos P (xs @ ys) = (algunos P xs ∨ algunos P ys)"<br />
have "algunos P ((a # xs) @ ys) = (algunos P [a] ∨ ( algunos P (xs @ ys)))" by simp<br />
also have "... = (algunos P [a] ∨ (algunos P xs ∨ algunos P ys))" using HI by simp<br />
finally show "algunos P ((a # xs) @ ys) = (algunos P (a # xs) ∨ algunos P ys)" by simp<br />
qed<br />
<br />
(* fraortmoy *)<br />
lemma algunos_append4:<br />
"algunos P (xs @ ys) = (algunos P xs ∨ algunos P ys)"<br />
proof (induct xs)<br />
show " algunos P ([] @ ys) = (algunos P [] ∨ algunos P ys)" by simp<br />
next <br />
fix a xs<br />
assume H1: "algunos P (xs @ ys) = (algunos P xs ∨ algunos P ys)"<br />
have "algunos P ((a # xs) @ ys) = (algunos P [a] ∨ algunos P (xs @ ys))" by simp<br />
also have "… = (algunos P [a] ∨ algunos P xs ∨ algunos P ys)" using H1 by simp<br />
also have "… = ((algunos P [a] ∨ algunos P xs) ∨ algunos P ys)" by simp<br />
also have "… = (algunos P (a#xs) ∨ algunos P ys)" by simp<br />
finally show "algunos P ((a # xs) @ ys) = (algunos P (a#xs) ∨ algunos P ys)" by simp<br />
qed<br />
<br />
{* danrodcha pablucoto anaprarod *}<br />
lemma algunos_append5:<br />
"algunos P (xs @ ys) = (algunos P xs ∨ algunos P ys)" (is "?Q xs")<br />
proof (induct xs)<br />
show "?Q []" by simp<br />
next<br />
fix a xs assume HI: "?Q xs"<br />
have "algunos P ((a#xs) @ ys) = algunos P (a#(xs @ ys))" by simp<br />
also have "… = (algunos P [a] ∨ algunos P (xs @ ys))" by simp<br />
also have "… = (algunos P [a] ∨ algunos P xs ∨ algunos P ys)" using HI by simp<br />
also have "… = (algunos P (a # xs) ∨ algunos P ys)" by simp<br />
finally show "?Q (a # xs)" by simp<br />
qed<br />
<br />
text {*<br />
--------------------------------------------------------------------- <br />
Ejercicio 9.1. Demostrar o refutar automáticamente<br />
algunos P (rev xs) = algunos P xs<br />
--------------------------------------------------------------------- <br />
*}<br />
<br />
(* ivamenjim migtermor marpoldia1 rubgonmar paupeddeg dancorgar anaprarod serrodcal *)<br />
<br />
lemma "algunos P (rev xs) = algunos P xs"<br />
apply (induct xs)<br />
apply simp<br />
apply (simp add: algunos_append)<br />
apply auto<br />
done<br />
<br />
(* ferrenseg crigomgom danrodcha pablucoto josgarsan pabrodmac lucnovdos*)<br />
lemma "algunos P (rev xs) = algunos P xs"<br />
by (induct xs) (auto simp add: algunos_append)<br />
<br />
(* wilmorort juacabsou *)<br />
lemma "algunos P (rev xs) = algunos P xs"<br />
by (induct xs,simp,simp add: algunos_append,auto) <br />
<br />
<br />
text {*<br />
--------------------------------------------------------------------- <br />
Ejercicio 9.2. Demostrar o refutar detalladamente<br />
algunos P (rev xs) = algunos P xs<br />
--------------------------------------------------------------------- <br />
*}<br />
<br />
lemma "algunos P (rev xs) = algunos P xs"<br />
oops<br />
<br />
(* migtermor *)<br />
lemma auxiliar1:<br />
"rev (a#xs) = rev xs @ [a]"<br />
by auto <br />
<br />
lemma "algunos P (rev xs) = algunos P xs" (is "?Q xs")<br />
proof (induct xs)<br />
show "?Q []" by simp<br />
next<br />
fix x xs<br />
assume HI: "?Q xs"<br />
have "algunos P (rev (x#xs)) = (algunos P (rev xs @ [x]))" using auxiliar1 by simp<br />
also have "… = ((algunos P (rev xs)) ∨ (algunos P [x]))" by (simp add: algunos_append)<br />
also have "… = ((P x) ∨ (algunos P (rev xs)))" by simp arith<br />
also have "… = ((P x) ∨ (algunos P xs))" using HI by simp<br />
finally show "algunos P (rev (x#xs)) = (algunos P (x#xs))" by simp<br />
qed<br />
<br />
(* ferrenseg *)<br />
<br />
lemma "algunos P (rev xs) = algunos P xs"<br />
proof (induct xs)<br />
show "algunos P (rev []) = algunos P []" by simp<br />
next<br />
fix x xs<br />
assume HI: "algunos P (rev xs) = algunos P xs"<br />
show "algunos P (rev (x # xs)) = algunos P (x # xs)"<br />
proof -<br />
have "algunos P (rev (x # xs)) = algunos P ((rev xs) @ [x])" by simp<br />
also have "… = (algunos P (rev xs) ∨ algunos P [x])" <br />
by (simp add: algunos_append)<br />
also have "… = (algunos P xs ∨ algunos P [x])" using HI by simp<br />
also have "… = (algunos P xs ∨ P x)" by simp<br />
also have "… = (P x ∨ algunos P xs)" by auto<br />
also have "… = algunos P (x # xs)" by simp<br />
finally show ?thesis<br />
qed<br />
qed<br />
<br />
(* ivamenjim marpoldia1*)<br />
<br />
lemma "algunos P (rev xs) = algunos P xs" (is "?P xs")<br />
proof (induct xs)<br />
show "?P []" by simp<br />
next <br />
fix a xs<br />
assume HI: "?P xs"<br />
have "algunos P (rev (a # xs)) = (algunos P ((rev xs)@[a]))" by simp<br />
also have "... = ((algunos P (rev xs)) ∨ algunos P [a])" by (simp add: algunos_append)<br />
also have "... = (algunos P xs ∨ algunos P [a])" using HI by simp<br />
also have "... = (algunos P [a] ∨ algunos P xs)" by arith<br />
finally show "algunos P (rev (a # xs)) = algunos P (a # xs)" by simp<br />
qed<br />
<br />
(* paupeddeg dancorgar serrodcal lucnovdos*)<br />
lemma "algunos P (rev xs) = algunos P xs"<br />
proof (induct xs)<br />
show "algunos P (rev []) = algunos P []" by simp<br />
next<br />
fix a xs<br />
assume HI:"algunos P (rev xs) = algunos P xs"<br />
have "algunos P (rev (a # xs)) = algunos P ((rev xs) @ [a])" by simp<br />
also have "... = ((algunos P (rev xs)) ∨ (algunos P [a]))" by (simp add: algunos_append)<br />
also have "... = ((algunos P xs) ∨ (algunos P [a]))" using HI by simp<br />
also have "... = ((algunos P [a]) ∨ (algunos P xs))" by arith<br />
finally show "algunos P (rev (a # xs)) = algunos P (a # xs)" by simp<br />
qed<br />
<br />
(*crigomgom pablucoto anaprarod rubgonmar juacabsou*)<br />
lemma "algunos P (rev xs) = algunos P xs"<br />
proof (induct xs)<br />
show "algunos P (rev []) = algunos P []" by simp<br />
next<br />
fix x xs<br />
assume HI: " algunos P (rev xs) = algunos P xs" <br />
have "algunos P (rev (x # xs)) = algunos P ((rev xs) @ [x])" by simp<br />
also have "... = (algunos P (rev xs) ∨ algunos P [x])" by (simp add: algunos_append)<br />
also have "... = (algunos P xs ∨ algunos P [x])" using HI by simp<br />
also have "... = (algunos P xs ∨ P x)" by simp<br />
also have "... = (P x ∨ algunos P xs)" by arith<br />
also have "... = algunos P (x#xs)" by simp<br />
finally show "algunos P (rev (x # xs)) = algunos P (x # xs)" by simp<br />
qed<br />
<br />
<br />
(*wilmorort*)<br />
lemma "algunos P (rev xs) = algunos P xs" (is "?P xs")<br />
proof (induct xs) <br />
show "?P []" by simp <br />
next<br />
fix a xs<br />
assume HI: "?P xs"<br />
have "algunos P (rev (a#xs)) = algunos P ((rev xs) @ [a])" by simp<br />
also have "... = (algunos P(rev xs) ∨ algunos P [a])" by (simp add: algunos_append)<br />
also have "... = (algunos P xs ∨ algunos P [a])" using HI by simp<br />
also have "... = (algunos P [a] ∨ algunos P xs)" by (simp add: HOL.disj_commute)<br />
also have "... = algunos P([a]@(xs))" by (simp)<br />
finally show "algunos P (rev (a#xs))= algunos P (a # xs)" by simp<br />
qed<br />
<br />
{*danrodcha *}<br />
lemma "algunos P (rev xs) = algunos P xs" (is "?Q xs")<br />
proof (induct xs)<br />
show "?Q []" by simp<br />
next<br />
fix a xs assume HI: "?Q xs"<br />
have "algunos P (rev (a#xs)) = algunos P ((rev xs) @ [a])" by simp<br />
also have "… = (algunos P (rev xs) ∨ algunos P [a])" by (simp add:algunos_append)<br />
also have "… = (algunos P xs ∨ algunos P [a])" using HI by simp<br />
also have "… = (algunos P [a] ∨ algunos P xs)" by (simp add: HOL.disj_commute)<br />
also have "… = algunos P (a # xs)" by simp<br />
finally show "?Q (a # xs)" by simp<br />
qed<br />
<br />
text {*<br />
--------------------------------------------------------------------- <br />
Ejercicio 10. Encontrar un término no trivial Z tal que sea cierta la <br />
siguiente ecuación:<br />
algunos (λx. P x ∨ Q x) xs = Z<br />
y demostrar la equivalencia de forma automática y detallada.<br />
--------------------------------------------------------------------- <br />
*}<br />
<br />
(* migtermor *)<br />
<br />
lemma "algunos (λx. P x ∨ Q x) xs = ((algunos (λx. P x) xs) ∨ (algunos (λx. Q x) xs))"<br />
by (induct xs) auto<br />
<br />
lemma "algunos (λx. P x ∨ Q x) xs = ((algunos (λx. P x) xs) ∨ (algunos (λx. Q x) xs))" (is "?R xs")<br />
proof (induct xs)<br />
show "?R []" by simp<br />
next<br />
fix x xs<br />
assume HI: "?R xs"<br />
have "algunos (λx. P x ∨ Q x) (x#xs) = (((λx. P x ∨ Q x) x) ∨ (algunos (λx. P x ∨ Q x) xs))"<br />
by simp<br />
also have "… = ((P x ∨ Q x) ∨ (algunos (λx. P x ∨ Q x) xs))" by simp<br />
also have H1: "… = ((((λx. P x) x) ∨ (algunos (λx. P x) xs)) ∨ (((λx. Q x) x) ∨ (algunos (λx. Q x) xs)))"<br />
using HI by simp arith<br />
have H2: "… = ((algunos (λx. P x) (x#xs)) ∨ (algunos (λx. Q x) (x#xs)))" <br />
by simp<br />
have C: "(algunos (λx. P x ∨ Q x) (x#xs)) = <br />
((algunos (λx. P x) (x#xs)) ∨ (algunos (λx. Q x) (x#xs)))" <br />
using H1 H2 by simp<br />
finally show "(algunos (λx. P x ∨ Q x) (x#xs)) = <br />
((algunos (λx. P x) (x#xs)) ∨ (algunos (λx. Q x) (x#xs)))" <br />
using C by simp<br />
qed<br />
<br />
<br />
(* ferrenseg *)<br />
<br />
lemma "algunos (λx. P x ∨ Q x) xs = (algunos P xs ∨ algunos Q xs)"<br />
proof (induct xs)<br />
show "algunos (λx. P x ∨ Q x) [] = (algunos P [] ∨ algunos Q [])" by simp<br />
next<br />
fix x xs<br />
assume HI: "algunos (λx. (P x ∨ Q x)) xs = (algunos P xs ∨ algunos Q xs)"<br />
show "algunos (λx. P x ∨ Q x) (x # xs) = (algunos P (x # xs) ∨ algunos Q (x # xs))"<br />
proof -<br />
have "algunos (λx. P x ∨ Q x) (x # xs) = <br />
((P x) ∨ (Q x) ∨ algunos (λx. P x ∨ Q x) xs)" by simp<br />
also have "… = ((P x) ∨ (Q x) ∨ algunos P xs ∨ algunos Q xs)" <br />
using HI by simp<br />
also have "… = (((P x) ∨ algunos P xs) ∨ ((Q x) ∨ algunos Q xs))" by auto<br />
also have "… = (algunos P (x # xs) ∨ algunos Q (x # xs))" by simp<br />
finally show ?thesis <br />
qed<br />
qed<br />
<br />
(* ivamenjim marpoldia1 paupeddeg *)<br />
<br />
lemma "algunos (λx. P x ∨ Q x) xs = (algunos P xs ∨ algunos Q xs)" <br />
by (induct xs) auto<br />
<br />
lemma "algunos (λx. P x ∨ Q x) xs = (algunos P xs ∨ algunos Q xs)"<br />
proof (induct xs)<br />
show "algunos (λx. P x ∨ Q x) [] = (algunos P [] ∨ algunos Q [])" by simp<br />
next <br />
fix a xs<br />
assume HI: "algunos (λx. P x ∨ Q x) xs = (algunos P xs ∨ algunos Q xs)"<br />
have "algunos (λx. P x ∨ Q x) (a # xs) = (P a ∨ Q a ∨ algunos (λx. P x ∨ Q x) xs)" by simp<br />
also have "... = (P a ∨ Q a ∨ algunos P xs ∨ algunos Q xs)" using HI by simp<br />
finally show "algunos (λx. P x ∨ Q x) (a # xs) = (algunos P (a # xs) ∨ algunos Q (a # xs))" by auto<br />
qed<br />
<br />
<br />
(* wilmorort danrodcha anaprarod juacabsou*)<br />
lemma "algunos (λx. P x ∨ Q x) xs = (¬todos(λx. ¬P x)xs ∨ ¬todos(λx. ¬Q x)xs)"<br />
by (induct xs) auto<br />
<br />
(* wilmorort *)<br />
lemma "algunos (λx. P x ∨ Q x) xs = (¬todos(λx. ¬P x)xs ∨ ¬todos(λx. ¬Q x)xs)"<br />
(is "?P xs")<br />
proof (induct xs)<br />
show "?P []" by simp<br />
next<br />
fix a xs<br />
assume HI: "?P xs"<br />
have "algunos (λx. P x ∨ Q x) (a # xs) = (P a ∨ Q a ∨ algunos (λx. P x ∨ Q x) xs)" by simp<br />
also have "... = (P a ∨ Q a ∨ (¬ todos (λx. ¬ P x) xs ∨ ¬ todos (λx. ¬ Q x) xs))" using HI by simp<br />
also have "... = (P a ∨ ¬ todos (λx. ¬ P x) xs ∨ Q a ∨ ¬ todos (λx. ¬ Q x) xs )" by arith<br />
finally show "algunos (λx. P x ∨ Q x) (a # xs) = (~ todos (λx. ¬ P x) (a # xs) ∨ ~ todos (λx. ¬ Q x) (a # xs))" by simp<br />
qed<br />
<br />
{* danrodcha *}<br />
lemma "algunos (λx. P x ∨ Q x) xs = (algunos P xs ∨ algunos Q xs)" (is "?R xs")<br />
proof (induct xs)<br />
show "?R []" by simp<br />
next<br />
fix a xs assume HI: "?R xs"<br />
have 1:" (Q a ∨ algunos P xs) = (algunos P xs ∨ Q a)" by (simp add: HOL.disj_commute)<br />
have "algunos (λx. P x ∨ Q x) (a # xs) = <br />
(algunos P [a] ∨ algunos Q [a] ∨ algunos (λx. P x ∨ Q x) xs)" by simp<br />
also have "… = (P a ∨ (Q a ∨ algunos P xs) ∨ algunos Q xs)" using HI by simp<br />
also have "… = (P a ∨ algunos P xs ∨ Q a ∨ algunos Q xs)" using 1 by simp<br />
also have "… = (algunos P (a # xs) ∨ algunos Q (a # xs))" by simp<br />
finally show "?R (a # xs)" by simp<br />
qed<br />
<br />
(* dancorgar *)<br />
lemma "algunos (λx. P x ∧ Q x) xs ⟹ (algunos P xs ∧ algunos Q xs)"<br />
apply (induct xs)<br />
apply auto<br />
done<br />
<br />
(* dancorgar *)<br />
lemma "algunos (λx. P x ∧ Q x) xs ⟹ (algunos P xs ∧ algunos Q xs)"<br />
proof (induct xs)<br />
show "algunos (λx. P x ∧ Q x) [] ⟹ (algunos P [] ∧ algunos Q [])" by simp<br />
next<br />
fix xf xs<br />
assume HI: "algunos (λx. P x ∧ Q x) xs ⟹ (algunos P xs ∧ algunos Q xs)"<br />
have F1: "algunos (λx. P x ∧ Q x) (xf#xs) ⟹ ((P xf ∧ Q xf) ∨ algunos (λx. P x ∧ Q x) xs)" by simp<br />
also have F2: "… ⟹ (P xf ∧ Q xf ∨ (algunos P xs ∧ algunos Q xs))" using HI by simp<br />
have F3: "((P xf ∧ Q xf) ∨ (algunos P xs ∧ algunos Q xs)) ⟹ <br />
((P xf ∨ algunos P xs) ∧ (Q xf ∨ algunos Q xs))" by blast<br />
have F4: "((P xf ∨ algunos P xs) ∧ (Q xf ∨ algunos Q xs)) ⟹ <br />
(algunos P (xf#xs) ∧ algunos Q (xf#xs))" by simp<br />
show "algunos (λx. P x ∧ Q x) (xf#xs) ⟹ (algunos P (xf#xs) ∧ algunos Q (xf#xs))"<br />
using F1 F2 F3 F4 by blast<br />
qed<br />
<br />
(*pablucoto crigomgom serrodcal*)<br />
-- "Automatica"<br />
lemma "algunos (λx. P x ∨ Q x) xs = (algunos P xs ∨ algunos Q xs)"<br />
by (induct xs) auto<br />
-- "Detallada"<br />
lemma "algunos (λx. P x ∨ Q x) xs = (algunos P xs ∨ algunos Q xs) "<br />
proof (induct xs)<br />
show "algunos (λx. P x ∨ Q x) [] = (algunos P [] ∨ algunos Q []) " by simp<br />
next<br />
fix a xs<br />
assume HI: " algunos (λx. P x ∨ Q x) xs = (algunos P xs ∨ algunos Q xs)"<br />
have " algunos (λx. P x ∨ Q x) (a # xs) = ( P a ∨ Q a ∨ algunos (λx. P x ∨ Q x) xs)" by simp<br />
also have "... = (P a ∨ Q a ∨ (algunos P xs ∨ algunos Q xs))" using HI by simp<br />
also have "… = (((P a) ∨ algunos P xs) ∨ ((Q a) ∨ algunos Q xs))" by arith <br />
also have "… = (algunos P (a#xs) ∨ algunos Q (a#xs))" by simp<br />
finally show "algunos (λx. P x ∨ Q x) (a # xs) = (algunos P (a # xs) ∨ algunos Q (a # xs)) " by simp<br />
qed<br />
<br />
(* anaprarod juacabsou *)<br />
lemma "algunos (λx. P x ∨ Q x) xs = (algunos P xs ∨ algunos Q xs)" (is "?P xs")<br />
proof (induct xs)<br />
show "?P []" by simp<br />
next<br />
fix x xs<br />
assume HI: "?P xs"<br />
have "algunos (λx. P x ∨ Q x) (x#xs) =(P x ∨ Q x ∨ algunos (λx. P x ∨ Q x) xs)" by simp<br />
also have "… = (P x ∨ Q x ∨ algunos P xs ∨ algunos Q xs)" using HI by simp<br />
also have "… = (P x ∨ algunos P xs ∨ Q x ∨ algunos Q xs)" by arith<br />
finally show "?P (x#xs)" by simp<br />
qed<br />
<br />
text {*<br />
--------------------------------------------------------------------- <br />
Ejercicio 11.1. Demostrar o refutar automáticamente<br />
algunos P xs = (¬ todos (λx. (¬ P x)) xs)<br />
--------------------------------------------------------------------- <br />
*}<br />
<br />
(* ivamenjim wilmorort migtermor marpoldia1 crigomgom rubgonmar paupeddeg danrodcha pablucoto dancorgar josgarsan anaprarod juacabsou lucnovdos serrodcal *)<br />
<br />
lemma "algunos P xs = (¬ todos (λx. (¬ P x)) xs)"<br />
by (induct xs) auto<br />
<br />
(* ferrenseg *)<br />
<br />
lemma "algunos P xs = (¬ todos (λx. (¬ P x)) xs)"<br />
by (induct xs) simp_all<br />
<br />
(* anaprarod *)<br />
lemma "algunos P xs = (¬ todos (λx. (¬ P x)) xs)" <br />
apply (induct xs)<br />
apply auto<br />
done<br />
<br />
<br />
text {*<br />
--------------------------------------------------------------------- <br />
Ejercicio 11.2. Demostrar o refutar datalladamente<br />
algunos P xs = (¬ todos (λx. (¬ P x)) xs)<br />
--------------------------------------------------------------------- <br />
*}<br />
<br />
(* ivamenjim migtermor marpoldia1 crigomgom paupeddeg josgarsan juacabsou serrodcal *)<br />
<br />
lemma "algunos P xs = (¬ todos (λx. (¬ P x)) xs)"<br />
proof (induct xs)<br />
show "algunos P [] = (¬ todos (λx. ¬ P x) [])" by simp<br />
next <br />
fix a xs<br />
assume HI: "algunos P xs = (¬ todos (λx. ¬ P x) xs)"<br />
have "algunos P (a # xs) = (P a ∨ (algunos P xs))" by simp<br />
also have "... = (P a ∨ (¬ todos (λx. ¬ P x) xs))" using HI by simp<br />
finally show "algunos P (a # xs) = (¬ todos (λx. ¬ P x) (a # xs))" by simp<br />
qed<br />
<br />
(* ferrenseg *)<br />
<br />
lemma "algunos P xs = (¬ todos (λx. (¬ P x)) xs)"<br />
proof (induct xs)<br />
show "algunos P [] = (¬ todos (λx. (¬ P x)) [])" by simp<br />
next<br />
fix x xs<br />
assume HI: "algunos P xs = (¬ todos (λx. (¬ P x)) xs)"<br />
show "algunos P (x # xs) = (¬ todos (λx. (¬ P x)) (x # xs))"<br />
proof - <br />
have "algunos P (x # xs) = ((P x) ∨ algunos P xs)" by simp<br />
also have "… = ((P x) ∨ ¬ todos (λx. (¬ P x)) xs)" using HI by simp<br />
also have "… = (¬ (¬ (P x) ∧ todos (λx. (¬ P x)) xs))" by simp<br />
also have "… = (¬ todos (λx. (¬ P x)) (x # xs))" by simp<br />
finally show ?thesis<br />
qed<br />
qed<br />
<br />
{* danrodcha pablucoto anaprarod rubgonmar *}<br />
lemma "algunos P xs = (¬ todos (λx. (¬ P x)) xs)" (is "?Q xs")<br />
proof (induct xs)<br />
show "?Q []" by simp<br />
next<br />
fix a xs assume HI: "?Q xs"<br />
have "algunos P (a # xs) = (P a ∨ algunos P xs)" by simp<br />
also have "… = (P a ∨ (¬ todos (λx. (¬ P x)) xs))" using HI by simp<br />
also have "… = (¬(¬(P a) ∧ todos(λx. (¬ P x)) xs))" by simp<br />
also have "… = (¬ todos(λx. (¬ P x)) (a # xs))" by simp<br />
finally show "?Q (a # xs)" by simp<br />
qed<br />
<br />
(* dancorgar*)<br />
lemma "algunos P xs = (¬ todos (λx. (¬ P x)) xs)"<br />
proof (induct xs)<br />
show "algunos P [] = (¬ todos (λx. (¬ P x)) [])" by simp<br />
next<br />
fix xf xs<br />
assume HI: "algunos P xs = (¬ todos (λx. (¬ P x)) xs)"<br />
have "algunos P (xf#xs) = ((P xf) ∨ algunos P xs)" by simp<br />
also have "… = ( (P xf) ∨ (¬ todos (λx. (¬ P x)) xs) )" using HI by simp<br />
also have "… = (¬((¬P xf) ∧ (todos (λx. (¬ P x)) xs)))" by simp<br />
finally show "algunos P (xf#xs) = (¬todos (λx. (¬ P x)) (xf#xs))" by simp<br />
qed<br />
<br />
(* lucnovdos*)<br />
lemma "algunos P xs = (¬ todos (λx. (¬ P x)) xs)"<br />
proof (induct xs)<br />
show "algunos P [] = (¬ todos (λx. (¬ P x)) [])" by simp<br />
next<br />
fix n xs<br />
assume HI: "algunos P xs = (¬ todos (λx. (¬ P x)) xs)"<br />
have "algunos P (n#xs) = ((P n) ∨ (algunos P xs))" by simp<br />
also have "… = ((P n) ∨ (¬ todos (λx. (¬ P x)) xs))" using HI by simp<br />
also have "… = (¬((¬P n) ∧ todos (λx. (¬ P x)) xs))" by simp<br />
also have "… = (¬ todos (λx. (¬ P x)) (n#xs))" by simp<br />
finally show "algunos P (n#xs) = (¬ todos (λx. (¬ P x)) (n#xs))" by simp<br />
qed<br />
<br />
text {*<br />
--------------------------------------------------------------------- <br />
Ejercicio 12. Definir la funcion primitiva recursiva <br />
estaEn :: 'a ⇒ 'a list ⇒ bool<br />
tal que (estaEn x xs) se verifica si el elemento x está en la lista<br />
xs. Por ejemplo, <br />
estaEn (2::nat) [3,2,4] = True<br />
estaEn (1::nat) [3,2,4] = False<br />
--------------------------------------------------------------------- <br />
*}<br />
<br />
(* ivamenjim ferrenseg migtermor serrodcal crigomgom rubgonmar marpoldia1 paupeddeg danrodcha dancorgar pablucoto anaprarod juacabsou lucnovdos*)<br />
<br />
fun estaEn :: "'a ⇒ 'a list ⇒ bool" where<br />
"estaEn x [] = False"<br />
| "estaEn x (a#xs) = ((a = x) ∨ (estaEn x xs))"<br />
<br />
value "estaEn (2::nat) [3,2,4] = True"<br />
value "estaEn (1::nat) [3,2,4] = False"<br />
<br />
<br />
text {*<br />
--------------------------------------------------------------------- <br />
Ejercicio 13. Expresar la relación existente entre estaEn y algunos. <br />
Demostrar dicha relación de forma automática y detallada.<br />
--------------------------------------------------------------------- <br />
*}<br />
<br />
(* migtermor crigomgom *)<br />
<br />
lemma "estaEn x xs = (algunos (λa. a=x) xs)"<br />
by (induct xs) auto<br />
<br />
lemma auxiliar13:<br />
"(x=a) = ((λa. a=x) a)"<br />
by auto<br />
<br />
lemma "estaEn x xs = (algunos (λa. a=x) xs)" (is "?P xs")<br />
proof (induct xs)<br />
show "?P []" by simp<br />
next<br />
fix a xs<br />
assume HI: "?P xs"<br />
have "estaEn x (a#xs) = ((a=x) ∨ (estaEn x xs))" by simp<br />
also have H: "… = ((a=x) ∨ (algunos (λa. a=x) xs))" using HI by simp<br />
also have "… = (((λa. a=x) a) ∨ (algunos (λa. a=x) xs))" using auxiliar13 by simp<br />
also have "… = (algunos (λa. a=x) (a#xs))" by simp<br />
have C: "estaEn x (a#xs) = (algunos (λa. a=x) (a#xs))" using H by simp<br />
finally show "estaEn x (a#xs) = (algunos (λa. a=x) (a#xs))" using C by simp<br />
qed<br />
<br />
<br />
(* ferrenseg *)<br />
<br />
lemma estaEn_algunos:<br />
"estaEn y xs = algunos (λx. x=y) xs"<br />
by (induct xs) auto<br />
<br />
lemma estaEn_algunos_2:<br />
"estaEn y xs = algunos (λx. x=y) xs"<br />
proof (induct xs)<br />
show "estaEn y [] = algunos (λx. x=y) []" by simp<br />
next<br />
fix x xs<br />
assume HI: "estaEn y xs = algunos (λx. x=y) xs"<br />
show "estaEn y (x # xs) = algunos (λx. x=y) (x # xs)"<br />
proof -<br />
have "estaEn y (x # xs) = (y=x ∨ estaEn y xs)" by simp<br />
also have "… = (y=x ∨ algunos (λx. x=y) xs)" using HI by simp<br />
also have "… = (x=y ∨ algunos (λx. x=y) xs)" by auto<br />
also have "… = algunos (λx. x=y) (x # xs)" by simp<br />
finally show ?thesis<br />
qed<br />
qed<br />
<br />
(* crigomgom paupeddeg dancorgar juacabsou serrodcal lucnovdos *)<br />
<br />
lemma "estaEn a xs = algunos (λx. x = a) xs"<br />
by (induct xs) auto<br />
<br />
lemma "estaEn x xs = (algunos (λa. a=x) xs)" <br />
proof (induct xs)<br />
show " estaEn x [] = algunos (λa. a = x) []" by simp<br />
next<br />
fix a xs<br />
assume HI: "estaEn x xs = algunos (λa. a = x) xs"<br />
have "estaEn x (a # xs) = (a = x ∨ estaEn x xs)" by simp<br />
also have "... = (a = x ∨ algunos (λa. a = x) xs)" using HI by simp<br />
finally show " estaEn x (a # xs) = algunos (λa. a = x) (a # xs)" by simp<br />
qed<br />
<br />
(* ivamenjim marpoldia1 *)<br />
<br />
lemma "estaEn x xs = algunos (λy. x=y) xs"<br />
by (induct xs) auto<br />
<br />
lemma "estaEn x xs = algunos (λy. x=y) xs"<br />
proof (induct xs)<br />
show "estaEn x [] = algunos (op = x) []" by simp<br />
next <br />
fix a xs<br />
assume HI: "estaEn x xs = algunos (op = x) xs"<br />
have "estaEn x (a # xs) = ((a = x) ∨ (estaEn x xs))" by simp<br />
also have "... = (a = x ∨ algunos (op = x) xs)" using HI by simp<br />
finally show " estaEn x (a # xs) = algunos (op = x) (a # xs)" by auto<br />
qed<br />
<br />
{* danrodcha *}<br />
lemma "estaEn a xs = algunos (op = a) xs"<br />
by (induct xs) auto<br />
<br />
{* danrodcha *}<br />
lemma "estaEn a xs = algunos (op = a) xs"<br />
proof (induct xs)<br />
show "estaEn a [] = algunos (op = a) []" by simp<br />
next<br />
fix x xs <br />
assume HI: "estaEn a xs = algunos (op = a) xs"<br />
have "estaEn a (x # xs) = ((x = a) ∨ (estaEn a xs))" by simp<br />
also have "… = (x = a ∨ algunos (op = a) xs)" using HI by simp<br />
also have "… = ((op = a) x ∨ algunos (op = a) xs)" by (simp add:HOL.eq_commute)<br />
also have "… = algunos (op = a) (x # xs)" by simp<br />
finally show "estaEn a (x # xs) = algunos (op = a) (x # xs)" by simp<br />
qed<br />
<br />
(* pablucoto *)<br />
<br />
-- "Automatica"<br />
lemma " estaEn y xs = algunos (λx. x=y) xs"<br />
by (induct xs) auto<br />
<br />
-- "Detallada"<br />
lemma " estaEn y xs = algunos (λx. x=y) xs"<br />
proof (induct xs)<br />
show "estaEn y [] = algunos (λx. x = y) []" by simp<br />
next<br />
fix a xs<br />
assume HI: " estaEn y xs = algunos (λx. x = y) xs "<br />
have " estaEn y (a # xs) = (y = a ∨ estaEn y xs)" by simp <br />
also have "... = (y = a ∨ algunos (λx. x = y) xs) " using HI by simp<br />
also have "… = (a=y ∨ algunos (λx. x=y) xs)" by auto<br />
also have "... = (algunos (λx. x = y) (a#xs))" by simp<br />
finally show "estaEn y (a # xs) = algunos (λx. x = y) (a # xs) " by simp<br />
qed<br />
<br />
<br />
(* anaprarod *)<br />
(* Automática*)<br />
lemma "estaEn a xs = algunos (λx. a=x) xs" (is "?P xs")<br />
by (induct xs) auto<br />
<br />
(* Detallada (igual que las anteriores pero obviando el paso previo al finally) *)<br />
lemma "estaEn a xs = algunos (λx. a=x) xs" (is "?P xs")<br />
proof (induct xs)<br />
show "?P []" by simp<br />
next<br />
fix x xs<br />
assume HI : "?P xs"<br />
have "estaEn a (x#xs) = (x=a ∨ estaEn a xs)" by simp<br />
also have "… = (x=a ∨ algunos (λx. a=x) xs)" using HI by simp<br />
also have "… = (a=x ∨ algunos (λx. a=x) xs)" using HI by auto<br />
finally show "?P (x#xs)" by simp<br />
qed<br />
<br />
<br />
end<br />
</source></div>Lucnovdoshttps://www.glc.us.es/~jalonso/RA2016/index.php?title=Relaci%C3%B3n_4&diff=653Relación 42016-11-23T19:55:17Z<p>Lucnovdos: </p>
<hr />
<div><source lang="isar"><br />
chapter {* R4: Cuantificadores sobre listas *}<br />
<br />
theory R4_Cuantificadores_sobre_listas<br />
imports Main <br />
begin<br />
<br />
text {* <br />
--------------------------------------------------------------------- <br />
Ejercicio 1. Definir la función <br />
todos :: ('a ⇒ bool) ⇒ 'a list ⇒ bool<br />
tal que (todos p xs) se verifica si todos los elementos de la lista <br />
xs cumplen la propiedad p. Por ejemplo, se verifica <br />
todos (λx. 1<length x) [[2,1,4],[1,3]]<br />
¬todos (λx. 1<length x) [[2,1,4],[3]]<br />
<br />
Nota: La función todos es equivalente a la predefinida list_all. <br />
--------------------------------------------------------------------- <br />
*}<br />
<br />
{*danrodcha*}<br />
fun todos :: "('a ⇒ bool) ⇒ 'a list ⇒ bool" where<br />
"todos p xs = "todos p [] = True"<br />
| "todos p (x#xs) = (p x ∧ todos p xs)""<br />
<br />
(* ivamenjim migtermor dancorgar wilmorort marpoldia1 ferrenseg paupeddeg pablucoto crigomgom anaprarod serrodcal juacabsou rubgonmar josgarsan fraortmoy lucnovdos pabrodmac fracorjim1*)<br />
fun todos :: "('a ⇒ bool) ⇒ 'a list ⇒ bool" where<br />
"todos p [] = True"<br />
| "todos p (x#xs) = (p x ∧ todos p xs)"<br />
<br />
text {* <br />
--------------------------------------------------------------------- <br />
Ejercicio 2. Definir la función <br />
algunos :: ('a ⇒ bool) ⇒ 'a list ⇒ bool<br />
tal que (algunos p xs) se verifica si algunos elementos de la lista <br />
xs cumplen la propiedad p. Por ejemplo, se verifica <br />
algunos (λx. 1<length x) [[2,1,4],[3]]<br />
¬algunos (λx. 1<length x) [[],[3]]"<br />
<br />
Nota: La función algunos es equivalente a la predefinida list_ex. <br />
--------------------------------------------------------------------- <br />
*}<br />
<br />
{*danrodcha ivamenjim migtermor dancorgar marpoldia1 ferrenseg wilmorort paupeddeg pablucoto crigomgom anaprarod serrodcal juacabsou rubgonmar josgarsan fraortmoy lucnovdos pabrodmac*}<br />
fun algunos :: "('a ⇒ bool) ⇒ 'a list ⇒ bool" where<br />
"algunos p [] = False"<br />
| "algunos p (x#xs) = (p x ∨ algunos p xs)"<br />
<br />
(* fracorjim1. En esta versión el procesamiento se detiene al encontrar una coincidencia *)<br />
<br />
fun algunos :: "('a ⇒ bool) ⇒ 'a list ⇒ bool" where<br />
"algunos p [] = False"<br />
| "algunos p (x#xs) = (if p x then True else algunos p xs)"<br />
<br />
<br />
text {*<br />
--------------------------------------------------------------------- <br />
Ejercicio 3.1. Demostrar o refutar automáticamente <br />
todos (λx. P x ∧ Q x) xs = (todos P xs ∧ todos Q xs)<br />
--------------------------------------------------------------------- <br />
*}<br />
<br />
{*danrodcha ivamenjim migtermor dancorgar marpoldia1 ferrenseg wilmorort paupeddeg pablucoto anaprarod serrodcal juacabsou rubgonmar josgarsan fraortmoy lucnovdos pabrodmac *}<br />
lemma "todos (λx. P x ∧ Q x) xs = (todos P xs ∧ todos Q xs)"<br />
by (induct xs) auto<br />
<br />
(* fracorjim1 *)<br />
<br />
lemma "todos (λx. P x ∧ Q x) xs = (todos P xs ∧ todos Q xs)"<br />
apply (induct xs)<br />
apply auto<br />
done<br />
<br />
text {*<br />
--------------------------------------------------------------------- <br />
Ejercicio 3.2. Demostrar o refutar detalladamente<br />
todos (λx. P x ∧ Q x) xs = (todos P xs ∧ todos Q xs)<br />
--------------------------------------------------------------------- <br />
*}<br />
<br />
{*danrodcha fracorjim1*}<br />
lemma "todos (λx. P x ∧ Q x) xs = (todos P xs ∧ todos Q xs)"<br />
proof (induct xs)<br />
show "?R []" by simp<br />
next<br />
fix a xs <br />
assume HI: "?R xs"<br />
have "todos (λx. P x ∧ Q x) (a#xs) = (P a ∧ Q a ∧ todos (λx. P x ∧ Q x) xs)" by simp<br />
also have "… = (P a ∧ Q a ∧ (todos P xs ∧ todos Q xs))" using HI by simp<br />
also have "… = ((P a ∧ todos P xs) ∧ (Q a ∧ todos Q xs))" by blast<br />
also have "… = (todos P (a#xs) ∧ todos Q (a#xs))" by simp<br />
finally show "?R (a#xs)" by simp<br />
qed<br />
<br />
(* ivamenjim wilmorort serrodcal josgarsan*)<br />
<br />
lemma "todos (λx. P x ∧ Q x) xs = (todos P xs ∧ todos Q xs)"<br />
proof (induct xs)<br />
show "todos (λx. P x ∧ Q x) [] = (todos P [] ∧ todos Q [])" by simp<br />
next <br />
fix a xs<br />
assume HI: "todos (λx. P x ∧ Q x) xs = (todos P xs ∧ todos Q xs)"<br />
have "todos (λx. P x ∧ Q x) (a # xs) = (P a ∧ Q a ∧ todos (λx. P x ∧ Q x) xs)" by simp<br />
also have "... = (P a ∧ Q a ∧ todos P xs ∧ todos Q xs)" using HI by simp<br />
finally show "todos (λx. P x ∧ Q x) (a # xs) = (todos P (a # xs) ∧ todos Q (a # xs))" by auto<br />
qed<br />
<br />
(* dancorgar paupeddeg *)<br />
lemma "todos (λx. P x ∧ Q x) xs = (todos P xs ∧ todos Q xs)"<br />
proof (induct xs)<br />
show "todos (λx. P x ∧ Q x) [] = (todos P [] ∧ todos Q [])" by simp<br />
next<br />
fix y xs<br />
assume HI: "todos (λx. P x ∧ Q x) xs = (todos P xs ∧ todos Q xs)"<br />
have "todos (λx. P x ∧ Q x) (y#xs) = ((P y ∧ Q y) ∧ (todos (λx. P x ∧ Q x) xs))" by simp<br />
also have "… = ((P y ∧ Q y) ∧ (todos P xs ∧ todos Q xs))" using HI by simp<br />
also have "… = ((P y ∧ todos P xs) ∧ (Q y ∧ todos Q xs))" by blast<br />
also have "… = ((todos P (y#xs)) ∧ (todos Q (y#xs)))" by simp<br />
finally show "todos (λx. P x ∧ Q x) (y#xs) = (todos P (y#xs) ∧ todos Q (y#xs))" by simp<br />
qed<br />
<br />
<br />
(* migtermor *)<br />
lemma "todos (λx. P x ∧ Q x) xs = (todos P xs ∧ todos Q xs)"<br />
proof (induct xs)<br />
show "todos (λx. P x ∧ Q x) [] = (todos P [] ∧ todos Q [])" by simp<br />
next <br />
fix x xs<br />
assume HI: "todos (λx. P x ∧ Q x) xs = (todos P xs ∧ todos Q xs)"<br />
have "(todos (λx. P x ∧ Q x) (x#xs)) = (( P x ∧ Q x) ∧ (todos (λx. P x ∧ Q x) xs))"<br />
by (simp only: todos.simps(2))<br />
also have "… = ((P x ∧ Q x) ∧ (todos P xs ∧ todos Q xs))" using HI by simp<br />
also have "… = (((P x)∧(todos P xs)) ∧ ((Q x) ∧ (todos Q xs)))" by arith<br />
also have "((P x ∧ Q x) ∧ (todos P xs ∧ todos Q xs)) = (((P x)∧(todos P xs)) ∧ ((Q x) ∧ (todos Q xs)))"<br />
by arith (* Este paso es exactamente el mismo que el anterior, pero sin cualquiera de los dos no funciona el "finally show" *)<br />
have "… = (((P x)∧(todos P xs))∧((Q x)∧(todos Q xs)))" by simp<br />
have "… = ((todos P (x#xs))∧(todos Q (x#xs)))" by simp<br />
finally show "(todos (λx. P x ∧ Q x) (x#xs)) = ((todos P (x#xs)) ∧ (todos Q (x#xs)))" by simp <br />
qed<br />
<br />
(* marpoldia1 *)<br />
lemma "todos (λx. P x ∧ Q x) xs = (todos P xs ∧ todos Q xs)"<br />
proof (induct xs)<br />
show "todos (λx. P x ∧ Q x) [] = (todos P [] ∧ todos Q [])" by simp<br />
next<br />
fix a xs<br />
assume HI: "todos (λx. P x ∧ Q x) xs = (todos P xs ∧ todos Q xs)" <br />
have "todos (λx. P x ∧ Q x) (a # xs) = ((P a ∧ Q a) ∧ (todos P xs ∧ todos Q xs))" using HI by simp<br />
also have "... = ((P a ∧ todos P xs) ∧ (Q a ∧ todos Q xs))" by blast<br />
also have "... = (todos P (a#xs) ∧ todos Q (a#xs)) " by simp<br />
finally show "todos (λx. P x ∧ Q x) (a#xs) = (todos P (a#xs) ∧ todos Q (a#xs))" by simp <br />
qed<br />
<br />
(* ferrenseg *)<br />
<br />
lemma "todos (λx. P x ∧ Q x) xs = (todos P xs ∧ todos Q xs)"<br />
proof (induct xs)<br />
show "todos (λx. P x ∧ Q x) [] = (todos P [] ∧ todos Q [])" by simp<br />
next <br />
fix n xs<br />
assume HI: "todos (λx. P x ∧ Q x) xs = (todos P xs ∧ todos Q xs)"<br />
have "todos (λx. P x ∧ Q x) (n # xs) = (P n ∧ Q n ∧ todos P xs ∧ todos Q xs)" using HI by simp<br />
also have "… = P n ∧ Q n ∧ todos P xs ∧ todos Q xs" by blast<br />
also have "⋯ = todos P (n # xs) ∧ todos Q (n # xs)" by simp <br />
finally show "todos (λx. P x ∧ Q x) (n # xs) = todos P (n # xs) ∧ todos Q (n # xs)" by simp<br />
qed<br />
<br />
(* lucnovdos *)<br />
<br />
lemma "todos (λx. P x ∧ Q x) xs = (todos P xs ∧ todos Q xs)"<br />
proof (induct xs)<br />
show "todos (λx. P x ∧ Q x) [] = (todos P [] ∧ todos Q [])" by simp<br />
next<br />
fix n xs<br />
assume HI: "todos (λx. P x ∧ Q x) xs = (todos P xs ∧ todos Q xs)" <br />
have "todos (λx. P x ∧ Q x) (n # xs) = ((P n ∧ Q n) ∧ (todos P xs ∧ todos Q xs))" using HI by simp<br />
also have "... = ((P n ∧ todos P xs) ∧ (Q n ∧ todos Q xs))" by arith<br />
also have "... = ((todos P(n#xs)) ∧ (todos Q(n#xs)))" by simp<br />
<br />
finally show "todos (λx. P x ∧ Q x) (n#xs) = (todos P (n#xs) ∧ todos Q (n#xs))" by simp <br />
qed<br />
<br />
{* wilmorort pablucoto crigomgom anaprarod juacabsou rubgonmar fraortmoy *}<br />
<br />
lemma "todos (λx. P x ∧ Q x) xs = (todos P xs ∧ todos Q xs)"<br />
proof (induct xs)<br />
show "todos (λx. P x ∧ Q x) [] = (todos P [] ∧ todos Q [])" by simp <br />
next<br />
fix a xs<br />
assume HI: "todos (λx. P x ∧ Q x) xs = (todos P xs ∧ todos Q xs)"<br />
have "todos (λx. P x ∧ Q x) (a # xs) = (P a ∧ Q a ∧ todos (λx. P x ∧ Q x) xs)" by simp<br />
also have "... = (P a ∧ Q a ∧ todos P xs ∧ todos Q xs)" using HI by simp<br />
also have "... = (P a ∧ todos P xs ∧ Q a ∧ todos Q xs)" by arith<br />
also have "... = (todos P (a # xs) ∧ todos Q (a # xs))" by simp (* Este paso se puede obviar*)<br />
finally show "todos (λx. P x ∧ Q x) (a # xs) = (todos P (a # xs) ∧ todos Q (a # xs))" by simp<br />
qed<br />
<br />
(* pabrodmac *)<br />
<br />
lemma "todos (λx. P x ∧ Q x) xs = (todos P xs ∧ todos Q xs)"<br />
proof (induct xs)<br />
show "todos (λx. P x ∧ Q x) [] = (todos P [] ∧ todos Q [])" by simp<br />
next <br />
fix x xs<br />
assume HI: "todos (λx. P x ∧ Q x) xs = (todos P xs ∧ todos Q xs)" <br />
have "todos (λx. P x ∧ Q x) (x#xs) = ( P x ∧ Q x ∧ todos (λx. P x ∧ Q x) xs)" by simp<br />
also have "... = (P x ∧ Q x ∧ (todos P xs ∧ todos Q xs))" using HI by simp<br />
also have "... = (P x ∧ todos P xs ∧ Q x ∧ todos Q xs)" by arith<br />
finally show "todos (λx. P x ∧ Q x) (x#xs) = (todos P (x#xs) ∧ todos Q (x#xs))" by simp<br />
qed<br />
<br />
text {*<br />
--------------------------------------------------------------------- <br />
Ejercicio 4.1. Demostrar o refutar automáticamente<br />
todos P (x @ y) = (todos P x ∧ todos P y)<br />
--------------------------------------------------------------------- <br />
*}<br />
{*danrodcha ivamenjim marpoldia1 migtermor ferrenseg wilmorort paupeddeg crigomgom anaprarod serrodcal juacabsou rubgonmar pablucoto fraortmoy josgarsan lucnovdos pabrodmac*}<br />
lemma "todos P (x @ y) = (todos P x ∧ todos P y)"<br />
by (induct x) auto<br />
<br />
{* anaprarod dancorgar fracorjim1*}<br />
lemma "todos P (x @ y) = (todos P x ∧ todos P y)"<br />
apply (induct x)<br />
apply auto<br />
done<br />
<br />
text {*<br />
--------------------------------------------------------------------- <br />
Ejercicio 4.2. Demostrar o refutar detalladamente<br />
todos P (x @ y) = (todos P x ∧ todos P y)<br />
--------------------------------------------------------------------- <br />
*}<br />
<br />
(* ivamenjim ferrenseg wilmorort dancorgar fraortmoy josgarsan lucnovdos*)<br />
<br />
lemma todos_append:<br />
"todos P (x @ y) = (todos P x ∧ todos P y)"<br />
proof (induct x)<br />
show "todos P ([] @ y) = (todos P [] ∧ todos P y)" by simp<br />
next <br />
fix a x<br />
assume HI: "todos P (x @ y) = (todos P x ∧ todos P y)"<br />
have "todos P ((a#x) @ y) = (P a ∧ (todos P (x @ y)))" by simp<br />
also have "... = (P a ∧ (todos P x ∧ todos P y))" using HI by simp<br />
finally show "todos P ((a # x) @ y) = (todos P (a # x) ∧ todos P y)" by simp<br />
qed<br />
<br />
(* migtermor serrodcal *)<br />
<br />
lemma todos_append1:<br />
"todos P (x @ y) = (todos P x ∧ todos P y)" (is "?P x")<br />
proof (induct x)<br />
show "?P []" by simp<br />
next <br />
fix a x<br />
assume HI: "?P x"<br />
have "todos P ((a#x) @ y) = (P a ∧ (todos P (x @ y)))" by simp<br />
also have "… = (P a ∧ (todos P x ∧ todos P y))" using HI by simp<br />
finally show "?P (a#x)" by simp<br />
qed <br />
<br />
(* marpoldia1 *)<br />
lemma todos_append:<br />
"todos P (x @ y) = (todos P x ∧ todos P y)"<br />
proof (induct x)<br />
show "todos P ([] @ y) = (todos P [] ∧ todos P y)" by simp<br />
next<br />
fix a x<br />
assume HI: "todos P (x @ y) = (todos P x ∧ todos P y)"<br />
have "todos P ((a # x) @ y) = todos P (a#(x@y))" by simp<br />
also have "... = (P a ∧ todos P (x @ y))" by simp<br />
also have "... = (P a ∧ (todos P x ∧ todos P y))" using HI by simp<br />
finally show "todos P ((a # x) @ y) = (todos P (a # x) ∧ todos P y)" by simp <br />
qed<br />
<br />
(* paupeddeg *)<br />
lemma todos_append3:<br />
"todos P (x @ y) = (todos P x ∧ todos P y)"<br />
proof (induct x)<br />
show "todos P ([] @ y) = (todos P [] ∧ todos P y)" by simp<br />
next<br />
fix a x<br />
assume HI: "todos P (x @ y) = (todos P x ∧ todos P y)"<br />
have "todos P ((a # x) @ y) = todos P (a # (x @ y))" by simp<br />
also have "... = (todos P [a] ∧ todos P (x @ y)) " by simp<br />
also have "... = (todos P [a] ∧ (todos P x ∧ todos P y))" using HI by simp<br />
finally show "todos P ((a # x) @ y) = (todos P (a # x) ∧ todos P y)" by simp<br />
qed<br />
<br />
(* crigomgom juacabsou anaprarod*)<br />
<br />
lemma todos_append4:<br />
"todos P (x @ y) = (todos P x ∧ todos P y)"<br />
proof (induct x)<br />
show "todos P ([] @ y) = (todos P [] ∧ todos P y)" by simp<br />
next<br />
fix x xs<br />
assume HI:"todos P (xs @ y) = (todos P xs ∧ todos P y)"<br />
have "todos P ((x # xs) @ y) = (P x ∧ todos P (xs @ y ))" by simp<br />
also have "... = (P x ∧ (todos P xs ∧ todos P y)) " using HI by simp<br />
also have "... = ((P x ∧ todos P xs) ∧ todos P y)" by simp<br />
finally show "todos P ((x # xs) @ y) = (todos P (x # xs) ∧ todos P y)" by simp<br />
qed<br />
<br />
(* pablucoto *)<br />
lemma todos_append5:<br />
"todos P (x @ y) = (todos P x ∧ todos P y)"<br />
proof (induct x )<br />
show "todos P ([] @ y) = (todos P [] ∧ todos P y)" by simp<br />
next<br />
fix a x<br />
assume HI:"todos P (x @ y) = (todos P x ∧ todos P y) " <br />
have "todos P ((a # x) @ y) = ( P a ∧ todos P (x @ y))" by simp<br />
also have " ... =( P a ∧ (todos P x ∧ todos P y))" using HI by simp<br />
also have "... = (todos P (a#x) ∧ todos P y)" by simp<br />
finally show "todos P ((a # x) @ y) = (todos P (a # x) ∧ todos P y)" by auto<br />
qed<br />
<br />
{*danrodcha*}<br />
lemma todos_append6:<br />
"todos P (x @ y) = (todos P x ∧ todos P y)" (is "?Q x")<br />
proof (induct x)<br />
show "?Q []" by simp<br />
next<br />
fix x a assume HI: "?Q x"<br />
have "todos P ((a # x) @ y) = todos P (a # x @ y)" by simp<br />
also have "… = ((P a) ∧ todos P (x @ y))" by simp<br />
also have "… = ((P a) ∧ (todos P x ∧ todos P y))" using HI by simp<br />
also have "… = (todos P (a # x) ∧ todos P y)" by simp<br />
finally show "?Q (a # x)" by simp<br />
qed<br />
<br />
{*pabrodmac*}<br />
lemma todos_append7:<br />
"todos P (x @ y) = (todos P x ∧ todos P y)"<br />
proof (induct x)<br />
show "todos P ([] @ y) = (todos P [] ∧ todos P y)" by simp<br />
next <br />
fix a x<br />
assume HI: "todos P (x @ y) = (todos P x ∧ todos P y)" <br />
have "todos P ((a#x) @ y) = (P a ∧ todos P (x @ y))" by simp<br />
also have "... = (P a ∧ (todos P x ∧ todos P y))" using HI by simp<br />
finally show "todos P ((a#x) @ y) = (todos P (a#x) ∧ todos P y)" by simp<br />
qed<br />
<br />
<br />
text {*<br />
--------------------------------------------------------------------- <br />
Ejercicio 5.1. Demostrar o refutar automáticamente <br />
todos P (rev xs) = todos P xs<br />
--------------------------------------------------------------------- <br />
*}<br />
<br />
lemma "todos P (rev xs) = todos P xs"<br />
oops<br />
<br />
(* migtermor ivamenjim marpoldia1 serrodcal anaprarod paupeddeg dancorgar *)<br />
lemma "todos P (rev xs) = todos P xs" <br />
apply (induct xs)<br />
apply simp<br />
apply (simp add: todos_append)<br />
apply auto<br />
done<br />
<br />
(* ferrenseg crigomgom rubgonmar fraortmoy josgarsan danrodcha lucnovdos pabrodmac*)<br />
<br />
lemma "todos P (rev xs) = todos P xs"<br />
by (induct xs) (auto simp add: todos_append)<br />
<br />
(* wilmorort *)<br />
lemma "todos P (rev xs) = todos P xs"<br />
by (induct xs,simp,simp add: todos_append,auto)<br />
<br />
(* juacabsou *)<br />
lemma "todos P (rev xs) = todos P xs"<br />
apply (induct xs,simp,simp add: todos_append,auto) done<br />
<br />
(* pablucoto * )<br />
lemma "todos P (rev xs) = todos P xs"<br />
by (induct xs) (auto, simp_all add: todos_append) <br />
lemma "todos P (rev xs) = todos P xs"<br />
by (induct xs) ( simp_all add: todos_append, auto)<br />
<br />
text {*<br />
--------------------------------------------------------------------- <br />
Ejercicio 5.2. Demostrar o refutar detalladamente<br />
todos P (rev xs) = todos P xs<br />
--------------------------------------------------------------------- <br />
*}<br />
<br />
lemma "todos P (rev xs) = todos P xs"<br />
oops<br />
<br />
(* migtermor *)<br />
<br />
lemma auxiliar:<br />
"rev (a#xs) = rev xs @ [a]"<br />
by auto<br />
<br />
lemma "todos P (rev xs) = todos P xs" (is "?Q xs")<br />
proof (induct xs)<br />
show "?Q []" by (simp only: rev.simps(1))<br />
next <br />
fix a xs<br />
assume HI: "?Q xs"<br />
have "todos P (rev (a#xs)) = (todos P (rev xs @ [a]))" by (simp add: auxiliar)<br />
have "… = ((todos P (rev xs)) ∧ (todos P [a]))" by (simp add: todos_append)<br />
have "… = (todos P (rev xs) ∧ P a)" by simp<br />
also have Aux: "… = (todos P xs ∧ P a)" using HI by simp<br />
have Aux1: "… = (P a ∧ todos P xs)" by arith<br />
have "(todos P (rev xs) ∧ P a) = (P a ∧ todos P xs)" using Aux Aux1 by simp<br />
finally show "todos P (rev (a#xs)) = todos P (a#xs)" by (simp add: todos_append)<br />
qed<br />
<br />
<br />
(* marpoldia1 ferrenseg crigomgom serrodcal juacabsou rubgonmar josgarsan pablucoto pabrodmac lucnovdos*)<br />
lemma "todos P (rev xs) = todos P xs"<br />
proof (induct xs)<br />
show "todos P (rev []) = todos P []" by simp<br />
next<br />
fix a xs<br />
assume HI: "todos P (rev xs) = todos P xs"<br />
have "todos P (rev (a # xs)) = (todos P ((rev xs)@[a]))" by simp<br />
also have "... = (todos P (rev xs) ∧ todos P [a])" by (simp add:todos_append)<br />
also have "... = (todos P xs ∧ P a)" using HI by simp<br />
also have "... = (P a ∧ todos P xs)" by arith<br />
also have "... = (todos P (a#xs))" by simp<br />
finally show "todos P (rev (a # xs)) = (todos P (a#xs))" by simp <br />
qed<br />
<br />
(* ivamenjim *)<br />
<br />
lemma "todos P (rev xs) = todos P xs" (is "?P xs")<br />
proof (induct xs)<br />
show "?P []" by simp<br />
next <br />
fix a xs<br />
assume HI: "?P xs"<br />
have "todos P (rev (a # xs)) = (todos P ((rev xs)@[a]))" by simp<br />
also have "... = ((todos P (rev xs)) ∧ todos P [a])" by (simp add: todos_append)<br />
also have "... = (todos P xs ∧ todos P [a])" using HI by simp<br />
also have "... = (todos P [a] ∧ todos P xs)" by arith<br />
finally show "todos P (rev (a # xs)) = todos P (a # xs)" by simp<br />
qed<br />
<br />
(* fraortmoy serrodcal *)<br />
<br />
lemma "todos P (rev xs) = todos P xs"<br />
proof (induct xs)<br />
show "todos P (rev []) = todos P []" by simp<br />
next<br />
fix a xs <br />
assume H1: "todos P (rev xs) = todos P xs"<br />
have " todos P (rev (a # xs)) = todos P (rev xs @ [a])" by simp<br />
also have "… = (todos P (rev xs) ∧ todos P [a])" by (simp add:todos_append)<br />
also have "… = (todos P xs ∧ todos P [a])" using H1 by simp<br />
also have "… = (todos P [a] ∧ todos P xs)" by arith<br />
finally show "todos P (rev (a # xs)) = todos P (a#xs)" by simp<br />
qed<br />
<br />
(* paupeddeg dancorgar *)<br />
lemma "todos P (rev xs) = todos P xs"<br />
proof (induct xs)<br />
show " todos P (rev []) = todos P []" by simp<br />
next<br />
fix a xs<br />
assume HI: "todos P (rev xs) = todos P xs "<br />
have "todos P (rev (a # xs)) = todos P (rev(xs) @ [a])" by simp<br />
also have "... = (todos P (rev xs) ∧ todos P [a]) " by (simp add: todos_append)<br />
also have "... = (todos P xs ∧ todos P [a])" using HI by simp<br />
also have "... = (todos P [a] ∧ todos P xs)" by arith<br />
also have "... = todos P ([a] @ xs)" by (simp add: todos_append)<br />
finally show "todos P (rev (a # xs)) = todos P (a # xs)" by simp<br />
qed <br />
<br />
(* wilmorort *)<br />
lemma "todos P (rev xs) = todos P xs" (is "?P xs")<br />
proof (induct xs) <br />
show "?P []" by simp <br />
next<br />
fix a xs<br />
assume HI: "?P xs"<br />
have "todos P (rev (a#xs)) = todos P ((rev xs) @ [a])" by simp<br />
also have "... = (todos P(rev xs) ∧ todos P [a])" by (simp add: todos_append)<br />
also have "... = (todos P xs ∧ todos P [a])" using HI by simp<br />
also have "... = (todos P [a] ∧ todos P xs)" by (simp add: HOL.conj_commute)<br />
also have "... = todos P([a]@(xs))" by (simp)<br />
finally show "todos P (rev (a#xs))= todos P (a # xs)" by simp<br />
qed<br />
<br />
(* anaprarod *) <br />
(* es igual que las anteriores pero con el final también con patrones *)<br />
<br />
lemma "todos P (rev xs) = todos P xs" (is "?P xs")<br />
proof (induct xs)<br />
show "?P []" by auto<br />
next<br />
fix a xs<br />
assume HI: "?P xs"<br />
have "todos P (rev (a#xs)) = todos P (rev xs @ [a])" by simp<br />
also have "… = (todos P (rev xs) ∧ todos P [a])" by (simp add: todos_append)<br />
also have "… = (todos P xs ∧ todos P [a])" using HI by simp<br />
also have "… = (todos P [a] ∧ todos P xs)" by arith<br />
also have "… = todos P (a#xs)" by simp<br />
finally show "?P (a # xs)" by simp<br />
qed<br />
<br />
{*danrodcha*}<br />
lemma "todos P (rev xs) = todos P xs" (is "?Q xs")<br />
proof (induct xs)<br />
show "?Q []" by simp<br />
next<br />
fix a xs assume HI: "?Q xs"<br />
have "todos P (rev (a#xs)) = todos P ((rev xs) @ [a])" by simp<br />
also have "… = (todos P (rev xs) ∧ todos P [a])" by (simp add:todos_append)<br />
also have "… = (todos P xs ∧ todos P [a])" using HI by simp<br />
also have "… = (todos P [a] ∧ todos P xs)" by (simp add: HOL.conj_commute)<br />
also have "… = todos P (a # xs)" by simp<br />
finally show "?Q (a # xs)" by simp<br />
qed<br />
<br />
text {*<br />
--------------------------------------------------------------------- <br />
Ejercicio 6. Demostrar o refutar:<br />
algunos (λx. P x ∧ Q x) xs = (algunos P xs ∧ algunos Q xs)<br />
--------------------------------------------------------------------- <br />
*}<br />
<br />
lemma "algunos (λx. P x ∧ Q x) xs = (algunos P xs ∧ algunos Q xs)"<br />
oops<br />
<br />
(* migtermor ivamenjim ferrenseg paupeddeg crigomgom serrodcal wilmorort juacabsou rubgonmar anaprarod marpoldia1 fraortmoy josgarsan danrodcha pablucoto lucnovdos*)<br />
lemma "algunos (λx. P x ∧ Q x) xs = (algunos P xs ∧ algunos Q xs)" <br />
quickcheck<br />
oops<br />
(* Quickcheck encuentra el siguiente contraejemplo: P={a1}, Q={a2}, xs={a1,a2}. En este ejemplo:<br />
· "algunos (λx. P x ∧ Q x) xs = False"<br />
· "(algunos P xs ∧ algunos Q xs) = True" *)<br />
<br />
(* dancorgar *)<br />
lemma "algunos (λx. P x ∧ Q x) xs = (algunos P xs ∧ algunos Q xs)"<br />
proof -<br />
assume H1: "xs = [a, b]"<br />
assume H2: "P a = True"<br />
assume H3: "Q a = False"<br />
assume H4: "P b = False"<br />
assume H5: "Q b = True" <br />
have F1: "(algunos P xs ∧ algunos Q xs) = True" using H1 H2 H3 H4 H5 by simp<br />
have F2: "algunos (λx. P x ∧ Q x) xs = False" using H1 H2 H3 H4 H5 by simp<br />
have "algunos (λx. P x ∧ Q x) xs ≠ (algunos P xs ∧ algunos Q xs)" using F1 F2 by simp<br />
oops<br />
<br />
text {*<br />
--------------------------------------------------------------------- <br />
Ejercicio 7.1. Demostrar o refutar automáticamente <br />
algunos P (map f xs) = algunos (P ∘ f) xs<br />
--------------------------------------------------------------------- <br />
*}<br />
<br />
(* ivamenjim migtermor marpoldia1 crigomgom rubgonmar wilmorort anaprarod <br />
fraortmoy juacabsou paupeddeg josgarsan danrodcha pablucoto lucnovdos *)<br />
<br />
lemma "algunos P (map f xs) = algunos (P o f) xs"<br />
by (induct xs) auto<br />
<br />
(* ferrenseg *)<br />
<br />
lemma "algunos P (map f xs) = algunos (P o f) xs"<br />
by (induct xs) simp_all<br />
<br />
(* anaprarod dancorgar serrodcal *)<br />
lemma "algunos P (map f xs) = algunos (P o f) xs"<br />
apply (induct xs)<br />
apply auto<br />
done<br />
<br />
text {*<br />
--------------------------------------------------------------------- <br />
Ejercicio 7.2. Demostrar o refutar detalladamente<br />
algunos P (map f xs) = algunos (P ∘ f) xs<br />
--------------------------------------------------------------------- <br />
*}<br />
<br />
lemma "algunos P (map f xs) = algunos (P o f) xs"<br />
oops<br />
<br />
(* migtermor *)<br />
<br />
lemma AUX: "algunos (λa. P (f a)) xs = algunos (P o f) xs"<br />
by (induct xs) auto<br />
<br />
lemma "algunos P (map f xs) = algunos (P o f) xs" (is "?Q xs")<br />
proof (induct xs)<br />
show "?Q []" by simp<br />
next<br />
fix x xs<br />
assume HI: "?Q xs"<br />
have "algunos P (map f (x#xs)) = (algunos P ((f x)#(map f xs)))" by simp<br />
also have "… = ((P (f x)) ∨ (algunos P (map f xs)))" by (simp only: algunos.simps(2))<br />
also have "… = ((P (f x)) ∨ (algunos (P o f) xs))" <br />
proof (cases)<br />
assume C1: "(P (f x))"<br />
have Aux: "((P (f x)) ∨ (algunos P (map f xs))) = True" using C1 by simp<br />
have Aux1: "… = ((P (f x)) ∨ (algunos (P o f) xs))" using C1 by simp<br />
then show "((P (f x)) ∨ (algunos P (map f xs))) = ((P (f x)) ∨ (algunos (P o f) xs))" <br />
using Aux Aux1 by simp<br />
next<br />
assume C2: "¬(P (f x))"<br />
have Aux2: "((P (f x)) ∨ (algunos P (map f xs))) = (algunos P (map f xs))" using C2 by simp<br />
have Aux3: "… = (algunos (P o f) xs)" using HI by (simp add: AUX)<br />
also have "… = ((P (f x)) ∨ (algunos (P o f) xs))" using C2 by simp<br />
then show "((P (f x)) ∨ (algunos P (map f xs))) = ((P (f x)) ∨ (algunos (P o f) xs))" <br />
using Aux2 Aux3 by simp<br />
qed<br />
also have "… = (((P o f) x) ∨ (algunos (P o f) xs))" by simp<br />
also have "… = (algunos (P o f) (x#xs))" by simp<br />
finally show "algunos P (map f (x#xs)) = (algunos (P o f) (x#xs))" by simp<br />
qed<br />
<br />
(* ferrenseg *)<br />
<br />
lemma "algunos P (map f xs) = algunos (P ∘ f) xs"<br />
proof (induct xs)<br />
show "algunos P (map f []) = algunos (P ∘ f) []" by simp<br />
next<br />
fix x xs<br />
assume HI: "algunos P (map f xs) = algunos (P ∘ f) xs"<br />
show "algunos P (map f (x # xs)) = algunos (P ∘ f) (x # xs)"<br />
proof -<br />
have "algunos P (map f (x # xs)) = algunos P ((f x) # (map f xs))" by simp<br />
also have "… = ((P (f x)) ∨ (algunos P (map f xs)))" by simp<br />
also have "… = (((P ∘ f) x) ∨ (algunos (P ∘ f) xs))" using HI by simp<br />
also have "… = algunos (P ∘ f) (x # xs)" by simp<br />
finally show ?thesis<br />
qed<br />
qed<br />
<br />
(* ivamenjim *)<br />
<br />
lemma "algunos P (map f xs) = algunos (P o f) xs"<br />
proof (induct xs)<br />
show "algunos P (map f []) = algunos (P ∘ f) []" by simp<br />
next <br />
fix a xs<br />
assume HI: "algunos P (map f xs) = algunos (P ∘ f) xs"<br />
have "algunos P (map f (a # xs)) = algunos P ((f a)#map f xs)" by simp<br />
also have "... = (algunos P (map f [a]) ∨ algunos P (map f xs))" by simp <br />
also have "... = (algunos P (map f [a]) ∨ algunos (P ∘ f) xs)" using HI by simp <br />
finally show "algunos P (map f (a # xs)) = algunos (P ∘ f) (a # xs)" by simp<br />
qed<br />
<br />
(* crigomgom rubgonmar anaprarod marpoldia1 juacabsou danrodcha paupeddeg josgarsan lucnovdos*)<br />
lemma "algunos P (map f xs) = algunos (P o f) xs"<br />
proof (induct xs)<br />
show "algunos P (map f []) = algunos (P ∘ f) []" by simp<br />
next<br />
fix x xs<br />
assume HI: "algunos P (map f xs) = algunos (P ∘ f) xs"<br />
have "algunos P (map f (x # xs)) = algunos P ((f x) # (map f xs))" by simp<br />
also have "... = (P (f x) ∨ algunos P (map f xs))" by simp<br />
also have "... = (P (f x) ∨ algunos (P ∘ f) xs)" using HI by simp<br />
also have "... = ((P ∘ f) x ∨ algunos (P ∘ f) xs)" by simp<br />
finally show "algunos P (map f (x # xs)) = algunos (P ∘ f) (x # xs)" by simp<br />
qed<br />
<br />
(* wilmorort pablucoto*)<br />
lemma "algunos P (map f xs) = algunos (P o f) xs" (is "?P xs" )<br />
proof (induct xs)<br />
show "?P []" by simp<br />
next<br />
fix a xs<br />
assume HI: "algunos P (map f xs) = algunos (P ∘ f) xs"<br />
have "algunos P (map f (a # xs)) = (P (f a) ∨ algunos P (map f xs))" by simp<br />
also have "... = (P (f a) ∨ algunos (P ∘ f) xs )" using HI by simp<br />
also have "... = ((P ∘ f) a ∨ algunos (P ∘ f)xs)" by simp<br />
finally show " algunos P (map f (a # xs)) = algunos (P ∘ f) (a # xs)" by simp<br />
qed<br />
<br />
(* fraortmoy serrodcal *)<br />
<br />
lemma "algunos P (map f xs) = algunos (P o f) xs"<br />
proof (induct xs)<br />
show "algunos P (map f []) = algunos (P ∘ f) []" by simp<br />
next<br />
fix a xs<br />
assume H1: "algunos P (map f xs) = algunos (P ∘ f) xs"<br />
have "algunos P (map f (a # xs)) = (algunos P (map f [a]) ∨ algunos P (map f xs))" by simp<br />
also have "… = (algunos (P ∘ f) [a] ∨ algunos (P ∘ f) xs )" using H1 by simp<br />
also have "… = algunos (P ∘ f) (a#xs)" by simp<br />
finally show "algunos P (map f (a # xs)) = algunos (P ∘ f) (a#xs)" by simp<br />
qed<br />
<br />
{*danrodcha*}<br />
lemma "algunos P (map f xs) = algunos (P o f) xs" (is "?Q xs")<br />
proof (induct xs)<br />
show "?Q []" by simp<br />
next<br />
fix a xs assume HI: "?Q xs"<br />
have " algunos P (map f (a # xs)) = algunos P ((f a)#(map f xs))" by simp<br />
also have "… = (P (f a) ∨ algunos P (map f xs))" by simp<br />
also have "… = (P (f a) ∨ algunos (P ∘ f) xs)" using HI by blast<br />
also have "… = ((P ∘ f) a ∨ algunos (P ∘ f) xs)" by simp<br />
also have "… = algunos (P ∘ f) (a # xs)" by simp<br />
finally show "?Q (a # xs)" by blast<br />
qed<br />
<br />
(* dancorgar *)<br />
lemma "algunos P (map f xs) = algunos (P o f) xs"<br />
proof (induct xs)<br />
show "algunos P (map f []) = algunos (P o f) []" by simp<br />
next<br />
fix x xs<br />
assume HI: "algunos P (map f xs) = algunos (P o f) xs"<br />
have "algunos P (map f (x#xs)) = ((P (f x)) ∨ (algunos P (map f xs)))" by simp<br />
also have "… = ((P (f x)) ∨ (algunos (P o f) xs))" using HI by simp<br />
finally show "algunos P (map f (x#xs)) = algunos (P o f) (x#xs)" by simp<br />
qed<br />
<br />
text {*<br />
--------------------------------------------------------------------- <br />
Ejercicio 8.1. Demostrar o refutar automáticamente <br />
algunos P (xs @ ys) = (algunos P xs ∨ algunos P ys)<br />
--------------------------------------------------------------------- <br />
*}<br />
<br />
(* ivamenjim marpoldia1 paupeddeg crigomgom rubgonmar wilmorort fraortmoy danrodcha pablucoto dancorgar josgarsan anaprarod juacabsou lucnovdos*)<br />
<br />
lemma "algunos P (xs @ ys) = (algunos P xs ∨ algunos P ys)"<br />
by (induct xs) auto<br />
<br />
(* ferrenseg *)<br />
lemma "algunos P (xs @ ys) = (algunos P xs ∨ algunos P ys)"<br />
by (induct xs) simp_all<br />
<br />
(* anaprarod *)<br />
lemma "algunos P (xs @ ys) = (algunos P xs ∨ algunos P ys)"<br />
apply (induct xs)<br />
apply auto<br />
done<br />
<br />
text {*<br />
--------------------------------------------------------------------- <br />
Ejercicio 8.2. Demostrar o refutar detalladamente<br />
algunos P (xs @ ys) = (algunos P xs ∨ algunos P ys)<br />
--------------------------------------------------------------------- <br />
*}<br />
<br />
(* ivamenjim ferrenseg marpoldia1 wilmorort dancorgar josgarsan juacabsou serrodcal lucnovdos*)<br />
<br />
lemma algunos_append:<br />
"algunos P (xs @ ys) = (algunos P xs ∨ algunos P ys)"<br />
proof (induct xs)<br />
show "algunos P ([] @ ys) = (algunos P [] ∨ algunos P ys)" by simp<br />
next <br />
fix a xs<br />
assume HI: "algunos P (xs @ ys) = (algunos P xs ∨ algunos P ys)"<br />
have "algunos P ((a # xs) @ ys) = (P a ∨ (algunos P (xs @ ys)))" by simp<br />
also have "... = (P a ∨ (algunos P xs ∨ algunos P ys))" using HI by simp<br />
finally show "algunos P ((a # xs) @ ys) = (algunos P (a # xs) ∨ algunos P ys)" by simp<br />
qed<br />
<br />
(* migtermor crigomgom *)<br />
<br />
lemma algunos_append2:<br />
"algunos P (xs @ ys) = (algunos P xs ∨ algunos P ys)" (is "?Q xs")<br />
proof (induct xs)<br />
show "?Q []" by simp<br />
next <br />
fix x xs<br />
assume HI: "?Q xs"<br />
have "algunos P ((x#xs) @ ys) = algunos P (x#(xs @ ys))" by simp<br />
also have "… = ((P x) ∨ (algunos P (xs @ ys)))" by simp<br />
also have "… = ((P x) ∨ (algunos P xs) ∨ (algunos P ys))" using HI by simp<br />
also have "… = ((algunos P (x#xs)) ∨ (algunos P ys))" by simp<br />
finally show "algunos P ((x#xs) @ ys) = (algunos P (x#xs) ∨ algunos P ys)" by simp<br />
qed<br />
<br />
(* paupeddeg*)<br />
lemma algunos_append3:<br />
"algunos P (xs @ ys) = (algunos P xs ∨ algunos P ys)"<br />
proof (induct xs)<br />
show "algunos P ([] @ ys) = (algunos P [] ∨ algunos P ys)" by simp<br />
next<br />
fix a xs<br />
assume HI: "algunos P (xs @ ys) = (algunos P xs ∨ algunos P ys)"<br />
have "algunos P ((a # xs) @ ys) = (algunos P [a] ∨ ( algunos P (xs @ ys)))" by simp<br />
also have "... = (algunos P [a] ∨ (algunos P xs ∨ algunos P ys))" using HI by simp<br />
finally show "algunos P ((a # xs) @ ys) = (algunos P (a # xs) ∨ algunos P ys)" by simp<br />
qed<br />
<br />
(* fraortmoy *)<br />
lemma algunos_append4:<br />
"algunos P (xs @ ys) = (algunos P xs ∨ algunos P ys)"<br />
proof (induct xs)<br />
show " algunos P ([] @ ys) = (algunos P [] ∨ algunos P ys)" by simp<br />
next <br />
fix a xs<br />
assume H1: "algunos P (xs @ ys) = (algunos P xs ∨ algunos P ys)"<br />
have "algunos P ((a # xs) @ ys) = (algunos P [a] ∨ algunos P (xs @ ys))" by simp<br />
also have "… = (algunos P [a] ∨ algunos P xs ∨ algunos P ys)" using H1 by simp<br />
also have "… = ((algunos P [a] ∨ algunos P xs) ∨ algunos P ys)" by simp<br />
also have "… = (algunos P (a#xs) ∨ algunos P ys)" by simp<br />
finally show "algunos P ((a # xs) @ ys) = (algunos P (a#xs) ∨ algunos P ys)" by simp<br />
qed<br />
<br />
{* danrodcha pablucoto anaprarod *}<br />
lemma algunos_append5:<br />
"algunos P (xs @ ys) = (algunos P xs ∨ algunos P ys)" (is "?Q xs")<br />
proof (induct xs)<br />
show "?Q []" by simp<br />
next<br />
fix a xs assume HI: "?Q xs"<br />
have "algunos P ((a#xs) @ ys) = algunos P (a#(xs @ ys))" by simp<br />
also have "… = (algunos P [a] ∨ algunos P (xs @ ys))" by simp<br />
also have "… = (algunos P [a] ∨ algunos P xs ∨ algunos P ys)" using HI by simp<br />
also have "… = (algunos P (a # xs) ∨ algunos P ys)" by simp<br />
finally show "?Q (a # xs)" by simp<br />
qed<br />
<br />
text {*<br />
--------------------------------------------------------------------- <br />
Ejercicio 9.1. Demostrar o refutar automáticamente<br />
algunos P (rev xs) = algunos P xs<br />
--------------------------------------------------------------------- <br />
*}<br />
<br />
(* ivamenjim migtermor marpoldia1 rubgonmar paupeddeg dancorgar anaprarod serrodcal *)<br />
<br />
lemma "algunos P (rev xs) = algunos P xs"<br />
apply (induct xs)<br />
apply simp<br />
apply (simp add: algunos_append)<br />
apply auto<br />
done<br />
<br />
(* ferrenseg crigomgom danrodcha pablucoto josgarsan pabrodmac lucnovdos*)<br />
lemma "algunos P (rev xs) = algunos P xs"<br />
by (induct xs) (auto simp add: algunos_append)<br />
<br />
(* wilmorort juacabsou *)<br />
lemma "algunos P (rev xs) = algunos P xs"<br />
by (induct xs,simp,simp add: algunos_append,auto) <br />
<br />
<br />
text {*<br />
--------------------------------------------------------------------- <br />
Ejercicio 9.2. Demostrar o refutar detalladamente<br />
algunos P (rev xs) = algunos P xs<br />
--------------------------------------------------------------------- <br />
*}<br />
<br />
lemma "algunos P (rev xs) = algunos P xs"<br />
oops<br />
<br />
(* migtermor *)<br />
lemma auxiliar1:<br />
"rev (a#xs) = rev xs @ [a]"<br />
by auto <br />
<br />
lemma "algunos P (rev xs) = algunos P xs" (is "?Q xs")<br />
proof (induct xs)<br />
show "?Q []" by simp<br />
next<br />
fix x xs<br />
assume HI: "?Q xs"<br />
have "algunos P (rev (x#xs)) = (algunos P (rev xs @ [x]))" using auxiliar1 by simp<br />
also have "… = ((algunos P (rev xs)) ∨ (algunos P [x]))" by (simp add: algunos_append)<br />
also have "… = ((P x) ∨ (algunos P (rev xs)))" by simp arith<br />
also have "… = ((P x) ∨ (algunos P xs))" using HI by simp<br />
finally show "algunos P (rev (x#xs)) = (algunos P (x#xs))" by simp<br />
qed<br />
<br />
(* ferrenseg *)<br />
<br />
lemma "algunos P (rev xs) = algunos P xs"<br />
proof (induct xs)<br />
show "algunos P (rev []) = algunos P []" by simp<br />
next<br />
fix x xs<br />
assume HI: "algunos P (rev xs) = algunos P xs"<br />
show "algunos P (rev (x # xs)) = algunos P (x # xs)"<br />
proof -<br />
have "algunos P (rev (x # xs)) = algunos P ((rev xs) @ [x])" by simp<br />
also have "… = (algunos P (rev xs) ∨ algunos P [x])" <br />
by (simp add: algunos_append)<br />
also have "… = (algunos P xs ∨ algunos P [x])" using HI by simp<br />
also have "… = (algunos P xs ∨ P x)" by simp<br />
also have "… = (P x ∨ algunos P xs)" by auto<br />
also have "… = algunos P (x # xs)" by simp<br />
finally show ?thesis<br />
qed<br />
qed<br />
<br />
(* ivamenjim marpoldia1*)<br />
<br />
lemma "algunos P (rev xs) = algunos P xs" (is "?P xs")<br />
proof (induct xs)<br />
show "?P []" by simp<br />
next <br />
fix a xs<br />
assume HI: "?P xs"<br />
have "algunos P (rev (a # xs)) = (algunos P ((rev xs)@[a]))" by simp<br />
also have "... = ((algunos P (rev xs)) ∨ algunos P [a])" by (simp add: algunos_append)<br />
also have "... = (algunos P xs ∨ algunos P [a])" using HI by simp<br />
also have "... = (algunos P [a] ∨ algunos P xs)" by arith<br />
finally show "algunos P (rev (a # xs)) = algunos P (a # xs)" by simp<br />
qed<br />
<br />
(* paupeddeg dancorgar serrodcal lucnovdos*)<br />
lemma "algunos P (rev xs) = algunos P xs"<br />
proof (induct xs)<br />
show "algunos P (rev []) = algunos P []" by simp<br />
next<br />
fix a xs<br />
assume HI:"algunos P (rev xs) = algunos P xs"<br />
have "algunos P (rev (a # xs)) = algunos P ((rev xs) @ [a])" by simp<br />
also have "... = ((algunos P (rev xs)) ∨ (algunos P [a]))" by (simp add: algunos_append)<br />
also have "... = ((algunos P xs) ∨ (algunos P [a]))" using HI by simp<br />
also have "... = ((algunos P [a]) ∨ (algunos P xs))" by arith<br />
finally show "algunos P (rev (a # xs)) = algunos P (a # xs)" by simp<br />
qed<br />
<br />
(*crigomgom pablucoto anaprarod rubgonmar juacabsou*)<br />
lemma "algunos P (rev xs) = algunos P xs"<br />
proof (induct xs)<br />
show "algunos P (rev []) = algunos P []" by simp<br />
next<br />
fix x xs<br />
assume HI: " algunos P (rev xs) = algunos P xs" <br />
have "algunos P (rev (x # xs)) = algunos P ((rev xs) @ [x])" by simp<br />
also have "... = (algunos P (rev xs) ∨ algunos P [x])" by (simp add: algunos_append)<br />
also have "... = (algunos P xs ∨ algunos P [x])" using HI by simp<br />
also have "... = (algunos P xs ∨ P x)" by simp<br />
also have "... = (P x ∨ algunos P xs)" by arith<br />
also have "... = algunos P (x#xs)" by simp<br />
finally show "algunos P (rev (x # xs)) = algunos P (x # xs)" by simp<br />
qed<br />
<br />
<br />
(*wilmorort*)<br />
lemma "algunos P (rev xs) = algunos P xs" (is "?P xs")<br />
proof (induct xs) <br />
show "?P []" by simp <br />
next<br />
fix a xs<br />
assume HI: "?P xs"<br />
have "algunos P (rev (a#xs)) = algunos P ((rev xs) @ [a])" by simp<br />
also have "... = (algunos P(rev xs) ∨ algunos P [a])" by (simp add: algunos_append)<br />
also have "... = (algunos P xs ∨ algunos P [a])" using HI by simp<br />
also have "... = (algunos P [a] ∨ algunos P xs)" by (simp add: HOL.disj_commute)<br />
also have "... = algunos P([a]@(xs))" by (simp)<br />
finally show "algunos P (rev (a#xs))= algunos P (a # xs)" by simp<br />
qed<br />
<br />
{*danrodcha *}<br />
lemma "algunos P (rev xs) = algunos P xs" (is "?Q xs")<br />
proof (induct xs)<br />
show "?Q []" by simp<br />
next<br />
fix a xs assume HI: "?Q xs"<br />
have "algunos P (rev (a#xs)) = algunos P ((rev xs) @ [a])" by simp<br />
also have "… = (algunos P (rev xs) ∨ algunos P [a])" by (simp add:algunos_append)<br />
also have "… = (algunos P xs ∨ algunos P [a])" using HI by simp<br />
also have "… = (algunos P [a] ∨ algunos P xs)" by (simp add: HOL.disj_commute)<br />
also have "… = algunos P (a # xs)" by simp<br />
finally show "?Q (a # xs)" by simp<br />
qed<br />
<br />
text {*<br />
--------------------------------------------------------------------- <br />
Ejercicio 10. Encontrar un término no trivial Z tal que sea cierta la <br />
siguiente ecuación:<br />
algunos (λx. P x ∨ Q x) xs = Z<br />
y demostrar la equivalencia de forma automática y detallada.<br />
--------------------------------------------------------------------- <br />
*}<br />
<br />
(* migtermor *)<br />
<br />
lemma "algunos (λx. P x ∨ Q x) xs = ((algunos (λx. P x) xs) ∨ (algunos (λx. Q x) xs))"<br />
by (induct xs) auto<br />
<br />
lemma "algunos (λx. P x ∨ Q x) xs = ((algunos (λx. P x) xs) ∨ (algunos (λx. Q x) xs))" (is "?R xs")<br />
proof (induct xs)<br />
show "?R []" by simp<br />
next<br />
fix x xs<br />
assume HI: "?R xs"<br />
have "algunos (λx. P x ∨ Q x) (x#xs) = (((λx. P x ∨ Q x) x) ∨ (algunos (λx. P x ∨ Q x) xs))"<br />
by simp<br />
also have "… = ((P x ∨ Q x) ∨ (algunos (λx. P x ∨ Q x) xs))" by simp<br />
also have H1: "… = ((((λx. P x) x) ∨ (algunos (λx. P x) xs)) ∨ (((λx. Q x) x) ∨ (algunos (λx. Q x) xs)))"<br />
using HI by simp arith<br />
have H2: "… = ((algunos (λx. P x) (x#xs)) ∨ (algunos (λx. Q x) (x#xs)))" <br />
by simp<br />
have C: "(algunos (λx. P x ∨ Q x) (x#xs)) = <br />
((algunos (λx. P x) (x#xs)) ∨ (algunos (λx. Q x) (x#xs)))" <br />
using H1 H2 by simp<br />
finally show "(algunos (λx. P x ∨ Q x) (x#xs)) = <br />
((algunos (λx. P x) (x#xs)) ∨ (algunos (λx. Q x) (x#xs)))" <br />
using C by simp<br />
qed<br />
<br />
<br />
(* ferrenseg *)<br />
<br />
lemma "algunos (λx. P x ∨ Q x) xs = (algunos P xs ∨ algunos Q xs)"<br />
proof (induct xs)<br />
show "algunos (λx. P x ∨ Q x) [] = (algunos P [] ∨ algunos Q [])" by simp<br />
next<br />
fix x xs<br />
assume HI: "algunos (λx. (P x ∨ Q x)) xs = (algunos P xs ∨ algunos Q xs)"<br />
show "algunos (λx. P x ∨ Q x) (x # xs) = (algunos P (x # xs) ∨ algunos Q (x # xs))"<br />
proof -<br />
have "algunos (λx. P x ∨ Q x) (x # xs) = <br />
((P x) ∨ (Q x) ∨ algunos (λx. P x ∨ Q x) xs)" by simp<br />
also have "… = ((P x) ∨ (Q x) ∨ algunos P xs ∨ algunos Q xs)" <br />
using HI by simp<br />
also have "… = (((P x) ∨ algunos P xs) ∨ ((Q x) ∨ algunos Q xs))" by auto<br />
also have "… = (algunos P (x # xs) ∨ algunos Q (x # xs))" by simp<br />
finally show ?thesis <br />
qed<br />
qed<br />
<br />
(* ivamenjim marpoldia1 paupeddeg *)<br />
<br />
lemma "algunos (λx. P x ∨ Q x) xs = (algunos P xs ∨ algunos Q xs)" <br />
by (induct xs) auto<br />
<br />
lemma "algunos (λx. P x ∨ Q x) xs = (algunos P xs ∨ algunos Q xs)"<br />
proof (induct xs)<br />
show "algunos (λx. P x ∨ Q x) [] = (algunos P [] ∨ algunos Q [])" by simp<br />
next <br />
fix a xs<br />
assume HI: "algunos (λx. P x ∨ Q x) xs = (algunos P xs ∨ algunos Q xs)"<br />
have "algunos (λx. P x ∨ Q x) (a # xs) = (P a ∨ Q a ∨ algunos (λx. P x ∨ Q x) xs)" by simp<br />
also have "... = (P a ∨ Q a ∨ algunos P xs ∨ algunos Q xs)" using HI by simp<br />
finally show "algunos (λx. P x ∨ Q x) (a # xs) = (algunos P (a # xs) ∨ algunos Q (a # xs))" by auto<br />
qed<br />
<br />
<br />
(* wilmorort danrodcha anaprarod juacabsou*)<br />
lemma "algunos (λx. P x ∨ Q x) xs = (¬todos(λx. ¬P x)xs ∨ ¬todos(λx. ¬Q x)xs)"<br />
by (induct xs) auto<br />
<br />
(* wilmorort *)<br />
lemma "algunos (λx. P x ∨ Q x) xs = (¬todos(λx. ¬P x)xs ∨ ¬todos(λx. ¬Q x)xs)"<br />
(is "?P xs")<br />
proof (induct xs)<br />
show "?P []" by simp<br />
next<br />
fix a xs<br />
assume HI: "?P xs"<br />
have "algunos (λx. P x ∨ Q x) (a # xs) = (P a ∨ Q a ∨ algunos (λx. P x ∨ Q x) xs)" by simp<br />
also have "... = (P a ∨ Q a ∨ (¬ todos (λx. ¬ P x) xs ∨ ¬ todos (λx. ¬ Q x) xs))" using HI by simp<br />
also have "... = (P a ∨ ¬ todos (λx. ¬ P x) xs ∨ Q a ∨ ¬ todos (λx. ¬ Q x) xs )" by arith<br />
finally show "algunos (λx. P x ∨ Q x) (a # xs) = (~ todos (λx. ¬ P x) (a # xs) ∨ ~ todos (λx. ¬ Q x) (a # xs))" by simp<br />
qed<br />
<br />
{* danrodcha *}<br />
lemma "algunos (λx. P x ∨ Q x) xs = (algunos P xs ∨ algunos Q xs)" (is "?R xs")<br />
proof (induct xs)<br />
show "?R []" by simp<br />
next<br />
fix a xs assume HI: "?R xs"<br />
have 1:" (Q a ∨ algunos P xs) = (algunos P xs ∨ Q a)" by (simp add: HOL.disj_commute)<br />
have "algunos (λx. P x ∨ Q x) (a # xs) = <br />
(algunos P [a] ∨ algunos Q [a] ∨ algunos (λx. P x ∨ Q x) xs)" by simp<br />
also have "… = (P a ∨ (Q a ∨ algunos P xs) ∨ algunos Q xs)" using HI by simp<br />
also have "… = (P a ∨ algunos P xs ∨ Q a ∨ algunos Q xs)" using 1 by simp<br />
also have "… = (algunos P (a # xs) ∨ algunos Q (a # xs))" by simp<br />
finally show "?R (a # xs)" by simp<br />
qed<br />
<br />
(* dancorgar *)<br />
lemma "algunos (λx. P x ∧ Q x) xs ⟹ (algunos P xs ∧ algunos Q xs)"<br />
apply (induct xs)<br />
apply auto<br />
done<br />
<br />
(* dancorgar *)<br />
lemma "algunos (λx. P x ∧ Q x) xs ⟹ (algunos P xs ∧ algunos Q xs)"<br />
proof (induct xs)<br />
show "algunos (λx. P x ∧ Q x) [] ⟹ (algunos P [] ∧ algunos Q [])" by simp<br />
next<br />
fix xf xs<br />
assume HI: "algunos (λx. P x ∧ Q x) xs ⟹ (algunos P xs ∧ algunos Q xs)"<br />
have F1: "algunos (λx. P x ∧ Q x) (xf#xs) ⟹ ((P xf ∧ Q xf) ∨ algunos (λx. P x ∧ Q x) xs)" by simp<br />
also have F2: "… ⟹ (P xf ∧ Q xf ∨ (algunos P xs ∧ algunos Q xs))" using HI by simp<br />
have F3: "((P xf ∧ Q xf) ∨ (algunos P xs ∧ algunos Q xs)) ⟹ <br />
((P xf ∨ algunos P xs) ∧ (Q xf ∨ algunos Q xs))" by blast<br />
have F4: "((P xf ∨ algunos P xs) ∧ (Q xf ∨ algunos Q xs)) ⟹ <br />
(algunos P (xf#xs) ∧ algunos Q (xf#xs))" by simp<br />
show "algunos (λx. P x ∧ Q x) (xf#xs) ⟹ (algunos P (xf#xs) ∧ algunos Q (xf#xs))"<br />
using F1 F2 F3 F4 by blast<br />
qed<br />
<br />
(*pablucoto crigomgom serrodcal*)<br />
-- "Automatica"<br />
lemma "algunos (λx. P x ∨ Q x) xs = (algunos P xs ∨ algunos Q xs)"<br />
by (induct xs) auto<br />
-- "Detallada"<br />
lemma "algunos (λx. P x ∨ Q x) xs = (algunos P xs ∨ algunos Q xs) "<br />
proof (induct xs)<br />
show "algunos (λx. P x ∨ Q x) [] = (algunos P [] ∨ algunos Q []) " by simp<br />
next<br />
fix a xs<br />
assume HI: " algunos (λx. P x ∨ Q x) xs = (algunos P xs ∨ algunos Q xs)"<br />
have " algunos (λx. P x ∨ Q x) (a # xs) = ( P a ∨ Q a ∨ algunos (λx. P x ∨ Q x) xs)" by simp<br />
also have "... = (P a ∨ Q a ∨ (algunos P xs ∨ algunos Q xs))" using HI by simp<br />
also have "… = (((P a) ∨ algunos P xs) ∨ ((Q a) ∨ algunos Q xs))" by arith <br />
also have "… = (algunos P (a#xs) ∨ algunos Q (a#xs))" by simp<br />
finally show "algunos (λx. P x ∨ Q x) (a # xs) = (algunos P (a # xs) ∨ algunos Q (a # xs)) " by simp<br />
qed<br />
<br />
(* anaprarod juacabsou *)<br />
lemma "algunos (λx. P x ∨ Q x) xs = (algunos P xs ∨ algunos Q xs)" (is "?P xs")<br />
proof (induct xs)<br />
show "?P []" by simp<br />
next<br />
fix x xs<br />
assume HI: "?P xs"<br />
have "algunos (λx. P x ∨ Q x) (x#xs) =(P x ∨ Q x ∨ algunos (λx. P x ∨ Q x) xs)" by simp<br />
also have "… = (P x ∨ Q x ∨ algunos P xs ∨ algunos Q xs)" using HI by simp<br />
also have "… = (P x ∨ algunos P xs ∨ Q x ∨ algunos Q xs)" by arith<br />
finally show "?P (x#xs)" by simp<br />
qed<br />
<br />
text {*<br />
--------------------------------------------------------------------- <br />
Ejercicio 11.1. Demostrar o refutar automáticamente<br />
algunos P xs = (¬ todos (λx. (¬ P x)) xs)<br />
--------------------------------------------------------------------- <br />
*}<br />
<br />
(* ivamenjim wilmorort migtermor marpoldia1 crigomgom rubgonmar paupeddeg danrodcha pablucoto dancorgar josgarsan anaprarod juacabsou lucnovdos serrodcal *)<br />
<br />
lemma "algunos P xs = (¬ todos (λx. (¬ P x)) xs)"<br />
by (induct xs) auto<br />
<br />
(* ferrenseg *)<br />
<br />
lemma "algunos P xs = (¬ todos (λx. (¬ P x)) xs)"<br />
by (induct xs) simp_all<br />
<br />
(* anaprarod *)<br />
lemma "algunos P xs = (¬ todos (λx. (¬ P x)) xs)" <br />
apply (induct xs)<br />
apply auto<br />
done<br />
<br />
<br />
text {*<br />
--------------------------------------------------------------------- <br />
Ejercicio 11.2. Demostrar o refutar datalladamente<br />
algunos P xs = (¬ todos (λx. (¬ P x)) xs)<br />
--------------------------------------------------------------------- <br />
*}<br />
<br />
(* ivamenjim migtermor marpoldia1 crigomgom paupeddeg josgarsan juacabsou serrodcal *)<br />
<br />
lemma "algunos P xs = (¬ todos (λx. (¬ P x)) xs)"<br />
proof (induct xs)<br />
show "algunos P [] = (¬ todos (λx. ¬ P x) [])" by simp<br />
next <br />
fix a xs<br />
assume HI: "algunos P xs = (¬ todos (λx. ¬ P x) xs)"<br />
have "algunos P (a # xs) = (P a ∨ (algunos P xs))" by simp<br />
also have "... = (P a ∨ (¬ todos (λx. ¬ P x) xs))" using HI by simp<br />
finally show "algunos P (a # xs) = (¬ todos (λx. ¬ P x) (a # xs))" by simp<br />
qed<br />
<br />
(* ferrenseg *)<br />
<br />
lemma "algunos P xs = (¬ todos (λx. (¬ P x)) xs)"<br />
proof (induct xs)<br />
show "algunos P [] = (¬ todos (λx. (¬ P x)) [])" by simp<br />
next<br />
fix x xs<br />
assume HI: "algunos P xs = (¬ todos (λx. (¬ P x)) xs)"<br />
show "algunos P (x # xs) = (¬ todos (λx. (¬ P x)) (x # xs))"<br />
proof - <br />
have "algunos P (x # xs) = ((P x) ∨ algunos P xs)" by simp<br />
also have "… = ((P x) ∨ ¬ todos (λx. (¬ P x)) xs)" using HI by simp<br />
also have "… = (¬ (¬ (P x) ∧ todos (λx. (¬ P x)) xs))" by simp<br />
also have "… = (¬ todos (λx. (¬ P x)) (x # xs))" by simp<br />
finally show ?thesis<br />
qed<br />
qed<br />
<br />
{* danrodcha pablucoto anaprarod rubgonmar *}<br />
lemma "algunos P xs = (¬ todos (λx. (¬ P x)) xs)" (is "?Q xs")<br />
proof (induct xs)<br />
show "?Q []" by simp<br />
next<br />
fix a xs assume HI: "?Q xs"<br />
have "algunos P (a # xs) = (P a ∨ algunos P xs)" by simp<br />
also have "… = (P a ∨ (¬ todos (λx. (¬ P x)) xs))" using HI by simp<br />
also have "… = (¬(¬(P a) ∧ todos(λx. (¬ P x)) xs))" by simp<br />
also have "… = (¬ todos(λx. (¬ P x)) (a # xs))" by simp<br />
finally show "?Q (a # xs)" by simp<br />
qed<br />
<br />
(* dancorgar*)<br />
lemma "algunos P xs = (¬ todos (λx. (¬ P x)) xs)"<br />
proof (induct xs)<br />
show "algunos P [] = (¬ todos (λx. (¬ P x)) [])" by simp<br />
next<br />
fix xf xs<br />
assume HI: "algunos P xs = (¬ todos (λx. (¬ P x)) xs)"<br />
have "algunos P (xf#xs) = ((P xf) ∨ algunos P xs)" by simp<br />
also have "… = ( (P xf) ∨ (¬ todos (λx. (¬ P x)) xs) )" using HI by simp<br />
also have "… = (¬((¬P xf) ∧ (todos (λx. (¬ P x)) xs)))" by simp<br />
finally show "algunos P (xf#xs) = (¬todos (λx. (¬ P x)) (xf#xs))" by simp<br />
qed<br />
<br />
(* lucnovdos*)<br />
lemma "algunos P xs = (¬ todos (λx. (¬ P x)) xs)"<br />
proof (induct xs)<br />
show "algunos P [] = (¬ todos (λx. (¬ P x)) [])" by simp<br />
next<br />
fix n xs<br />
assume HI: "algunos P xs = (¬ todos (λx. (¬ P x)) xs)"<br />
have "algunos P (n#xs) = ((P n) ∨ (algunos P xs))" by simp<br />
also have "… = ((P n) ∨ (¬ todos (λx. (¬ P x)) xs))" using HI by simp<br />
also have "… = (¬((¬P n) ∧ todos (λx. (¬ P x)) xs))" by simp<br />
also have "… = (¬ todos (λx. (¬ P x)) (n#xs))" by simp<br />
finally show "algunos P (n#xs) = (¬ todos (λx. (¬ P x)) (n#xs))" by simp<br />
qed<br />
<br />
text {*<br />
--------------------------------------------------------------------- <br />
Ejercicio 12. Definir la funcion primitiva recursiva <br />
estaEn :: 'a ⇒ 'a list ⇒ bool<br />
tal que (estaEn x xs) se verifica si el elemento x está en la lista<br />
xs. Por ejemplo, <br />
estaEn (2::nat) [3,2,4] = True<br />
estaEn (1::nat) [3,2,4] = False<br />
--------------------------------------------------------------------- <br />
*}<br />
<br />
(* ivamenjim ferrenseg migtermor serrodcal crigomgom rubgonmar marpoldia1 paupeddeg danrodcha dancorgar pablucoto anaprarod juacabsou lucnovdos*)<br />
<br />
fun estaEn :: "'a ⇒ 'a list ⇒ bool" where<br />
"estaEn x [] = False"<br />
| "estaEn x (a#xs) = ((a = x) ∨ (estaEn x xs))"<br />
<br />
value "estaEn (2::nat) [3,2,4] = True"<br />
value "estaEn (1::nat) [3,2,4] = False"<br />
<br />
<br />
text {*<br />
--------------------------------------------------------------------- <br />
Ejercicio 13. Expresar la relación existente entre estaEn y algunos. <br />
Demostrar dicha relación de forma automática y detallada.<br />
--------------------------------------------------------------------- <br />
*}<br />
<br />
(* migtermor crigomgom *)<br />
<br />
lemma "estaEn x xs = (algunos (λa. a=x) xs)"<br />
by (induct xs) auto<br />
<br />
lemma auxiliar13:<br />
"(x=a) = ((λa. a=x) a)"<br />
by auto<br />
<br />
lemma "estaEn x xs = (algunos (λa. a=x) xs)" (is "?P xs")<br />
proof (induct xs)<br />
show "?P []" by simp<br />
next<br />
fix a xs<br />
assume HI: "?P xs"<br />
have "estaEn x (a#xs) = ((a=x) ∨ (estaEn x xs))" by simp<br />
also have H: "… = ((a=x) ∨ (algunos (λa. a=x) xs))" using HI by simp<br />
also have "… = (((λa. a=x) a) ∨ (algunos (λa. a=x) xs))" using auxiliar13 by simp<br />
also have "… = (algunos (λa. a=x) (a#xs))" by simp<br />
have C: "estaEn x (a#xs) = (algunos (λa. a=x) (a#xs))" using H by simp<br />
finally show "estaEn x (a#xs) = (algunos (λa. a=x) (a#xs))" using C by simp<br />
qed<br />
<br />
<br />
(* ferrenseg *)<br />
<br />
lemma estaEn_algunos:<br />
"estaEn y xs = algunos (λx. x=y) xs"<br />
by (induct xs) auto<br />
<br />
lemma estaEn_algunos_2:<br />
"estaEn y xs = algunos (λx. x=y) xs"<br />
proof (induct xs)<br />
show "estaEn y [] = algunos (λx. x=y) []" by simp<br />
next<br />
fix x xs<br />
assume HI: "estaEn y xs = algunos (λx. x=y) xs"<br />
show "estaEn y (x # xs) = algunos (λx. x=y) (x # xs)"<br />
proof -<br />
have "estaEn y (x # xs) = (y=x ∨ estaEn y xs)" by simp<br />
also have "… = (y=x ∨ algunos (λx. x=y) xs)" using HI by simp<br />
also have "… = (x=y ∨ algunos (λx. x=y) xs)" by auto<br />
also have "… = algunos (λx. x=y) (x # xs)" by simp<br />
finally show ?thesis<br />
qed<br />
qed<br />
<br />
(* crigomgom paupeddeg dancorgar juacabsou serrodcal *)<br />
<br />
lemma "estaEn a xs = algunos (λx. x = a) xs"<br />
by (induct xs) auto<br />
<br />
lemma "estaEn x xs = (algunos (λa. a=x) xs)" <br />
proof (induct xs)<br />
show " estaEn x [] = algunos (λa. a = x) []" by simp<br />
next<br />
fix a xs<br />
assume HI: "estaEn x xs = algunos (λa. a = x) xs"<br />
have "estaEn x (a # xs) = (a = x ∨ estaEn x xs)" by simp<br />
also have "... = (a = x ∨ algunos (λa. a = x) xs)" using HI by simp<br />
finally show " estaEn x (a # xs) = algunos (λa. a = x) (a # xs)" by simp<br />
qed<br />
<br />
(* ivamenjim marpoldia1 *)<br />
<br />
lemma "estaEn x xs = algunos (λy. x=y) xs"<br />
by (induct xs) auto<br />
<br />
lemma "estaEn x xs = algunos (λy. x=y) xs"<br />
proof (induct xs)<br />
show "estaEn x [] = algunos (op = x) []" by simp<br />
next <br />
fix a xs<br />
assume HI: "estaEn x xs = algunos (op = x) xs"<br />
have "estaEn x (a # xs) = ((a = x) ∨ (estaEn x xs))" by simp<br />
also have "... = (a = x ∨ algunos (op = x) xs)" using HI by simp<br />
finally show " estaEn x (a # xs) = algunos (op = x) (a # xs)" by auto<br />
qed<br />
<br />
{* danrodcha *}<br />
lemma "estaEn a xs = algunos (op = a) xs"<br />
by (induct xs) auto<br />
<br />
{* danrodcha *}<br />
lemma "estaEn a xs = algunos (op = a) xs"<br />
proof (induct xs)<br />
show "estaEn a [] = algunos (op = a) []" by simp<br />
next<br />
fix x xs <br />
assume HI: "estaEn a xs = algunos (op = a) xs"<br />
have "estaEn a (x # xs) = ((x = a) ∨ (estaEn a xs))" by simp<br />
also have "… = (x = a ∨ algunos (op = a) xs)" using HI by simp<br />
also have "… = ((op = a) x ∨ algunos (op = a) xs)" by (simp add:HOL.eq_commute)<br />
also have "… = algunos (op = a) (x # xs)" by simp<br />
finally show "estaEn a (x # xs) = algunos (op = a) (x # xs)" by simp<br />
qed<br />
<br />
(* pablucoto *)<br />
<br />
-- "Automatica"<br />
lemma " estaEn y xs = algunos (λx. x=y) xs"<br />
by (induct xs) auto<br />
<br />
-- "Detallada"<br />
lemma " estaEn y xs = algunos (λx. x=y) xs"<br />
proof (induct xs)<br />
show "estaEn y [] = algunos (λx. x = y) []" by simp<br />
next<br />
fix a xs<br />
assume HI: " estaEn y xs = algunos (λx. x = y) xs "<br />
have " estaEn y (a # xs) = (y = a ∨ estaEn y xs)" by simp <br />
also have "... = (y = a ∨ algunos (λx. x = y) xs) " using HI by simp<br />
also have "… = (a=y ∨ algunos (λx. x=y) xs)" by auto<br />
also have "... = (algunos (λx. x = y) (a#xs))" by simp<br />
finally show "estaEn y (a # xs) = algunos (λx. x = y) (a # xs) " by simp<br />
qed<br />
<br />
<br />
(* anaprarod *)<br />
(* Automática*)<br />
lemma "estaEn a xs = algunos (λx. a=x) xs" (is "?P xs")<br />
by (induct xs) auto<br />
<br />
(* Detallada (igual que las anteriores pero obviando el paso previo al finally) *)<br />
lemma "estaEn a xs = algunos (λx. a=x) xs" (is "?P xs")<br />
proof (induct xs)<br />
show "?P []" by simp<br />
next<br />
fix x xs<br />
assume HI : "?P xs"<br />
have "estaEn a (x#xs) = (x=a ∨ estaEn a xs)" by simp<br />
also have "… = (x=a ∨ algunos (λx. a=x) xs)" using HI by simp<br />
also have "… = (a=x ∨ algunos (λx. a=x) xs)" using HI by auto<br />
finally show "?P (x#xs)" by simp<br />
qed<br />
<br />
<br />
end<br />
</source></div>Lucnovdoshttps://www.glc.us.es/~jalonso/RA2016/index.php?title=Relaci%C3%B3n_4&diff=640Relación 42016-11-23T11:49:16Z<p>Lucnovdos: </p>
<hr />
<div><source lang="isar"><br />
chapter {* R4: Cuantificadores sobre listas *}<br />
<br />
theory R4_Cuantificadores_sobre_listas<br />
imports Main <br />
begin<br />
<br />
text {* <br />
--------------------------------------------------------------------- <br />
Ejercicio 1. Definir la función <br />
todos :: ('a ⇒ bool) ⇒ 'a list ⇒ bool<br />
tal que (todos p xs) se verifica si todos los elementos de la lista <br />
xs cumplen la propiedad p. Por ejemplo, se verifica <br />
todos (λx. 1<length x) [[2,1,4],[1,3]]<br />
¬todos (λx. 1<length x) [[2,1,4],[3]]<br />
<br />
Nota: La función todos es equivalente a la predefinida list_all. <br />
--------------------------------------------------------------------- <br />
*}<br />
<br />
{*danrodcha*}<br />
fun todos :: "('a ⇒ bool) ⇒ 'a list ⇒ bool" where<br />
"todos p xs = "todos p [] = True"<br />
| "todos p (x#xs) = (p x ∧ todos p xs)""<br />
<br />
(* ivamenjim migtermor dancorgar wilmorort marpoldia1 ferrenseg paupeddeg pablucoto crigomgom anaprarod serrodcal juacabsou rubgonmar josgarsan fraortmoy lucnovdos pabrodmac*)<br />
fun todos :: "('a ⇒ bool) ⇒ 'a list ⇒ bool" where<br />
"todos p [] = True"<br />
| "todos p (x#xs) = (p x ∧ todos p xs)"<br />
<br />
text {* <br />
--------------------------------------------------------------------- <br />
Ejercicio 2. Definir la función <br />
algunos :: ('a ⇒ bool) ⇒ 'a list ⇒ bool<br />
tal que (algunos p xs) se verifica si algunos elementos de la lista <br />
xs cumplen la propiedad p. Por ejemplo, se verifica <br />
algunos (λx. 1<length x) [[2,1,4],[3]]<br />
¬algunos (λx. 1<length x) [[],[3]]"<br />
<br />
Nota: La función algunos es equivalente a la predefinida list_ex. <br />
--------------------------------------------------------------------- <br />
*}<br />
<br />
{*danrodcha ivamenjim migtermor dancorgar marpoldia1 ferrenseg wilmorort paupeddeg pablucoto crigomgom anaprarod serrodcal juacabsou rubgonmar josgarsan fraortmoy lucnovdos pabrodmac*}<br />
fun algunos :: "('a ⇒ bool) ⇒ 'a list ⇒ bool" where<br />
"algunos p [] = False"<br />
| "algunos p (x#xs) = (p x ∨ algunos p xs)"<br />
<br />
text {*<br />
--------------------------------------------------------------------- <br />
Ejercicio 3.1. Demostrar o refutar automáticamente <br />
todos (λx. P x ∧ Q x) xs = (todos P xs ∧ todos Q xs)<br />
--------------------------------------------------------------------- <br />
*}<br />
<br />
{*danrodcha ivamenjim migtermor dancorgar marpoldia1 ferrenseg wilmorort paupeddeg pablucoto anaprarod serrodcal juacabsou rubgonmar josgarsan fraortmoy lucnovdos pabrodmac*}<br />
lemma "todos (λx. P x ∧ Q x) xs = (todos P xs ∧ todos Q xs)"<br />
by (induct xs) auto<br />
<br />
text {*<br />
--------------------------------------------------------------------- <br />
Ejercicio 3.2. Demostrar o refutar detalladamente<br />
todos (λx. P x ∧ Q x) xs = (todos P xs ∧ todos Q xs)<br />
--------------------------------------------------------------------- <br />
*}<br />
<br />
{*danrodcha*}<br />
lemma "todos (λx. P x ∧ Q x) xs = (todos P xs ∧ todos Q xs)"<br />
proof (induct xs)<br />
show "?R []" by simp<br />
next<br />
fix a xs <br />
assume HI: "?R xs"<br />
have "todos (λx. P x ∧ Q x) (a#xs) = (P a ∧ Q a ∧ todos (λx. P x ∧ Q x) xs)" by simp<br />
also have "… = (P a ∧ Q a ∧ (todos P xs ∧ todos Q xs))" using HI by simp<br />
also have "… = ((P a ∧ todos P xs) ∧ (Q a ∧ todos Q xs))" by blast<br />
also have "… = (todos P (a#xs) ∧ todos Q (a#xs))" by simp<br />
finally show "?R (a#xs)" by simp<br />
qed<br />
<br />
(* ivamenjim wilmorort serrodcal josgarsan*)<br />
<br />
lemma "todos (λx. P x ∧ Q x) xs = (todos P xs ∧ todos Q xs)"<br />
proof (induct xs)<br />
show "todos (λx. P x ∧ Q x) [] = (todos P [] ∧ todos Q [])" by simp<br />
next <br />
fix a xs<br />
assume HI: "todos (λx. P x ∧ Q x) xs = (todos P xs ∧ todos Q xs)"<br />
have "todos (λx. P x ∧ Q x) (a # xs) = (P a ∧ Q a ∧ todos (λx. P x ∧ Q x) xs)" by simp<br />
also have "... = (P a ∧ Q a ∧ todos P xs ∧ todos Q xs)" using HI by simp<br />
finally show "todos (λx. P x ∧ Q x) (a # xs) = (todos P (a # xs) ∧ todos Q (a # xs))" by auto<br />
qed<br />
<br />
(* dancorgar paupeddeg *)<br />
lemma "todos (λx. P x ∧ Q x) xs = (todos P xs ∧ todos Q xs)"<br />
proof (induct xs)<br />
show "todos (λx. P x ∧ Q x) [] = (todos P [] ∧ todos Q [])" by simp<br />
next<br />
fix y xs<br />
assume HI: "todos (λx. P x ∧ Q x) xs = (todos P xs ∧ todos Q xs)"<br />
have "todos (λx. P x ∧ Q x) (y#xs) = ((P y ∧ Q y) ∧ (todos (λx. P x ∧ Q x) xs))" by simp<br />
also have "… = ((P y ∧ Q y) ∧ (todos P xs ∧ todos Q xs))" using HI by simp<br />
also have "… = ((P y ∧ todos P xs) ∧ (Q y ∧ todos Q xs))" by blast<br />
also have "… = ((todos P (y#xs)) ∧ (todos Q (y#xs)))" by simp<br />
finally show "todos (λx. P x ∧ Q x) (y#xs) = (todos P (y#xs) ∧ todos Q (y#xs))" by simp<br />
qed<br />
<br />
<br />
(* migtermor *)<br />
lemma "todos (λx. P x ∧ Q x) xs = (todos P xs ∧ todos Q xs)"<br />
proof (induct xs)<br />
show "todos (λx. P x ∧ Q x) [] = (todos P [] ∧ todos Q [])" by simp<br />
next <br />
fix x xs<br />
assume HI: "todos (λx. P x ∧ Q x) xs = (todos P xs ∧ todos Q xs)"<br />
have "(todos (λx. P x ∧ Q x) (x#xs)) = (( P x ∧ Q x) ∧ (todos (λx. P x ∧ Q x) xs))"<br />
by (simp only: todos.simps(2))<br />
also have "… = ((P x ∧ Q x) ∧ (todos P xs ∧ todos Q xs))" using HI by simp<br />
also have "… = (((P x)∧(todos P xs)) ∧ ((Q x) ∧ (todos Q xs)))" by arith<br />
also have "((P x ∧ Q x) ∧ (todos P xs ∧ todos Q xs)) = (((P x)∧(todos P xs)) ∧ ((Q x) ∧ (todos Q xs)))"<br />
by arith (* Este paso es exactamente el mismo que el anterior, pero sin cualquiera de los dos no funciona el "finally show" *)<br />
have "… = (((P x)∧(todos P xs))∧((Q x)∧(todos Q xs)))" by simp<br />
have "… = ((todos P (x#xs))∧(todos Q (x#xs)))" by simp<br />
finally show "(todos (λx. P x ∧ Q x) (x#xs)) = ((todos P (x#xs)) ∧ (todos Q (x#xs)))" by simp <br />
qed<br />
<br />
(* marpoldia1 *)<br />
lemma "todos (λx. P x ∧ Q x) xs = (todos P xs ∧ todos Q xs)"<br />
proof (induct xs)<br />
show "todos (λx. P x ∧ Q x) [] = (todos P [] ∧ todos Q [])" by simp<br />
next<br />
fix a xs<br />
assume HI: "todos (λx. P x ∧ Q x) xs = (todos P xs ∧ todos Q xs)" <br />
have "todos (λx. P x ∧ Q x) (a # xs) = ((P a ∧ Q a) ∧ (todos P xs ∧ todos Q xs))" using HI by simp<br />
also have "... = ((P a ∧ todos P xs) ∧ (Q a ∧ todos Q xs))" by blast<br />
also have "... = (todos P (a#xs) ∧ todos Q (a#xs)) " by simp<br />
finally show "todos (λx. P x ∧ Q x) (a#xs) = (todos P (a#xs) ∧ todos Q (a#xs))" by simp <br />
qed<br />
<br />
(* ferrenseg *)<br />
<br />
lemma "todos (λx. P x ∧ Q x) xs = (todos P xs ∧ todos Q xs)"<br />
proof (induct xs)<br />
show "todos (λx. P x ∧ Q x) [] = (todos P [] ∧ todos Q [])" by simp<br />
next <br />
fix n xs<br />
assume HI: "todos (λx. P x ∧ Q x) xs = (todos P xs ∧ todos Q xs)"<br />
have "todos (λx. P x ∧ Q x) (n # xs) = (P n ∧ Q n ∧ todos P xs ∧ todos Q xs)" using HI by simp<br />
also have "… = P n ∧ Q n ∧ todos P xs ∧ todos Q xs" by blast<br />
also have "⋯ = todos P (n # xs) ∧ todos Q (n # xs)" by simp <br />
finally show "todos (λx. P x ∧ Q x) (n # xs) = todos P (n # xs) ∧ todos Q (n # xs)" by simp<br />
qed<br />
<br />
(* lucnovdos *)<br />
<br />
lemma "todos (λx. P x ∧ Q x) xs = (todos P xs ∧ todos Q xs)"<br />
proof (induct xs)<br />
show "todos (λx. P x ∧ Q x) [] = (todos P [] ∧ todos Q [])" by simp<br />
next<br />
fix n xs<br />
assume HI: "todos (λx. P x ∧ Q x) xs = (todos P xs ∧ todos Q xs)" <br />
have "todos (λx. P x ∧ Q x) (n # xs) = ((P n ∧ Q n) ∧ (todos P xs ∧ todos Q xs))" using HI by simp<br />
also have "... = ((P n ∧ todos P xs) ∧ (Q n ∧ todos Q xs))" by arith<br />
also have "... = ((todos P(n#xs)) ∧ (todos Q(n#xs)))" by simp<br />
<br />
finally show "todos (λx. P x ∧ Q x) (n#xs) = (todos P (n#xs) ∧ todos Q (n#xs))" by simp <br />
qed<br />
<br />
{* wilmorort pablucoto crigomgom anaprarod juacabsou rubgonmar fraortmoy *}<br />
<br />
lemma "todos (λx. P x ∧ Q x) xs = (todos P xs ∧ todos Q xs)"<br />
proof (induct xs)<br />
show "todos (λx. P x ∧ Q x) [] = (todos P [] ∧ todos Q [])" by simp <br />
next<br />
fix a xs<br />
assume HI: "todos (λx. P x ∧ Q x) xs = (todos P xs ∧ todos Q xs)"<br />
have "todos (λx. P x ∧ Q x) (a # xs) = (P a ∧ Q a ∧ todos (λx. P x ∧ Q x) xs)" by simp<br />
also have "... = (P a ∧ Q a ∧ todos P xs ∧ todos Q xs)" using HI by simp<br />
also have "... = (P a ∧ todos P xs ∧ Q a ∧ todos Q xs)" by arith<br />
also have "... = (todos P (a # xs) ∧ todos Q (a # xs))" by simp (* Este paso se puede obviar*)<br />
finally show "todos (λx. P x ∧ Q x) (a # xs) = (todos P (a # xs) ∧ todos Q (a # xs))" by simp<br />
qed<br />
<br />
(* pabrodmac *)<br />
<br />
lemma "todos (λx. P x ∧ Q x) xs = (todos P xs ∧ todos Q xs)"<br />
proof (induct xs)<br />
show "todos (λx. P x ∧ Q x) [] = (todos P [] ∧ todos Q [])" by simp<br />
next <br />
fix x xs<br />
assume HI: "todos (λx. P x ∧ Q x) xs = (todos P xs ∧ todos Q xs)" <br />
have "todos (λx. P x ∧ Q x) (x#xs) = ( P x ∧ Q x ∧ todos (λx. P x ∧ Q x) xs)" by simp<br />
also have "... = (P x ∧ Q x ∧ (todos P xs ∧ todos Q xs))" using HI by simp<br />
also have "... = (P x ∧ todos P xs ∧ Q x ∧ todos Q xs)" by arith<br />
finally show "todos (λx. P x ∧ Q x) (x#xs) = (todos P (x#xs) ∧ todos Q (x#xs))" by simp<br />
qed<br />
<br />
text {*<br />
--------------------------------------------------------------------- <br />
Ejercicio 4.1. Demostrar o refutar automáticamente<br />
todos P (x @ y) = (todos P x ∧ todos P y)<br />
--------------------------------------------------------------------- <br />
*}<br />
{*danrodcha ivamenjim marpoldia1 migtermor ferrenseg wilmorort paupeddeg crigomgom anaprarod serrodcal juacabsou rubgonmar pablucoto fraortmoy josgarsan lucnovdos pabrodmac*}<br />
lemma "todos P (x @ y) = (todos P x ∧ todos P y)"<br />
by (induct x) auto<br />
<br />
{* anaprarod dancorgar *}<br />
lemma "todos P (x @ y) = (todos P x ∧ todos P y)"<br />
apply (induct x)<br />
apply auto<br />
done<br />
<br />
text {*<br />
--------------------------------------------------------------------- <br />
Ejercicio 4.2. Demostrar o refutar detalladamente<br />
todos P (x @ y) = (todos P x ∧ todos P y)<br />
--------------------------------------------------------------------- <br />
*}<br />
<br />
(* ivamenjim ferrenseg wilmorort dancorgar fraortmoy josgarsan lucnovdos*)<br />
<br />
lemma todos_append:<br />
"todos P (x @ y) = (todos P x ∧ todos P y)"<br />
proof (induct x)<br />
show "todos P ([] @ y) = (todos P [] ∧ todos P y)" by simp<br />
next <br />
fix a x<br />
assume HI: "todos P (x @ y) = (todos P x ∧ todos P y)"<br />
have "todos P ((a#x) @ y) = (P a ∧ (todos P (x @ y)))" by simp<br />
also have "... = (P a ∧ (todos P x ∧ todos P y))" using HI by simp<br />
finally show "todos P ((a # x) @ y) = (todos P (a # x) ∧ todos P y)" by simp<br />
qed<br />
<br />
(* migtermor serrodcal *)<br />
<br />
lemma todos_append1:<br />
"todos P (x @ y) = (todos P x ∧ todos P y)" (is "?P x")<br />
proof (induct x)<br />
show "?P []" by simp<br />
next <br />
fix a x<br />
assume HI: "?P x"<br />
have "todos P ((a#x) @ y) = (P a ∧ (todos P (x @ y)))" by simp<br />
also have "… = (P a ∧ (todos P x ∧ todos P y))" using HI by simp<br />
finally show "?P (a#x)" by simp<br />
qed <br />
<br />
(* marpoldia1 *)<br />
lemma todos_append:<br />
"todos P (x @ y) = (todos P x ∧ todos P y)"<br />
proof (induct x)<br />
show "todos P ([] @ y) = (todos P [] ∧ todos P y)" by simp<br />
next<br />
fix a x<br />
assume HI: "todos P (x @ y) = (todos P x ∧ todos P y)"<br />
have "todos P ((a # x) @ y) = todos P (a#(x@y))" by simp<br />
also have "... = (P a ∧ todos P (x @ y))" by simp<br />
also have "... = (P a ∧ (todos P x ∧ todos P y))" using HI by simp<br />
finally show "todos P ((a # x) @ y) = (todos P (a # x) ∧ todos P y)" by simp <br />
qed<br />
<br />
(* paupeddeg *)<br />
lemma todos_append3:<br />
"todos P (x @ y) = (todos P x ∧ todos P y)"<br />
proof (induct x)<br />
show "todos P ([] @ y) = (todos P [] ∧ todos P y)" by simp<br />
next<br />
fix a x<br />
assume HI: "todos P (x @ y) = (todos P x ∧ todos P y)"<br />
have "todos P ((a # x) @ y) = todos P (a # (x @ y))" by simp<br />
also have "... = (todos P [a] ∧ todos P (x @ y)) " by simp<br />
also have "... = (todos P [a] ∧ (todos P x ∧ todos P y))" using HI by simp<br />
finally show "todos P ((a # x) @ y) = (todos P (a # x) ∧ todos P y)" by simp<br />
qed<br />
<br />
(* crigomgom juacabsou anaprarod*)<br />
<br />
lemma todos_append4:<br />
"todos P (x @ y) = (todos P x ∧ todos P y)"<br />
proof (induct x)<br />
show "todos P ([] @ y) = (todos P [] ∧ todos P y)" by simp<br />
next<br />
fix x xs<br />
assume HI:"todos P (xs @ y) = (todos P xs ∧ todos P y)"<br />
have "todos P ((x # xs) @ y) = (P x ∧ todos P (xs @ y ))" by simp<br />
also have "... = (P x ∧ (todos P xs ∧ todos P y)) " using HI by simp<br />
also have "... = ((P x ∧ todos P xs) ∧ todos P y)" by simp<br />
finally show "todos P ((x # xs) @ y) = (todos P (x # xs) ∧ todos P y)" by simp<br />
qed<br />
<br />
(* pablucoto *)<br />
lemma todos_append5:<br />
"todos P (x @ y) = (todos P x ∧ todos P y)"<br />
proof (induct x )<br />
show "todos P ([] @ y) = (todos P [] ∧ todos P y)" by simp<br />
next<br />
fix a x<br />
assume HI:"todos P (x @ y) = (todos P x ∧ todos P y) " <br />
have "todos P ((a # x) @ y) = ( P a ∧ todos P (x @ y))" by simp<br />
also have " ... =( P a ∧ (todos P x ∧ todos P y))" using HI by simp<br />
also have "... = (todos P (a#x) ∧ todos P y)" by simp<br />
finally show "todos P ((a # x) @ y) = (todos P (a # x) ∧ todos P y)" by auto<br />
qed<br />
<br />
{*danrodcha*}<br />
lemma todos_append6:<br />
"todos P (x @ y) = (todos P x ∧ todos P y)" (is "?Q x")<br />
proof (induct x)<br />
show "?Q []" by simp<br />
next<br />
fix x a assume HI: "?Q x"<br />
have "todos P ((a # x) @ y) = todos P (a # x @ y)" by simp<br />
also have "… = ((P a) ∧ todos P (x @ y))" by simp<br />
also have "… = ((P a) ∧ (todos P x ∧ todos P y))" using HI by simp<br />
also have "… = (todos P (a # x) ∧ todos P y)" by simp<br />
finally show "?Q (a # x)" by simp<br />
qed<br />
<br />
{*pabrodmac*}<br />
lemma todos_append7:<br />
"todos P (x @ y) = (todos P x ∧ todos P y)"<br />
proof (induct x)<br />
show "todos P ([] @ y) = (todos P [] ∧ todos P y)" by simp<br />
next <br />
fix a x<br />
assume HI: "todos P (x @ y) = (todos P x ∧ todos P y)" <br />
have "todos P ((a#x) @ y) = (P a ∧ todos P (x @ y))" by simp<br />
also have "... = (P a ∧ (todos P x ∧ todos P y))" using HI by simp<br />
finally show "todos P ((a#x) @ y) = (todos P (a#x) ∧ todos P y)" by simp<br />
qed<br />
<br />
<br />
text {*<br />
--------------------------------------------------------------------- <br />
Ejercicio 5.1. Demostrar o refutar automáticamente <br />
todos P (rev xs) = todos P xs<br />
--------------------------------------------------------------------- <br />
*}<br />
<br />
lemma "todos P (rev xs) = todos P xs"<br />
oops<br />
<br />
(* migtermor ivamenjim marpoldia1 serrodcal anaprarod paupeddeg dancorgar *)<br />
lemma "todos P (rev xs) = todos P xs" <br />
apply (induct xs)<br />
apply simp<br />
apply (simp add: todos_append)<br />
apply auto<br />
done<br />
<br />
(* ferrenseg crigomgom rubgonmar fraortmoy josgarsan danrodcha lucnovdos pabrodmac*)<br />
<br />
lemma "todos P (rev xs) = todos P xs"<br />
by (induct xs) (auto simp add: todos_append)<br />
<br />
(* wilmorort *)<br />
lemma "todos P (rev xs) = todos P xs"<br />
by (induct xs,simp,simp add: todos_append,auto)<br />
<br />
(* juacabsou *)<br />
lemma "todos P (rev xs) = todos P xs"<br />
apply (induct xs,simp,simp add: todos_append,auto) done<br />
<br />
(* pablucoto * )<br />
lemma "todos P (rev xs) = todos P xs"<br />
by (induct xs) (auto, simp_all add: todos_append) <br />
lemma "todos P (rev xs) = todos P xs"<br />
by (induct xs) ( simp_all add: todos_append, auto)<br />
<br />
text {*<br />
--------------------------------------------------------------------- <br />
Ejercicio 5.2. Demostrar o refutar detalladamente<br />
todos P (rev xs) = todos P xs<br />
--------------------------------------------------------------------- <br />
*}<br />
<br />
lemma "todos P (rev xs) = todos P xs"<br />
oops<br />
<br />
(* migtermor *)<br />
<br />
lemma auxiliar:<br />
"rev (a#xs) = rev xs @ [a]"<br />
by auto<br />
<br />
lemma "todos P (rev xs) = todos P xs" (is "?Q xs")<br />
proof (induct xs)<br />
show "?Q []" by (simp only: rev.simps(1))<br />
next <br />
fix a xs<br />
assume HI: "?Q xs"<br />
have "todos P (rev (a#xs)) = (todos P (rev xs @ [a]))" by (simp add: auxiliar)<br />
have "… = ((todos P (rev xs)) ∧ (todos P [a]))" by (simp add: todos_append)<br />
have "… = (todos P (rev xs) ∧ P a)" by simp<br />
also have Aux: "… = (todos P xs ∧ P a)" using HI by simp<br />
have Aux1: "… = (P a ∧ todos P xs)" by arith<br />
have "(todos P (rev xs) ∧ P a) = (P a ∧ todos P xs)" using Aux Aux1 by simp<br />
finally show "todos P (rev (a#xs)) = todos P (a#xs)" by (simp add: todos_append)<br />
qed<br />
<br />
<br />
(* marpoldia1 ferrenseg crigomgom serrodcal juacabsou rubgonmar josgarsan pablucoto pabrodmac lucnovdos*)<br />
lemma "todos P (rev xs) = todos P xs"<br />
proof (induct xs)<br />
show "todos P (rev []) = todos P []" by simp<br />
next<br />
fix a xs<br />
assume HI: "todos P (rev xs) = todos P xs"<br />
have "todos P (rev (a # xs)) = (todos P ((rev xs)@[a]))" by simp<br />
also have "... = (todos P (rev xs) ∧ todos P [a])" by (simp add:todos_append)<br />
also have "... = (todos P xs ∧ P a)" using HI by simp<br />
also have "... = (P a ∧ todos P xs)" by arith<br />
also have "... = (todos P (a#xs))" by simp<br />
finally show "todos P (rev (a # xs)) = (todos P (a#xs))" by simp <br />
qed<br />
<br />
(* ivamenjim *)<br />
<br />
lemma "todos P (rev xs) = todos P xs" (is "?P xs")<br />
proof (induct xs)<br />
show "?P []" by simp<br />
next <br />
fix a xs<br />
assume HI: "?P xs"<br />
have "todos P (rev (a # xs)) = (todos P ((rev xs)@[a]))" by simp<br />
also have "... = ((todos P (rev xs)) ∧ todos P [a])" by (simp add: todos_append)<br />
also have "... = (todos P xs ∧ todos P [a])" using HI by simp<br />
also have "... = (todos P [a] ∧ todos P xs)" by arith<br />
finally show "todos P (rev (a # xs)) = todos P (a # xs)" by simp<br />
qed<br />
<br />
(* fraortmoy *)<br />
<br />
lemma "todos P (rev xs) = todos P xs"<br />
proof (induct xs)<br />
show "todos P (rev []) = todos P []" by simp<br />
next<br />
fix a xs <br />
assume H1: "todos P (rev xs) = todos P xs"<br />
have " todos P (rev (a # xs)) = todos P (rev xs @ [a])" by simp<br />
also have "… = (todos P (rev xs) ∧ todos P [a])" by (simp add:todos_append)<br />
also have "… = (todos P xs ∧ todos P [a])" using H1 by simp<br />
also have "… = (todos P [a] ∧ todos P xs)" by arith<br />
finally show "todos P (rev (a # xs)) = todos P (a#xs)" by simp<br />
qed<br />
<br />
(* paupeddeg dancorgar *)<br />
lemma "todos P (rev xs) = todos P xs"<br />
proof (induct xs)<br />
show " todos P (rev []) = todos P []" by simp<br />
next<br />
fix a xs<br />
assume HI: "todos P (rev xs) = todos P xs "<br />
have "todos P (rev (a # xs)) = todos P (rev(xs) @ [a])" by simp<br />
also have "... = (todos P (rev xs) ∧ todos P [a]) " by (simp add: todos_append)<br />
also have "... = (todos P xs ∧ todos P [a])" using HI by simp<br />
also have "... = (todos P [a] ∧ todos P xs)" by arith<br />
also have "... = todos P ([a] @ xs)" by (simp add: todos_append)<br />
finally show "todos P (rev (a # xs)) = todos P (a # xs)" by simp<br />
qed <br />
<br />
(* wilmorort *)<br />
lemma "todos P (rev xs) = todos P xs" (is "?P xs")<br />
proof (induct xs) <br />
show "?P []" by simp <br />
next<br />
fix a xs<br />
assume HI: "?P xs"<br />
have "todos P (rev (a#xs)) = todos P ((rev xs) @ [a])" by simp<br />
also have "... = (todos P(rev xs) ∧ todos P [a])" by (simp add: todos_append)<br />
also have "... = (todos P xs ∧ todos P [a])" using HI by simp<br />
also have "... = (todos P [a] ∧ todos P xs)" by (simp add: HOL.conj_commute)<br />
also have "... = todos P([a]@(xs))" by (simp)<br />
finally show "todos P (rev (a#xs))= todos P (a # xs)" by simp<br />
qed<br />
<br />
(* anaprarod *) <br />
(* es igual que las anteriores pero con el final también con patrones *)<br />
<br />
lemma "todos P (rev xs) = todos P xs" (is "?P xs")<br />
proof (induct xs)<br />
show "?P []" by auto<br />
next<br />
fix a xs<br />
assume HI: "?P xs"<br />
have "todos P (rev (a#xs)) = todos P (rev xs @ [a])" by simp<br />
also have "… = (todos P (rev xs) ∧ todos P [a])" by (simp add: todos_append)<br />
also have "… = (todos P xs ∧ todos P [a])" using HI by simp<br />
also have "… = (todos P [a] ∧ todos P xs)" by arith<br />
also have "… = todos P (a#xs)" by simp<br />
finally show "?P (a # xs)" by simp<br />
qed<br />
<br />
{*danrodcha*}<br />
lemma "todos P (rev xs) = todos P xs" (is "?Q xs")<br />
proof (induct xs)<br />
show "?Q []" by simp<br />
next<br />
fix a xs assume HI: "?Q xs"<br />
have "todos P (rev (a#xs)) = todos P ((rev xs) @ [a])" by simp<br />
also have "… = (todos P (rev xs) ∧ todos P [a])" by (simp add:todos_append)<br />
also have "… = (todos P xs ∧ todos P [a])" using HI by simp<br />
also have "… = (todos P [a] ∧ todos P xs)" by (simp add: HOL.conj_commute)<br />
also have "… = todos P (a # xs)" by simp<br />
finally show "?Q (a # xs)" by simp<br />
qed<br />
<br />
text {*<br />
--------------------------------------------------------------------- <br />
Ejercicio 6. Demostrar o refutar:<br />
algunos (λx. P x ∧ Q x) xs = (algunos P xs ∧ algunos Q xs)<br />
--------------------------------------------------------------------- <br />
*}<br />
<br />
lemma "algunos (λx. P x ∧ Q x) xs = (algunos P xs ∧ algunos Q xs)"<br />
oops<br />
<br />
(* migtermor ivamenjim ferrenseg paupeddeg crigomgom serrodcal wilmorort juacabsou rubgonmar anaprarod marpoldia1 fraortmoy josgarsan danrodcha pablucoto lucnovdos*)<br />
lemma "algunos (λx. P x ∧ Q x) xs = (algunos P xs ∧ algunos Q xs)" <br />
quickcheck<br />
oops<br />
(* Quickcheck encuentra el siguiente contraejemplo: P={a1}, Q={a2}, xs={a1,a2}. En este ejemplo:<br />
· "algunos (λx. P x ∧ Q x) xs = False"<br />
· "(algunos P xs ∧ algunos Q xs) = True" *)<br />
<br />
(* dancorgar *)<br />
lemma "algunos (λx. P x ∧ Q x) xs = (algunos P xs ∧ algunos Q xs)"<br />
proof -<br />
assume H1: "xs = [a, b]"<br />
assume H2: "P a = True"<br />
assume H3: "Q a = False"<br />
assume H4: "P b = False"<br />
assume H5: "Q b = True" <br />
have F1: "(algunos P xs ∧ algunos Q xs) = True" using H1 H2 H3 H4 H5 by simp<br />
have F2: "algunos (λx. P x ∧ Q x) xs = False" using H1 H2 H3 H4 H5 by simp<br />
have "algunos (λx. P x ∧ Q x) xs ≠ (algunos P xs ∧ algunos Q xs)" using F1 F2 by simp<br />
oops<br />
<br />
text {*<br />
--------------------------------------------------------------------- <br />
Ejercicio 7.1. Demostrar o refutar automáticamente <br />
algunos P (map f xs) = algunos (P ∘ f) xs<br />
--------------------------------------------------------------------- <br />
*}<br />
<br />
(* ivamenjim migtermor marpoldia1 crigomgom rubgonmar wilmorort anaprarod <br />
fraortmoy juacabsou paupeddeg josgarsan danrodcha pablucoto lucnovdos *)<br />
<br />
lemma "algunos P (map f xs) = algunos (P o f) xs"<br />
by (induct xs) auto<br />
<br />
(* ferrenseg *)<br />
<br />
lemma "algunos P (map f xs) = algunos (P o f) xs"<br />
by (induct xs) simp_all<br />
<br />
(* anaprarod dancorgar *)<br />
lemma "algunos P (map f xs) = algunos (P o f) xs"<br />
apply (induct xs)<br />
apply auto<br />
done<br />
<br />
text {*<br />
--------------------------------------------------------------------- <br />
Ejercicio 7.2. Demostrar o refutar detalladamente<br />
algunos P (map f xs) = algunos (P ∘ f) xs<br />
--------------------------------------------------------------------- <br />
*}<br />
<br />
lemma "algunos P (map f xs) = algunos (P o f) xs"<br />
oops<br />
<br />
(* migtermor *)<br />
<br />
lemma AUX: "algunos (λa. P (f a)) xs = algunos (P o f) xs"<br />
by (induct xs) auto<br />
<br />
lemma "algunos P (map f xs) = algunos (P o f) xs" (is "?Q xs")<br />
proof (induct xs)<br />
show "?Q []" by simp<br />
next<br />
fix x xs<br />
assume HI: "?Q xs"<br />
have "algunos P (map f (x#xs)) = (algunos P ((f x)#(map f xs)))" by simp<br />
also have "… = ((P (f x)) ∨ (algunos P (map f xs)))" by (simp only: algunos.simps(2))<br />
also have "… = ((P (f x)) ∨ (algunos (P o f) xs))" <br />
proof (cases)<br />
assume C1: "(P (f x))"<br />
have Aux: "((P (f x)) ∨ (algunos P (map f xs))) = True" using C1 by simp<br />
have Aux1: "… = ((P (f x)) ∨ (algunos (P o f) xs))" using C1 by simp<br />
then show "((P (f x)) ∨ (algunos P (map f xs))) = ((P (f x)) ∨ (algunos (P o f) xs))" <br />
using Aux Aux1 by simp<br />
next<br />
assume C2: "¬(P (f x))"<br />
have Aux2: "((P (f x)) ∨ (algunos P (map f xs))) = (algunos P (map f xs))" using C2 by simp<br />
have Aux3: "… = (algunos (P o f) xs)" using HI by (simp add: AUX)<br />
also have "… = ((P (f x)) ∨ (algunos (P o f) xs))" using C2 by simp<br />
then show "((P (f x)) ∨ (algunos P (map f xs))) = ((P (f x)) ∨ (algunos (P o f) xs))" <br />
using Aux2 Aux3 by simp<br />
qed<br />
also have "… = (((P o f) x) ∨ (algunos (P o f) xs))" by simp<br />
also have "… = (algunos (P o f) (x#xs))" by simp<br />
finally show "algunos P (map f (x#xs)) = (algunos (P o f) (x#xs))" by simp<br />
qed<br />
<br />
(* ferrenseg *)<br />
<br />
lemma "algunos P (map f xs) = algunos (P ∘ f) xs"<br />
proof (induct xs)<br />
show "algunos P (map f []) = algunos (P ∘ f) []" by simp<br />
next<br />
fix x xs<br />
assume HI: "algunos P (map f xs) = algunos (P ∘ f) xs"<br />
show "algunos P (map f (x # xs)) = algunos (P ∘ f) (x # xs)"<br />
proof -<br />
have "algunos P (map f (x # xs)) = algunos P ((f x) # (map f xs))" by simp<br />
also have "… = ((P (f x)) ∨ (algunos P (map f xs)))" by simp<br />
also have "… = (((P ∘ f) x) ∨ (algunos (P ∘ f) xs))" using HI by simp<br />
also have "… = algunos (P ∘ f) (x # xs)" by simp<br />
finally show ?thesis<br />
qed<br />
qed<br />
<br />
(* ivamenjim *)<br />
<br />
lemma "algunos P (map f xs) = algunos (P o f) xs"<br />
proof (induct xs)<br />
show "algunos P (map f []) = algunos (P ∘ f) []" by simp<br />
next <br />
fix a xs<br />
assume HI: "algunos P (map f xs) = algunos (P ∘ f) xs"<br />
have "algunos P (map f (a # xs)) = algunos P ((f a)#map f xs)" by simp<br />
also have "... = (algunos P (map f [a]) ∨ algunos P (map f xs))" by simp <br />
also have "... = (algunos P (map f [a]) ∨ algunos (P ∘ f) xs)" using HI by simp <br />
finally show "algunos P (map f (a # xs)) = algunos (P ∘ f) (a # xs)" by simp<br />
qed<br />
<br />
(* crigomgom rubgonmar anaprarod marpoldia1 juacabsou danrodcha paupeddeg josgarsan*)<br />
lemma "algunos P (map f xs) = algunos (P o f) xs"<br />
proof (induct xs)<br />
show "algunos P (map f []) = algunos (P ∘ f) []" by simp<br />
next<br />
fix x xs<br />
assume HI: "algunos P (map f xs) = algunos (P ∘ f) xs"<br />
have "algunos P (map f (x # xs)) = algunos P ((f x) # (map f xs))" by simp<br />
also have "... = (P (f x) ∨ algunos P (map f xs))" by simp<br />
also have "... = (P (f x) ∨ algunos (P ∘ f) xs)" using HI by simp<br />
also have "... = ((P ∘ f) x ∨ algunos (P ∘ f) xs)" by simp<br />
finally show "algunos P (map f (x # xs)) = algunos (P ∘ f) (x # xs)" by simp<br />
qed<br />
<br />
(* wilmorort pablucoto*)<br />
lemma "algunos P (map f xs) = algunos (P o f) xs" (is "?P xs" )<br />
proof (induct xs)<br />
show "?P []" by simp<br />
next<br />
fix a xs<br />
assume HI: "algunos P (map f xs) = algunos (P ∘ f) xs"<br />
have "algunos P (map f (a # xs)) = (P (f a) ∨ algunos P (map f xs))" by simp<br />
also have "... = (P (f a) ∨ algunos (P ∘ f) xs )" using HI by simp<br />
also have "... = ((P ∘ f) a ∨ algunos (P ∘ f)xs)" by simp<br />
finally show " algunos P (map f (a # xs)) = algunos (P ∘ f) (a # xs)" by simp<br />
qed<br />
<br />
(* fraortmoy *)<br />
<br />
lemma "algunos P (map f xs) = algunos (P o f) xs"<br />
proof (induct xs)<br />
show "algunos P (map f []) = algunos (P ∘ f) []" by simp<br />
next<br />
fix a xs<br />
assume H1: "algunos P (map f xs) = algunos (P ∘ f) xs"<br />
have "algunos P (map f (a # xs)) = (algunos P (map f [a]) ∨ algunos P (map f xs))" by simp<br />
also have "… = (algunos (P ∘ f) [a] ∨ algunos (P ∘ f) xs )" using H1 by simp<br />
also have "… = algunos (P ∘ f) (a#xs)" by simp<br />
finally show "algunos P (map f (a # xs)) = algunos (P ∘ f) (a#xs)" by simp<br />
qed<br />
<br />
{*danrodcha*}<br />
lemma "algunos P (map f xs) = algunos (P o f) xs" (is "?Q xs")<br />
proof (induct xs)<br />
show "?Q []" by simp<br />
next<br />
fix a xs assume HI: "?Q xs"<br />
have " algunos P (map f (a # xs)) = algunos P ((f a)#(map f xs))" by simp<br />
also have "… = (P (f a) ∨ algunos P (map f xs))" by simp<br />
also have "… = (P (f a) ∨ algunos (P ∘ f) xs)" using HI by blast<br />
also have "… = ((P ∘ f) a ∨ algunos (P ∘ f) xs)" by simp<br />
also have "… = algunos (P ∘ f) (a # xs)" by simp<br />
finally show "?Q (a # xs)" by blast<br />
qed<br />
<br />
(* dancorgar *)<br />
lemma "algunos P (map f xs) = algunos (P o f) xs"<br />
proof (induct xs)<br />
show "algunos P (map f []) = algunos (P o f) []" by simp<br />
next<br />
fix x xs<br />
assume HI: "algunos P (map f xs) = algunos (P o f) xs"<br />
have "algunos P (map f (x#xs)) = ((P (f x)) ∨ (algunos P (map f xs)))" by simp<br />
also have "… = ((P (f x)) ∨ (algunos (P o f) xs))" using HI by simp<br />
finally show "algunos P (map f (x#xs)) = algunos (P o f) (x#xs)" by simp<br />
qed<br />
<br />
text {*<br />
--------------------------------------------------------------------- <br />
Ejercicio 8.1. Demostrar o refutar automáticamente <br />
algunos P (xs @ ys) = (algunos P xs ∨ algunos P ys)<br />
--------------------------------------------------------------------- <br />
*}<br />
<br />
(* ivamenjim marpoldia1 paupeddeg crigomgom rubgonmar wilmorort fraortmoy danrodcha pablucoto dancorgar josgarsan anaprarod*)<br />
<br />
lemma "algunos P (xs @ ys) = (algunos P xs ∨ algunos P ys)"<br />
by (induct xs) auto<br />
<br />
(* ferrenseg *)<br />
lemma "algunos P (xs @ ys) = (algunos P xs ∨ algunos P ys)"<br />
by (induct xs) simp_all<br />
<br />
(* anaprarod *)<br />
lemma "algunos P (xs @ ys) = (algunos P xs ∨ algunos P ys)"<br />
apply (induct xs)<br />
apply auto<br />
done<br />
<br />
text {*<br />
--------------------------------------------------------------------- <br />
Ejercicio 8.2. Demostrar o refutar detalladamente<br />
algunos P (xs @ ys) = (algunos P xs ∨ algunos P ys)<br />
--------------------------------------------------------------------- <br />
*}<br />
<br />
(* ivamenjim ferrenseg marpoldia1 wilmorort dancorgar josgarsan*)<br />
<br />
lemma algunos_append:<br />
"algunos P (xs @ ys) = (algunos P xs ∨ algunos P ys)"<br />
proof (induct xs)<br />
show "algunos P ([] @ ys) = (algunos P [] ∨ algunos P ys)" by simp<br />
next <br />
fix a xs<br />
assume HI: "algunos P (xs @ ys) = (algunos P xs ∨ algunos P ys)"<br />
have "algunos P ((a # xs) @ ys) = (P a ∨ (algunos P (xs @ ys)))" by simp<br />
also have "... = (P a ∨ (algunos P xs ∨ algunos P ys))" using HI by simp<br />
finally show "algunos P ((a # xs) @ ys) = (algunos P (a # xs) ∨ algunos P ys)" by simp<br />
qed<br />
<br />
(* migtermor crigomgom *)<br />
<br />
lemma algunos_append2:<br />
"algunos P (xs @ ys) = (algunos P xs ∨ algunos P ys)" (is "?Q xs")<br />
proof (induct xs)<br />
show "?Q []" by simp<br />
next <br />
fix x xs<br />
assume HI: "?Q xs"<br />
have "algunos P ((x#xs) @ ys) = algunos P (x#(xs @ ys))" by simp<br />
also have "… = ((P x) ∨ (algunos P (xs @ ys)))" by simp<br />
also have "… = ((P x) ∨ (algunos P xs) ∨ (algunos P ys))" using HI by simp<br />
also have "… = ((algunos P (x#xs)) ∨ (algunos P ys))" by simp<br />
finally show "algunos P ((x#xs) @ ys) = (algunos P (x#xs) ∨ algunos P ys)" by simp<br />
qed<br />
<br />
(* paupeddeg*)<br />
lemma algunos_append3:<br />
"algunos P (xs @ ys) = (algunos P xs ∨ algunos P ys)"<br />
proof (induct xs)<br />
show "algunos P ([] @ ys) = (algunos P [] ∨ algunos P ys)" by simp<br />
next<br />
fix a xs<br />
assume HI: "algunos P (xs @ ys) = (algunos P xs ∨ algunos P ys)"<br />
have "algunos P ((a # xs) @ ys) = (algunos P [a] ∨ ( algunos P (xs @ ys)))" by simp<br />
also have "... = (algunos P [a] ∨ (algunos P xs ∨ algunos P ys))" using HI by simp<br />
finally show "algunos P ((a # xs) @ ys) = (algunos P (a # xs) ∨ algunos P ys)" by simp<br />
qed<br />
<br />
(* fraortmoy *)<br />
lemma algunos_append4:<br />
"algunos P (xs @ ys) = (algunos P xs ∨ algunos P ys)"<br />
proof (induct xs)<br />
show " algunos P ([] @ ys) = (algunos P [] ∨ algunos P ys)" by simp<br />
next <br />
fix a xs<br />
assume H1: "algunos P (xs @ ys) = (algunos P xs ∨ algunos P ys)"<br />
have "algunos P ((a # xs) @ ys) = (algunos P [a] ∨ algunos P (xs @ ys))" by simp<br />
also have "… = (algunos P [a] ∨ algunos P xs ∨ algunos P ys)" using H1 by simp<br />
also have "… = ((algunos P [a] ∨ algunos P xs) ∨ algunos P ys)" by simp<br />
also have "… = (algunos P (a#xs) ∨ algunos P ys)" by simp<br />
finally show "algunos P ((a # xs) @ ys) = (algunos P (a#xs) ∨ algunos P ys)" by simp<br />
qed<br />
<br />
{* danrodcha pablucoto anaprarod *}<br />
lemma algunos_append5:<br />
"algunos P (xs @ ys) = (algunos P xs ∨ algunos P ys)" (is "?Q xs")<br />
proof (induct xs)<br />
show "?Q []" by simp<br />
next<br />
fix a xs assume HI: "?Q xs"<br />
have "algunos P ((a#xs) @ ys) = algunos P (a#(xs @ ys))" by simp<br />
also have "… = (algunos P [a] ∨ algunos P (xs @ ys))" by simp<br />
also have "… = (algunos P [a] ∨ algunos P xs ∨ algunos P ys)" using HI by simp<br />
also have "… = (algunos P (a # xs) ∨ algunos P ys)" by simp<br />
finally show "?Q (a # xs)" by simp<br />
qed<br />
<br />
text {*<br />
--------------------------------------------------------------------- <br />
Ejercicio 9.1. Demostrar o refutar automáticamente<br />
algunos P (rev xs) = algunos P xs<br />
--------------------------------------------------------------------- <br />
*}<br />
<br />
(* ivamenjim migtermor marpoldia1 rubgonmar paupeddeg dancorgar anaprarod *)<br />
<br />
lemma "algunos P (rev xs) = algunos P xs"<br />
apply (induct xs)<br />
apply simp<br />
apply (simp add: algunos_append)<br />
apply auto<br />
done<br />
<br />
(* ferrenseg crigomgom danrodcha pablucoto josgarsan*)<br />
lemma "algunos P (rev xs) = algunos P xs"<br />
by (induct xs) (auto simp add: algunos_append)<br />
<br />
(* wilmorort *)<br />
lemma "algunos P (rev xs) = algunos P xs"<br />
by (induct xs,simp,simp add: algunos_append,auto) <br />
<br />
<br />
text {*<br />
--------------------------------------------------------------------- <br />
Ejercicio 9.2. Demostrar o refutar detalladamente<br />
algunos P (rev xs) = algunos P xs<br />
--------------------------------------------------------------------- <br />
*}<br />
<br />
lemma "algunos P (rev xs) = algunos P xs"<br />
oops<br />
<br />
(* migtermor *)<br />
lemma auxiliar1:<br />
"rev (a#xs) = rev xs @ [a]"<br />
by auto <br />
<br />
lemma "algunos P (rev xs) = algunos P xs" (is "?Q xs")<br />
proof (induct xs)<br />
show "?Q []" by simp<br />
next<br />
fix x xs<br />
assume HI: "?Q xs"<br />
have "algunos P (rev (x#xs)) = (algunos P (rev xs @ [x]))" using auxiliar1 by simp<br />
also have "… = ((algunos P (rev xs)) ∨ (algunos P [x]))" by (simp add: algunos_append)<br />
also have "… = ((P x) ∨ (algunos P (rev xs)))" by simp arith<br />
also have "… = ((P x) ∨ (algunos P xs))" using HI by simp<br />
finally show "algunos P (rev (x#xs)) = (algunos P (x#xs))" by simp<br />
qed<br />
<br />
(* ferrenseg *)<br />
<br />
lemma "algunos P (rev xs) = algunos P xs"<br />
proof (induct xs)<br />
show "algunos P (rev []) = algunos P []" by simp<br />
next<br />
fix x xs<br />
assume HI: "algunos P (rev xs) = algunos P xs"<br />
show "algunos P (rev (x # xs)) = algunos P (x # xs)"<br />
proof -<br />
have "algunos P (rev (x # xs)) = algunos P ((rev xs) @ [x])" by simp<br />
also have "… = (algunos P (rev xs) ∨ algunos P [x])" <br />
by (simp add: algunos_append)<br />
also have "… = (algunos P xs ∨ algunos P [x])" using HI by simp<br />
also have "… = (algunos P xs ∨ P x)" by simp<br />
also have "… = (P x ∨ algunos P xs)" by auto<br />
also have "… = algunos P (x # xs)" by simp<br />
finally show ?thesis<br />
qed<br />
qed<br />
<br />
(* ivamenjim marpoldia1*)<br />
<br />
lemma "algunos P (rev xs) = algunos P xs" (is "?P xs")<br />
proof (induct xs)<br />
show "?P []" by simp<br />
next <br />
fix a xs<br />
assume HI: "?P xs"<br />
have "algunos P (rev (a # xs)) = (algunos P ((rev xs)@[a]))" by simp<br />
also have "... = ((algunos P (rev xs)) ∨ algunos P [a])" by (simp add: algunos_append)<br />
also have "... = (algunos P xs ∨ algunos P [a])" using HI by simp<br />
also have "... = (algunos P [a] ∨ algunos P xs)" by arith<br />
finally show "algunos P (rev (a # xs)) = algunos P (a # xs)" by simp<br />
qed<br />
<br />
(* paupeddeg dancorgar *)<br />
lemma "algunos P (rev xs) = algunos P xs"<br />
proof (induct xs)<br />
show "algunos P (rev []) = algunos P []" by simp<br />
next<br />
fix a xs<br />
assume HI:"algunos P (rev xs) = algunos P xs"<br />
have "algunos P (rev (a # xs)) = algunos P ((rev xs) @ [a])" by simp<br />
also have "... = ((algunos P (rev xs)) ∨ (algunos P [a]))" by (simp add: algunos_append)<br />
also have "... = ((algunos P xs) ∨ (algunos P [a]))" using HI by simp<br />
also have "... = ((algunos P [a]) ∨ (algunos P xs))" by arith<br />
finally show "algunos P (rev (a # xs)) = algunos P (a # xs)" by simp<br />
qed<br />
<br />
(*crigomgom pablucoto anaprarod *)<br />
lemma "algunos P (rev xs) = algunos P xs"<br />
proof (induct xs)<br />
show "algunos P (rev []) = algunos P []" by simp<br />
next<br />
fix x xs<br />
assume HI: " algunos P (rev xs) = algunos P xs" <br />
have "algunos P (rev (x # xs)) = algunos P ((rev xs) @ [x])" by simp<br />
also have "... = (algunos P (rev xs) ∨ algunos P [x])" by (simp add: algunos_append)<br />
also have "... = (algunos P xs ∨ algunos P [x])" using HI by simp<br />
also have "... = (algunos P xs ∨ P x)" by simp<br />
also have "... = (P x ∨ algunos P xs)" by arith<br />
also have "... = algunos P (x#xs)" by simp<br />
finally show "algunos P (rev (x # xs)) = algunos P (x # xs)" by simp<br />
qed<br />
<br />
<br />
(*wilmorort*)<br />
lemma "algunos P (rev xs) = algunos P xs" (is "?P xs")<br />
proof (induct xs) <br />
show "?P []" by simp <br />
next<br />
fix a xs<br />
assume HI: "?P xs"<br />
have "algunos P (rev (a#xs)) = algunos P ((rev xs) @ [a])" by simp<br />
also have "... = (algunos P(rev xs) ∨ algunos P [a])" by (simp add: algunos_append)<br />
also have "... = (algunos P xs ∨ algunos P [a])" using HI by simp<br />
also have "... = (algunos P [a] ∨ algunos P xs)" by (simp add: HOL.disj_commute)<br />
also have "... = algunos P([a]@(xs))" by (simp)<br />
finally show "algunos P (rev (a#xs))= algunos P (a # xs)" by simp<br />
qed<br />
<br />
{*danrodcha *}<br />
lemma "algunos P (rev xs) = algunos P xs" (is "?Q xs")<br />
proof (induct xs)<br />
show "?Q []" by simp<br />
next<br />
fix a xs assume HI: "?Q xs"<br />
have "algunos P (rev (a#xs)) = algunos P ((rev xs) @ [a])" by simp<br />
also have "… = (algunos P (rev xs) ∨ algunos P [a])" by (simp add:algunos_append)<br />
also have "… = (algunos P xs ∨ algunos P [a])" using HI by simp<br />
also have "… = (algunos P [a] ∨ algunos P xs)" by (simp add: HOL.disj_commute)<br />
also have "… = algunos P (a # xs)" by simp<br />
finally show "?Q (a # xs)" by simp<br />
qed<br />
<br />
text {*<br />
--------------------------------------------------------------------- <br />
Ejercicio 10. Encontrar un término no trivial Z tal que sea cierta la <br />
siguiente ecuación:<br />
algunos (λx. P x ∨ Q x) xs = Z<br />
y demostrar la equivalencia de forma automática y detallada.<br />
--------------------------------------------------------------------- <br />
*}<br />
<br />
(* migtermor *)<br />
<br />
lemma "algunos (λx. P x ∨ Q x) xs = ((algunos (λx. P x) xs) ∨ (algunos (λx. Q x) xs))"<br />
by (induct xs) auto<br />
<br />
lemma "algunos (λx. P x ∨ Q x) xs = ((algunos (λx. P x) xs) ∨ (algunos (λx. Q x) xs))" (is "?R xs")<br />
proof (induct xs)<br />
show "?R []" by simp<br />
next<br />
fix x xs<br />
assume HI: "?R xs"<br />
have "algunos (λx. P x ∨ Q x) (x#xs) = (((λx. P x ∨ Q x) x) ∨ (algunos (λx. P x ∨ Q x) xs))"<br />
by simp<br />
also have "… = ((P x ∨ Q x) ∨ (algunos (λx. P x ∨ Q x) xs))" by simp<br />
also have H1: "… = ((((λx. P x) x) ∨ (algunos (λx. P x) xs)) ∨ (((λx. Q x) x) ∨ (algunos (λx. Q x) xs)))"<br />
using HI by simp arith<br />
have H2: "… = ((algunos (λx. P x) (x#xs)) ∨ (algunos (λx. Q x) (x#xs)))" <br />
by simp<br />
have C: "(algunos (λx. P x ∨ Q x) (x#xs)) = <br />
((algunos (λx. P x) (x#xs)) ∨ (algunos (λx. Q x) (x#xs)))" <br />
using H1 H2 by simp<br />
finally show "(algunos (λx. P x ∨ Q x) (x#xs)) = <br />
((algunos (λx. P x) (x#xs)) ∨ (algunos (λx. Q x) (x#xs)))" <br />
using C by simp<br />
qed<br />
<br />
<br />
(* ferrenseg *)<br />
<br />
lemma "algunos (λx. P x ∨ Q x) xs = (algunos P xs ∨ algunos Q xs)"<br />
proof (induct xs)<br />
show "algunos (λx. P x ∨ Q x) [] = (algunos P [] ∨ algunos Q [])" by simp<br />
next<br />
fix x xs<br />
assume HI: "algunos (λx. (P x ∨ Q x)) xs = (algunos P xs ∨ algunos Q xs)"<br />
show "algunos (λx. P x ∨ Q x) (x # xs) = (algunos P (x # xs) ∨ algunos Q (x # xs))"<br />
proof -<br />
have "algunos (λx. P x ∨ Q x) (x # xs) = <br />
((P x) ∨ (Q x) ∨ algunos (λx. P x ∨ Q x) xs)" by simp<br />
also have "… = ((P x) ∨ (Q x) ∨ algunos P xs ∨ algunos Q xs)" <br />
using HI by simp<br />
also have "… = (((P x) ∨ algunos P xs) ∨ ((Q x) ∨ algunos Q xs))" by auto<br />
also have "… = (algunos P (x # xs) ∨ algunos Q (x # xs))" by simp<br />
finally show ?thesis <br />
qed<br />
qed<br />
<br />
(* ivamenjim marpoldia1 paupeddeg *)<br />
<br />
lemma "algunos (λx. P x ∨ Q x) xs = (algunos P xs ∨ algunos Q xs)" <br />
by (induct xs) auto<br />
<br />
lemma "algunos (λx. P x ∨ Q x) xs = (algunos P xs ∨ algunos Q xs)"<br />
proof (induct xs)<br />
show "algunos (λx. P x ∨ Q x) [] = (algunos P [] ∨ algunos Q [])" by simp<br />
next <br />
fix a xs<br />
assume HI: "algunos (λx. P x ∨ Q x) xs = (algunos P xs ∨ algunos Q xs)"<br />
have "algunos (λx. P x ∨ Q x) (a # xs) = (P a ∨ Q a ∨ algunos (λx. P x ∨ Q x) xs)" by simp<br />
also have "... = (P a ∨ Q a ∨ algunos P xs ∨ algunos Q xs)" using HI by simp<br />
finally show "algunos (λx. P x ∨ Q x) (a # xs) = (algunos P (a # xs) ∨ algunos Q (a # xs))" by auto<br />
qed<br />
<br />
<br />
(* wilmorort danrodcha anaprarod*)<br />
lemma "algunos (λx. P x ∨ Q x) xs = (¬todos(λx. ¬P x)xs ∨ ¬todos(λx. ¬Q x)xs)"<br />
by (induct xs) auto<br />
<br />
(* wilmorort *)<br />
lemma "algunos (λx. P x ∨ Q x) xs = (¬todos(λx. ¬P x)xs ∨ ¬todos(λx. ¬Q x)xs)"<br />
(is "?P xs")<br />
proof (induct xs)<br />
show "?P []" by simp<br />
next<br />
fix a xs<br />
assume HI: "?P xs"<br />
have "algunos (λx. P x ∨ Q x) (a # xs) = (P a ∨ Q a ∨ algunos (λx. P x ∨ Q x) xs)" by simp<br />
also have "... = (P a ∨ Q a ∨ (¬ todos (λx. ¬ P x) xs ∨ ¬ todos (λx. ¬ Q x) xs))" using HI by simp<br />
also have "... = (P a ∨ ¬ todos (λx. ¬ P x) xs ∨ Q a ∨ ¬ todos (λx. ¬ Q x) xs )" by arith<br />
finally show "algunos (λx. P x ∨ Q x) (a # xs) = (~ todos (λx. ¬ P x) (a # xs) ∨ ~ todos (λx. ¬ Q x) (a # xs))" by simp<br />
qed<br />
<br />
{* danrodcha *}<br />
lemma "algunos (λx. P x ∨ Q x) xs = (algunos P xs ∨ algunos Q xs)" (is "?R xs")<br />
proof (induct xs)<br />
show "?R []" by simp<br />
next<br />
fix a xs assume HI: "?R xs"<br />
have 1:" (Q a ∨ algunos P xs) = (algunos P xs ∨ Q a)" by (simp add: HOL.disj_commute)<br />
have "algunos (λx. P x ∨ Q x) (a # xs) = <br />
(algunos P [a] ∨ algunos Q [a] ∨ algunos (λx. P x ∨ Q x) xs)" by simp<br />
also have "… = (P a ∨ (Q a ∨ algunos P xs) ∨ algunos Q xs)" using HI by simp<br />
also have "… = (P a ∨ algunos P xs ∨ Q a ∨ algunos Q xs)" using 1 by simp<br />
also have "… = (algunos P (a # xs) ∨ algunos Q (a # xs))" by simp<br />
finally show "?R (a # xs)" by simp<br />
qed<br />
<br />
(* dancorgar *)<br />
lemma "algunos (λx. P x ∧ Q x) xs ⟹ (algunos P xs ∧ algunos Q xs)"<br />
apply (induct xs)<br />
apply auto<br />
done<br />
<br />
(* dancorgar *)<br />
lemma "algunos (λx. P x ∧ Q x) xs ⟹ (algunos P xs ∧ algunos Q xs)"<br />
proof (induct xs)<br />
show "algunos (λx. P x ∧ Q x) [] ⟹ (algunos P [] ∧ algunos Q [])" by simp<br />
next<br />
fix xf xs<br />
assume HI: "algunos (λx. P x ∧ Q x) xs ⟹ (algunos P xs ∧ algunos Q xs)"<br />
have F1: "algunos (λx. P x ∧ Q x) (xf#xs) ⟹ ((P xf ∧ Q xf) ∨ algunos (λx. P x ∧ Q x) xs)" by simp<br />
also have F2: "… ⟹ (P xf ∧ Q xf ∨ (algunos P xs ∧ algunos Q xs))" using HI by simp<br />
have F3: "((P xf ∧ Q xf) ∨ (algunos P xs ∧ algunos Q xs)) ⟹ <br />
((P xf ∨ algunos P xs) ∧ (Q xf ∨ algunos Q xs))" by blast<br />
have F4: "((P xf ∨ algunos P xs) ∧ (Q xf ∨ algunos Q xs)) ⟹ <br />
(algunos P (xf#xs) ∧ algunos Q (xf#xs))" by simp<br />
show "algunos (λx. P x ∧ Q x) (xf#xs) ⟹ (algunos P (xf#xs) ∧ algunos Q (xf#xs))"<br />
using F1 F2 F3 F4 by blast<br />
qed<br />
<br />
(*pablucoto crigomgom*)<br />
-- "Automatica"<br />
lemma "algunos (λx. P x ∨ Q x) xs = (algunos P xs ∨ algunos Q xs)"<br />
by (induct xs) auto<br />
-- "Detallada"<br />
lemma "algunos (λx. P x ∨ Q x) xs = (algunos P xs ∨ algunos Q xs) "<br />
proof (induct xs)<br />
show "algunos (λx. P x ∨ Q x) [] = (algunos P [] ∨ algunos Q []) " by simp<br />
next<br />
fix a xs<br />
assume HI: " algunos (λx. P x ∨ Q x) xs = (algunos P xs ∨ algunos Q xs)"<br />
have " algunos (λx. P x ∨ Q x) (a # xs) = ( P a ∨ Q a ∨ algunos (λx. P x ∨ Q x) xs)" by simp<br />
also have "... = (P a ∨ Q a ∨ (algunos P xs ∨ algunos Q xs))" using HI by simp<br />
also have "… = (((P a) ∨ algunos P xs) ∨ ((Q a) ∨ algunos Q xs))" by arith <br />
also have "… = (algunos P (a#xs) ∨ algunos Q (a#xs))" by simp<br />
finally show "algunos (λx. P x ∨ Q x) (a # xs) = (algunos P (a # xs) ∨ algunos Q (a # xs)) " by simp<br />
qed<br />
<br />
(* anaprarod *)<br />
lemma "algunos (λx. P x ∨ Q x) xs = (algunos P xs ∨ algunos Q xs)" (is "?P xs")<br />
proof (induct xs)<br />
show "?P []" by simp<br />
next<br />
fix x xs<br />
assume HI: "?P xs"<br />
have "algunos (λx. P x ∨ Q x) (x#xs) =(P x ∨ Q x ∨ algunos (λx. P x ∨ Q x) xs)" by simp<br />
also have "… = (P x ∨ Q x ∨ algunos P xs ∨ algunos Q xs)" using HI by simp<br />
also have "… = (P x ∨ algunos P xs ∨ Q x ∨ algunos Q xs)" by arith<br />
finally show "?P (x#xs)" by simp<br />
qed<br />
<br />
text {*<br />
--------------------------------------------------------------------- <br />
Ejercicio 11.1. Demostrar o refutar automáticamente<br />
algunos P xs = (¬ todos (λx. (¬ P x)) xs)<br />
--------------------------------------------------------------------- <br />
*}<br />
<br />
(* ivamenjim wilmorort migtermor marpoldia1 crigomgom rubgonmar paupeddeg danrodcha pablucoto dancorgar josgarsan anaprarod*)<br />
<br />
lemma "algunos P xs = (¬ todos (λx. (¬ P x)) xs)"<br />
by (induct xs) auto<br />
<br />
(* ferrenseg *)<br />
<br />
lemma "algunos P xs = (¬ todos (λx. (¬ P x)) xs)"<br />
by (induct xs) simp_all<br />
<br />
(* anaprarod *)<br />
lemma "algunos P xs = (¬ todos (λx. (¬ P x)) xs)" <br />
apply (induct xs)<br />
apply auto<br />
done<br />
<br />
<br />
text {*<br />
--------------------------------------------------------------------- <br />
Ejercicio 11.2. Demostrar o refutar datalladamente<br />
algunos P xs = (¬ todos (λx. (¬ P x)) xs)<br />
--------------------------------------------------------------------- <br />
*}<br />
<br />
(* ivamenjim migtermor marpoldia1 crigomgom paupeddeg josgarsan*)<br />
<br />
lemma "algunos P xs = (¬ todos (λx. (¬ P x)) xs)"<br />
proof (induct xs)<br />
show "algunos P [] = (¬ todos (λx. ¬ P x) [])" by simp<br />
next <br />
fix a xs<br />
assume HI: "algunos P xs = (¬ todos (λx. ¬ P x) xs)"<br />
have "algunos P (a # xs) = (P a ∨ (algunos P xs))" by simp<br />
also have "... = (P a ∨ (¬ todos (λx. ¬ P x) xs))" using HI by simp<br />
finally show "algunos P (a # xs) = (¬ todos (λx. ¬ P x) (a # xs))" by simp<br />
qed<br />
<br />
(* ferrenseg *)<br />
<br />
lemma "algunos P xs = (¬ todos (λx. (¬ P x)) xs)"<br />
proof (induct xs)<br />
show "algunos P [] = (¬ todos (λx. (¬ P x)) [])" by simp<br />
next<br />
fix x xs<br />
assume HI: "algunos P xs = (¬ todos (λx. (¬ P x)) xs)"<br />
show "algunos P (x # xs) = (¬ todos (λx. (¬ P x)) (x # xs))"<br />
proof - <br />
have "algunos P (x # xs) = ((P x) ∨ algunos P xs)" by simp<br />
also have "… = ((P x) ∨ ¬ todos (λx. (¬ P x)) xs)" using HI by simp<br />
also have "… = (¬ (¬ (P x) ∧ todos (λx. (¬ P x)) xs))" by simp<br />
also have "… = (¬ todos (λx. (¬ P x)) (x # xs))" by simp<br />
finally show ?thesis<br />
qed<br />
qed<br />
<br />
{* danrodcha pablucoto anaprarod *}<br />
lemma "algunos P xs = (¬ todos (λx. (¬ P x)) xs)" (is "?Q xs")<br />
proof (induct xs)<br />
show "?Q []" by simp<br />
next<br />
fix a xs assume HI: "?Q xs"<br />
have "algunos P (a # xs) = (P a ∨ algunos P xs)" by simp<br />
also have "… = (P a ∨ (¬ todos (λx. (¬ P x)) xs))" using HI by simp<br />
also have "… = (¬(¬(P a) ∧ todos(λx. (¬ P x)) xs))" by simp<br />
also have "… = (¬ todos(λx. (¬ P x)) (a # xs))" by simp<br />
finally show "?Q (a # xs)" by simp<br />
qed<br />
<br />
(* dancorgar *)<br />
lemma "algunos P xs = (¬ todos (λx. (¬ P x)) xs)"<br />
proof (induct xs)<br />
show "algunos P [] = (¬ todos (λx. (¬ P x)) [])" by simp<br />
next<br />
fix xf xs<br />
assume HI: "algunos P xs = (¬ todos (λx. (¬ P x)) xs)"<br />
have "algunos P (xf#xs) = ((P xf) ∨ algunos P xs)" by simp<br />
also have "… = ( (P xf) ∨ (¬ todos (λx. (¬ P x)) xs) )" using HI by simp<br />
also have "… = (¬((¬P xf) ∧ (todos (λx. (¬ P x)) xs)))" by simp<br />
finally show "algunos P (xf#xs) = (¬todos (λx. (¬ P x)) (xf#xs))" by simp<br />
qed<br />
<br />
text {*<br />
--------------------------------------------------------------------- <br />
Ejercicio 12. Definir la funcion primitiva recursiva <br />
estaEn :: 'a ⇒ 'a list ⇒ bool<br />
tal que (estaEn x xs) se verifica si el elemento x está en la lista<br />
xs. Por ejemplo, <br />
estaEn (2::nat) [3,2,4] = True<br />
estaEn (1::nat) [3,2,4] = False<br />
--------------------------------------------------------------------- <br />
*}<br />
<br />
(* ivamenjim ferrenseg migtermor serrodcal crigomgom rubgonmar marpoldia1 paupeddeg danrodcha dancorgar pablucoto anaprarod*)<br />
<br />
fun estaEn :: "'a ⇒ 'a list ⇒ bool" where<br />
"estaEn x [] = False"<br />
| "estaEn x (a#xs) = ((a = x) ∨ (estaEn x xs))"<br />
<br />
value "estaEn (2::nat) [3,2,4] = True"<br />
value "estaEn (1::nat) [3,2,4] = False"<br />
<br />
<br />
text {*<br />
--------------------------------------------------------------------- <br />
Ejercicio 13. Expresar la relación existente entre estaEn y algunos. <br />
Demostrar dicha relación de forma automática y detallada.<br />
--------------------------------------------------------------------- <br />
*}<br />
<br />
(* migtermor crigomgom *)<br />
<br />
lemma "estaEn x xs = (algunos (λa. a=x) xs)"<br />
by (induct xs) auto<br />
<br />
lemma auxiliar13:<br />
"(x=a) = ((λa. a=x) a)"<br />
by auto<br />
<br />
lemma "estaEn x xs = (algunos (λa. a=x) xs)" (is "?P xs")<br />
proof (induct xs)<br />
show "?P []" by simp<br />
next<br />
fix a xs<br />
assume HI: "?P xs"<br />
have "estaEn x (a#xs) = ((a=x) ∨ (estaEn x xs))" by simp<br />
also have H: "… = ((a=x) ∨ (algunos (λa. a=x) xs))" using HI by simp<br />
also have "… = (((λa. a=x) a) ∨ (algunos (λa. a=x) xs))" using auxiliar13 by simp<br />
also have "… = (algunos (λa. a=x) (a#xs))" by simp<br />
have C: "estaEn x (a#xs) = (algunos (λa. a=x) (a#xs))" using H by simp<br />
finally show "estaEn x (a#xs) = (algunos (λa. a=x) (a#xs))" using C by simp<br />
qed<br />
<br />
<br />
(* ferrenseg *)<br />
<br />
lemma estaEn_algunos:<br />
"estaEn y xs = algunos (λx. x=y) xs"<br />
by (induct xs) auto<br />
<br />
lemma estaEn_algunos_2:<br />
"estaEn y xs = algunos (λx. x=y) xs"<br />
proof (induct xs)<br />
show "estaEn y [] = algunos (λx. x=y) []" by simp<br />
next<br />
fix x xs<br />
assume HI: "estaEn y xs = algunos (λx. x=y) xs"<br />
show "estaEn y (x # xs) = algunos (λx. x=y) (x # xs)"<br />
proof -<br />
have "estaEn y (x # xs) = (y=x ∨ estaEn y xs)" by simp<br />
also have "… = (y=x ∨ algunos (λx. x=y) xs)" using HI by simp<br />
also have "… = (x=y ∨ algunos (λx. x=y) xs)" by auto<br />
also have "… = algunos (λx. x=y) (x # xs)" by simp<br />
finally show ?thesis<br />
qed<br />
qed<br />
<br />
(* crigomgom paupeddeg dancorgar *)<br />
<br />
lemma "estaEn a xs = algunos (λx. x = a) xs"<br />
by (induct xs) auto<br />
<br />
lemma "estaEn x xs = (algunos (λa. a=x) xs)" <br />
proof (induct xs)<br />
show " estaEn x [] = algunos (λa. a = x) []" by simp<br />
next<br />
fix a xs<br />
assume HI: "estaEn x xs = algunos (λa. a = x) xs"<br />
have "estaEn x (a # xs) = (a = x ∨ estaEn x xs)" by simp<br />
also have "... = (a = x ∨ algunos (λa. a = x) xs)" using HI by simp<br />
finally show " estaEn x (a # xs) = algunos (λa. a = x) (a # xs)" by simp<br />
qed<br />
<br />
(* ivamenjim marpoldia1 *)<br />
<br />
lemma "estaEn x xs = algunos (λy. x=y) xs"<br />
by (induct xs) auto<br />
<br />
lemma "estaEn x xs = algunos (λy. x=y) xs"<br />
proof (induct xs)<br />
show "estaEn x [] = algunos (op = x) []" by simp<br />
next <br />
fix a xs<br />
assume HI: "estaEn x xs = algunos (op = x) xs"<br />
have "estaEn x (a # xs) = ((a = x) ∨ (estaEn x xs))" by simp<br />
also have "... = (a = x ∨ algunos (op = x) xs)" using HI by simp<br />
finally show " estaEn x (a # xs) = algunos (op = x) (a # xs)" by auto<br />
qed<br />
<br />
{* danrodcha *}<br />
lemma "estaEn a xs = algunos (op = a) xs"<br />
by (induct xs) auto<br />
<br />
{* danrodcha *}<br />
lemma "estaEn a xs = algunos (op = a) xs"<br />
proof (induct xs)<br />
show "estaEn a [] = algunos (op = a) []" by simp<br />
next<br />
fix x xs <br />
assume HI: "estaEn a xs = algunos (op = a) xs"<br />
have "estaEn a (x # xs) = ((x = a) ∨ (estaEn a xs))" by simp<br />
also have "… = (x = a ∨ algunos (op = a) xs)" using HI by simp<br />
also have "… = ((op = a) x ∨ algunos (op = a) xs)" by (simp add:HOL.eq_commute)<br />
also have "… = algunos (op = a) (x # xs)" by simp<br />
finally show "estaEn a (x # xs) = algunos (op = a) (x # xs)" by simp<br />
qed<br />
<br />
(* pablucoto *)<br />
<br />
-- "Automatica"<br />
lemma " estaEn y xs = algunos (λx. x=y) xs"<br />
by (induct xs) auto<br />
<br />
-- "Detallada"<br />
lemma " estaEn y xs = algunos (λx. x=y) xs"<br />
proof (induct xs)<br />
show "estaEn y [] = algunos (λx. x = y) []" by simp<br />
next<br />
fix a xs<br />
assume HI: " estaEn y xs = algunos (λx. x = y) xs "<br />
have " estaEn y (a # xs) = (y = a ∨ estaEn y xs)" by simp <br />
also have "... = (y = a ∨ algunos (λx. x = y) xs) " using HI by simp<br />
also have "… = (a=y ∨ algunos (λx. x=y) xs)" by auto<br />
also have "... = (algunos (λx. x = y) (a#xs))" by simp<br />
finally show "estaEn y (a # xs) = algunos (λx. x = y) (a # xs) " by simp<br />
qed<br />
<br />
<br />
(* anaprarod *)<br />
(* Automática*)<br />
lemma "estaEn a xs = algunos (λx. a=x) xs" (is "?P xs")<br />
by (induct xs) auto<br />
<br />
(* Detallada (igual que las anteriores pero obviando el paso previo al finally) *)<br />
lemma "estaEn a xs = algunos (λx. a=x) xs" (is "?P xs")<br />
proof (induct xs)<br />
show "?P []" by simp<br />
next<br />
fix x xs<br />
assume HI : "?P xs"<br />
have "estaEn a (x#xs) = (x=a ∨ estaEn a xs)" by simp<br />
also have "… = (x=a ∨ algunos (λx. a=x) xs)" using HI by simp<br />
also have "… = (a=x ∨ algunos (λx. a=x) xs)" using HI by auto<br />
finally show "?P (x#xs)" by simp<br />
qed<br />
<br />
<br />
end<br />
</source></div>Lucnovdoshttps://www.glc.us.es/~jalonso/RA2016/index.php?title=Relaci%C3%B3n_4&diff=610Relación 42016-11-22T23:27:29Z<p>Lucnovdos: </p>
<hr />
<div><source lang="isar"><br />
chapter {* R4: Cuantificadores sobre listas *}<br />
<br />
theory R4_Cuantificadores_sobre_listas<br />
imports Main <br />
begin<br />
<br />
text {* <br />
--------------------------------------------------------------------- <br />
Ejercicio 1. Definir la función <br />
todos :: ('a ⇒ bool) ⇒ 'a list ⇒ bool<br />
tal que (todos p xs) se verifica si todos los elementos de la lista <br />
xs cumplen la propiedad p. Por ejemplo, se verifica <br />
todos (λx. 1<length x) [[2,1,4],[1,3]]<br />
¬todos (λx. 1<length x) [[2,1,4],[3]]<br />
<br />
Nota: La función todos es equivalente a la predefinida list_all. <br />
--------------------------------------------------------------------- <br />
*}<br />
<br />
{*danrodcha*}<br />
fun todos :: "('a ⇒ bool) ⇒ 'a list ⇒ bool" where<br />
"todos p xs = "todos p [] = True"<br />
| "todos p (x#xs) = (p x ∧ todos p xs)""<br />
<br />
(* ivamenjim migtermor dancorgar wilmorort marpoldia1 ferrenseg paupeddeg pablucoto crigomgom anaprarod serrodcal juacabsou rubgonmar josgarsan fraortmoy lucnovdos*)<br />
fun todos :: "('a ⇒ bool) ⇒ 'a list ⇒ bool" where<br />
"todos p [] = True"<br />
| "todos p (x#xs) = (p x ∧ todos p xs)"<br />
<br />
text {* <br />
--------------------------------------------------------------------- <br />
Ejercicio 2. Definir la función <br />
algunos :: ('a ⇒ bool) ⇒ 'a list ⇒ bool<br />
tal que (algunos p xs) se verifica si algunos elementos de la lista <br />
xs cumplen la propiedad p. Por ejemplo, se verifica <br />
algunos (λx. 1<length x) [[2,1,4],[3]]<br />
¬algunos (λx. 1<length x) [[],[3]]"<br />
<br />
Nota: La función algunos es equivalente a la predefinida list_ex. <br />
--------------------------------------------------------------------- <br />
*}<br />
<br />
{*danrodcha ivamenjim migtermor dancorgar marpoldia1 ferrenseg wilmorort paupeddeg pablucoto crigomgom anaprarod serrodcal juacabsou rubgonmar josgarsan fraortmoy lucnovdos*}<br />
fun algunos :: "('a ⇒ bool) ⇒ 'a list ⇒ bool" where<br />
"algunos p [] = False"<br />
| "algunos p (x#xs) = (p x ∨ algunos p xs)"<br />
<br />
text {*<br />
--------------------------------------------------------------------- <br />
Ejercicio 3.1. Demostrar o refutar automáticamente <br />
todos (λx. P x ∧ Q x) xs = (todos P xs ∧ todos Q xs)<br />
--------------------------------------------------------------------- <br />
*}<br />
<br />
{*danrodcha ivamenjim migtermor dancorgar marpoldia1 ferrenseg wilmorort paupeddeg pablucoto anaprarod serrodcal juacabsou rubgonmar josgarsan fraortmoy lucnovdos*}<br />
lemma "todos (λx. P x ∧ Q x) xs = (todos P xs ∧ todos Q xs)"<br />
by (induct xs) auto<br />
<br />
text {*<br />
--------------------------------------------------------------------- <br />
Ejercicio 3.2. Demostrar o refutar detalladamente<br />
todos (λx. P x ∧ Q x) xs = (todos P xs ∧ todos Q xs)<br />
--------------------------------------------------------------------- <br />
*}<br />
<br />
{*danrodcha*}<br />
lemma "todos (λx. P x ∧ Q x) xs = (todos P xs ∧ todos Q xs)"<br />
proof (induct xs)<br />
show "?R []" by simp<br />
next<br />
fix a xs <br />
assume HI: "?R xs"<br />
have "todos (λx. P x ∧ Q x) (a#xs) = (P a ∧ Q a ∧ todos (λx. P x ∧ Q x) xs)" by simp<br />
also have "… = (P a ∧ Q a ∧ (todos P xs ∧ todos Q xs))" using HI by simp<br />
also have "… = ((P a ∧ todos P xs) ∧ (Q a ∧ todos Q xs))" by blast<br />
also have "… = (todos P (a#xs) ∧ todos Q (a#xs))" by simp<br />
finally show "?R (a#xs)" by simp<br />
qed<br />
<br />
(* ivamenjim wilmorort serrodcal josgarsan*)<br />
<br />
lemma "todos (λx. P x ∧ Q x) xs = (todos P xs ∧ todos Q xs)"<br />
proof (induct xs)<br />
show "todos (λx. P x ∧ Q x) [] = (todos P [] ∧ todos Q [])" by simp<br />
next <br />
fix a xs<br />
assume HI: "todos (λx. P x ∧ Q x) xs = (todos P xs ∧ todos Q xs)"<br />
have "todos (λx. P x ∧ Q x) (a # xs) = (P a ∧ Q a ∧ todos (λx. P x ∧ Q x) xs)" by simp<br />
also have "... = (P a ∧ Q a ∧ todos P xs ∧ todos Q xs)" using HI by simp<br />
finally show "todos (λx. P x ∧ Q x) (a # xs) = (todos P (a # xs) ∧ todos Q (a # xs))" by auto<br />
qed<br />
<br />
(* dancorgar paupeddeg *)<br />
lemma "todos (λx. P x ∧ Q x) xs = (todos P xs ∧ todos Q xs)"<br />
proof (induct xs)<br />
show "todos (λx. P x ∧ Q x) [] = (todos P [] ∧ todos Q [])" by simp<br />
next<br />
fix y xs<br />
assume HI: "todos (λx. P x ∧ Q x) xs = (todos P xs ∧ todos Q xs)"<br />
have "todos (λx. P x ∧ Q x) (y#xs) = ((P y ∧ Q y) ∧ (todos (λx. P x ∧ Q x) xs))" by simp<br />
also have "… = ((P y ∧ Q y) ∧ (todos P xs ∧ todos Q xs))" using HI by simp<br />
also have "… = ((P y ∧ todos P xs) ∧ (Q y ∧ todos Q xs))" by blast<br />
also have "… = ((todos P (y#xs)) ∧ (todos Q (y#xs)))" by simp<br />
finally show "todos (λx. P x ∧ Q x) (y#xs) = (todos P (y#xs) ∧ todos Q (y#xs))" by simp<br />
qed<br />
<br />
<br />
(* migtermor *)<br />
lemma "todos (λx. P x ∧ Q x) xs = (todos P xs ∧ todos Q xs)"<br />
proof (induct xs)<br />
show "todos (λx. P x ∧ Q x) [] = (todos P [] ∧ todos Q [])" by simp<br />
next <br />
fix x xs<br />
assume HI: "todos (λx. P x ∧ Q x) xs = (todos P xs ∧ todos Q xs)"<br />
have "(todos (λx. P x ∧ Q x) (x#xs)) = (( P x ∧ Q x) ∧ (todos (λx. P x ∧ Q x) xs))"<br />
by (simp only: todos.simps(2))<br />
also have "… = ((P x ∧ Q x) ∧ (todos P xs ∧ todos Q xs))" using HI by simp<br />
also have "… = (((P x)∧(todos P xs)) ∧ ((Q x) ∧ (todos Q xs)))" by arith<br />
also have "((P x ∧ Q x) ∧ (todos P xs ∧ todos Q xs)) = (((P x)∧(todos P xs)) ∧ ((Q x) ∧ (todos Q xs)))"<br />
by arith (* Este paso es exactamente el mismo que el anterior, pero sin cualquiera de los dos no funciona el "finally show" *)<br />
have "… = (((P x)∧(todos P xs))∧((Q x)∧(todos Q xs)))" by simp<br />
have "… = ((todos P (x#xs))∧(todos Q (x#xs)))" by simp<br />
finally show "(todos (λx. P x ∧ Q x) (x#xs)) = ((todos P (x#xs)) ∧ (todos Q (x#xs)))" by simp <br />
qed<br />
<br />
(* marpoldia1 *)<br />
lemma "todos (λx. P x ∧ Q x) xs = (todos P xs ∧ todos Q xs)"<br />
proof (induct xs)<br />
show "todos (λx. P x ∧ Q x) [] = (todos P [] ∧ todos Q [])" by simp<br />
next<br />
fix a xs<br />
assume HI: "todos (λx. P x ∧ Q x) xs = (todos P xs ∧ todos Q xs)" <br />
have "todos (λx. P x ∧ Q x) (a # xs) = ((P a ∧ Q a) ∧ (todos P xs ∧ todos Q xs))" using HI by simp<br />
also have "... = ((P a ∧ todos P xs) ∧ (Q a ∧ todos Q xs))" by blast<br />
also have "... = (todos P (a#xs) ∧ todos Q (a#xs)) " by simp<br />
finally show "todos (λx. P x ∧ Q x) (a#xs) = (todos P (a#xs) ∧ todos Q (a#xs))" by simp <br />
qed<br />
<br />
(* ferrenseg *)<br />
<br />
lemma "todos (λx. P x ∧ Q x) xs = (todos P xs ∧ todos Q xs)"<br />
proof (induct xs)<br />
show "todos (λx. P x ∧ Q x) [] = (todos P [] ∧ todos Q [])" by simp<br />
next <br />
fix n xs<br />
assume HI: "todos (λx. P x ∧ Q x) xs = (todos P xs ∧ todos Q xs)"<br />
have "todos (λx. P x ∧ Q x) (n # xs) = (P n ∧ Q n ∧ todos P xs ∧ todos Q xs)" using HI by simp<br />
also have "… = P n ∧ Q n ∧ todos P xs ∧ todos Q xs" by blast<br />
also have "⋯ = todos P (n # xs) ∧ todos Q (n # xs)" by simp <br />
finally show "todos (λx. P x ∧ Q x) (n # xs) = todos P (n # xs) ∧ todos Q (n # xs)" by simp<br />
qed<br />
<br />
(* lucnovdos *)<br />
<br />
lemma "todos (λx. P x ∧ Q x) xs = (todos P xs ∧ todos Q xs)"<br />
proof (induct xs)<br />
show "todos (λx. P x ∧ Q x) [] = (todos P [] ∧ todos Q [])" by simp<br />
next<br />
fix n xs<br />
assume HI: "todos (λx. P x ∧ Q x) xs = (todos P xs ∧ todos Q xs)" <br />
have "todos (λx. P x ∧ Q x) (n # xs) = ((P n ∧ Q n) ∧ (todos P xs ∧ todos Q xs))" using HI by simp<br />
also have "... = ((P n ∧ todos P xs) ∧ (Q n ∧ todos Q xs))" by arith<br />
also have "... = ((todos P(n#xs)) ∧ (todos Q(n#xs)))" by simp<br />
<br />
finally show "todos (λx. P x ∧ Q x) (n#xs) = (todos P (n#xs) ∧ todos Q (n#xs))" by simp <br />
qed<br />
<br />
{* wilmorort pablucoto crigomgom anaprarod juacabsou rubgonmar fraortmoy *}<br />
<br />
lemma "todos (λx. P x ∧ Q x) xs = (todos P xs ∧ todos Q xs)"<br />
proof (induct xs)<br />
show "todos (λx. P x ∧ Q x) [] = (todos P [] ∧ todos Q [])" by simp <br />
next<br />
fix a xs<br />
assume HI: "todos (λx. P x ∧ Q x) xs = (todos P xs ∧ todos Q xs)"<br />
have "todos (λx. P x ∧ Q x) (a # xs) = (P a ∧ Q a ∧ todos (λx. P x ∧ Q x) xs)" by simp<br />
also have "... = (P a ∧ Q a ∧ todos P xs ∧ todos Q xs)" using HI by simp<br />
also have "... = (P a ∧ todos P xs ∧ Q a ∧ todos Q xs)" by arith<br />
also have "... = (todos P (a # xs) ∧ todos Q (a # xs))" by simp (* Este paso se puede obviar*)<br />
finally show "todos (λx. P x ∧ Q x) (a # xs) = (todos P (a # xs) ∧ todos Q (a # xs))" by simp<br />
qed<br />
<br />
text {*<br />
--------------------------------------------------------------------- <br />
Ejercicio 4.1. Demostrar o refutar automáticamente<br />
todos P (x @ y) = (todos P x ∧ todos P y)<br />
--------------------------------------------------------------------- <br />
*}<br />
{*danrodcha ivamenjim marpoldia1 migtermor ferrenseg wilmorort paupeddeg crigomgom anaprarod serrodcal juacabsou rubgonmar pablucoto fraortmoy josgarsan lucnovdos*}<br />
lemma "todos P (x @ y) = (todos P x ∧ todos P y)"<br />
by (induct x) auto<br />
<br />
{* anaprarod dancorgar *}<br />
lemma "todos P (x @ y) = (todos P x ∧ todos P y)"<br />
apply (induct x)<br />
apply auto<br />
done<br />
<br />
text {*<br />
--------------------------------------------------------------------- <br />
Ejercicio 4.2. Demostrar o refutar detalladamente<br />
todos P (x @ y) = (todos P x ∧ todos P y)<br />
--------------------------------------------------------------------- <br />
*}<br />
<br />
(* ivamenjim ferrenseg wilmorort dancorgar fraortmoy josgarsan lucnovdos*)<br />
<br />
lemma todos_append:<br />
"todos P (x @ y) = (todos P x ∧ todos P y)"<br />
proof (induct x)<br />
show "todos P ([] @ y) = (todos P [] ∧ todos P y)" by simp<br />
next <br />
fix a x<br />
assume HI: "todos P (x @ y) = (todos P x ∧ todos P y)"<br />
have "todos P ((a#x) @ y) = (P a ∧ (todos P (x @ y)))" by simp<br />
also have "... = (P a ∧ (todos P x ∧ todos P y))" using HI by simp<br />
finally show "todos P ((a # x) @ y) = (todos P (a # x) ∧ todos P y)" by simp<br />
qed<br />
<br />
(* migtermor serrodcal *)<br />
<br />
lemma todos_append1:<br />
"todos P (x @ y) = (todos P x ∧ todos P y)" (is "?P x")<br />
proof (induct x)<br />
show "?P []" by simp<br />
next <br />
fix a x<br />
assume HI: "?P x"<br />
have "todos P ((a#x) @ y) = (P a ∧ (todos P (x @ y)))" by simp<br />
also have "… = (P a ∧ (todos P x ∧ todos P y))" using HI by simp<br />
finally show "?P (a#x)" by simp<br />
qed <br />
<br />
(* marpoldia1 *)<br />
lemma todos_append:<br />
"todos P (x @ y) = (todos P x ∧ todos P y)"<br />
proof (induct x)<br />
show "todos P ([] @ y) = (todos P [] ∧ todos P y)" by simp<br />
next<br />
fix a x<br />
assume HI: "todos P (x @ y) = (todos P x ∧ todos P y)"<br />
have "todos P ((a # x) @ y) = todos P (a#(x@y))" by simp<br />
also have "... = (P a ∧ todos P (x @ y))" by simp<br />
also have "... = (P a ∧ (todos P x ∧ todos P y))" using HI by simp<br />
finally show "todos P ((a # x) @ y) = (todos P (a # x) ∧ todos P y)" by simp <br />
qed<br />
<br />
(* paupeddeg *)<br />
lemma todos_append3:<br />
"todos P (x @ y) = (todos P x ∧ todos P y)"<br />
proof (induct x)<br />
show "todos P ([] @ y) = (todos P [] ∧ todos P y)" by simp<br />
next<br />
fix a x<br />
assume HI: "todos P (x @ y) = (todos P x ∧ todos P y)"<br />
have "todos P ((a # x) @ y) = todos P (a # (x @ y))" by simp<br />
also have "... = (todos P [a] ∧ todos P (x @ y)) " by simp<br />
also have "... = (todos P [a] ∧ (todos P x ∧ todos P y))" using HI by simp<br />
finally show "todos P ((a # x) @ y) = (todos P (a # x) ∧ todos P y)" by simp<br />
qed<br />
<br />
(* crigomgom juacabsou anaprarod*)<br />
<br />
lemma todos_append4:<br />
"todos P (x @ y) = (todos P x ∧ todos P y)"<br />
proof (induct x)<br />
show "todos P ([] @ y) = (todos P [] ∧ todos P y)" by simp<br />
next<br />
fix x xs<br />
assume HI:"todos P (xs @ y) = (todos P xs ∧ todos P y)"<br />
have "todos P ((x # xs) @ y) = (P x ∧ todos P (xs @ y ))" by simp<br />
also have "... = (P x ∧ (todos P xs ∧ todos P y)) " using HI by simp<br />
also have "... = ((P x ∧ todos P xs) ∧ todos P y)" by simp<br />
finally show "todos P ((x # xs) @ y) = (todos P (x # xs) ∧ todos P y)" by simp<br />
qed<br />
<br />
(* pablucoto *)<br />
lemma todos_append5:<br />
"todos P (x @ y) = (todos P x ∧ todos P y)"<br />
proof (induct x )<br />
show "todos P ([] @ y) = (todos P [] ∧ todos P y)" by simp<br />
next<br />
fix a x<br />
assume HI:"todos P (x @ y) = (todos P x ∧ todos P y) " <br />
have "todos P ((a # x) @ y) = ( P a ∧ todos P (x @ y))" by simp<br />
also have " ... =( P a ∧ (todos P x ∧ todos P y))" using HI by simp<br />
also have "... = (todos P (a#x) ∧ todos P y)" by simp<br />
finally show "todos P ((a # x) @ y) = (todos P (a # x) ∧ todos P y)" by auto<br />
qed<br />
<br />
{*danrodcha*}<br />
lemma todos_append6:<br />
"todos P (x @ y) = (todos P x ∧ todos P y)" (is "?Q x")<br />
proof (induct x)<br />
show "?Q []" by simp<br />
next<br />
fix x a assume HI: "?Q x"<br />
have "todos P ((a # x) @ y) = todos P (a # x @ y)" by simp<br />
also have "… = ((P a) ∧ todos P (x @ y))" by simp<br />
also have "… = ((P a) ∧ (todos P x ∧ todos P y))" using HI by simp<br />
also have "… = (todos P (a # x) ∧ todos P y)" by simp<br />
finally show "?Q (a # x)" by simp<br />
qed<br />
<br />
text {*<br />
--------------------------------------------------------------------- <br />
Ejercicio 5.1. Demostrar o refutar automáticamente <br />
todos P (rev xs) = todos P xs<br />
--------------------------------------------------------------------- <br />
*}<br />
<br />
lemma "todos P (rev xs) = todos P xs"<br />
oops<br />
<br />
(* migtermor ivamenjim marpoldia1 serrodcal anaprarod paupeddeg *)<br />
lemma "todos P (rev xs) = todos P xs" <br />
apply (induct xs)<br />
apply simp<br />
apply (simp add: todos_append)<br />
apply auto<br />
done<br />
<br />
(* ferrenseg crigomgom rubgonmar fraortmoy josgarsan danrodcha lucnovdos*)<br />
<br />
lemma "todos P (rev xs) = todos P xs"<br />
by (induct xs) (auto simp add: todos_append)<br />
<br />
(* wilmorort *)<br />
lemma "todos P (rev xs) = todos P xs"<br />
by (induct xs,simp,simp add: todos_append,auto)<br />
<br />
(* juacabsou *)<br />
lemma "todos P (rev xs) = todos P xs"<br />
apply (induct xs,simp,simp add: todos_append,auto) done<br />
<br />
(* dancorgar *)<br />
lemma "todos P (rev xs) = todos P xs"<br />
apply (induct xs)<br />
apply (simp add: todos_append)<br />
apply (simp add: todos_append)<br />
apply auto<br />
done<br />
<br />
(* pablucoto * )<br />
lemma "todos P (rev xs) = todos P xs"<br />
by (induct xs) (auto, simp_all add: todos_append) <br />
lemma "todos P (rev xs) = todos P xs"<br />
by (induct xs) ( simp_all add: todos_append, auto)<br />
<br />
text {*<br />
--------------------------------------------------------------------- <br />
Ejercicio 5.2. Demostrar o refutar detalladamente<br />
todos P (rev xs) = todos P xs<br />
--------------------------------------------------------------------- <br />
*}<br />
<br />
lemma "todos P (rev xs) = todos P xs"<br />
oops<br />
<br />
(* migtermor *)<br />
<br />
lemma auxiliar:<br />
"rev (a#xs) = rev xs @ [a]"<br />
by auto<br />
<br />
lemma "todos P (rev xs) = todos P xs" (is "?Q xs")<br />
proof (induct xs)<br />
show "?Q []" by (simp only: rev.simps(1))<br />
next <br />
fix a xs<br />
assume HI: "?Q xs"<br />
have "todos P (rev (a#xs)) = (todos P (rev xs @ [a]))" by (simp add: auxiliar)<br />
have "… = ((todos P (rev xs)) ∧ (todos P [a]))" by (simp add: todos_append)<br />
have "… = (todos P (rev xs) ∧ P a)" by simp<br />
also have Aux: "… = (todos P xs ∧ P a)" using HI by simp<br />
have Aux1: "… = (P a ∧ todos P xs)" by arith<br />
have "(todos P (rev xs) ∧ P a) = (P a ∧ todos P xs)" using Aux Aux1 by simp<br />
finally show "todos P (rev (a#xs)) = todos P (a#xs)" by (simp add: todos_append)<br />
qed<br />
<br />
<br />
(* marpoldia1 ferrenseg crigomgom serrodcal juacabsou rubgonmar josgarsan pablucoto*)<br />
lemma "todos P (rev xs) = todos P xs"<br />
proof (induct xs)<br />
show "todos P (rev []) = todos P []" by simp<br />
next<br />
fix a xs<br />
assume HI: "todos P (rev xs) = todos P xs"<br />
have "todos P (rev (a # xs)) = (todos P ((rev xs)@[a]))" by simp<br />
also have "... = (todos P (rev xs) ∧ todos P [a])" by (simp add:todos_append)<br />
also have "... = (todos P xs ∧ P a)" using HI by simp<br />
also have "... = (P a ∧ todos P xs)" by arith<br />
also have "... = (todos P (a#xs))" by simp<br />
finally show "todos P (rev (a # xs)) = (todos P (a#xs))" by simp <br />
qed<br />
<br />
(* ivamenjim *)<br />
<br />
lemma "todos P (rev xs) = todos P xs" (is "?P xs")<br />
proof (induct xs)<br />
show "?P []" by simp<br />
next <br />
fix a xs<br />
assume HI: "?P xs"<br />
have "todos P (rev (a # xs)) = (todos P ((rev xs)@[a]))" by simp<br />
also have "... = ((todos P (rev xs)) ∧ todos P [a])" by (simp add: todos_append)<br />
also have "... = (todos P xs ∧ todos P [a])" using HI by simp<br />
also have "... = (todos P [a] ∧ todos P xs)" by arith<br />
finally show "todos P (rev (a # xs)) = todos P (a # xs)" by simp<br />
qed<br />
<br />
(* fraortmoy *)<br />
<br />
lemma "todos P (rev xs) = todos P xs"<br />
proof (induct xs)<br />
show "todos P (rev []) = todos P []" by simp<br />
next<br />
fix a xs <br />
assume H1: "todos P (rev xs) = todos P xs"<br />
have " todos P (rev (a # xs)) = todos P (rev xs @ [a])" by simp<br />
also have "… = (todos P (rev xs) ∧ todos P [a])" by (simp add:todos_append)<br />
also have "… = (todos P xs ∧ todos P [a])" using H1 by simp<br />
also have "… = (todos P [a] ∧ todos P xs)" by arith<br />
finally show "todos P (rev (a # xs)) = todos P (a#xs)" by simp<br />
qed<br />
<br />
(* paupeddeg dancorgar *)<br />
lemma "todos P (rev xs) = todos P xs"<br />
proof (induct xs)<br />
show " todos P (rev []) = todos P []" by simp<br />
next<br />
fix a xs<br />
assume HI: "todos P (rev xs) = todos P xs "<br />
have "todos P (rev (a # xs)) = todos P (rev(xs) @ [a])" by simp<br />
also have "... = (todos P (rev xs) ∧ todos P [a]) " by (simp add: todos_append)<br />
also have "... = (todos P xs ∧ todos P [a])" using HI by simp<br />
also have "... = (todos P [a] ∧ todos P xs)" by arith<br />
also have "... = todos P ([a] @ xs)" by (simp add: todos_append)<br />
finally show "todos P (rev (a # xs)) = todos P (a # xs)" by simp<br />
qed <br />
<br />
(* wilmorort *)<br />
lemma "todos P (rev xs) = todos P xs" (is "?P xs")<br />
proof (induct xs) <br />
show "?P []" by simp <br />
next<br />
fix a xs<br />
assume HI: "?P xs"<br />
have "todos P (rev (a#xs)) = todos P ((rev xs) @ [a])" by simp<br />
also have "... = (todos P(rev xs) ∧ todos P [a])" by (simp add: todos_append)<br />
also have "... = (todos P xs ∧ todos P [a])" using HI by simp<br />
also have "... = (todos P [a] ∧ todos P xs)" by (simp add: HOL.conj_commute)<br />
also have "... = todos P([a]@(xs))" by (simp)<br />
finally show "todos P (rev (a#xs))= todos P (a # xs)" by simp<br />
qed<br />
<br />
(* anaprarod *) <br />
(* es igual que las anteriores pero con el final también con patrones *)<br />
<br />
lemma "todos P (rev xs) = todos P xs" (is "?P xs")<br />
proof (induct xs)<br />
show "?P []" by auto<br />
next<br />
fix a xs<br />
assume HI: "?P xs"<br />
have "todos P (rev (a#xs)) = todos P (rev xs @ [a])" by simp<br />
also have "… = (todos P (rev xs) ∧ todos P [a])" by (simp add: todos_append)<br />
also have "… = (todos P xs ∧ todos P [a])" using HI by simp<br />
also have "… = (todos P [a] ∧ todos P xs)" by arith<br />
also have "… = todos P (a#xs)" by simp<br />
finally show "?P (a # xs)" by simp<br />
qed<br />
<br />
{*danrodcha*}<br />
lemma "todos P (rev xs) = todos P xs" (is "?Q xs")<br />
proof (induct xs)<br />
show "?Q []" by simp<br />
next<br />
fix a xs assume HI: "?Q xs"<br />
have "todos P (rev (a#xs)) = todos P ((rev xs) @ [a])" by simp<br />
also have "… = (todos P (rev xs) ∧ todos P [a])" by (simp add:todos_append)<br />
also have "… = (todos P xs ∧ todos P [a])" using HI by simp<br />
also have "… = (todos P [a] ∧ todos P xs)" by (simp add: HOL.conj_commute)<br />
also have "… = todos P (a # xs)" by simp<br />
finally show "?Q (a # xs)" by simp<br />
qed<br />
<br />
text {*<br />
--------------------------------------------------------------------- <br />
Ejercicio 6. Demostrar o refutar:<br />
algunos (λx. P x ∧ Q x) xs = (algunos P xs ∧ algunos Q xs)<br />
--------------------------------------------------------------------- <br />
*}<br />
<br />
lemma "algunos (λx. P x ∧ Q x) xs = (algunos P xs ∧ algunos Q xs)"<br />
oops<br />
<br />
(* migtermor ivamenjim ferrenseg paupeddeg crigomgom serrodcal wilmorort juacabsou rubgonmar anaprarod marpoldia1 fraortmoy josgarsan danrodcha pablucoto*)<br />
lemma "algunos (λx. P x ∧ Q x) xs = (algunos P xs ∧ algunos Q xs)" <br />
quickcheck<br />
oops<br />
(* Quickcheck encuentra el siguiente contraejemplo: P={a1}, Q={a2}, xs={a1,a2}. En este ejemplo:<br />
· "algunos (λx. P x ∧ Q x) xs = False"<br />
· "(algunos P xs ∧ algunos Q xs) = True" *)<br />
<br />
text {*<br />
--------------------------------------------------------------------- <br />
Ejercicio 7.1. Demostrar o refutar automáticamente <br />
algunos P (map f xs) = algunos (P ∘ f) xs<br />
--------------------------------------------------------------------- <br />
*}<br />
<br />
(* ivamenjim migtermor marpoldia1 crigomgom rubgonmar wilmorort anaprarod <br />
fraortmoy juacabsou paupeddeg josgarsan danrodcha*)<br />
<br />
lemma "algunos P (map f xs) = algunos (P o f) xs"<br />
by (induct xs) auto<br />
<br />
(* ferrenseg *)<br />
<br />
lemma "algunos P (map f xs) = algunos (P o f) xs"<br />
by (induct xs) simp_all<br />
<br />
(* anaprarod *)<br />
lemma "algunos P (map f xs) = algunos (P o f) xs"<br />
apply (induct xs)<br />
apply auto<br />
done<br />
<br />
text {*<br />
--------------------------------------------------------------------- <br />
Ejercicio 7.2. Demostrar o refutar datalladamente<br />
algunos P (map f xs) = algunos (P ∘ f) xs<br />
--------------------------------------------------------------------- <br />
*}<br />
<br />
lemma "algunos P (map f xs) = algunos (P o f) xs"<br />
oops<br />
<br />
(* migtermor *)<br />
<br />
lemma AUX: "algunos (λa. P (f a)) xs = algunos (P o f) xs"<br />
by (induct xs) auto<br />
<br />
lemma "algunos P (map f xs) = algunos (P o f) xs" (is "?Q xs")<br />
proof (induct xs)<br />
show "?Q []" by simp<br />
next<br />
fix x xs<br />
assume HI: "?Q xs"<br />
have "algunos P (map f (x#xs)) = (algunos P ((f x)#(map f xs)))" by simp<br />
also have "… = ((P (f x)) ∨ (algunos P (map f xs)))" by (simp only: algunos.simps(2))<br />
also have "… = ((P (f x)) ∨ (algunos (P o f) xs))" <br />
proof (cases)<br />
assume C1: "(P (f x))"<br />
have Aux: "((P (f x)) ∨ (algunos P (map f xs))) = True" using C1 by simp<br />
have Aux1: "… = ((P (f x)) ∨ (algunos (P o f) xs))" using C1 by simp<br />
then show "((P (f x)) ∨ (algunos P (map f xs))) = ((P (f x)) ∨ (algunos (P o f) xs))" <br />
using Aux Aux1 by simp<br />
next<br />
assume C2: "¬(P (f x))"<br />
have Aux2: "((P (f x)) ∨ (algunos P (map f xs))) = (algunos P (map f xs))" using C2 by simp<br />
have Aux3: "… = (algunos (P o f) xs)" using HI by (simp add: AUX)<br />
also have "… = ((P (f x)) ∨ (algunos (P o f) xs))" using C2 by simp<br />
then show "((P (f x)) ∨ (algunos P (map f xs))) = ((P (f x)) ∨ (algunos (P o f) xs))" <br />
using Aux2 Aux3 by simp<br />
qed<br />
also have "… = (((P o f) x) ∨ (algunos (P o f) xs))" by simp<br />
also have "… = (algunos (P o f) (x#xs))" by simp<br />
finally show "algunos P (map f (x#xs)) = (algunos (P o f) (x#xs))" by simp<br />
qed<br />
<br />
(* ferrenseg *)<br />
<br />
lemma "algunos P (map f xs) = algunos (P ∘ f) xs"<br />
proof (induct xs)<br />
show "algunos P (map f []) = algunos (P ∘ f) []" by simp<br />
next<br />
fix x xs<br />
assume HI: "algunos P (map f xs) = algunos (P ∘ f) xs"<br />
show "algunos P (map f (x # xs)) = algunos (P ∘ f) (x # xs)"<br />
proof -<br />
have "algunos P (map f (x # xs)) = algunos P ((f x) # (map f xs))" by simp<br />
also have "… = ((P (f x)) ∨ (algunos P (map f xs)))" by simp<br />
also have "… = (((P ∘ f) x) ∨ (algunos (P ∘ f) xs))" using HI by simp<br />
also have "… = algunos (P ∘ f) (x # xs)" by simp<br />
finally show ?thesis<br />
qed<br />
qed<br />
<br />
(* ivamenjim *)<br />
<br />
lemma "algunos P (map f xs) = algunos (P o f) xs"<br />
proof (induct xs)<br />
show "algunos P (map f []) = algunos (P ∘ f) []" by simp<br />
next <br />
fix a xs<br />
assume HI: "algunos P (map f xs) = algunos (P ∘ f) xs"<br />
have "algunos P (map f (a # xs)) = algunos P ((f a)#map f xs)" by simp<br />
also have "... = (algunos P (map f [a]) ∨ algunos P (map f xs))" by simp <br />
also have "... = (algunos P (map f [a]) ∨ algunos (P ∘ f) xs)" using HI by simp <br />
finally show "algunos P (map f (a # xs)) = algunos (P ∘ f) (a # xs)" by simp<br />
qed<br />
<br />
(* crigomgom rubgonmar anaprarod marpoldia1 juacabsou danrodcha paupeddeg*)<br />
lemma "algunos P (map f xs) = algunos (P o f) xs"<br />
proof (induct xs)<br />
show "algunos P (map f []) = algunos (P ∘ f) []" by simp<br />
next<br />
fix x xs<br />
assume HI: "algunos P (map f xs) = algunos (P ∘ f) xs"<br />
have "algunos P (map f (x # xs)) = algunos P ((f x) # (map f xs))" by simp<br />
also have "... = (P (f x) ∨ algunos P (map f xs))" by simp<br />
also have "... = (P (f x) ∨ algunos (P ∘ f) xs)" using HI by simp<br />
also have "... = ((P ∘ f) x ∨ algunos (P ∘ f) xs)" by simp<br />
finally show "algunos P (map f (x # xs)) = algunos (P ∘ f) (x # xs)" by simp<br />
qed<br />
<br />
(* wilmorort *)<br />
lemma "algunos P (map f xs) = algunos (P o f) xs" (is "?P xs" )<br />
proof (induct xs)<br />
show "?P []" by simp<br />
next<br />
fix a xs<br />
assume HI: "algunos P (map f xs) = algunos (P ∘ f) xs"<br />
have "algunos P (map f (a # xs)) = (P (f a) ∨ algunos P (map f xs))" by simp<br />
also have "... = (P (f a) ∨ algunos (P ∘ f) xs )" using HI by simp<br />
also have "... = ((P ∘ f) a ∨ algunos (P ∘ f)xs)" by simp<br />
finally show " algunos P (map f (a # xs)) = algunos (P ∘ f) (a # xs)" by simp<br />
qed<br />
<br />
(* fraortmoy *)<br />
<br />
lemma "algunos P (map f xs) = algunos (P o f) xs"<br />
proof (induct xs)<br />
show "algunos P (map f []) = algunos (P ∘ f) []" by simp<br />
next<br />
fix a xs<br />
assume H1: "algunos P (map f xs) = algunos (P ∘ f) xs"<br />
have "algunos P (map f (a # xs)) = (algunos P (map f [a]) ∨ algunos P (map f xs))" by simp<br />
also have "… = (algunos (P ∘ f) [a] ∨ algunos (P ∘ f) xs )" using H1 by simp<br />
also have "… = algunos (P ∘ f) (a#xs)" by simp<br />
finally show "algunos P (map f (a # xs)) = algunos (P ∘ f) (a#xs)" by simp<br />
qed<br />
<br />
{*danrodcha*}<br />
lemma "algunos P (map f xs) = algunos (P o f) xs" (is "?Q xs")<br />
proof (induct xs)<br />
show "?Q []" by simp<br />
next<br />
fix a xs assume HI: "?Q xs"<br />
have " algunos P (map f (a # xs)) = algunos P ((f a)#(map f xs))" by simp<br />
also have "… = (P (f a) ∨ algunos P (map f xs))" by simp<br />
also have "… = (P (f a) ∨ algunos (P ∘ f) xs)" using HI by blast<br />
also have "… = ((P ∘ f) a ∨ algunos (P ∘ f) xs)" by simp<br />
also have "… = algunos (P ∘ f) (a # xs)" by simp<br />
finally show "?Q (a # xs)" by blast<br />
qed<br />
<br />
text {*<br />
--------------------------------------------------------------------- <br />
Ejercicio 8.1. Demostrar o refutar automáticamente <br />
algunos P (xs @ ys) = (algunos P xs ∨ algunos P ys)<br />
--------------------------------------------------------------------- <br />
*}<br />
<br />
(* ivamenjim marpoldia1 paupeddeg crigomgom rubgonmar wilmorort fraortmoy danrodcha *)<br />
<br />
lemma "algunos P (xs @ ys) = (algunos P xs ∨ algunos P ys)"<br />
by (induct xs) auto<br />
<br />
(* ferrenseg *)<br />
lemma "algunos P (xs @ ys) = (algunos P xs ∨ algunos P ys)"<br />
by (induct xs) simp_all<br />
<br />
text {*<br />
--------------------------------------------------------------------- <br />
Ejercicio 8.2. Demostrar o refutar detalladamente<br />
algunos P (xs @ ys) = (algunos P xs ∨ algunos P ys)<br />
--------------------------------------------------------------------- <br />
*}<br />
<br />
(* ivamenjim ferrenseg marpoldia1 wilmorort*)<br />
<br />
lemma algunos_append:<br />
"algunos P (xs @ ys) = (algunos P xs ∨ algunos P ys)"<br />
proof (induct xs)<br />
show "algunos P ([] @ ys) = (algunos P [] ∨ algunos P ys)" by simp<br />
next <br />
fix a xs<br />
assume HI: "algunos P (xs @ ys) = (algunos P xs ∨ algunos P ys)"<br />
have "algunos P ((a # xs) @ ys) = (P a ∨ (algunos P (xs @ ys)))" by simp<br />
also have "... = (P a ∨ (algunos P xs ∨ algunos P ys))" using HI by simp<br />
finally show "algunos P ((a # xs) @ ys) = (algunos P (a # xs) ∨ algunos P ys)" by simp<br />
qed<br />
<br />
(* migtermor crigomgom *)<br />
<br />
lemma algunos_append2:<br />
"algunos P (xs @ ys) = (algunos P xs ∨ algunos P ys)" (is "?Q xs")<br />
proof (induct xs)<br />
show "?Q []" by simp<br />
next <br />
fix x xs<br />
assume HI: "?Q xs"<br />
have "algunos P ((x#xs) @ ys) = algunos P (x#(xs @ ys))" by simp<br />
also have "… = ((P x) ∨ (algunos P (xs @ ys)))" by simp<br />
also have "… = ((P x) ∨ (algunos P xs) ∨ (algunos P ys))" using HI by simp<br />
also have "… = ((algunos P (x#xs)) ∨ (algunos P ys))" by simp<br />
finally show "algunos P ((x#xs) @ ys) = (algunos P (x#xs) ∨ algunos P ys)" by simp<br />
qed<br />
<br />
(* paupeddeg*)<br />
lemma algunos_append3:<br />
"algunos P (xs @ ys) = (algunos P xs ∨ algunos P ys)"<br />
proof (induct xs)<br />
show "algunos P ([] @ ys) = (algunos P [] ∨ algunos P ys)" by simp<br />
next<br />
fix a xs<br />
assume HI: "algunos P (xs @ ys) = (algunos P xs ∨ algunos P ys)"<br />
have "algunos P ((a # xs) @ ys) = (algunos P [a] ∨ ( algunos P (xs @ ys)))" by simp<br />
also have "... = (algunos P [a] ∨ (algunos P xs ∨ algunos P ys))" using HI by simp<br />
finally show "algunos P ((a # xs) @ ys) = (algunos P (a # xs) ∨ algunos P ys)" by simp<br />
qed<br />
<br />
(* fraortmoy *)<br />
lemma algunos_append4:<br />
"algunos P (xs @ ys) = (algunos P xs ∨ algunos P ys)"<br />
proof (induct xs)<br />
show " algunos P ([] @ ys) = (algunos P [] ∨ algunos P ys)" by simp<br />
next <br />
fix a xs<br />
assume H1: "algunos P (xs @ ys) = (algunos P xs ∨ algunos P ys)"<br />
have "algunos P ((a # xs) @ ys) = (algunos P [a] ∨ algunos P (xs @ ys))" by simp<br />
also have "… = (algunos P [a] ∨ algunos P xs ∨ algunos P ys)" using H1 by simp<br />
also have "… = ((algunos P [a] ∨ algunos P xs) ∨ algunos P ys)" by simp<br />
also have "… = (algunos P (a#xs) ∨ algunos P ys)" by simp<br />
finally show "algunos P ((a # xs) @ ys) = (algunos P (a#xs) ∨ algunos P ys)" by simp<br />
qed<br />
<br />
{* danrodcha *}<br />
lemma algunos_append5:<br />
"algunos P (xs @ ys) = (algunos P xs ∨ algunos P ys)" (is "?Q xs")<br />
proof (induct xs)<br />
show "?Q []" by simp<br />
next<br />
fix a xs assume HI: "?Q xs"<br />
have "algunos P ((a#xs) @ ys) = algunos P (a#(xs @ ys))" by simp<br />
also have "… = (algunos P [a] ∨ algunos P (xs @ ys))" by simp<br />
also have "… = (algunos P [a] ∨ algunos P xs ∨ algunos P ys)" using HI by simp<br />
also have "… = (algunos P (a # xs) ∨ algunos P ys)" by simp<br />
finally show "?Q (a # xs)" by simp<br />
qed<br />
<br />
text {*<br />
--------------------------------------------------------------------- <br />
Ejercicio 9.1. Demostrar o refutar automáticamente<br />
algunos P (rev xs) = algunos P xs<br />
--------------------------------------------------------------------- <br />
*}<br />
<br />
(* ivamenjim migtermor marpoldia1 rubgonmar paupeddeg *)<br />
<br />
lemma "algunos P (rev xs) = algunos P xs"<br />
apply (induct xs)<br />
apply simp<br />
apply (simp add: algunos_append)<br />
apply auto<br />
done<br />
<br />
(* ferrenseg crigomgom danrodcha*)<br />
lemma "algunos P (rev xs) = algunos P xs"<br />
by (induct xs) (auto simp add: algunos_append)<br />
<br />
(* wilmorort *)<br />
lemma "algunos P (rev xs) = algunos P xs"<br />
by (induct xs,simp,simp add: algunos_append,auto) <br />
<br />
<br />
text {*<br />
--------------------------------------------------------------------- <br />
Ejercicio 9.2. Demostrar o refutar detalladamente<br />
algunos P (rev xs) = algunos P xs<br />
--------------------------------------------------------------------- <br />
*}<br />
<br />
lemma "algunos P (rev xs) = algunos P xs"<br />
oops<br />
<br />
(* migtermor *)<br />
lemma auxiliar1:<br />
"rev (a#xs) = rev xs @ [a]"<br />
by auto <br />
<br />
lemma "algunos P (rev xs) = algunos P xs" (is "?Q xs")<br />
proof (induct xs)<br />
show "?Q []" by simp<br />
next<br />
fix x xs<br />
assume HI: "?Q xs"<br />
have "algunos P (rev (x#xs)) = (algunos P (rev xs @ [x]))" using auxiliar1 by simp<br />
also have "… = ((algunos P (rev xs)) ∨ (algunos P [x]))" by (simp add: algunos_append)<br />
also have "… = ((P x) ∨ (algunos P (rev xs)))" by simp arith<br />
also have "… = ((P x) ∨ (algunos P xs))" using HI by simp<br />
finally show "algunos P (rev (x#xs)) = (algunos P (x#xs))" by simp<br />
qed<br />
<br />
(* ferrenseg *)<br />
<br />
lemma "algunos P (rev xs) = algunos P xs"<br />
proof (induct xs)<br />
show "algunos P (rev []) = algunos P []" by simp<br />
next<br />
fix x xs<br />
assume HI: "algunos P (rev xs) = algunos P xs"<br />
show "algunos P (rev (x # xs)) = algunos P (x # xs)"<br />
proof -<br />
have "algunos P (rev (x # xs)) = algunos P ((rev xs) @ [x])" by simp<br />
also have "… = (algunos P (rev xs) ∨ algunos P [x])" <br />
by (simp add: algunos_append)<br />
also have "… = (algunos P xs ∨ algunos P [x])" using HI by simp<br />
also have "… = (algunos P xs ∨ P x)" by simp<br />
also have "… = (P x ∨ algunos P xs)" by auto<br />
also have "… = algunos P (x # xs)" by simp<br />
finally show ?thesis<br />
qed<br />
qed<br />
<br />
(* ivamenjim marpoldia1*)<br />
<br />
lemma "algunos P (rev xs) = algunos P xs" (is "?P xs")<br />
proof (induct xs)<br />
show "?P []" by simp<br />
next <br />
fix a xs<br />
assume HI: "?P xs"<br />
have "algunos P (rev (a # xs)) = (algunos P ((rev xs)@[a]))" by simp<br />
also have "... = ((algunos P (rev xs)) ∨ algunos P [a])" by (simp add: algunos_append)<br />
also have "... = (algunos P xs ∨ algunos P [a])" using HI by simp<br />
also have "... = (algunos P [a] ∨ algunos P xs)" by arith<br />
finally show "algunos P (rev (a # xs)) = algunos P (a # xs)" by simp<br />
qed<br />
<br />
(*paupeddeg*)<br />
lemma "algunos P (rev xs) = algunos P xs"<br />
proof (induct xs)<br />
show "algunos P (rev []) = algunos P []" by simp<br />
next<br />
fix a xs<br />
assume HI:"algunos P (rev xs) = algunos P xs"<br />
have "algunos P (rev (a # xs)) = algunos P ((rev xs) @ [a])" by simp<br />
also have "... = ((algunos P (rev xs)) ∨ (algunos P [a]))" by (simp add: algunos_append)<br />
also have "... = ((algunos P xs) ∨ (algunos P [a]))" using HI by simp<br />
also have "... = ((algunos P [a]) ∨ (algunos P xs))" by arith<br />
finally show "algunos P (rev (a # xs)) = algunos P (a # xs)" by simp<br />
qed<br />
<br />
(*crigomgom *)<br />
lemma "algunos P (rev xs) = algunos P xs"<br />
proof (induct xs)<br />
show "algunos P (rev []) = algunos P []" by simp<br />
next<br />
fix x xs<br />
assume HI: " algunos P (rev xs) = algunos P xs" <br />
have "algunos P (rev (x # xs)) = algunos P ((rev xs) @ [x])" by simp<br />
also have "... = (algunos P (rev xs) ∨ algunos P [x])" by (simp add: algunos_append)<br />
also have "... = (algunos P xs ∨ algunos P [x])" using HI by simp<br />
also have "... = (algunos P xs ∨ P x)" by simp<br />
also have "... = (P x ∨ algunos P xs)" by arith<br />
also have "... = algunos P (x#xs)" by simp<br />
finally show "algunos P (rev (x # xs)) = algunos P (x # xs)" by simp<br />
qed<br />
<br />
<br />
(*wilmorort*)<br />
lemma "algunos P (rev xs) = algunos P xs" (is "?P xs")<br />
proof (induct xs) <br />
show "?P []" by simp <br />
next<br />
fix a xs<br />
assume HI: "?P xs"<br />
have "algunos P (rev (a#xs)) = algunos P ((rev xs) @ [a])" by simp<br />
also have "... = (algunos P(rev xs) ∨ algunos P [a])" by (simp add: algunos_append)<br />
also have "... = (algunos P xs ∨ algunos P [a])" using HI by simp<br />
also have "... = (algunos P [a] ∨ algunos P xs)" by (simp add: HOL.disj_commute)<br />
also have "... = algunos P([a]@(xs))" by (simp)<br />
finally show "algunos P (rev (a#xs))= algunos P (a # xs)" by simp<br />
qed<br />
<br />
{*danrodcha *}<br />
lemma "algunos P (rev xs) = algunos P xs" (is "?Q xs")<br />
proof (induct xs)<br />
show "?Q []" by simp<br />
next<br />
fix a xs assume HI: "?Q xs"<br />
have "algunos P (rev (a#xs)) = algunos P ((rev xs) @ [a])" by simp<br />
also have "… = (algunos P (rev xs) ∨ algunos P [a])" by (simp add:algunos_append)<br />
also have "… = (algunos P xs ∨ algunos P [a])" using HI by simp<br />
also have "… = (algunos P [a] ∨ algunos P xs)" by (simp add: HOL.disj_commute)<br />
also have "… = algunos P (a # xs)" by simp<br />
finally show "?Q (a # xs)" by simp<br />
qed<br />
<br />
text {*<br />
--------------------------------------------------------------------- <br />
Ejercicio 10. Encontrar un término no trivial Z tal que sea cierta la <br />
siguiente ecuación:<br />
algunos (λx. P x ∨ Q x) xs = Z<br />
y demostrar la equivalencia de forma automática y detallada.<br />
--------------------------------------------------------------------- <br />
*}<br />
<br />
(* migtermor *)<br />
<br />
lemma "algunos (λx. P x ∨ Q x) xs = ((algunos (λx. P x) xs) ∨ (algunos (λx. Q x) xs))"<br />
by (induct xs) auto<br />
<br />
lemma "algunos (λx. P x ∨ Q x) xs = ((algunos (λx. P x) xs) ∨ (algunos (λx. Q x) xs))" (is "?R xs")<br />
proof (induct xs)<br />
show "?R []" by simp<br />
next<br />
fix x xs<br />
assume HI: "?R xs"<br />
have "algunos (λx. P x ∨ Q x) (x#xs) = (((λx. P x ∨ Q x) x) ∨ (algunos (λx. P x ∨ Q x) xs))"<br />
by simp<br />
also have "… = ((P x ∨ Q x) ∨ (algunos (λx. P x ∨ Q x) xs))" by simp<br />
also have H1: "… = ((((λx. P x) x) ∨ (algunos (λx. P x) xs)) ∨ (((λx. Q x) x) ∨ (algunos (λx. Q x) xs)))"<br />
using HI by simp arith<br />
have H2: "… = ((algunos (λx. P x) (x#xs)) ∨ (algunos (λx. Q x) (x#xs)))" <br />
by simp<br />
have C: "(algunos (λx. P x ∨ Q x) (x#xs)) = <br />
((algunos (λx. P x) (x#xs)) ∨ (algunos (λx. Q x) (x#xs)))" <br />
using H1 H2 by simp<br />
finally show "(algunos (λx. P x ∨ Q x) (x#xs)) = <br />
((algunos (λx. P x) (x#xs)) ∨ (algunos (λx. Q x) (x#xs)))" <br />
using C by simp<br />
qed<br />
<br />
<br />
(* ferrenseg *)<br />
<br />
lemma "algunos (λx. P x ∨ Q x) xs = (algunos P xs ∨ algunos Q xs)"<br />
proof (induct xs)<br />
show "algunos (λx. P x ∨ Q x) [] = (algunos P [] ∨ algunos Q [])" by simp<br />
next<br />
fix x xs<br />
assume HI: "algunos (λx. (P x ∨ Q x)) xs = (algunos P xs ∨ algunos Q xs)"<br />
show "algunos (λx. P x ∨ Q x) (x # xs) = (algunos P (x # xs) ∨ algunos Q (x # xs))"<br />
proof -<br />
have "algunos (λx. P x ∨ Q x) (x # xs) = <br />
((P x) ∨ (Q x) ∨ algunos (λx. P x ∨ Q x) xs)" by simp<br />
also have "… = ((P x) ∨ (Q x) ∨ algunos P xs ∨ algunos Q xs)" <br />
using HI by simp<br />
also have "… = (((P x) ∨ algunos P xs) ∨ ((Q x) ∨ algunos Q xs))" by auto<br />
also have "… = (algunos P (x # xs) ∨ algunos Q (x # xs))" by simp<br />
finally show ?thesis <br />
qed<br />
qed<br />
<br />
(* ivamenjim marpoldia1 paupeddeg *)<br />
<br />
lemma "algunos (λx. P x ∨ Q x) xs = (algunos P xs ∨ algunos Q xs)" <br />
by (induct xs) auto<br />
<br />
lemma "algunos (λx. P x ∨ Q x) xs = (algunos P xs ∨ algunos Q xs)"<br />
proof (induct xs)<br />
show "algunos (λx. P x ∨ Q x) [] = (algunos P [] ∨ algunos Q [])" by simp<br />
next <br />
fix a xs<br />
assume HI: "algunos (λx. P x ∨ Q x) xs = (algunos P xs ∨ algunos Q xs)"<br />
have "algunos (λx. P x ∨ Q x) (a # xs) = (P a ∨ Q a ∨ algunos (λx. P x ∨ Q x) xs)" by simp<br />
also have "... = (P a ∨ Q a ∨ algunos P xs ∨ algunos Q xs)" using HI by simp<br />
finally show "algunos (λx. P x ∨ Q x) (a # xs) = (algunos P (a # xs) ∨ algunos Q (a # xs))" by auto<br />
qed<br />
<br />
<br />
(* wilmorort danrodcha*)<br />
lemma "algunos (λx. P x ∨ Q x) xs = (¬todos(λx. ¬P x)xs ∨ ¬todos(λx. ¬Q x)xs)"<br />
by (induct xs) auto<br />
<br />
(* wilmorort *)<br />
lemma "algunos (λx. P x ∨ Q x) xs = (¬todos(λx. ¬P x)xs ∨ ¬todos(λx. ¬Q x)xs)"<br />
(is "?P xs")<br />
proof (induct xs)<br />
show "?P []" by simp<br />
next<br />
fix a xs<br />
assume HI: "?P xs"<br />
have "algunos (λx. P x ∨ Q x) (a # xs) = (P a ∨ Q a ∨ algunos (λx. P x ∨ Q x) xs)" by simp<br />
also have "... = (P a ∨ Q a ∨ (¬ todos (λx. ¬ P x) xs ∨ ¬ todos (λx. ¬ Q x) xs))" using HI by simp<br />
also have "... = (P a ∨ ¬ todos (λx. ¬ P x) xs ∨ Q a ∨ ¬ todos (λx. ¬ Q x) xs )" by arith<br />
finally show "algunos (λx. P x ∨ Q x) (a # xs) = (~ todos (λx. ¬ P x) (a # xs) ∨ ~ todos (λx. ¬ Q x) (a # xs))" by simp<br />
qed<br />
<br />
{* danrodcha *}<br />
lemma "algunos (λx. P x ∨ Q x) xs = (algunos P xs ∨ algunos Q xs)" (is "?R xs")<br />
proof (induct xs)<br />
show "?R []" by simp<br />
next<br />
fix a xs assume HI: "?R xs"<br />
have 1:" (Q a ∨ algunos P xs) = (algunos P xs ∨ Q a)" by (simp add: HOL.disj_commute)<br />
have "algunos (λx. P x ∨ Q x) (a # xs) = <br />
(algunos P [a] ∨ algunos Q [a] ∨ algunos (λx. P x ∨ Q x) xs)" by simp<br />
also have "… = (P a ∨ (Q a ∨ algunos P xs) ∨ algunos Q xs)" using HI by simp<br />
also have "… = (P a ∨ algunos P xs ∨ Q a ∨ algunos Q xs)" using 1 by simp<br />
also have "… = (algunos P (a # xs) ∨ algunos Q (a # xs))" by simp<br />
finally show "?R (a # xs)" by simp<br />
qed<br />
<br />
<br />
<br />
text {*<br />
--------------------------------------------------------------------- <br />
Ejercicio 11.1. Demostrar o refutar automáticamente<br />
algunos P xs = (¬ todos (λx. (¬ P x)) xs)<br />
--------------------------------------------------------------------- <br />
*}<br />
<br />
(* ivamenjim wilmorort migtermor marpoldia1 crigomgom rubgonmar paupeddeg danrodcha*)<br />
<br />
lemma "algunos P xs = (¬ todos (λx. (¬ P x)) xs)"<br />
by (induct xs) auto<br />
<br />
(* ferrenseg *)<br />
<br />
lemma "algunos P xs = (¬ todos (λx. (¬ P x)) xs)"<br />
by (induct xs) simp_all<br />
<br />
text {*<br />
--------------------------------------------------------------------- <br />
Ejercicio 11.2. Demostrar o refutar datalladamente<br />
algunos P xs = (¬ todos (λx. (¬ P x)) xs)<br />
--------------------------------------------------------------------- <br />
*}<br />
<br />
(* ivamenjim migtermor marpoldia1 crigomgom paupeddeg*)<br />
<br />
lemma "algunos P xs = (¬ todos (λx. (¬ P x)) xs)"<br />
proof (induct xs)<br />
show "algunos P [] = (¬ todos (λx. ¬ P x) [])" by simp<br />
next <br />
fix a xs<br />
assume HI: "algunos P xs = (¬ todos (λx. ¬ P x) xs)"<br />
have "algunos P (a # xs) = (P a ∨ (algunos P xs))" by simp<br />
also have "... = (P a ∨ (¬ todos (λx. ¬ P x) xs))" using HI by simp<br />
finally show "algunos P (a # xs) = (¬ todos (λx. ¬ P x) (a # xs))" by simp<br />
qed<br />
<br />
(* ferrenseg *)<br />
<br />
lemma "algunos P xs = (¬ todos (λx. (¬ P x)) xs)"<br />
proof (induct xs)<br />
show "algunos P [] = (¬ todos (λx. (¬ P x)) [])" by simp<br />
next<br />
fix x xs<br />
assume HI: "algunos P xs = (¬ todos (λx. (¬ P x)) xs)"<br />
show "algunos P (x # xs) = (¬ todos (λx. (¬ P x)) (x # xs))"<br />
proof - <br />
have "algunos P (x # xs) = ((P x) ∨ algunos P xs)" by simp<br />
also have "… = ((P x) ∨ ¬ todos (λx. (¬ P x)) xs)" using HI by simp<br />
also have "… = (¬ (¬ (P x) ∧ todos (λx. (¬ P x)) xs))" by simp<br />
also have "… = (¬ todos (λx. (¬ P x)) (x # xs))" by simp<br />
finally show ?thesis<br />
qed<br />
qed<br />
<br />
{* danrodcha *}<br />
lemma "algunos P xs = (¬ todos (λx. (¬ P x)) xs)" (is "?Q xs")<br />
proof (induct xs)<br />
show "?Q []" by simp<br />
next<br />
fix a xs assume HI: "?Q xs"<br />
have "algunos P (a # xs) = (P a ∨ algunos P xs)" by simp<br />
also have "… = (P a ∨ (¬ todos (λx. (¬ P x)) xs))" using HI by simp<br />
also have "… = (¬(¬(P a) ∧ todos(λx. (¬ P x)) xs))" by simp<br />
also have "… = (¬ todos(λx. (¬ P x)) (a # xs))" by simp<br />
finally show "?Q (a # xs)" by simp<br />
qed<br />
<br />
text {*<br />
--------------------------------------------------------------------- <br />
Ejercicio 12. Definir la funcion primitiva recursiva <br />
estaEn :: 'a ⇒ 'a list ⇒ bool<br />
tal que (estaEn x xs) se verifica si el elemento x está en la lista<br />
xs. Por ejemplo, <br />
estaEn (2::nat) [3,2,4] = True<br />
estaEn (1::nat) [3,2,4] = False<br />
--------------------------------------------------------------------- <br />
*}<br />
<br />
(* ivamenjim ferrenseg migtermor serrodcal crigomgom rubgonmar marpoldia1 paupeddeg danrodcha*)<br />
<br />
fun estaEn :: "'a ⇒ 'a list ⇒ bool" where<br />
"estaEn x [] = False"<br />
| "estaEn x (a#xs) = ((a = x) ∨ (estaEn x xs))"<br />
<br />
value "estaEn (2::nat) [3,2,4] = True"<br />
value "estaEn (1::nat) [3,2,4] = False"<br />
<br />
<br />
text {*<br />
--------------------------------------------------------------------- <br />
Ejercicio 13. Expresar la relación existente entre estaEn y algunos. <br />
Demostrar dicha relación de forma automática y detallada.<br />
--------------------------------------------------------------------- <br />
*}<br />
<br />
(* migtermor crigomgom *)<br />
<br />
lemma "estaEn x xs = (algunos (λa. a=x) xs)"<br />
by (induct xs) auto<br />
<br />
lemma auxiliar13:<br />
"(x=a) = ((λa. a=x) a)"<br />
by auto<br />
<br />
lemma "estaEn x xs = (algunos (λa. a=x) xs)" (is "?P xs")<br />
proof (induct xs)<br />
show "?P []" by simp<br />
next<br />
fix a xs<br />
assume HI: "?P xs"<br />
have "estaEn x (a#xs) = ((a=x) ∨ (estaEn x xs))" by simp<br />
also have H: "… = ((a=x) ∨ (algunos (λa. a=x) xs))" using HI by simp<br />
also have "… = (((λa. a=x) a) ∨ (algunos (λa. a=x) xs))" using auxiliar13 by simp<br />
also have "… = (algunos (λa. a=x) (a#xs))" by simp<br />
have C: "estaEn x (a#xs) = (algunos (λa. a=x) (a#xs))" using H by simp<br />
finally show "estaEn x (a#xs) = (algunos (λa. a=x) (a#xs))" using C by simp<br />
qed<br />
<br />
<br />
(* ferrenseg *)<br />
<br />
lemma estaEn_algunos:<br />
"estaEn y xs = algunos (λx. x=y) xs"<br />
by (induct xs) auto<br />
<br />
lemma estaEn_algunos_2:<br />
"estaEn y xs = algunos (λx. x=y) xs"<br />
proof (induct xs)<br />
show "estaEn y [] = algunos (λx. x=y) []" by simp<br />
next<br />
fix x xs<br />
assume HI: "estaEn y xs = algunos (λx. x=y) xs"<br />
show "estaEn y (x # xs) = algunos (λx. x=y) (x # xs)"<br />
proof -<br />
have "estaEn y (x # xs) = (y=x ∨ estaEn y xs)" by simp<br />
also have "… = (y=x ∨ algunos (λx. x=y) xs)" using HI by simp<br />
also have "… = (x=y ∨ algunos (λx. x=y) xs)" by auto<br />
also have "… = algunos (λx. x=y) (x # xs)" by simp<br />
finally show ?thesis<br />
qed<br />
qed<br />
<br />
(*crigomgom paupeddeg*)<br />
<br />
lemma "estaEn a xs = algunos (λx. x = a) xs"<br />
by (induct xs) auto<br />
<br />
lemma "estaEn x xs = (algunos (λa. a=x) xs)" <br />
proof (induct xs)<br />
show " estaEn x [] = algunos (λa. a = x) []" by simp<br />
next<br />
fix a xs<br />
assume HI: "estaEn x xs = algunos (λa. a = x) xs"<br />
have "estaEn x (a # xs) = (a = x ∨ estaEn x xs)" by simp<br />
also have "... = (a = x ∨ algunos (λa. a = x) xs)" using HI by simp<br />
finally show " estaEn x (a # xs) = algunos (λa. a = x) (a # xs)" by simp<br />
qed<br />
<br />
(* ivamenjim marpoldia1 *)<br />
<br />
lemma "estaEn x xs = algunos (λy. x=y) xs"<br />
by (induct xs) auto<br />
<br />
lemma "estaEn x xs = algunos (λy. x=y) xs"<br />
proof (induct xs)<br />
show "estaEn x [] = algunos (op = x) []" by simp<br />
next <br />
fix a xs<br />
assume HI: "estaEn x xs = algunos (op = x) xs"<br />
have "estaEn x (a # xs) = ((a = x) ∨ (estaEn x xs))" by simp<br />
also have "... = (a = x ∨ algunos (op = x) xs)" using HI by simp<br />
finally show " estaEn x (a # xs) = algunos (op = x) (a # xs)" by auto<br />
qed<br />
<br />
{* danrodcha *}<br />
lemma "estaEn a xs = algunos (op = a) xs"<br />
by (induct xs) auto<br />
<br />
{* danrodcha *}<br />
lemma "estaEn a xs = algunos (op = a) xs"<br />
proof (induct xs)<br />
show "estaEn a [] = algunos (op = a) []" by simp<br />
next<br />
fix x xs <br />
assume HI: "estaEn a xs = algunos (op = a) xs"<br />
have "estaEn a (x # xs) = ((x = a) ∨ (estaEn a xs))" by simp<br />
also have "… = (x = a ∨ algunos (op = a) xs)" using HI by simp<br />
also have "… = ((op = a) x ∨ algunos (op = a) xs)" by (simp add:HOL.eq_commute)<br />
also have "… = algunos (op = a) (x # xs)" by simp<br />
finally show "estaEn a (x # xs) = algunos (op = a) (x # xs)" by simp<br />
qed<br />
<br />
end<br />
</source></div>Lucnovdoshttps://www.glc.us.es/~jalonso/RA2016/index.php?title=Relaci%C3%B3n_4&diff=607Relación 42016-11-22T22:58:01Z<p>Lucnovdos: </p>
<hr />
<div><source lang="isar"><br />
chapter {* R4: Cuantificadores sobre listas *}<br />
<br />
theory R4_Cuantificadores_sobre_listas<br />
imports Main <br />
begin<br />
<br />
text {* <br />
--------------------------------------------------------------------- <br />
Ejercicio 1. Definir la función <br />
todos :: ('a ⇒ bool) ⇒ 'a list ⇒ bool<br />
tal que (todos p xs) se verifica si todos los elementos de la lista <br />
xs cumplen la propiedad p. Por ejemplo, se verifica <br />
todos (λx. 1<length x) [[2,1,4],[1,3]]<br />
¬todos (λx. 1<length x) [[2,1,4],[3]]<br />
<br />
Nota: La función todos es equivalente a la predefinida list_all. <br />
--------------------------------------------------------------------- <br />
*}<br />
<br />
{*danrodcha*}<br />
fun todos :: "('a ⇒ bool) ⇒ 'a list ⇒ bool" where<br />
"todos p xs = "todos p [] = True"<br />
| "todos p (x#xs) = (p x ∧ todos p xs)""<br />
<br />
(* ivamenjim migtermor dancorgar wilmorort marpoldia1 ferrenseg paupeddeg pablucoto crigomgom anaprarod serrodcal juacabsou rubgonmar josgarsan fraortmoy lucnovdos*)<br />
fun todos :: "('a ⇒ bool) ⇒ 'a list ⇒ bool" where<br />
"todos p [] = True"<br />
| "todos p (x#xs) = (p x ∧ todos p xs)"<br />
<br />
text {* <br />
--------------------------------------------------------------------- <br />
Ejercicio 2. Definir la función <br />
algunos :: ('a ⇒ bool) ⇒ 'a list ⇒ bool<br />
tal que (algunos p xs) se verifica si algunos elementos de la lista <br />
xs cumplen la propiedad p. Por ejemplo, se verifica <br />
algunos (λx. 1<length x) [[2,1,4],[3]]<br />
¬algunos (λx. 1<length x) [[],[3]]"<br />
<br />
Nota: La función algunos es equivalente a la predefinida list_ex. <br />
--------------------------------------------------------------------- <br />
*}<br />
<br />
{*danrodcha ivamenjim migtermor dancorgar marpoldia1 ferrenseg wilmorort paupeddeg pablucoto crigomgom anaprarod serrodcal juacabsou rubgonmar josgarsan fraortmoy lucnovdos*}<br />
fun algunos :: "('a ⇒ bool) ⇒ 'a list ⇒ bool" where<br />
"algunos p [] = False"<br />
| "algunos p (x#xs) = (p x ∨ algunos p xs)"<br />
<br />
text {*<br />
--------------------------------------------------------------------- <br />
Ejercicio 3.1. Demostrar o refutar automáticamente <br />
todos (λx. P x ∧ Q x) xs = (todos P xs ∧ todos Q xs)<br />
--------------------------------------------------------------------- <br />
*}<br />
<br />
{*danrodcha ivamenjim migtermor dancorgar marpoldia1 ferrenseg wilmorort paupeddeg pablucoto anaprarod serrodcal juacabsou rubgonmar josgarsan fraortmoy lucnovdos*}<br />
lemma "todos (λx. P x ∧ Q x) xs = (todos P xs ∧ todos Q xs)"<br />
by (induct xs) auto<br />
<br />
text {*<br />
--------------------------------------------------------------------- <br />
Ejercicio 3.2. Demostrar o refutar detalladamente<br />
todos (λx. P x ∧ Q x) xs = (todos P xs ∧ todos Q xs)<br />
--------------------------------------------------------------------- <br />
*}<br />
<br />
{*danrodcha*}<br />
lemma "todos (λx. P x ∧ Q x) xs = (todos P xs ∧ todos Q xs)"<br />
proof (induct xs)<br />
show "?R []" by simp<br />
next<br />
fix a xs <br />
assume HI: "?R xs"<br />
have "todos (λx. P x ∧ Q x) (a#xs) = (P a ∧ Q a ∧ todos (λx. P x ∧ Q x) xs)" by simp<br />
also have "… = (P a ∧ Q a ∧ (todos P xs ∧ todos Q xs))" using HI by simp<br />
also have "… = ((P a ∧ todos P xs) ∧ (Q a ∧ todos Q xs))" by blast<br />
also have "… = (todos P (a#xs) ∧ todos Q (a#xs))" by simp<br />
finally show "?R (a#xs)" by simp<br />
qed<br />
<br />
(* ivamenjim wilmorort serrodcal josgarsan*)<br />
<br />
lemma "todos (λx. P x ∧ Q x) xs = (todos P xs ∧ todos Q xs)"<br />
proof (induct xs)<br />
show "todos (λx. P x ∧ Q x) [] = (todos P [] ∧ todos Q [])" by simp<br />
next <br />
fix a xs<br />
assume HI: "todos (λx. P x ∧ Q x) xs = (todos P xs ∧ todos Q xs)"<br />
have "todos (λx. P x ∧ Q x) (a # xs) = (P a ∧ Q a ∧ todos (λx. P x ∧ Q x) xs)" by simp<br />
also have "... = (P a ∧ Q a ∧ todos P xs ∧ todos Q xs)" using HI by simp<br />
finally show "todos (λx. P x ∧ Q x) (a # xs) = (todos P (a # xs) ∧ todos Q (a # xs))" by auto<br />
qed<br />
<br />
(* dancorgar paupeddeg *)<br />
lemma "todos (λx. P x ∧ Q x) xs = (todos P xs ∧ todos Q xs)"<br />
proof (induct xs)<br />
show "todos (λx. P x ∧ Q x) [] = (todos P [] ∧ todos Q [])" by simp<br />
next<br />
fix y xs<br />
assume HI: "todos (λx. P x ∧ Q x) xs = (todos P xs ∧ todos Q xs)"<br />
have "todos (λx. P x ∧ Q x) (y#xs) = ((P y ∧ Q y) ∧ (todos (λx. P x ∧ Q x) xs))" by simp<br />
also have "… = ((P y ∧ Q y) ∧ (todos P xs ∧ todos Q xs))" using HI by simp<br />
also have "… = ((P y ∧ todos P xs) ∧ (Q y ∧ todos Q xs))" by blast<br />
also have "… = ((todos P (y#xs)) ∧ (todos Q (y#xs)))" by simp<br />
finally show "todos (λx. P x ∧ Q x) (y#xs) = (todos P (y#xs) ∧ todos Q (y#xs))" by simp<br />
qed<br />
<br />
<br />
(* migtermor *)<br />
lemma "todos (λx. P x ∧ Q x) xs = (todos P xs ∧ todos Q xs)"<br />
proof (induct xs)<br />
show "todos (λx. P x ∧ Q x) [] = (todos P [] ∧ todos Q [])" by simp<br />
next <br />
fix x xs<br />
assume HI: "todos (λx. P x ∧ Q x) xs = (todos P xs ∧ todos Q xs)"<br />
have "(todos (λx. P x ∧ Q x) (x#xs)) = (( P x ∧ Q x) ∧ (todos (λx. P x ∧ Q x) xs))"<br />
by (simp only: todos.simps(2))<br />
also have "… = ((P x ∧ Q x) ∧ (todos P xs ∧ todos Q xs))" using HI by simp<br />
also have "… = (((P x)∧(todos P xs)) ∧ ((Q x) ∧ (todos Q xs)))" by arith<br />
also have "((P x ∧ Q x) ∧ (todos P xs ∧ todos Q xs)) = (((P x)∧(todos P xs)) ∧ ((Q x) ∧ (todos Q xs)))"<br />
by arith (* Este paso es exactamente el mismo que el anterior, pero sin cualquiera de los dos no funciona el "finally show" *)<br />
have "… = (((P x)∧(todos P xs))∧((Q x)∧(todos Q xs)))" by simp<br />
have "… = ((todos P (x#xs))∧(todos Q (x#xs)))" by simp<br />
finally show "(todos (λx. P x ∧ Q x) (x#xs)) = ((todos P (x#xs)) ∧ (todos Q (x#xs)))" by simp <br />
qed<br />
<br />
(* marpoldia1 *)<br />
lemma "todos (λx. P x ∧ Q x) xs = (todos P xs ∧ todos Q xs)"<br />
proof (induct xs)<br />
show "todos (λx. P x ∧ Q x) [] = (todos P [] ∧ todos Q [])" by simp<br />
next<br />
fix a xs<br />
assume HI: "todos (λx. P x ∧ Q x) xs = (todos P xs ∧ todos Q xs)" <br />
have "todos (λx. P x ∧ Q x) (a # xs) = ((P a ∧ Q a) ∧ (todos P xs ∧ todos Q xs))" using HI by simp<br />
also have "... = ((P a ∧ todos P xs) ∧ (Q a ∧ todos Q xs))" by blast<br />
also have "... = (todos P (a#xs) ∧ todos Q (a#xs)) " by simp<br />
finally show "todos (λx. P x ∧ Q x) (a#xs) = (todos P (a#xs) ∧ todos Q (a#xs))" by simp <br />
qed<br />
<br />
(* ferrenseg *)<br />
<br />
lemma "todos (λx. P x ∧ Q x) xs = (todos P xs ∧ todos Q xs)"<br />
proof (induct xs)<br />
show "todos (λx. P x ∧ Q x) [] = (todos P [] ∧ todos Q [])" by simp<br />
next <br />
fix n xs<br />
assume HI: "todos (λx. P x ∧ Q x) xs = (todos P xs ∧ todos Q xs)"<br />
have "todos (λx. P x ∧ Q x) (n # xs) = (P n ∧ Q n ∧ todos P xs ∧ todos Q xs)" using HI by simp<br />
also have "… = P n ∧ Q n ∧ todos P xs ∧ todos Q xs" by blast<br />
also have "⋯ = todos P (n # xs) ∧ todos Q (n # xs)" by simp <br />
finally show "todos (λx. P x ∧ Q x) (n # xs) = todos P (n # xs) ∧ todos Q (n # xs)" by simp<br />
qed<br />
<br />
(* lucnovdos *)<br />
<br />
lemma "todos (λx. P x ∧ Q x) xs = (todos P xs ∧ todos Q xs)"<br />
proof (induct xs)<br />
show "todos (λx. P x ∧ Q x) [] = (todos P [] ∧ todos Q [])" by simp<br />
next<br />
fix n xs<br />
assume HI: "todos (λx. P x ∧ Q x) xs = (todos P xs ∧ todos Q xs)" <br />
have "todos (λx. P x ∧ Q x) (n # xs) = ((P n ∧ Q n) ∧ (todos P xs ∧ todos Q xs))" using HI by simp<br />
also have "... = ((P n ∧ todos P xs) ∧ (Q n ∧ todos Q xs))" by arith<br />
also have "... = ((todos P(n#xs)) ∧ (todos Q(n#xs)))" by simp<br />
<br />
finally show "todos (λx. P x ∧ Q x) (n#xs) = (todos P (n#xs) ∧ todos Q (n#xs))" by simp <br />
qed<br />
<br />
{* wilmorort pablucoto crigomgom anaprarod juacabsou rubgonmar fraortmoy *}<br />
<br />
lemma "todos (λx. P x ∧ Q x) xs = (todos P xs ∧ todos Q xs)"<br />
proof (induct xs)<br />
show "todos (λx. P x ∧ Q x) [] = (todos P [] ∧ todos Q [])" by simp <br />
next<br />
fix a xs<br />
assume HI: "todos (λx. P x ∧ Q x) xs = (todos P xs ∧ todos Q xs)"<br />
have "todos (λx. P x ∧ Q x) (a # xs) = (P a ∧ Q a ∧ todos (λx. P x ∧ Q x) xs)" by simp<br />
also have "... = (P a ∧ Q a ∧ todos P xs ∧ todos Q xs)" using HI by simp<br />
also have "... = (P a ∧ todos P xs ∧ Q a ∧ todos Q xs)" by arith<br />
also have "... = (todos P (a # xs) ∧ todos Q (a # xs))" by simp (* Este paso se puede obviar*)<br />
finally show "todos (λx. P x ∧ Q x) (a # xs) = (todos P (a # xs) ∧ todos Q (a # xs))" by simp<br />
qed<br />
<br />
text {*<br />
--------------------------------------------------------------------- <br />
Ejercicio 4.1. Demostrar o refutar automáticamente<br />
todos P (x @ y) = (todos P x ∧ todos P y)<br />
--------------------------------------------------------------------- <br />
*}<br />
{*danrodcha ivamenjim marpoldia1 migtermor ferrenseg wilmorort paupeddeg crigomgom anaprarod serrodcal juacabsou rubgonmar pablucoto fraortmoy josgarsan lucnovdos*}<br />
lemma "todos P (x @ y) = (todos P x ∧ todos P y)"<br />
by (induct x) auto<br />
<br />
{* anaprarod dancorgar *}<br />
lemma "todos P (x @ y) = (todos P x ∧ todos P y)"<br />
apply (induct x)<br />
apply auto<br />
done<br />
<br />
text {*<br />
--------------------------------------------------------------------- <br />
Ejercicio 4.2. Demostrar o refutar detalladamente<br />
todos P (x @ y) = (todos P x ∧ todos P y)<br />
--------------------------------------------------------------------- <br />
*}<br />
<br />
(* ivamenjim ferrenseg wilmorort dancorgar fraortmoy josgarsan*)<br />
<br />
lemma todos_append:<br />
"todos P (x @ y) = (todos P x ∧ todos P y)"<br />
proof (induct x)<br />
show "todos P ([] @ y) = (todos P [] ∧ todos P y)" by simp<br />
next <br />
fix a x<br />
assume HI: "todos P (x @ y) = (todos P x ∧ todos P y)"<br />
have "todos P ((a#x) @ y) = (P a ∧ (todos P (x @ y)))" by simp<br />
also have "... = (P a ∧ (todos P x ∧ todos P y))" using HI by simp<br />
finally show "todos P ((a # x) @ y) = (todos P (a # x) ∧ todos P y)" by simp<br />
qed<br />
<br />
(* migtermor serrodcal *)<br />
<br />
lemma todos_append1:<br />
"todos P (x @ y) = (todos P x ∧ todos P y)" (is "?P x")<br />
proof (induct x)<br />
show "?P []" by simp<br />
next <br />
fix a x<br />
assume HI: "?P x"<br />
have "todos P ((a#x) @ y) = (P a ∧ (todos P (x @ y)))" by simp<br />
also have "… = (P a ∧ (todos P x ∧ todos P y))" using HI by simp<br />
finally show "?P (a#x)" by simp<br />
qed <br />
<br />
(* marpoldia1 *)<br />
lemma todos_append:<br />
"todos P (x @ y) = (todos P x ∧ todos P y)"<br />
proof (induct x)<br />
show "todos P ([] @ y) = (todos P [] ∧ todos P y)" by simp<br />
next<br />
fix a x<br />
assume HI: "todos P (x @ y) = (todos P x ∧ todos P y)"<br />
have "todos P ((a # x) @ y) = todos P (a#(x@y))" by simp<br />
also have "... = (P a ∧ todos P (x @ y))" by simp<br />
also have "... = (P a ∧ (todos P x ∧ todos P y))" using HI by simp<br />
finally show "todos P ((a # x) @ y) = (todos P (a # x) ∧ todos P y)" by simp <br />
qed<br />
<br />
(* paupeddeg *)<br />
lemma todos_append3:<br />
"todos P (x @ y) = (todos P x ∧ todos P y)"<br />
proof (induct x)<br />
show "todos P ([] @ y) = (todos P [] ∧ todos P y)" by simp<br />
next<br />
fix a x<br />
assume HI: "todos P (x @ y) = (todos P x ∧ todos P y)"<br />
have "todos P ((a # x) @ y) = todos P (a # (x @ y))" by simp<br />
also have "... = (todos P [a] ∧ todos P (x @ y)) " by simp<br />
also have "... = (todos P [a] ∧ (todos P x ∧ todos P y))" using HI by simp<br />
finally show "todos P ((a # x) @ y) = (todos P (a # x) ∧ todos P y)" by simp<br />
qed<br />
<br />
(* crigomgom juacabsou anaprarod*)<br />
<br />
lemma todos_append4:<br />
"todos P (x @ y) = (todos P x ∧ todos P y)"<br />
proof (induct x)<br />
show "todos P ([] @ y) = (todos P [] ∧ todos P y)" by simp<br />
next<br />
fix x xs<br />
assume HI:"todos P (xs @ y) = (todos P xs ∧ todos P y)"<br />
have "todos P ((x # xs) @ y) = (P x ∧ todos P (xs @ y ))" by simp<br />
also have "... = (P x ∧ (todos P xs ∧ todos P y)) " using HI by simp<br />
also have "... = ((P x ∧ todos P xs) ∧ todos P y)" by simp<br />
finally show "todos P ((x # xs) @ y) = (todos P (x # xs) ∧ todos P y)" by simp<br />
qed<br />
<br />
(* pablucoto *)<br />
lemma todos_append5:<br />
"todos P (x @ y) = (todos P x ∧ todos P y)"<br />
proof (induct x )<br />
show "todos P ([] @ y) = (todos P [] ∧ todos P y)" by simp<br />
next<br />
fix a x<br />
assume HI:"todos P (x @ y) = (todos P x ∧ todos P y) " <br />
have "todos P ((a # x) @ y) = ( P a ∧ todos P (x @ y))" by simp<br />
also have " ... =( P a ∧ (todos P x ∧ todos P y))" using HI by simp<br />
also have "... = (todos P (a#x) ∧ todos P y)" by simp<br />
finally show "todos P ((a # x) @ y) = (todos P (a # x) ∧ todos P y)" by auto<br />
qed<br />
<br />
{*danrodcha*}<br />
lemma todos_append6:<br />
"todos P (x @ y) = (todos P x ∧ todos P y)" (is "?Q x")<br />
proof (induct x)<br />
show "?Q []" by simp<br />
next<br />
fix x a assume HI: "?Q x"<br />
have "todos P ((a # x) @ y) = todos P (a # x @ y)" by simp<br />
also have "… = ((P a) ∧ todos P (x @ y))" by simp<br />
also have "… = ((P a) ∧ (todos P x ∧ todos P y))" using HI by simp<br />
also have "… = (todos P (a # x) ∧ todos P y)" by simp<br />
finally show "?Q (a # x)" by simp<br />
qed<br />
<br />
text {*<br />
--------------------------------------------------------------------- <br />
Ejercicio 5.1. Demostrar o refutar automáticamente <br />
todos P (rev xs) = todos P xs<br />
--------------------------------------------------------------------- <br />
*}<br />
<br />
lemma "todos P (rev xs) = todos P xs"<br />
oops<br />
<br />
(* migtermor ivamenjim marpoldia1 serrodcal anaprarod paupeddeg *)<br />
lemma "todos P (rev xs) = todos P xs" <br />
apply (induct xs)<br />
apply simp<br />
apply (simp add: todos_append)<br />
apply auto<br />
done<br />
<br />
(* ferrenseg crigomgom rubgonmar fraortmoy josgarsan danrodcha*)<br />
<br />
lemma "todos P (rev xs) = todos P xs"<br />
by (induct xs) (auto simp add: todos_append)<br />
<br />
(* wilmorort *)<br />
lemma "todos P (rev xs) = todos P xs"<br />
by (induct xs,simp,simp add: todos_append,auto)<br />
<br />
(* juacabsou *)<br />
lemma "todos P (rev xs) = todos P xs"<br />
apply (induct xs,simp,simp add: todos_append,auto) done<br />
<br />
(* dancorgar *)<br />
lemma "todos P (rev xs) = todos P xs"<br />
apply (induct xs)<br />
apply (simp add: todos_append)<br />
apply (simp add: todos_append)<br />
apply auto<br />
done<br />
<br />
(* pablucoto * )<br />
lemma "todos P (rev xs) = todos P xs"<br />
by (induct xs) (auto, simp_all add: todos_append) <br />
lemma "todos P (rev xs) = todos P xs"<br />
by (induct xs) ( simp_all add: todos_append, auto)<br />
<br />
text {*<br />
--------------------------------------------------------------------- <br />
Ejercicio 5.2. Demostrar o refutar detalladamente<br />
todos P (rev xs) = todos P xs<br />
--------------------------------------------------------------------- <br />
*}<br />
<br />
lemma "todos P (rev xs) = todos P xs"<br />
oops<br />
<br />
(* migtermor *)<br />
<br />
lemma auxiliar:<br />
"rev (a#xs) = rev xs @ [a]"<br />
by auto<br />
<br />
lemma "todos P (rev xs) = todos P xs" (is "?Q xs")<br />
proof (induct xs)<br />
show "?Q []" by (simp only: rev.simps(1))<br />
next <br />
fix a xs<br />
assume HI: "?Q xs"<br />
have "todos P (rev (a#xs)) = (todos P (rev xs @ [a]))" by (simp add: auxiliar)<br />
have "… = ((todos P (rev xs)) ∧ (todos P [a]))" by (simp add: todos_append)<br />
have "… = (todos P (rev xs) ∧ P a)" by simp<br />
also have Aux: "… = (todos P xs ∧ P a)" using HI by simp<br />
have Aux1: "… = (P a ∧ todos P xs)" by arith<br />
have "(todos P (rev xs) ∧ P a) = (P a ∧ todos P xs)" using Aux Aux1 by simp<br />
finally show "todos P (rev (a#xs)) = todos P (a#xs)" by (simp add: todos_append)<br />
qed<br />
<br />
<br />
(* marpoldia1 ferrenseg crigomgom serrodcal juacabsou rubgonmar josgarsan pablucoto*)<br />
lemma "todos P (rev xs) = todos P xs"<br />
proof (induct xs)<br />
show "todos P (rev []) = todos P []" by simp<br />
next<br />
fix a xs<br />
assume HI: "todos P (rev xs) = todos P xs"<br />
have "todos P (rev (a # xs)) = (todos P ((rev xs)@[a]))" by simp<br />
also have "... = (todos P (rev xs) ∧ todos P [a])" by (simp add:todos_append)<br />
also have "... = (todos P xs ∧ P a)" using HI by simp<br />
also have "... = (P a ∧ todos P xs)" by arith<br />
also have "... = (todos P (a#xs))" by simp<br />
finally show "todos P (rev (a # xs)) = (todos P (a#xs))" by simp <br />
qed<br />
<br />
(* ivamenjim *)<br />
<br />
lemma "todos P (rev xs) = todos P xs" (is "?P xs")<br />
proof (induct xs)<br />
show "?P []" by simp<br />
next <br />
fix a xs<br />
assume HI: "?P xs"<br />
have "todos P (rev (a # xs)) = (todos P ((rev xs)@[a]))" by simp<br />
also have "... = ((todos P (rev xs)) ∧ todos P [a])" by (simp add: todos_append)<br />
also have "... = (todos P xs ∧ todos P [a])" using HI by simp<br />
also have "... = (todos P [a] ∧ todos P xs)" by arith<br />
finally show "todos P (rev (a # xs)) = todos P (a # xs)" by simp<br />
qed<br />
<br />
(* fraortmoy *)<br />
<br />
lemma "todos P (rev xs) = todos P xs"<br />
proof (induct xs)<br />
show "todos P (rev []) = todos P []" by simp<br />
next<br />
fix a xs <br />
assume H1: "todos P (rev xs) = todos P xs"<br />
have " todos P (rev (a # xs)) = todos P (rev xs @ [a])" by simp<br />
also have "… = (todos P (rev xs) ∧ todos P [a])" by (simp add:todos_append)<br />
also have "… = (todos P xs ∧ todos P [a])" using H1 by simp<br />
also have "… = (todos P [a] ∧ todos P xs)" by arith<br />
finally show "todos P (rev (a # xs)) = todos P (a#xs)" by simp<br />
qed<br />
<br />
(* paupeddeg dancorgar *)<br />
lemma "todos P (rev xs) = todos P xs"<br />
proof (induct xs)<br />
show " todos P (rev []) = todos P []" by simp<br />
next<br />
fix a xs<br />
assume HI: "todos P (rev xs) = todos P xs "<br />
have "todos P (rev (a # xs)) = todos P (rev(xs) @ [a])" by simp<br />
also have "... = (todos P (rev xs) ∧ todos P [a]) " by (simp add: todos_append)<br />
also have "... = (todos P xs ∧ todos P [a])" using HI by simp<br />
also have "... = (todos P [a] ∧ todos P xs)" by arith<br />
also have "... = todos P ([a] @ xs)" by (simp add: todos_append)<br />
finally show "todos P (rev (a # xs)) = todos P (a # xs)" by simp<br />
qed <br />
<br />
(* wilmorort *)<br />
lemma "todos P (rev xs) = todos P xs" (is "?P xs")<br />
proof (induct xs) <br />
show "?P []" by simp <br />
next<br />
fix a xs<br />
assume HI: "?P xs"<br />
have "todos P (rev (a#xs)) = todos P ((rev xs) @ [a])" by simp<br />
also have "... = (todos P(rev xs) ∧ todos P [a])" by (simp add: todos_append)<br />
also have "... = (todos P xs ∧ todos P [a])" using HI by simp<br />
also have "... = (todos P [a] ∧ todos P xs)" by (simp add: HOL.conj_commute)<br />
also have "... = todos P([a]@(xs))" by (simp)<br />
finally show "todos P (rev (a#xs))= todos P (a # xs)" by simp<br />
qed<br />
<br />
(* anaprarod *) <br />
(* es igual que las anteriores pero con el final también con patrones *)<br />
<br />
lemma "todos P (rev xs) = todos P xs" (is "?P xs")<br />
proof (induct xs)<br />
show "?P []" by auto<br />
next<br />
fix a xs<br />
assume HI: "?P xs"<br />
have "todos P (rev (a#xs)) = todos P (rev xs @ [a])" by simp<br />
also have "… = (todos P (rev xs) ∧ todos P [a])" by (simp add: todos_append)<br />
also have "… = (todos P xs ∧ todos P [a])" using HI by simp<br />
also have "… = (todos P [a] ∧ todos P xs)" by arith<br />
also have "… = todos P (a#xs)" by simp<br />
finally show "?P (a # xs)" by simp<br />
qed<br />
<br />
{*danrodcha*}<br />
lemma "todos P (rev xs) = todos P xs" (is "?Q xs")<br />
proof (induct xs)<br />
show "?Q []" by simp<br />
next<br />
fix a xs assume HI: "?Q xs"<br />
have "todos P (rev (a#xs)) = todos P ((rev xs) @ [a])" by simp<br />
also have "… = (todos P (rev xs) ∧ todos P [a])" by (simp add:todos_append)<br />
also have "… = (todos P xs ∧ todos P [a])" using HI by simp<br />
also have "… = (todos P [a] ∧ todos P xs)" by (simp add: HOL.conj_commute)<br />
also have "… = todos P (a # xs)" by simp<br />
finally show "?Q (a # xs)" by simp<br />
qed<br />
<br />
text {*<br />
--------------------------------------------------------------------- <br />
Ejercicio 6. Demostrar o refutar:<br />
algunos (λx. P x ∧ Q x) xs = (algunos P xs ∧ algunos Q xs)<br />
--------------------------------------------------------------------- <br />
*}<br />
<br />
lemma "algunos (λx. P x ∧ Q x) xs = (algunos P xs ∧ algunos Q xs)"<br />
oops<br />
<br />
(* migtermor ivamenjim ferrenseg paupeddeg crigomgom serrodcal wilmorort juacabsou rubgonmar anaprarod marpoldia1 fraortmoy josgarsan danrodcha*)<br />
lemma "algunos (λx. P x ∧ Q x) xs = (algunos P xs ∧ algunos Q xs)" <br />
quickcheck<br />
oops<br />
(* Quickcheck encuentra el siguiente contraejemplo: P={a1}, Q={a2}, xs={a1,a2}. En este ejemplo:<br />
· "algunos (λx. P x ∧ Q x) xs = False"<br />
· "(algunos P xs ∧ algunos Q xs) = True" *)<br />
<br />
text {*<br />
--------------------------------------------------------------------- <br />
Ejercicio 7.1. Demostrar o refutar automáticamente <br />
algunos P (map f xs) = algunos (P ∘ f) xs<br />
--------------------------------------------------------------------- <br />
*}<br />
<br />
(* ivamenjim migtermor marpoldia1 crigomgom rubgonmar wilmorort anaprarod <br />
fraortmoy juacabsou paupeddeg josgarsan danrodcha*)<br />
<br />
lemma "algunos P (map f xs) = algunos (P o f) xs"<br />
by (induct xs) auto<br />
<br />
(* ferrenseg *)<br />
<br />
lemma "algunos P (map f xs) = algunos (P o f) xs"<br />
by (induct xs) simp_all<br />
<br />
(* anaprarod *)<br />
lemma "algunos P (map f xs) = algunos (P o f) xs"<br />
apply (induct xs)<br />
apply auto<br />
done<br />
<br />
text {*<br />
--------------------------------------------------------------------- <br />
Ejercicio 7.2. Demostrar o refutar datalladamente<br />
algunos P (map f xs) = algunos (P ∘ f) xs<br />
--------------------------------------------------------------------- <br />
*}<br />
<br />
lemma "algunos P (map f xs) = algunos (P o f) xs"<br />
oops<br />
<br />
(* migtermor *)<br />
<br />
lemma AUX: "algunos (λa. P (f a)) xs = algunos (P o f) xs"<br />
by (induct xs) auto<br />
<br />
lemma "algunos P (map f xs) = algunos (P o f) xs" (is "?Q xs")<br />
proof (induct xs)<br />
show "?Q []" by simp<br />
next<br />
fix x xs<br />
assume HI: "?Q xs"<br />
have "algunos P (map f (x#xs)) = (algunos P ((f x)#(map f xs)))" by simp<br />
also have "… = ((P (f x)) ∨ (algunos P (map f xs)))" by (simp only: algunos.simps(2))<br />
also have "… = ((P (f x)) ∨ (algunos (P o f) xs))" <br />
proof (cases)<br />
assume C1: "(P (f x))"<br />
have Aux: "((P (f x)) ∨ (algunos P (map f xs))) = True" using C1 by simp<br />
have Aux1: "… = ((P (f x)) ∨ (algunos (P o f) xs))" using C1 by simp<br />
then show "((P (f x)) ∨ (algunos P (map f xs))) = ((P (f x)) ∨ (algunos (P o f) xs))" <br />
using Aux Aux1 by simp<br />
next<br />
assume C2: "¬(P (f x))"<br />
have Aux2: "((P (f x)) ∨ (algunos P (map f xs))) = (algunos P (map f xs))" using C2 by simp<br />
have Aux3: "… = (algunos (P o f) xs)" using HI by (simp add: AUX)<br />
also have "… = ((P (f x)) ∨ (algunos (P o f) xs))" using C2 by simp<br />
then show "((P (f x)) ∨ (algunos P (map f xs))) = ((P (f x)) ∨ (algunos (P o f) xs))" <br />
using Aux2 Aux3 by simp<br />
qed<br />
also have "… = (((P o f) x) ∨ (algunos (P o f) xs))" by simp<br />
also have "… = (algunos (P o f) (x#xs))" by simp<br />
finally show "algunos P (map f (x#xs)) = (algunos (P o f) (x#xs))" by simp<br />
qed<br />
<br />
(* ferrenseg *)<br />
<br />
lemma "algunos P (map f xs) = algunos (P ∘ f) xs"<br />
proof (induct xs)<br />
show "algunos P (map f []) = algunos (P ∘ f) []" by simp<br />
next<br />
fix x xs<br />
assume HI: "algunos P (map f xs) = algunos (P ∘ f) xs"<br />
show "algunos P (map f (x # xs)) = algunos (P ∘ f) (x # xs)"<br />
proof -<br />
have "algunos P (map f (x # xs)) = algunos P ((f x) # (map f xs))" by simp<br />
also have "… = ((P (f x)) ∨ (algunos P (map f xs)))" by simp<br />
also have "… = (((P ∘ f) x) ∨ (algunos (P ∘ f) xs))" using HI by simp<br />
also have "… = algunos (P ∘ f) (x # xs)" by simp<br />
finally show ?thesis<br />
qed<br />
qed<br />
<br />
(* ivamenjim *)<br />
<br />
lemma "algunos P (map f xs) = algunos (P o f) xs"<br />
proof (induct xs)<br />
show "algunos P (map f []) = algunos (P ∘ f) []" by simp<br />
next <br />
fix a xs<br />
assume HI: "algunos P (map f xs) = algunos (P ∘ f) xs"<br />
have "algunos P (map f (a # xs)) = algunos P ((f a)#map f xs)" by simp<br />
also have "... = (algunos P (map f [a]) ∨ algunos P (map f xs))" by simp <br />
also have "... = (algunos P (map f [a]) ∨ algunos (P ∘ f) xs)" using HI by simp <br />
finally show "algunos P (map f (a # xs)) = algunos (P ∘ f) (a # xs)" by simp<br />
qed<br />
<br />
(* crigomgom rubgonmar anaprarod marpoldia1 juacabsou danrodcha paupeddeg*)<br />
lemma "algunos P (map f xs) = algunos (P o f) xs"<br />
proof (induct xs)<br />
show "algunos P (map f []) = algunos (P ∘ f) []" by simp<br />
next<br />
fix x xs<br />
assume HI: "algunos P (map f xs) = algunos (P ∘ f) xs"<br />
have "algunos P (map f (x # xs)) = algunos P ((f x) # (map f xs))" by simp<br />
also have "... = (P (f x) ∨ algunos P (map f xs))" by simp<br />
also have "... = (P (f x) ∨ algunos (P ∘ f) xs)" using HI by simp<br />
also have "... = ((P ∘ f) x ∨ algunos (P ∘ f) xs)" by simp<br />
finally show "algunos P (map f (x # xs)) = algunos (P ∘ f) (x # xs)" by simp<br />
qed<br />
<br />
(* wilmorort *)<br />
lemma "algunos P (map f xs) = algunos (P o f) xs" (is "?P xs" )<br />
proof (induct xs)<br />
show "?P []" by simp<br />
next<br />
fix a xs<br />
assume HI: "algunos P (map f xs) = algunos (P ∘ f) xs"<br />
have "algunos P (map f (a # xs)) = (P (f a) ∨ algunos P (map f xs))" by simp<br />
also have "... = (P (f a) ∨ algunos (P ∘ f) xs )" using HI by simp<br />
also have "... = ((P ∘ f) a ∨ algunos (P ∘ f)xs)" by simp<br />
finally show " algunos P (map f (a # xs)) = algunos (P ∘ f) (a # xs)" by simp<br />
qed<br />
<br />
(* fraortmoy *)<br />
<br />
lemma "algunos P (map f xs) = algunos (P o f) xs"<br />
proof (induct xs)<br />
show "algunos P (map f []) = algunos (P ∘ f) []" by simp<br />
next<br />
fix a xs<br />
assume H1: "algunos P (map f xs) = algunos (P ∘ f) xs"<br />
have "algunos P (map f (a # xs)) = (algunos P (map f [a]) ∨ algunos P (map f xs))" by simp<br />
also have "… = (algunos (P ∘ f) [a] ∨ algunos (P ∘ f) xs )" using H1 by simp<br />
also have "… = algunos (P ∘ f) (a#xs)" by simp<br />
finally show "algunos P (map f (a # xs)) = algunos (P ∘ f) (a#xs)" by simp<br />
qed<br />
<br />
{*danrodcha*}<br />
lemma "algunos P (map f xs) = algunos (P o f) xs" (is "?Q xs")<br />
proof (induct xs)<br />
show "?Q []" by simp<br />
next<br />
fix a xs assume HI: "?Q xs"<br />
have " algunos P (map f (a # xs)) = algunos P ((f a)#(map f xs))" by simp<br />
also have "… = (P (f a) ∨ algunos P (map f xs))" by simp<br />
also have "… = (P (f a) ∨ algunos (P ∘ f) xs)" using HI by blast<br />
also have "… = ((P ∘ f) a ∨ algunos (P ∘ f) xs)" by simp<br />
also have "… = algunos (P ∘ f) (a # xs)" by simp<br />
finally show "?Q (a # xs)" by blast<br />
qed<br />
<br />
text {*<br />
--------------------------------------------------------------------- <br />
Ejercicio 8.1. Demostrar o refutar automáticamente <br />
algunos P (xs @ ys) = (algunos P xs ∨ algunos P ys)<br />
--------------------------------------------------------------------- <br />
*}<br />
<br />
(* ivamenjim marpoldia1 paupeddeg crigomgom rubgonmar wilmorort fraortmoy danrodcha *)<br />
<br />
lemma "algunos P (xs @ ys) = (algunos P xs ∨ algunos P ys)"<br />
by (induct xs) auto<br />
<br />
(* ferrenseg *)<br />
lemma "algunos P (xs @ ys) = (algunos P xs ∨ algunos P ys)"<br />
by (induct xs) simp_all<br />
<br />
text {*<br />
--------------------------------------------------------------------- <br />
Ejercicio 8.2. Demostrar o refutar detalladamente<br />
algunos P (xs @ ys) = (algunos P xs ∨ algunos P ys)<br />
--------------------------------------------------------------------- <br />
*}<br />
<br />
(* ivamenjim ferrenseg marpoldia1 wilmorort*)<br />
<br />
lemma algunos_append:<br />
"algunos P (xs @ ys) = (algunos P xs ∨ algunos P ys)"<br />
proof (induct xs)<br />
show "algunos P ([] @ ys) = (algunos P [] ∨ algunos P ys)" by simp<br />
next <br />
fix a xs<br />
assume HI: "algunos P (xs @ ys) = (algunos P xs ∨ algunos P ys)"<br />
have "algunos P ((a # xs) @ ys) = (P a ∨ (algunos P (xs @ ys)))" by simp<br />
also have "... = (P a ∨ (algunos P xs ∨ algunos P ys))" using HI by simp<br />
finally show "algunos P ((a # xs) @ ys) = (algunos P (a # xs) ∨ algunos P ys)" by simp<br />
qed<br />
<br />
(* migtermor crigomgom *)<br />
<br />
lemma algunos_append2:<br />
"algunos P (xs @ ys) = (algunos P xs ∨ algunos P ys)" (is "?Q xs")<br />
proof (induct xs)<br />
show "?Q []" by simp<br />
next <br />
fix x xs<br />
assume HI: "?Q xs"<br />
have "algunos P ((x#xs) @ ys) = algunos P (x#(xs @ ys))" by simp<br />
also have "… = ((P x) ∨ (algunos P (xs @ ys)))" by simp<br />
also have "… = ((P x) ∨ (algunos P xs) ∨ (algunos P ys))" using HI by simp<br />
also have "… = ((algunos P (x#xs)) ∨ (algunos P ys))" by simp<br />
finally show "algunos P ((x#xs) @ ys) = (algunos P (x#xs) ∨ algunos P ys)" by simp<br />
qed<br />
<br />
(* paupeddeg*)<br />
lemma algunos_append3:<br />
"algunos P (xs @ ys) = (algunos P xs ∨ algunos P ys)"<br />
proof (induct xs)<br />
show "algunos P ([] @ ys) = (algunos P [] ∨ algunos P ys)" by simp<br />
next<br />
fix a xs<br />
assume HI: "algunos P (xs @ ys) = (algunos P xs ∨ algunos P ys)"<br />
have "algunos P ((a # xs) @ ys) = (algunos P [a] ∨ ( algunos P (xs @ ys)))" by simp<br />
also have "... = (algunos P [a] ∨ (algunos P xs ∨ algunos P ys))" using HI by simp<br />
finally show "algunos P ((a # xs) @ ys) = (algunos P (a # xs) ∨ algunos P ys)" by simp<br />
qed<br />
<br />
(* fraortmoy *)<br />
lemma algunos_append4:<br />
"algunos P (xs @ ys) = (algunos P xs ∨ algunos P ys)"<br />
proof (induct xs)<br />
show " algunos P ([] @ ys) = (algunos P [] ∨ algunos P ys)" by simp<br />
next <br />
fix a xs<br />
assume H1: "algunos P (xs @ ys) = (algunos P xs ∨ algunos P ys)"<br />
have "algunos P ((a # xs) @ ys) = (algunos P [a] ∨ algunos P (xs @ ys))" by simp<br />
also have "… = (algunos P [a] ∨ algunos P xs ∨ algunos P ys)" using H1 by simp<br />
also have "… = ((algunos P [a] ∨ algunos P xs) ∨ algunos P ys)" by simp<br />
also have "… = (algunos P (a#xs) ∨ algunos P ys)" by simp<br />
finally show "algunos P ((a # xs) @ ys) = (algunos P (a#xs) ∨ algunos P ys)" by simp<br />
qed<br />
<br />
{* danrodcha *}<br />
lemma algunos_append5:<br />
"algunos P (xs @ ys) = (algunos P xs ∨ algunos P ys)" (is "?Q xs")<br />
proof (induct xs)<br />
show "?Q []" by simp<br />
next<br />
fix a xs assume HI: "?Q xs"<br />
have "algunos P ((a#xs) @ ys) = algunos P (a#(xs @ ys))" by simp<br />
also have "… = (algunos P [a] ∨ algunos P (xs @ ys))" by simp<br />
also have "… = (algunos P [a] ∨ algunos P xs ∨ algunos P ys)" using HI by simp<br />
also have "… = (algunos P (a # xs) ∨ algunos P ys)" by simp<br />
finally show "?Q (a # xs)" by simp<br />
qed<br />
<br />
text {*<br />
--------------------------------------------------------------------- <br />
Ejercicio 9.1. Demostrar o refutar automáticamente<br />
algunos P (rev xs) = algunos P xs<br />
--------------------------------------------------------------------- <br />
*}<br />
<br />
(* ivamenjim migtermor marpoldia1 rubgonmar paupeddeg *)<br />
<br />
lemma "algunos P (rev xs) = algunos P xs"<br />
apply (induct xs)<br />
apply simp<br />
apply (simp add: algunos_append)<br />
apply auto<br />
done<br />
<br />
(* ferrenseg crigomgom danrodcha*)<br />
lemma "algunos P (rev xs) = algunos P xs"<br />
by (induct xs) (auto simp add: algunos_append)<br />
<br />
(* wilmorort *)<br />
lemma "algunos P (rev xs) = algunos P xs"<br />
by (induct xs,simp,simp add: algunos_append,auto) <br />
<br />
<br />
text {*<br />
--------------------------------------------------------------------- <br />
Ejercicio 9.2. Demostrar o refutar detalladamente<br />
algunos P (rev xs) = algunos P xs<br />
--------------------------------------------------------------------- <br />
*}<br />
<br />
lemma "algunos P (rev xs) = algunos P xs"<br />
oops<br />
<br />
(* migtermor *)<br />
lemma auxiliar1:<br />
"rev (a#xs) = rev xs @ [a]"<br />
by auto <br />
<br />
lemma "algunos P (rev xs) = algunos P xs" (is "?Q xs")<br />
proof (induct xs)<br />
show "?Q []" by simp<br />
next<br />
fix x xs<br />
assume HI: "?Q xs"<br />
have "algunos P (rev (x#xs)) = (algunos P (rev xs @ [x]))" using auxiliar1 by simp<br />
also have "… = ((algunos P (rev xs)) ∨ (algunos P [x]))" by (simp add: algunos_append)<br />
also have "… = ((P x) ∨ (algunos P (rev xs)))" by simp arith<br />
also have "… = ((P x) ∨ (algunos P xs))" using HI by simp<br />
finally show "algunos P (rev (x#xs)) = (algunos P (x#xs))" by simp<br />
qed<br />
<br />
(* ferrenseg *)<br />
<br />
lemma "algunos P (rev xs) = algunos P xs"<br />
proof (induct xs)<br />
show "algunos P (rev []) = algunos P []" by simp<br />
next<br />
fix x xs<br />
assume HI: "algunos P (rev xs) = algunos P xs"<br />
show "algunos P (rev (x # xs)) = algunos P (x # xs)"<br />
proof -<br />
have "algunos P (rev (x # xs)) = algunos P ((rev xs) @ [x])" by simp<br />
also have "… = (algunos P (rev xs) ∨ algunos P [x])" <br />
by (simp add: algunos_append)<br />
also have "… = (algunos P xs ∨ algunos P [x])" using HI by simp<br />
also have "… = (algunos P xs ∨ P x)" by simp<br />
also have "… = (P x ∨ algunos P xs)" by auto<br />
also have "… = algunos P (x # xs)" by simp<br />
finally show ?thesis<br />
qed<br />
qed<br />
<br />
(* ivamenjim marpoldia1*)<br />
<br />
lemma "algunos P (rev xs) = algunos P xs" (is "?P xs")<br />
proof (induct xs)<br />
show "?P []" by simp<br />
next <br />
fix a xs<br />
assume HI: "?P xs"<br />
have "algunos P (rev (a # xs)) = (algunos P ((rev xs)@[a]))" by simp<br />
also have "... = ((algunos P (rev xs)) ∨ algunos P [a])" by (simp add: algunos_append)<br />
also have "... = (algunos P xs ∨ algunos P [a])" using HI by simp<br />
also have "... = (algunos P [a] ∨ algunos P xs)" by arith<br />
finally show "algunos P (rev (a # xs)) = algunos P (a # xs)" by simp<br />
qed<br />
<br />
(*paupeddeg*)<br />
lemma "algunos P (rev xs) = algunos P xs"<br />
proof (induct xs)<br />
show "algunos P (rev []) = algunos P []" by simp<br />
next<br />
fix a xs<br />
assume HI:"algunos P (rev xs) = algunos P xs"<br />
have "algunos P (rev (a # xs)) = algunos P ((rev xs) @ [a])" by simp<br />
also have "... = ((algunos P (rev xs)) ∨ (algunos P [a]))" by (simp add: algunos_append)<br />
also have "... = ((algunos P xs) ∨ (algunos P [a]))" using HI by simp<br />
also have "... = ((algunos P [a]) ∨ (algunos P xs))" by arith<br />
finally show "algunos P (rev (a # xs)) = algunos P (a # xs)" by simp<br />
qed<br />
<br />
(*crigomgom *)<br />
lemma "algunos P (rev xs) = algunos P xs"<br />
proof (induct xs)<br />
show "algunos P (rev []) = algunos P []" by simp<br />
next<br />
fix x xs<br />
assume HI: " algunos P (rev xs) = algunos P xs" <br />
have "algunos P (rev (x # xs)) = algunos P ((rev xs) @ [x])" by simp<br />
also have "... = (algunos P (rev xs) ∨ algunos P [x])" by (simp add: algunos_append)<br />
also have "... = (algunos P xs ∨ algunos P [x])" using HI by simp<br />
also have "... = (algunos P xs ∨ P x)" by simp<br />
also have "... = (P x ∨ algunos P xs)" by arith<br />
also have "... = algunos P (x#xs)" by simp<br />
finally show "algunos P (rev (x # xs)) = algunos P (x # xs)" by simp<br />
qed<br />
<br />
<br />
(*wilmorort*)<br />
lemma "algunos P (rev xs) = algunos P xs" (is "?P xs")<br />
proof (induct xs) <br />
show "?P []" by simp <br />
next<br />
fix a xs<br />
assume HI: "?P xs"<br />
have "algunos P (rev (a#xs)) = algunos P ((rev xs) @ [a])" by simp<br />
also have "... = (algunos P(rev xs) ∨ algunos P [a])" by (simp add: algunos_append)<br />
also have "... = (algunos P xs ∨ algunos P [a])" using HI by simp<br />
also have "... = (algunos P [a] ∨ algunos P xs)" by (simp add: HOL.disj_commute)<br />
also have "... = algunos P([a]@(xs))" by (simp)<br />
finally show "algunos P (rev (a#xs))= algunos P (a # xs)" by simp<br />
qed<br />
<br />
{*danrodcha *}<br />
lemma "algunos P (rev xs) = algunos P xs" (is "?Q xs")<br />
proof (induct xs)<br />
show "?Q []" by simp<br />
next<br />
fix a xs assume HI: "?Q xs"<br />
have "algunos P (rev (a#xs)) = algunos P ((rev xs) @ [a])" by simp<br />
also have "… = (algunos P (rev xs) ∨ algunos P [a])" by (simp add:algunos_append)<br />
also have "… = (algunos P xs ∨ algunos P [a])" using HI by simp<br />
also have "… = (algunos P [a] ∨ algunos P xs)" by (simp add: HOL.disj_commute)<br />
also have "… = algunos P (a # xs)" by simp<br />
finally show "?Q (a # xs)" by simp<br />
qed<br />
<br />
text {*<br />
--------------------------------------------------------------------- <br />
Ejercicio 10. Encontrar un término no trivial Z tal que sea cierta la <br />
siguiente ecuación:<br />
algunos (λx. P x ∨ Q x) xs = Z<br />
y demostrar la equivalencia de forma automática y detallada.<br />
--------------------------------------------------------------------- <br />
*}<br />
<br />
(* migtermor *)<br />
<br />
lemma "algunos (λx. P x ∨ Q x) xs = ((algunos (λx. P x) xs) ∨ (algunos (λx. Q x) xs))"<br />
by (induct xs) auto<br />
<br />
lemma "algunos (λx. P x ∨ Q x) xs = ((algunos (λx. P x) xs) ∨ (algunos (λx. Q x) xs))" (is "?R xs")<br />
proof (induct xs)<br />
show "?R []" by simp<br />
next<br />
fix x xs<br />
assume HI: "?R xs"<br />
have "algunos (λx. P x ∨ Q x) (x#xs) = (((λx. P x ∨ Q x) x) ∨ (algunos (λx. P x ∨ Q x) xs))"<br />
by simp<br />
also have "… = ((P x ∨ Q x) ∨ (algunos (λx. P x ∨ Q x) xs))" by simp<br />
also have H1: "… = ((((λx. P x) x) ∨ (algunos (λx. P x) xs)) ∨ (((λx. Q x) x) ∨ (algunos (λx. Q x) xs)))"<br />
using HI by simp arith<br />
have H2: "… = ((algunos (λx. P x) (x#xs)) ∨ (algunos (λx. Q x) (x#xs)))" <br />
by simp<br />
have C: "(algunos (λx. P x ∨ Q x) (x#xs)) = <br />
((algunos (λx. P x) (x#xs)) ∨ (algunos (λx. Q x) (x#xs)))" <br />
using H1 H2 by simp<br />
finally show "(algunos (λx. P x ∨ Q x) (x#xs)) = <br />
((algunos (λx. P x) (x#xs)) ∨ (algunos (λx. Q x) (x#xs)))" <br />
using C by simp<br />
qed<br />
<br />
<br />
(* ferrenseg *)<br />
<br />
lemma "algunos (λx. P x ∨ Q x) xs = (algunos P xs ∨ algunos Q xs)"<br />
proof (induct xs)<br />
show "algunos (λx. P x ∨ Q x) [] = (algunos P [] ∨ algunos Q [])" by simp<br />
next<br />
fix x xs<br />
assume HI: "algunos (λx. (P x ∨ Q x)) xs = (algunos P xs ∨ algunos Q xs)"<br />
show "algunos (λx. P x ∨ Q x) (x # xs) = (algunos P (x # xs) ∨ algunos Q (x # xs))"<br />
proof -<br />
have "algunos (λx. P x ∨ Q x) (x # xs) = <br />
((P x) ∨ (Q x) ∨ algunos (λx. P x ∨ Q x) xs)" by simp<br />
also have "… = ((P x) ∨ (Q x) ∨ algunos P xs ∨ algunos Q xs)" <br />
using HI by simp<br />
also have "… = (((P x) ∨ algunos P xs) ∨ ((Q x) ∨ algunos Q xs))" by auto<br />
also have "… = (algunos P (x # xs) ∨ algunos Q (x # xs))" by simp<br />
finally show ?thesis <br />
qed<br />
qed<br />
<br />
(* ivamenjim marpoldia1 *)<br />
<br />
lemma "algunos (λx. P x ∨ Q x) xs = (algunos P xs ∨ algunos Q xs)" <br />
by (induct xs) auto<br />
<br />
lemma "algunos (λx. P x ∨ Q x) xs = (algunos P xs ∨ algunos Q xs)"<br />
proof (induct xs)<br />
show "algunos (λx. P x ∨ Q x) [] = (algunos P [] ∨ algunos Q [])" by simp<br />
next <br />
fix a xs<br />
assume HI: "algunos (λx. P x ∨ Q x) xs = (algunos P xs ∨ algunos Q xs)"<br />
have "algunos (λx. P x ∨ Q x) (a # xs) = (P a ∨ Q a ∨ algunos (λx. P x ∨ Q x) xs)" by simp<br />
also have "... = (P a ∨ Q a ∨ algunos P xs ∨ algunos Q xs)" using HI by simp<br />
finally show "algunos (λx. P x ∨ Q x) (a # xs) = (algunos P (a # xs) ∨ algunos Q (a # xs))" by auto<br />
qed<br />
<br />
<br />
(* wilmorort danrodcha*)<br />
lemma "algunos (λx. P x ∨ Q x) xs = (¬todos(λx. ¬P x)xs ∨ ¬todos(λx. ¬Q x)xs)"<br />
by (induct xs) auto<br />
<br />
(* wilmorort *)<br />
lemma "algunos (λx. P x ∨ Q x) xs = (¬todos(λx. ¬P x)xs ∨ ¬todos(λx. ¬Q x)xs)"<br />
(is "?P xs")<br />
proof (induct xs)<br />
show "?P []" by simp<br />
next<br />
fix a xs<br />
assume HI: "?P xs"<br />
have "algunos (λx. P x ∨ Q x) (a # xs) = (P a ∨ Q a ∨ algunos (λx. P x ∨ Q x) xs)" by simp<br />
also have "... = (P a ∨ Q a ∨ (¬ todos (λx. ¬ P x) xs ∨ ¬ todos (λx. ¬ Q x) xs))" using HI by simp<br />
also have "... = (P a ∨ ¬ todos (λx. ¬ P x) xs ∨ Q a ∨ ¬ todos (λx. ¬ Q x) xs )" by arith<br />
finally show "algunos (λx. P x ∨ Q x) (a # xs) = (~ todos (λx. ¬ P x) (a # xs) ∨ ~ todos (λx. ¬ Q x) (a # xs))" by simp<br />
qed<br />
<br />
{* danrodcha *}<br />
lemma "algunos (λx. P x ∨ Q x) xs = (algunos P xs ∨ algunos Q xs)" (is "?R xs")<br />
proof (induct xs)<br />
show "?R []" by simp<br />
next<br />
fix a xs assume HI: "?R xs"<br />
have 1:" (Q a ∨ algunos P xs) = (algunos P xs ∨ Q a)" by (simp add: HOL.disj_commute)<br />
have "algunos (λx. P x ∨ Q x) (a # xs) = <br />
(algunos P [a] ∨ algunos Q [a] ∨ algunos (λx. P x ∨ Q x) xs)" by simp<br />
also have "… = (P a ∨ (Q a ∨ algunos P xs) ∨ algunos Q xs)" using HI by simp<br />
also have "… = (P a ∨ algunos P xs ∨ Q a ∨ algunos Q xs)" using 1 by simp<br />
also have "… = (algunos P (a # xs) ∨ algunos Q (a # xs))" by simp<br />
finally show "?R (a # xs)" by simp<br />
qed<br />
<br />
<br />
<br />
text {*<br />
--------------------------------------------------------------------- <br />
Ejercicio 11.1. Demostrar o refutar automáticamente<br />
algunos P xs = (¬ todos (λx. (¬ P x)) xs)<br />
--------------------------------------------------------------------- <br />
*}<br />
<br />
(* ivamenjim wilmorort migtermor marpoldia1 crigomgom rubgonmar paupeddeg danrodcha*)<br />
<br />
lemma "algunos P xs = (¬ todos (λx. (¬ P x)) xs)"<br />
by (induct xs) auto<br />
<br />
(* ferrenseg *)<br />
<br />
lemma "algunos P xs = (¬ todos (λx. (¬ P x)) xs)"<br />
by (induct xs) simp_all<br />
<br />
text {*<br />
--------------------------------------------------------------------- <br />
Ejercicio 11.2. Demostrar o refutar datalladamente<br />
algunos P xs = (¬ todos (λx. (¬ P x)) xs)<br />
--------------------------------------------------------------------- <br />
*}<br />
<br />
(* ivamenjim migtermor marpoldia1 crigomgom paupeddeg*)<br />
<br />
lemma "algunos P xs = (¬ todos (λx. (¬ P x)) xs)"<br />
proof (induct xs)<br />
show "algunos P [] = (¬ todos (λx. ¬ P x) [])" by simp<br />
next <br />
fix a xs<br />
assume HI: "algunos P xs = (¬ todos (λx. ¬ P x) xs)"<br />
have "algunos P (a # xs) = (P a ∨ (algunos P xs))" by simp<br />
also have "... = (P a ∨ (¬ todos (λx. ¬ P x) xs))" using HI by simp<br />
finally show "algunos P (a # xs) = (¬ todos (λx. ¬ P x) (a # xs))" by simp<br />
qed<br />
<br />
(* ferrenseg *)<br />
<br />
lemma "algunos P xs = (¬ todos (λx. (¬ P x)) xs)"<br />
proof (induct xs)<br />
show "algunos P [] = (¬ todos (λx. (¬ P x)) [])" by simp<br />
next<br />
fix x xs<br />
assume HI: "algunos P xs = (¬ todos (λx. (¬ P x)) xs)"<br />
show "algunos P (x # xs) = (¬ todos (λx. (¬ P x)) (x # xs))"<br />
proof - <br />
have "algunos P (x # xs) = ((P x) ∨ algunos P xs)" by simp<br />
also have "… = ((P x) ∨ ¬ todos (λx. (¬ P x)) xs)" using HI by simp<br />
also have "… = (¬ (¬ (P x) ∧ todos (λx. (¬ P x)) xs))" by simp<br />
also have "… = (¬ todos (λx. (¬ P x)) (x # xs))" by simp<br />
finally show ?thesis<br />
qed<br />
qed<br />
<br />
{* danrodcha *}<br />
lemma "algunos P xs = (¬ todos (λx. (¬ P x)) xs)" (is "?Q xs")<br />
proof (induct xs)<br />
show "?Q []" by simp<br />
next<br />
fix a xs assume HI: "?Q xs"<br />
have "algunos P (a # xs) = (P a ∨ algunos P xs)" by simp<br />
also have "… = (P a ∨ (¬ todos (λx. (¬ P x)) xs))" using HI by simp<br />
also have "… = (¬(¬(P a) ∧ todos(λx. (¬ P x)) xs))" by simp<br />
also have "… = (¬ todos(λx. (¬ P x)) (a # xs))" by simp<br />
finally show "?Q (a # xs)" by simp<br />
qed<br />
<br />
text {*<br />
--------------------------------------------------------------------- <br />
Ejercicio 12. Definir la funcion primitiva recursiva <br />
estaEn :: 'a ⇒ 'a list ⇒ bool<br />
tal que (estaEn x xs) se verifica si el elemento x está en la lista<br />
xs. Por ejemplo, <br />
estaEn (2::nat) [3,2,4] = True<br />
estaEn (1::nat) [3,2,4] = False<br />
--------------------------------------------------------------------- <br />
*}<br />
<br />
(* ivamenjim ferrenseg migtermor serrodcal crigomgom rubgonmar marpoldia1 paupeddeg danrodcha*)<br />
<br />
fun estaEn :: "'a ⇒ 'a list ⇒ bool" where<br />
"estaEn x [] = False"<br />
| "estaEn x (a#xs) = ((a = x) ∨ (estaEn x xs))"<br />
<br />
value "estaEn (2::nat) [3,2,4] = True"<br />
value "estaEn (1::nat) [3,2,4] = False"<br />
<br />
<br />
text {*<br />
--------------------------------------------------------------------- <br />
Ejercicio 13. Expresar la relación existente entre estaEn y algunos. <br />
Demostrar dicha relación de forma automática y detallada.<br />
--------------------------------------------------------------------- <br />
*}<br />
<br />
(* migtermor crigomgom *)<br />
<br />
lemma "estaEn x xs = (algunos (λa. a=x) xs)"<br />
by (induct xs) auto<br />
<br />
lemma auxiliar13:<br />
"(x=a) = ((λa. a=x) a)"<br />
by auto<br />
<br />
lemma "estaEn x xs = (algunos (λa. a=x) xs)" (is "?P xs")<br />
proof (induct xs)<br />
show "?P []" by simp<br />
next<br />
fix a xs<br />
assume HI: "?P xs"<br />
have "estaEn x (a#xs) = ((a=x) ∨ (estaEn x xs))" by simp<br />
also have H: "… = ((a=x) ∨ (algunos (λa. a=x) xs))" using HI by simp<br />
also have "… = (((λa. a=x) a) ∨ (algunos (λa. a=x) xs))" using auxiliar13 by simp<br />
also have "… = (algunos (λa. a=x) (a#xs))" by simp<br />
have C: "estaEn x (a#xs) = (algunos (λa. a=x) (a#xs))" using H by simp<br />
finally show "estaEn x (a#xs) = (algunos (λa. a=x) (a#xs))" using C by simp<br />
qed<br />
<br />
<br />
(* ferrenseg *)<br />
<br />
lemma estaEn_algunos:<br />
"estaEn y xs = algunos (λx. x=y) xs"<br />
by (induct xs) auto<br />
<br />
lemma estaEn_algunos_2:<br />
"estaEn y xs = algunos (λx. x=y) xs"<br />
proof (induct xs)<br />
show "estaEn y [] = algunos (λx. x=y) []" by simp<br />
next<br />
fix x xs<br />
assume HI: "estaEn y xs = algunos (λx. x=y) xs"<br />
show "estaEn y (x # xs) = algunos (λx. x=y) (x # xs)"<br />
proof -<br />
have "estaEn y (x # xs) = (y=x ∨ estaEn y xs)" by simp<br />
also have "… = (y=x ∨ algunos (λx. x=y) xs)" using HI by simp<br />
also have "… = (x=y ∨ algunos (λx. x=y) xs)" by auto<br />
also have "… = algunos (λx. x=y) (x # xs)" by simp<br />
finally show ?thesis<br />
qed<br />
qed<br />
<br />
(*crigomgom paupeddeg*)<br />
<br />
lemma "estaEn a xs = algunos (λx. x = a) xs"<br />
by (induct xs) auto<br />
<br />
lemma "estaEn x xs = (algunos (λa. a=x) xs)" <br />
proof (induct xs)<br />
show " estaEn x [] = algunos (λa. a = x) []" by simp<br />
next<br />
fix a xs<br />
assume HI: "estaEn x xs = algunos (λa. a = x) xs"<br />
have "estaEn x (a # xs) = (a = x ∨ estaEn x xs)" by simp<br />
also have "... = (a = x ∨ algunos (λa. a = x) xs)" using HI by simp<br />
finally show " estaEn x (a # xs) = algunos (λa. a = x) (a # xs)" by simp<br />
qed<br />
<br />
(* ivamenjim marpoldia1 *)<br />
<br />
lemma "estaEn x xs = algunos (λy. x=y) xs"<br />
by (induct xs) auto<br />
<br />
lemma "estaEn x xs = algunos (λy. x=y) xs"<br />
proof (induct xs)<br />
show "estaEn x [] = algunos (op = x) []" by simp<br />
next <br />
fix a xs<br />
assume HI: "estaEn x xs = algunos (op = x) xs"<br />
have "estaEn x (a # xs) = ((a = x) ∨ (estaEn x xs))" by simp<br />
also have "... = (a = x ∨ algunos (op = x) xs)" using HI by simp<br />
finally show " estaEn x (a # xs) = algunos (op = x) (a # xs)" by auto<br />
qed<br />
<br />
{* danrodcha *}<br />
lemma "estaEn a xs = algunos (op = a) xs"<br />
by (induct xs) auto<br />
<br />
{* danrodcha *}<br />
lemma "estaEn a xs = algunos (op = a) xs"<br />
proof (induct xs)<br />
show "estaEn a [] = algunos (op = a) []" by simp<br />
next<br />
fix x xs <br />
assume HI: "estaEn a xs = algunos (op = a) xs"<br />
have "estaEn a (x # xs) = ((x = a) ∨ (estaEn a xs))" by simp<br />
also have "… = (x = a ∨ algunos (op = a) xs)" using HI by simp<br />
also have "… = ((op = a) x ∨ algunos (op = a) xs)" by (simp add:HOL.eq_commute)<br />
also have "… = algunos (op = a) (x # xs)" by simp<br />
finally show "estaEn a (x # xs) = algunos (op = a) (x # xs)" by simp<br />
qed<br />
<br />
end<br />
</source></div>Lucnovdoshttps://www.glc.us.es/~jalonso/RA2016/index.php?title=Relaci%C3%B3n_4&diff=606Relación 42016-11-22T22:56:27Z<p>Lucnovdos: </p>
<hr />
<div><source lang="isar"><br />
chapter {* R4: Cuantificadores sobre listas *}<br />
<br />
theory R4_Cuantificadores_sobre_listas<br />
imports Main <br />
begin<br />
<br />
text {* <br />
--------------------------------------------------------------------- <br />
Ejercicio 1. Definir la función <br />
todos :: ('a ⇒ bool) ⇒ 'a list ⇒ bool<br />
tal que (todos p xs) se verifica si todos los elementos de la lista <br />
xs cumplen la propiedad p. Por ejemplo, se verifica <br />
todos (λx. 1<length x) [[2,1,4],[1,3]]<br />
¬todos (λx. 1<length x) [[2,1,4],[3]]<br />
<br />
Nota: La función todos es equivalente a la predefinida list_all. <br />
--------------------------------------------------------------------- <br />
*}<br />
<br />
{*danrodcha*}<br />
fun todos :: "('a ⇒ bool) ⇒ 'a list ⇒ bool" where<br />
"todos p xs = "todos p [] = True"<br />
| "todos p (x#xs) = (p x ∧ todos p xs)""<br />
<br />
(* ivamenjim migtermor dancorgar wilmorort marpoldia1 ferrenseg paupeddeg pablucoto crigomgom anaprarod serrodcal juacabsou rubgonmar josgarsan fraortmoy lucnovdos*)<br />
fun todos :: "('a ⇒ bool) ⇒ 'a list ⇒ bool" where<br />
"todos p [] = True"<br />
| "todos p (x#xs) = (p x ∧ todos p xs)"<br />
<br />
text {* <br />
--------------------------------------------------------------------- <br />
Ejercicio 2. Definir la función <br />
algunos :: ('a ⇒ bool) ⇒ 'a list ⇒ bool<br />
tal que (algunos p xs) se verifica si algunos elementos de la lista <br />
xs cumplen la propiedad p. Por ejemplo, se verifica <br />
algunos (λx. 1<length x) [[2,1,4],[3]]<br />
¬algunos (λx. 1<length x) [[],[3]]"<br />
<br />
Nota: La función algunos es equivalente a la predefinida list_ex. <br />
--------------------------------------------------------------------- <br />
*}<br />
<br />
{*danrodcha ivamenjim migtermor dancorgar marpoldia1 ferrenseg wilmorort paupeddeg pablucoto crigomgom anaprarod serrodcal juacabsou rubgonmar josgarsan fraortmoy lucnovdos*}<br />
fun algunos :: "('a ⇒ bool) ⇒ 'a list ⇒ bool" where<br />
"algunos p [] = False"<br />
| "algunos p (x#xs) = (p x ∨ algunos p xs)"<br />
<br />
text {*<br />
--------------------------------------------------------------------- <br />
Ejercicio 3.1. Demostrar o refutar automáticamente <br />
todos (λx. P x ∧ Q x) xs = (todos P xs ∧ todos Q xs)<br />
--------------------------------------------------------------------- <br />
*}<br />
<br />
{*danrodcha ivamenjim migtermor dancorgar marpoldia1 ferrenseg wilmorort paupeddeg pablucoto anaprarod serrodcal juacabsou rubgonmar josgarsan fraortmoy lucnovdos*}<br />
lemma "todos (λx. P x ∧ Q x) xs = (todos P xs ∧ todos Q xs)"<br />
by (induct xs) auto<br />
<br />
text {*<br />
--------------------------------------------------------------------- <br />
Ejercicio 3.2. Demostrar o refutar detalladamente<br />
todos (λx. P x ∧ Q x) xs = (todos P xs ∧ todos Q xs)<br />
--------------------------------------------------------------------- <br />
*}<br />
<br />
{*danrodcha*}<br />
lemma "todos (λx. P x ∧ Q x) xs = (todos P xs ∧ todos Q xs)"<br />
proof (induct xs)<br />
show "?R []" by simp<br />
next<br />
fix a xs <br />
assume HI: "?R xs"<br />
have "todos (λx. P x ∧ Q x) (a#xs) = (P a ∧ Q a ∧ todos (λx. P x ∧ Q x) xs)" by simp<br />
also have "… = (P a ∧ Q a ∧ (todos P xs ∧ todos Q xs))" using HI by simp<br />
also have "… = ((P a ∧ todos P xs) ∧ (Q a ∧ todos Q xs))" by blast<br />
also have "… = (todos P (a#xs) ∧ todos Q (a#xs))" by simp<br />
finally show "?R (a#xs)" by simp<br />
qed<br />
<br />
(* ivamenjim wilmorort serrodcal josgarsan*)<br />
<br />
lemma "todos (λx. P x ∧ Q x) xs = (todos P xs ∧ todos Q xs)"<br />
proof (induct xs)<br />
show "todos (λx. P x ∧ Q x) [] = (todos P [] ∧ todos Q [])" by simp<br />
next <br />
fix a xs<br />
assume HI: "todos (λx. P x ∧ Q x) xs = (todos P xs ∧ todos Q xs)"<br />
have "todos (λx. P x ∧ Q x) (a # xs) = (P a ∧ Q a ∧ todos (λx. P x ∧ Q x) xs)" by simp<br />
also have "... = (P a ∧ Q a ∧ todos P xs ∧ todos Q xs)" using HI by simp<br />
finally show "todos (λx. P x ∧ Q x) (a # xs) = (todos P (a # xs) ∧ todos Q (a # xs))" by auto<br />
qed<br />
<br />
(* dancorgar paupeddeg *)<br />
lemma "todos (λx. P x ∧ Q x) xs = (todos P xs ∧ todos Q xs)"<br />
proof (induct xs)<br />
show "todos (λx. P x ∧ Q x) [] = (todos P [] ∧ todos Q [])" by simp<br />
next<br />
fix y xs<br />
assume HI: "todos (λx. P x ∧ Q x) xs = (todos P xs ∧ todos Q xs)"<br />
have "todos (λx. P x ∧ Q x) (y#xs) = ((P y ∧ Q y) ∧ (todos (λx. P x ∧ Q x) xs))" by simp<br />
also have "… = ((P y ∧ Q y) ∧ (todos P xs ∧ todos Q xs))" using HI by simp<br />
also have "… = ((P y ∧ todos P xs) ∧ (Q y ∧ todos Q xs))" by blast<br />
also have "… = ((todos P (y#xs)) ∧ (todos Q (y#xs)))" by simp<br />
finally show "todos (λx. P x ∧ Q x) (y#xs) = (todos P (y#xs) ∧ todos Q (y#xs))" by simp<br />
qed<br />
<br />
<br />
(* migtermor *)<br />
lemma "todos (λx. P x ∧ Q x) xs = (todos P xs ∧ todos Q xs)"<br />
proof (induct xs)<br />
show "todos (λx. P x ∧ Q x) [] = (todos P [] ∧ todos Q [])" by simp<br />
next <br />
fix x xs<br />
assume HI: "todos (λx. P x ∧ Q x) xs = (todos P xs ∧ todos Q xs)"<br />
have "(todos (λx. P x ∧ Q x) (x#xs)) = (( P x ∧ Q x) ∧ (todos (λx. P x ∧ Q x) xs))"<br />
by (simp only: todos.simps(2))<br />
also have "… = ((P x ∧ Q x) ∧ (todos P xs ∧ todos Q xs))" using HI by simp<br />
also have "… = (((P x)∧(todos P xs)) ∧ ((Q x) ∧ (todos Q xs)))" by arith<br />
also have "((P x ∧ Q x) ∧ (todos P xs ∧ todos Q xs)) = (((P x)∧(todos P xs)) ∧ ((Q x) ∧ (todos Q xs)))"<br />
by arith (* Este paso es exactamente el mismo que el anterior, pero sin cualquiera de los dos no funciona el "finally show" *)<br />
have "… = (((P x)∧(todos P xs))∧((Q x)∧(todos Q xs)))" by simp<br />
have "… = ((todos P (x#xs))∧(todos Q (x#xs)))" by simp<br />
finally show "(todos (λx. P x ∧ Q x) (x#xs)) = ((todos P (x#xs)) ∧ (todos Q (x#xs)))" by simp <br />
qed<br />
<br />
(* marpoldia1 *)<br />
lemma "todos (λx. P x ∧ Q x) xs = (todos P xs ∧ todos Q xs)"<br />
proof (induct xs)<br />
show "todos (λx. P x ∧ Q x) [] = (todos P [] ∧ todos Q [])" by simp<br />
next<br />
fix a xs<br />
assume HI: "todos (λx. P x ∧ Q x) xs = (todos P xs ∧ todos Q xs)" <br />
have "todos (λx. P x ∧ Q x) (a # xs) = ((P a ∧ Q a) ∧ (todos P xs ∧ todos Q xs))" using HI by simp<br />
also have "... = ((P a ∧ todos P xs) ∧ (Q a ∧ todos Q xs))" by blast<br />
also have "... = (todos P (a#xs) ∧ todos Q (a#xs)) " by simp<br />
finally show "todos (λx. P x ∧ Q x) (a#xs) = (todos P (a#xs) ∧ todos Q (a#xs))" by simp <br />
qed<br />
<br />
(* ferrenseg *)<br />
<br />
lemma "todos (λx. P x ∧ Q x) xs = (todos P xs ∧ todos Q xs)"<br />
proof (induct xs)<br />
show "todos (λx. P x ∧ Q x) [] = (todos P [] ∧ todos Q [])" by simp<br />
next <br />
fix n xs<br />
assume HI: "todos (λx. P x ∧ Q x) xs = (todos P xs ∧ todos Q xs)"<br />
have "todos (λx. P x ∧ Q x) (n # xs) = (P n ∧ Q n ∧ todos P xs ∧ todos Q xs)" using HI by simp<br />
also have "… = P n ∧ Q n ∧ todos P xs ∧ todos Q xs" by blast<br />
also have "⋯ = todos P (n # xs) ∧ todos Q (n # xs)" by simp <br />
finally show "todos (λx. P x ∧ Q x) (n # xs) = todos P (n # xs) ∧ todos Q (n # xs)" by simp<br />
qed<br />
<br />
(* lucnovdos *)<br />
<br />
lemma "todos (λx. P x ∧ Q x) xs = (todos P xs ∧ todos Q xs)"<br />
proof (induct xs)<br />
show "todos (λx. P x ∧ Q x) [] = (todos P [] ∧ todos Q [])" by simp<br />
next<br />
fix n xs<br />
assume HI: "todos (λx. P x ∧ Q x) xs = (todos P xs ∧ todos Q xs)" <br />
have "todos (λx. P x ∧ Q x) (n # xs) = ((P n ∧ Q n) ∧ (todos P xs ∧ todos Q xs))" using HI by simp<br />
also have "... = ((P n ∧ todos P xs) ∧ (Q n ∧ todos Q xs))" by arith<br />
also have "... = ((todos P(n#xs)) ∧ (todos Q(n#xs)))" by simp<br />
<br />
finally show "todos (λx. P x ∧ Q x) (n#xs) = (todos P (n#xs) ∧ todos Q (n#xs))" by simp <br />
qed<br />
<br />
{* wilmorort pablucoto crigomgom anaprarod juacabsou rubgonmar fraortmoy *}<br />
<br />
lemma "todos (λx. P x ∧ Q x) xs = (todos P xs ∧ todos Q xs)"<br />
proof (induct xs)<br />
show "todos (λx. P x ∧ Q x) [] = (todos P [] ∧ todos Q [])" by simp <br />
next<br />
fix a xs<br />
assume HI: "todos (λx. P x ∧ Q x) xs = (todos P xs ∧ todos Q xs)"<br />
have "todos (λx. P x ∧ Q x) (a # xs) = (P a ∧ Q a ∧ todos (λx. P x ∧ Q x) xs)" by simp<br />
also have "... = (P a ∧ Q a ∧ todos P xs ∧ todos Q xs)" using HI by simp<br />
also have "... = (P a ∧ todos P xs ∧ Q a ∧ todos Q xs)" by arith<br />
also have "... = (todos P (a # xs) ∧ todos Q (a # xs))" by simp (* Este paso se puede obviar*)<br />
finally show "todos (λx. P x ∧ Q x) (a # xs) = (todos P (a # xs) ∧ todos Q (a # xs))" by simp<br />
qed<br />
<br />
text {*<br />
--------------------------------------------------------------------- <br />
Ejercicio 4.1. Demostrar o refutar automáticamente<br />
todos P (x @ y) = (todos P x ∧ todos P y)<br />
--------------------------------------------------------------------- <br />
*}<br />
{*danrodcha ivamenjim marpoldia1 migtermor ferrenseg wilmorort paupeddeg crigomgom anaprarod serrodcal juacabsou rubgonmar pablucoto fraortmoy josgarsan*}<br />
lemma "todos P (x @ y) = (todos P x ∧ todos P y)"<br />
by (induct x) auto<br />
<br />
{* anaprarod dancorgar *}<br />
lemma "todos P (x @ y) = (todos P x ∧ todos P y)"<br />
apply (induct x)<br />
apply auto<br />
done<br />
<br />
text {*<br />
--------------------------------------------------------------------- <br />
Ejercicio 4.2. Demostrar o refutar detalladamente<br />
todos P (x @ y) = (todos P x ∧ todos P y)<br />
--------------------------------------------------------------------- <br />
*}<br />
<br />
(* ivamenjim ferrenseg wilmorort dancorgar fraortmoy josgarsan*)<br />
<br />
lemma todos_append:<br />
"todos P (x @ y) = (todos P x ∧ todos P y)"<br />
proof (induct x)<br />
show "todos P ([] @ y) = (todos P [] ∧ todos P y)" by simp<br />
next <br />
fix a x<br />
assume HI: "todos P (x @ y) = (todos P x ∧ todos P y)"<br />
have "todos P ((a#x) @ y) = (P a ∧ (todos P (x @ y)))" by simp<br />
also have "... = (P a ∧ (todos P x ∧ todos P y))" using HI by simp<br />
finally show "todos P ((a # x) @ y) = (todos P (a # x) ∧ todos P y)" by simp<br />
qed<br />
<br />
(* migtermor serrodcal *)<br />
<br />
lemma todos_append1:<br />
"todos P (x @ y) = (todos P x ∧ todos P y)" (is "?P x")<br />
proof (induct x)<br />
show "?P []" by simp<br />
next <br />
fix a x<br />
assume HI: "?P x"<br />
have "todos P ((a#x) @ y) = (P a ∧ (todos P (x @ y)))" by simp<br />
also have "… = (P a ∧ (todos P x ∧ todos P y))" using HI by simp<br />
finally show "?P (a#x)" by simp<br />
qed <br />
<br />
(* marpoldia1 *)<br />
lemma todos_append:<br />
"todos P (x @ y) = (todos P x ∧ todos P y)"<br />
proof (induct x)<br />
show "todos P ([] @ y) = (todos P [] ∧ todos P y)" by simp<br />
next<br />
fix a x<br />
assume HI: "todos P (x @ y) = (todos P x ∧ todos P y)"<br />
have "todos P ((a # x) @ y) = todos P (a#(x@y))" by simp<br />
also have "... = (P a ∧ todos P (x @ y))" by simp<br />
also have "... = (P a ∧ (todos P x ∧ todos P y))" using HI by simp<br />
finally show "todos P ((a # x) @ y) = (todos P (a # x) ∧ todos P y)" by simp <br />
qed<br />
<br />
(* paupeddeg *)<br />
lemma todos_append3:<br />
"todos P (x @ y) = (todos P x ∧ todos P y)"<br />
proof (induct x)<br />
show "todos P ([] @ y) = (todos P [] ∧ todos P y)" by simp<br />
next<br />
fix a x<br />
assume HI: "todos P (x @ y) = (todos P x ∧ todos P y)"<br />
have "todos P ((a # x) @ y) = todos P (a # (x @ y))" by simp<br />
also have "... = (todos P [a] ∧ todos P (x @ y)) " by simp<br />
also have "... = (todos P [a] ∧ (todos P x ∧ todos P y))" using HI by simp<br />
finally show "todos P ((a # x) @ y) = (todos P (a # x) ∧ todos P y)" by simp<br />
qed<br />
<br />
(* crigomgom juacabsou anaprarod*)<br />
<br />
lemma todos_append4:<br />
"todos P (x @ y) = (todos P x ∧ todos P y)"<br />
proof (induct x)<br />
show "todos P ([] @ y) = (todos P [] ∧ todos P y)" by simp<br />
next<br />
fix x xs<br />
assume HI:"todos P (xs @ y) = (todos P xs ∧ todos P y)"<br />
have "todos P ((x # xs) @ y) = (P x ∧ todos P (xs @ y ))" by simp<br />
also have "... = (P x ∧ (todos P xs ∧ todos P y)) " using HI by simp<br />
also have "... = ((P x ∧ todos P xs) ∧ todos P y)" by simp<br />
finally show "todos P ((x # xs) @ y) = (todos P (x # xs) ∧ todos P y)" by simp<br />
qed<br />
<br />
(* pablucoto *)<br />
lemma todos_append5:<br />
"todos P (x @ y) = (todos P x ∧ todos P y)"<br />
proof (induct x )<br />
show "todos P ([] @ y) = (todos P [] ∧ todos P y)" by simp<br />
next<br />
fix a x<br />
assume HI:"todos P (x @ y) = (todos P x ∧ todos P y) " <br />
have "todos P ((a # x) @ y) = ( P a ∧ todos P (x @ y))" by simp<br />
also have " ... =( P a ∧ (todos P x ∧ todos P y))" using HI by simp<br />
also have "... = (todos P (a#x) ∧ todos P y)" by simp<br />
finally show "todos P ((a # x) @ y) = (todos P (a # x) ∧ todos P y)" by auto<br />
qed<br />
<br />
{*danrodcha*}<br />
lemma todos_append6:<br />
"todos P (x @ y) = (todos P x ∧ todos P y)" (is "?Q x")<br />
proof (induct x)<br />
show "?Q []" by simp<br />
next<br />
fix x a assume HI: "?Q x"<br />
have "todos P ((a # x) @ y) = todos P (a # x @ y)" by simp<br />
also have "… = ((P a) ∧ todos P (x @ y))" by simp<br />
also have "… = ((P a) ∧ (todos P x ∧ todos P y))" using HI by simp<br />
also have "… = (todos P (a # x) ∧ todos P y)" by simp<br />
finally show "?Q (a # x)" by simp<br />
qed<br />
<br />
text {*<br />
--------------------------------------------------------------------- <br />
Ejercicio 5.1. Demostrar o refutar automáticamente <br />
todos P (rev xs) = todos P xs<br />
--------------------------------------------------------------------- <br />
*}<br />
<br />
lemma "todos P (rev xs) = todos P xs"<br />
oops<br />
<br />
(* migtermor ivamenjim marpoldia1 serrodcal anaprarod paupeddeg *)<br />
lemma "todos P (rev xs) = todos P xs" <br />
apply (induct xs)<br />
apply simp<br />
apply (simp add: todos_append)<br />
apply auto<br />
done<br />
<br />
(* ferrenseg crigomgom rubgonmar fraortmoy josgarsan danrodcha*)<br />
<br />
lemma "todos P (rev xs) = todos P xs"<br />
by (induct xs) (auto simp add: todos_append)<br />
<br />
(* wilmorort *)<br />
lemma "todos P (rev xs) = todos P xs"<br />
by (induct xs,simp,simp add: todos_append,auto)<br />
<br />
(* juacabsou *)<br />
lemma "todos P (rev xs) = todos P xs"<br />
apply (induct xs,simp,simp add: todos_append,auto) done<br />
<br />
(* dancorgar *)<br />
lemma "todos P (rev xs) = todos P xs"<br />
apply (induct xs)<br />
apply (simp add: todos_append)<br />
apply (simp add: todos_append)<br />
apply auto<br />
done<br />
<br />
(* pablucoto * )<br />
lemma "todos P (rev xs) = todos P xs"<br />
by (induct xs) (auto, simp_all add: todos_append) <br />
lemma "todos P (rev xs) = todos P xs"<br />
by (induct xs) ( simp_all add: todos_append, auto)<br />
<br />
text {*<br />
--------------------------------------------------------------------- <br />
Ejercicio 5.2. Demostrar o refutar detalladamente<br />
todos P (rev xs) = todos P xs<br />
--------------------------------------------------------------------- <br />
*}<br />
<br />
lemma "todos P (rev xs) = todos P xs"<br />
oops<br />
<br />
(* migtermor *)<br />
<br />
lemma auxiliar:<br />
"rev (a#xs) = rev xs @ [a]"<br />
by auto<br />
<br />
lemma "todos P (rev xs) = todos P xs" (is "?Q xs")<br />
proof (induct xs)<br />
show "?Q []" by (simp only: rev.simps(1))<br />
next <br />
fix a xs<br />
assume HI: "?Q xs"<br />
have "todos P (rev (a#xs)) = (todos P (rev xs @ [a]))" by (simp add: auxiliar)<br />
have "… = ((todos P (rev xs)) ∧ (todos P [a]))" by (simp add: todos_append)<br />
have "… = (todos P (rev xs) ∧ P a)" by simp<br />
also have Aux: "… = (todos P xs ∧ P a)" using HI by simp<br />
have Aux1: "… = (P a ∧ todos P xs)" by arith<br />
have "(todos P (rev xs) ∧ P a) = (P a ∧ todos P xs)" using Aux Aux1 by simp<br />
finally show "todos P (rev (a#xs)) = todos P (a#xs)" by (simp add: todos_append)<br />
qed<br />
<br />
<br />
(* marpoldia1 ferrenseg crigomgom serrodcal juacabsou rubgonmar josgarsan pablucoto*)<br />
lemma "todos P (rev xs) = todos P xs"<br />
proof (induct xs)<br />
show "todos P (rev []) = todos P []" by simp<br />
next<br />
fix a xs<br />
assume HI: "todos P (rev xs) = todos P xs"<br />
have "todos P (rev (a # xs)) = (todos P ((rev xs)@[a]))" by simp<br />
also have "... = (todos P (rev xs) ∧ todos P [a])" by (simp add:todos_append)<br />
also have "... = (todos P xs ∧ P a)" using HI by simp<br />
also have "... = (P a ∧ todos P xs)" by arith<br />
also have "... = (todos P (a#xs))" by simp<br />
finally show "todos P (rev (a # xs)) = (todos P (a#xs))" by simp <br />
qed<br />
<br />
(* ivamenjim *)<br />
<br />
lemma "todos P (rev xs) = todos P xs" (is "?P xs")<br />
proof (induct xs)<br />
show "?P []" by simp<br />
next <br />
fix a xs<br />
assume HI: "?P xs"<br />
have "todos P (rev (a # xs)) = (todos P ((rev xs)@[a]))" by simp<br />
also have "... = ((todos P (rev xs)) ∧ todos P [a])" by (simp add: todos_append)<br />
also have "... = (todos P xs ∧ todos P [a])" using HI by simp<br />
also have "... = (todos P [a] ∧ todos P xs)" by arith<br />
finally show "todos P (rev (a # xs)) = todos P (a # xs)" by simp<br />
qed<br />
<br />
(* fraortmoy *)<br />
<br />
lemma "todos P (rev xs) = todos P xs"<br />
proof (induct xs)<br />
show "todos P (rev []) = todos P []" by simp<br />
next<br />
fix a xs <br />
assume H1: "todos P (rev xs) = todos P xs"<br />
have " todos P (rev (a # xs)) = todos P (rev xs @ [a])" by simp<br />
also have "… = (todos P (rev xs) ∧ todos P [a])" by (simp add:todos_append)<br />
also have "… = (todos P xs ∧ todos P [a])" using H1 by simp<br />
also have "… = (todos P [a] ∧ todos P xs)" by arith<br />
finally show "todos P (rev (a # xs)) = todos P (a#xs)" by simp<br />
qed<br />
<br />
(* paupeddeg dancorgar *)<br />
lemma "todos P (rev xs) = todos P xs"<br />
proof (induct xs)<br />
show " todos P (rev []) = todos P []" by simp<br />
next<br />
fix a xs<br />
assume HI: "todos P (rev xs) = todos P xs "<br />
have "todos P (rev (a # xs)) = todos P (rev(xs) @ [a])" by simp<br />
also have "... = (todos P (rev xs) ∧ todos P [a]) " by (simp add: todos_append)<br />
also have "... = (todos P xs ∧ todos P [a])" using HI by simp<br />
also have "... = (todos P [a] ∧ todos P xs)" by arith<br />
also have "... = todos P ([a] @ xs)" by (simp add: todos_append)<br />
finally show "todos P (rev (a # xs)) = todos P (a # xs)" by simp<br />
qed <br />
<br />
(* wilmorort *)<br />
lemma "todos P (rev xs) = todos P xs" (is "?P xs")<br />
proof (induct xs) <br />
show "?P []" by simp <br />
next<br />
fix a xs<br />
assume HI: "?P xs"<br />
have "todos P (rev (a#xs)) = todos P ((rev xs) @ [a])" by simp<br />
also have "... = (todos P(rev xs) ∧ todos P [a])" by (simp add: todos_append)<br />
also have "... = (todos P xs ∧ todos P [a])" using HI by simp<br />
also have "... = (todos P [a] ∧ todos P xs)" by (simp add: HOL.conj_commute)<br />
also have "... = todos P([a]@(xs))" by (simp)<br />
finally show "todos P (rev (a#xs))= todos P (a # xs)" by simp<br />
qed<br />
<br />
(* anaprarod *) <br />
(* es igual que las anteriores pero con el final también con patrones *)<br />
<br />
lemma "todos P (rev xs) = todos P xs" (is "?P xs")<br />
proof (induct xs)<br />
show "?P []" by auto<br />
next<br />
fix a xs<br />
assume HI: "?P xs"<br />
have "todos P (rev (a#xs)) = todos P (rev xs @ [a])" by simp<br />
also have "… = (todos P (rev xs) ∧ todos P [a])" by (simp add: todos_append)<br />
also have "… = (todos P xs ∧ todos P [a])" using HI by simp<br />
also have "… = (todos P [a] ∧ todos P xs)" by arith<br />
also have "… = todos P (a#xs)" by simp<br />
finally show "?P (a # xs)" by simp<br />
qed<br />
<br />
{*danrodcha*}<br />
lemma "todos P (rev xs) = todos P xs" (is "?Q xs")<br />
proof (induct xs)<br />
show "?Q []" by simp<br />
next<br />
fix a xs assume HI: "?Q xs"<br />
have "todos P (rev (a#xs)) = todos P ((rev xs) @ [a])" by simp<br />
also have "… = (todos P (rev xs) ∧ todos P [a])" by (simp add:todos_append)<br />
also have "… = (todos P xs ∧ todos P [a])" using HI by simp<br />
also have "… = (todos P [a] ∧ todos P xs)" by (simp add: HOL.conj_commute)<br />
also have "… = todos P (a # xs)" by simp<br />
finally show "?Q (a # xs)" by simp<br />
qed<br />
<br />
text {*<br />
--------------------------------------------------------------------- <br />
Ejercicio 6. Demostrar o refutar:<br />
algunos (λx. P x ∧ Q x) xs = (algunos P xs ∧ algunos Q xs)<br />
--------------------------------------------------------------------- <br />
*}<br />
<br />
lemma "algunos (λx. P x ∧ Q x) xs = (algunos P xs ∧ algunos Q xs)"<br />
oops<br />
<br />
(* migtermor ivamenjim ferrenseg paupeddeg crigomgom serrodcal wilmorort juacabsou rubgonmar anaprarod marpoldia1 fraortmoy josgarsan danrodcha*)<br />
lemma "algunos (λx. P x ∧ Q x) xs = (algunos P xs ∧ algunos Q xs)" <br />
quickcheck<br />
oops<br />
(* Quickcheck encuentra el siguiente contraejemplo: P={a1}, Q={a2}, xs={a1,a2}. En este ejemplo:<br />
· "algunos (λx. P x ∧ Q x) xs = False"<br />
· "(algunos P xs ∧ algunos Q xs) = True" *)<br />
<br />
text {*<br />
--------------------------------------------------------------------- <br />
Ejercicio 7.1. Demostrar o refutar automáticamente <br />
algunos P (map f xs) = algunos (P ∘ f) xs<br />
--------------------------------------------------------------------- <br />
*}<br />
<br />
(* ivamenjim migtermor marpoldia1 crigomgom rubgonmar wilmorort anaprarod <br />
fraortmoy juacabsou paupeddeg josgarsan danrodcha*)<br />
<br />
lemma "algunos P (map f xs) = algunos (P o f) xs"<br />
by (induct xs) auto<br />
<br />
(* ferrenseg *)<br />
<br />
lemma "algunos P (map f xs) = algunos (P o f) xs"<br />
by (induct xs) simp_all<br />
<br />
(* anaprarod *)<br />
lemma "algunos P (map f xs) = algunos (P o f) xs"<br />
apply (induct xs)<br />
apply auto<br />
done<br />
<br />
text {*<br />
--------------------------------------------------------------------- <br />
Ejercicio 7.2. Demostrar o refutar datalladamente<br />
algunos P (map f xs) = algunos (P ∘ f) xs<br />
--------------------------------------------------------------------- <br />
*}<br />
<br />
lemma "algunos P (map f xs) = algunos (P o f) xs"<br />
oops<br />
<br />
(* migtermor *)<br />
<br />
lemma AUX: "algunos (λa. P (f a)) xs = algunos (P o f) xs"<br />
by (induct xs) auto<br />
<br />
lemma "algunos P (map f xs) = algunos (P o f) xs" (is "?Q xs")<br />
proof (induct xs)<br />
show "?Q []" by simp<br />
next<br />
fix x xs<br />
assume HI: "?Q xs"<br />
have "algunos P (map f (x#xs)) = (algunos P ((f x)#(map f xs)))" by simp<br />
also have "… = ((P (f x)) ∨ (algunos P (map f xs)))" by (simp only: algunos.simps(2))<br />
also have "… = ((P (f x)) ∨ (algunos (P o f) xs))" <br />
proof (cases)<br />
assume C1: "(P (f x))"<br />
have Aux: "((P (f x)) ∨ (algunos P (map f xs))) = True" using C1 by simp<br />
have Aux1: "… = ((P (f x)) ∨ (algunos (P o f) xs))" using C1 by simp<br />
then show "((P (f x)) ∨ (algunos P (map f xs))) = ((P (f x)) ∨ (algunos (P o f) xs))" <br />
using Aux Aux1 by simp<br />
next<br />
assume C2: "¬(P (f x))"<br />
have Aux2: "((P (f x)) ∨ (algunos P (map f xs))) = (algunos P (map f xs))" using C2 by simp<br />
have Aux3: "… = (algunos (P o f) xs)" using HI by (simp add: AUX)<br />
also have "… = ((P (f x)) ∨ (algunos (P o f) xs))" using C2 by simp<br />
then show "((P (f x)) ∨ (algunos P (map f xs))) = ((P (f x)) ∨ (algunos (P o f) xs))" <br />
using Aux2 Aux3 by simp<br />
qed<br />
also have "… = (((P o f) x) ∨ (algunos (P o f) xs))" by simp<br />
also have "… = (algunos (P o f) (x#xs))" by simp<br />
finally show "algunos P (map f (x#xs)) = (algunos (P o f) (x#xs))" by simp<br />
qed<br />
<br />
(* ferrenseg *)<br />
<br />
lemma "algunos P (map f xs) = algunos (P ∘ f) xs"<br />
proof (induct xs)<br />
show "algunos P (map f []) = algunos (P ∘ f) []" by simp<br />
next<br />
fix x xs<br />
assume HI: "algunos P (map f xs) = algunos (P ∘ f) xs"<br />
show "algunos P (map f (x # xs)) = algunos (P ∘ f) (x # xs)"<br />
proof -<br />
have "algunos P (map f (x # xs)) = algunos P ((f x) # (map f xs))" by simp<br />
also have "… = ((P (f x)) ∨ (algunos P (map f xs)))" by simp<br />
also have "… = (((P ∘ f) x) ∨ (algunos (P ∘ f) xs))" using HI by simp<br />
also have "… = algunos (P ∘ f) (x # xs)" by simp<br />
finally show ?thesis<br />
qed<br />
qed<br />
<br />
(* ivamenjim *)<br />
<br />
lemma "algunos P (map f xs) = algunos (P o f) xs"<br />
proof (induct xs)<br />
show "algunos P (map f []) = algunos (P ∘ f) []" by simp<br />
next <br />
fix a xs<br />
assume HI: "algunos P (map f xs) = algunos (P ∘ f) xs"<br />
have "algunos P (map f (a # xs)) = algunos P ((f a)#map f xs)" by simp<br />
also have "... = (algunos P (map f [a]) ∨ algunos P (map f xs))" by simp <br />
also have "... = (algunos P (map f [a]) ∨ algunos (P ∘ f) xs)" using HI by simp <br />
finally show "algunos P (map f (a # xs)) = algunos (P ∘ f) (a # xs)" by simp<br />
qed<br />
<br />
(* crigomgom rubgonmar anaprarod marpoldia1 juacabsou danrodcha paupeddeg*)<br />
lemma "algunos P (map f xs) = algunos (P o f) xs"<br />
proof (induct xs)<br />
show "algunos P (map f []) = algunos (P ∘ f) []" by simp<br />
next<br />
fix x xs<br />
assume HI: "algunos P (map f xs) = algunos (P ∘ f) xs"<br />
have "algunos P (map f (x # xs)) = algunos P ((f x) # (map f xs))" by simp<br />
also have "... = (P (f x) ∨ algunos P (map f xs))" by simp<br />
also have "... = (P (f x) ∨ algunos (P ∘ f) xs)" using HI by simp<br />
also have "... = ((P ∘ f) x ∨ algunos (P ∘ f) xs)" by simp<br />
finally show "algunos P (map f (x # xs)) = algunos (P ∘ f) (x # xs)" by simp<br />
qed<br />
<br />
(* wilmorort *)<br />
lemma "algunos P (map f xs) = algunos (P o f) xs" (is "?P xs" )<br />
proof (induct xs)<br />
show "?P []" by simp<br />
next<br />
fix a xs<br />
assume HI: "algunos P (map f xs) = algunos (P ∘ f) xs"<br />
have "algunos P (map f (a # xs)) = (P (f a) ∨ algunos P (map f xs))" by simp<br />
also have "... = (P (f a) ∨ algunos (P ∘ f) xs )" using HI by simp<br />
also have "... = ((P ∘ f) a ∨ algunos (P ∘ f)xs)" by simp<br />
finally show " algunos P (map f (a # xs)) = algunos (P ∘ f) (a # xs)" by simp<br />
qed<br />
<br />
(* fraortmoy *)<br />
<br />
lemma "algunos P (map f xs) = algunos (P o f) xs"<br />
proof (induct xs)<br />
show "algunos P (map f []) = algunos (P ∘ f) []" by simp<br />
next<br />
fix a xs<br />
assume H1: "algunos P (map f xs) = algunos (P ∘ f) xs"<br />
have "algunos P (map f (a # xs)) = (algunos P (map f [a]) ∨ algunos P (map f xs))" by simp<br />
also have "… = (algunos (P ∘ f) [a] ∨ algunos (P ∘ f) xs )" using H1 by simp<br />
also have "… = algunos (P ∘ f) (a#xs)" by simp<br />
finally show "algunos P (map f (a # xs)) = algunos (P ∘ f) (a#xs)" by simp<br />
qed<br />
<br />
{*danrodcha*}<br />
lemma "algunos P (map f xs) = algunos (P o f) xs" (is "?Q xs")<br />
proof (induct xs)<br />
show "?Q []" by simp<br />
next<br />
fix a xs assume HI: "?Q xs"<br />
have " algunos P (map f (a # xs)) = algunos P ((f a)#(map f xs))" by simp<br />
also have "… = (P (f a) ∨ algunos P (map f xs))" by simp<br />
also have "… = (P (f a) ∨ algunos (P ∘ f) xs)" using HI by blast<br />
also have "… = ((P ∘ f) a ∨ algunos (P ∘ f) xs)" by simp<br />
also have "… = algunos (P ∘ f) (a # xs)" by simp<br />
finally show "?Q (a # xs)" by blast<br />
qed<br />
<br />
text {*<br />
--------------------------------------------------------------------- <br />
Ejercicio 8.1. Demostrar o refutar automáticamente <br />
algunos P (xs @ ys) = (algunos P xs ∨ algunos P ys)<br />
--------------------------------------------------------------------- <br />
*}<br />
<br />
(* ivamenjim marpoldia1 paupeddeg crigomgom rubgonmar wilmorort fraortmoy danrodcha *)<br />
<br />
lemma "algunos P (xs @ ys) = (algunos P xs ∨ algunos P ys)"<br />
by (induct xs) auto<br />
<br />
(* ferrenseg *)<br />
lemma "algunos P (xs @ ys) = (algunos P xs ∨ algunos P ys)"<br />
by (induct xs) simp_all<br />
<br />
text {*<br />
--------------------------------------------------------------------- <br />
Ejercicio 8.2. Demostrar o refutar detalladamente<br />
algunos P (xs @ ys) = (algunos P xs ∨ algunos P ys)<br />
--------------------------------------------------------------------- <br />
*}<br />
<br />
(* ivamenjim ferrenseg marpoldia1 wilmorort*)<br />
<br />
lemma algunos_append:<br />
"algunos P (xs @ ys) = (algunos P xs ∨ algunos P ys)"<br />
proof (induct xs)<br />
show "algunos P ([] @ ys) = (algunos P [] ∨ algunos P ys)" by simp<br />
next <br />
fix a xs<br />
assume HI: "algunos P (xs @ ys) = (algunos P xs ∨ algunos P ys)"<br />
have "algunos P ((a # xs) @ ys) = (P a ∨ (algunos P (xs @ ys)))" by simp<br />
also have "... = (P a ∨ (algunos P xs ∨ algunos P ys))" using HI by simp<br />
finally show "algunos P ((a # xs) @ ys) = (algunos P (a # xs) ∨ algunos P ys)" by simp<br />
qed<br />
<br />
(* migtermor crigomgom *)<br />
<br />
lemma algunos_append2:<br />
"algunos P (xs @ ys) = (algunos P xs ∨ algunos P ys)" (is "?Q xs")<br />
proof (induct xs)<br />
show "?Q []" by simp<br />
next <br />
fix x xs<br />
assume HI: "?Q xs"<br />
have "algunos P ((x#xs) @ ys) = algunos P (x#(xs @ ys))" by simp<br />
also have "… = ((P x) ∨ (algunos P (xs @ ys)))" by simp<br />
also have "… = ((P x) ∨ (algunos P xs) ∨ (algunos P ys))" using HI by simp<br />
also have "… = ((algunos P (x#xs)) ∨ (algunos P ys))" by simp<br />
finally show "algunos P ((x#xs) @ ys) = (algunos P (x#xs) ∨ algunos P ys)" by simp<br />
qed<br />
<br />
(* paupeddeg*)<br />
lemma algunos_append3:<br />
"algunos P (xs @ ys) = (algunos P xs ∨ algunos P ys)"<br />
proof (induct xs)<br />
show "algunos P ([] @ ys) = (algunos P [] ∨ algunos P ys)" by simp<br />
next<br />
fix a xs<br />
assume HI: "algunos P (xs @ ys) = (algunos P xs ∨ algunos P ys)"<br />
have "algunos P ((a # xs) @ ys) = (algunos P [a] ∨ ( algunos P (xs @ ys)))" by simp<br />
also have "... = (algunos P [a] ∨ (algunos P xs ∨ algunos P ys))" using HI by simp<br />
finally show "algunos P ((a # xs) @ ys) = (algunos P (a # xs) ∨ algunos P ys)" by simp<br />
qed<br />
<br />
(* fraortmoy *)<br />
lemma algunos_append4:<br />
"algunos P (xs @ ys) = (algunos P xs ∨ algunos P ys)"<br />
proof (induct xs)<br />
show " algunos P ([] @ ys) = (algunos P [] ∨ algunos P ys)" by simp<br />
next <br />
fix a xs<br />
assume H1: "algunos P (xs @ ys) = (algunos P xs ∨ algunos P ys)"<br />
have "algunos P ((a # xs) @ ys) = (algunos P [a] ∨ algunos P (xs @ ys))" by simp<br />
also have "… = (algunos P [a] ∨ algunos P xs ∨ algunos P ys)" using H1 by simp<br />
also have "… = ((algunos P [a] ∨ algunos P xs) ∨ algunos P ys)" by simp<br />
also have "… = (algunos P (a#xs) ∨ algunos P ys)" by simp<br />
finally show "algunos P ((a # xs) @ ys) = (algunos P (a#xs) ∨ algunos P ys)" by simp<br />
qed<br />
<br />
{* danrodcha *}<br />
lemma algunos_append5:<br />
"algunos P (xs @ ys) = (algunos P xs ∨ algunos P ys)" (is "?Q xs")<br />
proof (induct xs)<br />
show "?Q []" by simp<br />
next<br />
fix a xs assume HI: "?Q xs"<br />
have "algunos P ((a#xs) @ ys) = algunos P (a#(xs @ ys))" by simp<br />
also have "… = (algunos P [a] ∨ algunos P (xs @ ys))" by simp<br />
also have "… = (algunos P [a] ∨ algunos P xs ∨ algunos P ys)" using HI by simp<br />
also have "… = (algunos P (a # xs) ∨ algunos P ys)" by simp<br />
finally show "?Q (a # xs)" by simp<br />
qed<br />
<br />
text {*<br />
--------------------------------------------------------------------- <br />
Ejercicio 9.1. Demostrar o refutar automáticamente<br />
algunos P (rev xs) = algunos P xs<br />
--------------------------------------------------------------------- <br />
*}<br />
<br />
(* ivamenjim migtermor marpoldia1 rubgonmar paupeddeg *)<br />
<br />
lemma "algunos P (rev xs) = algunos P xs"<br />
apply (induct xs)<br />
apply simp<br />
apply (simp add: algunos_append)<br />
apply auto<br />
done<br />
<br />
(* ferrenseg crigomgom danrodcha*)<br />
lemma "algunos P (rev xs) = algunos P xs"<br />
by (induct xs) (auto simp add: algunos_append)<br />
<br />
(* wilmorort *)<br />
lemma "algunos P (rev xs) = algunos P xs"<br />
by (induct xs,simp,simp add: algunos_append,auto) <br />
<br />
<br />
text {*<br />
--------------------------------------------------------------------- <br />
Ejercicio 9.2. Demostrar o refutar detalladamente<br />
algunos P (rev xs) = algunos P xs<br />
--------------------------------------------------------------------- <br />
*}<br />
<br />
lemma "algunos P (rev xs) = algunos P xs"<br />
oops<br />
<br />
(* migtermor *)<br />
lemma auxiliar1:<br />
"rev (a#xs) = rev xs @ [a]"<br />
by auto <br />
<br />
lemma "algunos P (rev xs) = algunos P xs" (is "?Q xs")<br />
proof (induct xs)<br />
show "?Q []" by simp<br />
next<br />
fix x xs<br />
assume HI: "?Q xs"<br />
have "algunos P (rev (x#xs)) = (algunos P (rev xs @ [x]))" using auxiliar1 by simp<br />
also have "… = ((algunos P (rev xs)) ∨ (algunos P [x]))" by (simp add: algunos_append)<br />
also have "… = ((P x) ∨ (algunos P (rev xs)))" by simp arith<br />
also have "… = ((P x) ∨ (algunos P xs))" using HI by simp<br />
finally show "algunos P (rev (x#xs)) = (algunos P (x#xs))" by simp<br />
qed<br />
<br />
(* ferrenseg *)<br />
<br />
lemma "algunos P (rev xs) = algunos P xs"<br />
proof (induct xs)<br />
show "algunos P (rev []) = algunos P []" by simp<br />
next<br />
fix x xs<br />
assume HI: "algunos P (rev xs) = algunos P xs"<br />
show "algunos P (rev (x # xs)) = algunos P (x # xs)"<br />
proof -<br />
have "algunos P (rev (x # xs)) = algunos P ((rev xs) @ [x])" by simp<br />
also have "… = (algunos P (rev xs) ∨ algunos P [x])" <br />
by (simp add: algunos_append)<br />
also have "… = (algunos P xs ∨ algunos P [x])" using HI by simp<br />
also have "… = (algunos P xs ∨ P x)" by simp<br />
also have "… = (P x ∨ algunos P xs)" by auto<br />
also have "… = algunos P (x # xs)" by simp<br />
finally show ?thesis<br />
qed<br />
qed<br />
<br />
(* ivamenjim marpoldia1*)<br />
<br />
lemma "algunos P (rev xs) = algunos P xs" (is "?P xs")<br />
proof (induct xs)<br />
show "?P []" by simp<br />
next <br />
fix a xs<br />
assume HI: "?P xs"<br />
have "algunos P (rev (a # xs)) = (algunos P ((rev xs)@[a]))" by simp<br />
also have "... = ((algunos P (rev xs)) ∨ algunos P [a])" by (simp add: algunos_append)<br />
also have "... = (algunos P xs ∨ algunos P [a])" using HI by simp<br />
also have "... = (algunos P [a] ∨ algunos P xs)" by arith<br />
finally show "algunos P (rev (a # xs)) = algunos P (a # xs)" by simp<br />
qed<br />
<br />
(*paupeddeg*)<br />
lemma "algunos P (rev xs) = algunos P xs"<br />
proof (induct xs)<br />
show "algunos P (rev []) = algunos P []" by simp<br />
next<br />
fix a xs<br />
assume HI:"algunos P (rev xs) = algunos P xs"<br />
have "algunos P (rev (a # xs)) = algunos P ((rev xs) @ [a])" by simp<br />
also have "... = ((algunos P (rev xs)) ∨ (algunos P [a]))" by (simp add: algunos_append)<br />
also have "... = ((algunos P xs) ∨ (algunos P [a]))" using HI by simp<br />
also have "... = ((algunos P [a]) ∨ (algunos P xs))" by arith<br />
finally show "algunos P (rev (a # xs)) = algunos P (a # xs)" by simp<br />
qed<br />
<br />
(*crigomgom *)<br />
lemma "algunos P (rev xs) = algunos P xs"<br />
proof (induct xs)<br />
show "algunos P (rev []) = algunos P []" by simp<br />
next<br />
fix x xs<br />
assume HI: " algunos P (rev xs) = algunos P xs" <br />
have "algunos P (rev (x # xs)) = algunos P ((rev xs) @ [x])" by simp<br />
also have "... = (algunos P (rev xs) ∨ algunos P [x])" by (simp add: algunos_append)<br />
also have "... = (algunos P xs ∨ algunos P [x])" using HI by simp<br />
also have "... = (algunos P xs ∨ P x)" by simp<br />
also have "... = (P x ∨ algunos P xs)" by arith<br />
also have "... = algunos P (x#xs)" by simp<br />
finally show "algunos P (rev (x # xs)) = algunos P (x # xs)" by simp<br />
qed<br />
<br />
<br />
(*wilmorort*)<br />
lemma "algunos P (rev xs) = algunos P xs" (is "?P xs")<br />
proof (induct xs) <br />
show "?P []" by simp <br />
next<br />
fix a xs<br />
assume HI: "?P xs"<br />
have "algunos P (rev (a#xs)) = algunos P ((rev xs) @ [a])" by simp<br />
also have "... = (algunos P(rev xs) ∨ algunos P [a])" by (simp add: algunos_append)<br />
also have "... = (algunos P xs ∨ algunos P [a])" using HI by simp<br />
also have "... = (algunos P [a] ∨ algunos P xs)" by (simp add: HOL.disj_commute)<br />
also have "... = algunos P([a]@(xs))" by (simp)<br />
finally show "algunos P (rev (a#xs))= algunos P (a # xs)" by simp<br />
qed<br />
<br />
{*danrodcha *}<br />
lemma "algunos P (rev xs) = algunos P xs" (is "?Q xs")<br />
proof (induct xs)<br />
show "?Q []" by simp<br />
next<br />
fix a xs assume HI: "?Q xs"<br />
have "algunos P (rev (a#xs)) = algunos P ((rev xs) @ [a])" by simp<br />
also have "… = (algunos P (rev xs) ∨ algunos P [a])" by (simp add:algunos_append)<br />
also have "… = (algunos P xs ∨ algunos P [a])" using HI by simp<br />
also have "… = (algunos P [a] ∨ algunos P xs)" by (simp add: HOL.disj_commute)<br />
also have "… = algunos P (a # xs)" by simp<br />
finally show "?Q (a # xs)" by simp<br />
qed<br />
<br />
text {*<br />
--------------------------------------------------------------------- <br />
Ejercicio 10. Encontrar un término no trivial Z tal que sea cierta la <br />
siguiente ecuación:<br />
algunos (λx. P x ∨ Q x) xs = Z<br />
y demostrar la equivalencia de forma automática y detallada.<br />
--------------------------------------------------------------------- <br />
*}<br />
<br />
(* migtermor *)<br />
<br />
lemma "algunos (λx. P x ∨ Q x) xs = ((algunos (λx. P x) xs) ∨ (algunos (λx. Q x) xs))"<br />
by (induct xs) auto<br />
<br />
lemma "algunos (λx. P x ∨ Q x) xs = ((algunos (λx. P x) xs) ∨ (algunos (λx. Q x) xs))" (is "?R xs")<br />
proof (induct xs)<br />
show "?R []" by simp<br />
next<br />
fix x xs<br />
assume HI: "?R xs"<br />
have "algunos (λx. P x ∨ Q x) (x#xs) = (((λx. P x ∨ Q x) x) ∨ (algunos (λx. P x ∨ Q x) xs))"<br />
by simp<br />
also have "… = ((P x ∨ Q x) ∨ (algunos (λx. P x ∨ Q x) xs))" by simp<br />
also have H1: "… = ((((λx. P x) x) ∨ (algunos (λx. P x) xs)) ∨ (((λx. Q x) x) ∨ (algunos (λx. Q x) xs)))"<br />
using HI by simp arith<br />
have H2: "… = ((algunos (λx. P x) (x#xs)) ∨ (algunos (λx. Q x) (x#xs)))" <br />
by simp<br />
have C: "(algunos (λx. P x ∨ Q x) (x#xs)) = <br />
((algunos (λx. P x) (x#xs)) ∨ (algunos (λx. Q x) (x#xs)))" <br />
using H1 H2 by simp<br />
finally show "(algunos (λx. P x ∨ Q x) (x#xs)) = <br />
((algunos (λx. P x) (x#xs)) ∨ (algunos (λx. Q x) (x#xs)))" <br />
using C by simp<br />
qed<br />
<br />
<br />
(* ferrenseg *)<br />
<br />
lemma "algunos (λx. P x ∨ Q x) xs = (algunos P xs ∨ algunos Q xs)"<br />
proof (induct xs)<br />
show "algunos (λx. P x ∨ Q x) [] = (algunos P [] ∨ algunos Q [])" by simp<br />
next<br />
fix x xs<br />
assume HI: "algunos (λx. (P x ∨ Q x)) xs = (algunos P xs ∨ algunos Q xs)"<br />
show "algunos (λx. P x ∨ Q x) (x # xs) = (algunos P (x # xs) ∨ algunos Q (x # xs))"<br />
proof -<br />
have "algunos (λx. P x ∨ Q x) (x # xs) = <br />
((P x) ∨ (Q x) ∨ algunos (λx. P x ∨ Q x) xs)" by simp<br />
also have "… = ((P x) ∨ (Q x) ∨ algunos P xs ∨ algunos Q xs)" <br />
using HI by simp<br />
also have "… = (((P x) ∨ algunos P xs) ∨ ((Q x) ∨ algunos Q xs))" by auto<br />
also have "… = (algunos P (x # xs) ∨ algunos Q (x # xs))" by simp<br />
finally show ?thesis <br />
qed<br />
qed<br />
<br />
(* ivamenjim marpoldia1 *)<br />
<br />
lemma "algunos (λx. P x ∨ Q x) xs = (algunos P xs ∨ algunos Q xs)" <br />
by (induct xs) auto<br />
<br />
lemma "algunos (λx. P x ∨ Q x) xs = (algunos P xs ∨ algunos Q xs)"<br />
proof (induct xs)<br />
show "algunos (λx. P x ∨ Q x) [] = (algunos P [] ∨ algunos Q [])" by simp<br />
next <br />
fix a xs<br />
assume HI: "algunos (λx. P x ∨ Q x) xs = (algunos P xs ∨ algunos Q xs)"<br />
have "algunos (λx. P x ∨ Q x) (a # xs) = (P a ∨ Q a ∨ algunos (λx. P x ∨ Q x) xs)" by simp<br />
also have "... = (P a ∨ Q a ∨ algunos P xs ∨ algunos Q xs)" using HI by simp<br />
finally show "algunos (λx. P x ∨ Q x) (a # xs) = (algunos P (a # xs) ∨ algunos Q (a # xs))" by auto<br />
qed<br />
<br />
<br />
(* wilmorort danrodcha*)<br />
lemma "algunos (λx. P x ∨ Q x) xs = (¬todos(λx. ¬P x)xs ∨ ¬todos(λx. ¬Q x)xs)"<br />
by (induct xs) auto<br />
<br />
(* wilmorort *)<br />
lemma "algunos (λx. P x ∨ Q x) xs = (¬todos(λx. ¬P x)xs ∨ ¬todos(λx. ¬Q x)xs)"<br />
(is "?P xs")<br />
proof (induct xs)<br />
show "?P []" by simp<br />
next<br />
fix a xs<br />
assume HI: "?P xs"<br />
have "algunos (λx. P x ∨ Q x) (a # xs) = (P a ∨ Q a ∨ algunos (λx. P x ∨ Q x) xs)" by simp<br />
also have "... = (P a ∨ Q a ∨ (¬ todos (λx. ¬ P x) xs ∨ ¬ todos (λx. ¬ Q x) xs))" using HI by simp<br />
also have "... = (P a ∨ ¬ todos (λx. ¬ P x) xs ∨ Q a ∨ ¬ todos (λx. ¬ Q x) xs )" by arith<br />
finally show "algunos (λx. P x ∨ Q x) (a # xs) = (~ todos (λx. ¬ P x) (a # xs) ∨ ~ todos (λx. ¬ Q x) (a # xs))" by simp<br />
qed<br />
<br />
{* danrodcha *}<br />
lemma "algunos (λx. P x ∨ Q x) xs = (algunos P xs ∨ algunos Q xs)" (is "?R xs")<br />
proof (induct xs)<br />
show "?R []" by simp<br />
next<br />
fix a xs assume HI: "?R xs"<br />
have 1:" (Q a ∨ algunos P xs) = (algunos P xs ∨ Q a)" by (simp add: HOL.disj_commute)<br />
have "algunos (λx. P x ∨ Q x) (a # xs) = <br />
(algunos P [a] ∨ algunos Q [a] ∨ algunos (λx. P x ∨ Q x) xs)" by simp<br />
also have "… = (P a ∨ (Q a ∨ algunos P xs) ∨ algunos Q xs)" using HI by simp<br />
also have "… = (P a ∨ algunos P xs ∨ Q a ∨ algunos Q xs)" using 1 by simp<br />
also have "… = (algunos P (a # xs) ∨ algunos Q (a # xs))" by simp<br />
finally show "?R (a # xs)" by simp<br />
qed<br />
<br />
<br />
<br />
text {*<br />
--------------------------------------------------------------------- <br />
Ejercicio 11.1. Demostrar o refutar automáticamente<br />
algunos P xs = (¬ todos (λx. (¬ P x)) xs)<br />
--------------------------------------------------------------------- <br />
*}<br />
<br />
(* ivamenjim wilmorort migtermor marpoldia1 crigomgom rubgonmar paupeddeg danrodcha*)<br />
<br />
lemma "algunos P xs = (¬ todos (λx. (¬ P x)) xs)"<br />
by (induct xs) auto<br />
<br />
(* ferrenseg *)<br />
<br />
lemma "algunos P xs = (¬ todos (λx. (¬ P x)) xs)"<br />
by (induct xs) simp_all<br />
<br />
text {*<br />
--------------------------------------------------------------------- <br />
Ejercicio 11.2. Demostrar o refutar datalladamente<br />
algunos P xs = (¬ todos (λx. (¬ P x)) xs)<br />
--------------------------------------------------------------------- <br />
*}<br />
<br />
(* ivamenjim migtermor marpoldia1 crigomgom paupeddeg*)<br />
<br />
lemma "algunos P xs = (¬ todos (λx. (¬ P x)) xs)"<br />
proof (induct xs)<br />
show "algunos P [] = (¬ todos (λx. ¬ P x) [])" by simp<br />
next <br />
fix a xs<br />
assume HI: "algunos P xs = (¬ todos (λx. ¬ P x) xs)"<br />
have "algunos P (a # xs) = (P a ∨ (algunos P xs))" by simp<br />
also have "... = (P a ∨ (¬ todos (λx. ¬ P x) xs))" using HI by simp<br />
finally show "algunos P (a # xs) = (¬ todos (λx. ¬ P x) (a # xs))" by simp<br />
qed<br />
<br />
(* ferrenseg *)<br />
<br />
lemma "algunos P xs = (¬ todos (λx. (¬ P x)) xs)"<br />
proof (induct xs)<br />
show "algunos P [] = (¬ todos (λx. (¬ P x)) [])" by simp<br />
next<br />
fix x xs<br />
assume HI: "algunos P xs = (¬ todos (λx. (¬ P x)) xs)"<br />
show "algunos P (x # xs) = (¬ todos (λx. (¬ P x)) (x # xs))"<br />
proof - <br />
have "algunos P (x # xs) = ((P x) ∨ algunos P xs)" by simp<br />
also have "… = ((P x) ∨ ¬ todos (λx. (¬ P x)) xs)" using HI by simp<br />
also have "… = (¬ (¬ (P x) ∧ todos (λx. (¬ P x)) xs))" by simp<br />
also have "… = (¬ todos (λx. (¬ P x)) (x # xs))" by simp<br />
finally show ?thesis<br />
qed<br />
qed<br />
<br />
{* danrodcha *}<br />
lemma "algunos P xs = (¬ todos (λx. (¬ P x)) xs)" (is "?Q xs")<br />
proof (induct xs)<br />
show "?Q []" by simp<br />
next<br />
fix a xs assume HI: "?Q xs"<br />
have "algunos P (a # xs) = (P a ∨ algunos P xs)" by simp<br />
also have "… = (P a ∨ (¬ todos (λx. (¬ P x)) xs))" using HI by simp<br />
also have "… = (¬(¬(P a) ∧ todos(λx. (¬ P x)) xs))" by simp<br />
also have "… = (¬ todos(λx. (¬ P x)) (a # xs))" by simp<br />
finally show "?Q (a # xs)" by simp<br />
qed<br />
<br />
text {*<br />
--------------------------------------------------------------------- <br />
Ejercicio 12. Definir la funcion primitiva recursiva <br />
estaEn :: 'a ⇒ 'a list ⇒ bool<br />
tal que (estaEn x xs) se verifica si el elemento x está en la lista<br />
xs. Por ejemplo, <br />
estaEn (2::nat) [3,2,4] = True<br />
estaEn (1::nat) [3,2,4] = False<br />
--------------------------------------------------------------------- <br />
*}<br />
<br />
(* ivamenjim ferrenseg migtermor serrodcal crigomgom rubgonmar marpoldia1 paupeddeg danrodcha*)<br />
<br />
fun estaEn :: "'a ⇒ 'a list ⇒ bool" where<br />
"estaEn x [] = False"<br />
| "estaEn x (a#xs) = ((a = x) ∨ (estaEn x xs))"<br />
<br />
value "estaEn (2::nat) [3,2,4] = True"<br />
value "estaEn (1::nat) [3,2,4] = False"<br />
<br />
<br />
text {*<br />
--------------------------------------------------------------------- <br />
Ejercicio 13. Expresar la relación existente entre estaEn y algunos. <br />
Demostrar dicha relación de forma automática y detallada.<br />
--------------------------------------------------------------------- <br />
*}<br />
<br />
(* migtermor crigomgom *)<br />
<br />
lemma "estaEn x xs = (algunos (λa. a=x) xs)"<br />
by (induct xs) auto<br />
<br />
lemma auxiliar13:<br />
"(x=a) = ((λa. a=x) a)"<br />
by auto<br />
<br />
lemma "estaEn x xs = (algunos (λa. a=x) xs)" (is "?P xs")<br />
proof (induct xs)<br />
show "?P []" by simp<br />
next<br />
fix a xs<br />
assume HI: "?P xs"<br />
have "estaEn x (a#xs) = ((a=x) ∨ (estaEn x xs))" by simp<br />
also have H: "… = ((a=x) ∨ (algunos (λa. a=x) xs))" using HI by simp<br />
also have "… = (((λa. a=x) a) ∨ (algunos (λa. a=x) xs))" using auxiliar13 by simp<br />
also have "… = (algunos (λa. a=x) (a#xs))" by simp<br />
have C: "estaEn x (a#xs) = (algunos (λa. a=x) (a#xs))" using H by simp<br />
finally show "estaEn x (a#xs) = (algunos (λa. a=x) (a#xs))" using C by simp<br />
qed<br />
<br />
<br />
(* ferrenseg *)<br />
<br />
lemma estaEn_algunos:<br />
"estaEn y xs = algunos (λx. x=y) xs"<br />
by (induct xs) auto<br />
<br />
lemma estaEn_algunos_2:<br />
"estaEn y xs = algunos (λx. x=y) xs"<br />
proof (induct xs)<br />
show "estaEn y [] = algunos (λx. x=y) []" by simp<br />
next<br />
fix x xs<br />
assume HI: "estaEn y xs = algunos (λx. x=y) xs"<br />
show "estaEn y (x # xs) = algunos (λx. x=y) (x # xs)"<br />
proof -<br />
have "estaEn y (x # xs) = (y=x ∨ estaEn y xs)" by simp<br />
also have "… = (y=x ∨ algunos (λx. x=y) xs)" using HI by simp<br />
also have "… = (x=y ∨ algunos (λx. x=y) xs)" by auto<br />
also have "… = algunos (λx. x=y) (x # xs)" by simp<br />
finally show ?thesis<br />
qed<br />
qed<br />
<br />
(*crigomgom paupeddeg*)<br />
<br />
lemma "estaEn a xs = algunos (λx. x = a) xs"<br />
by (induct xs) auto<br />
<br />
lemma "estaEn x xs = (algunos (λa. a=x) xs)" <br />
proof (induct xs)<br />
show " estaEn x [] = algunos (λa. a = x) []" by simp<br />
next<br />
fix a xs<br />
assume HI: "estaEn x xs = algunos (λa. a = x) xs"<br />
have "estaEn x (a # xs) = (a = x ∨ estaEn x xs)" by simp<br />
also have "... = (a = x ∨ algunos (λa. a = x) xs)" using HI by simp<br />
finally show " estaEn x (a # xs) = algunos (λa. a = x) (a # xs)" by simp<br />
qed<br />
<br />
(* ivamenjim marpoldia1 *)<br />
<br />
lemma "estaEn x xs = algunos (λy. x=y) xs"<br />
by (induct xs) auto<br />
<br />
lemma "estaEn x xs = algunos (λy. x=y) xs"<br />
proof (induct xs)<br />
show "estaEn x [] = algunos (op = x) []" by simp<br />
next <br />
fix a xs<br />
assume HI: "estaEn x xs = algunos (op = x) xs"<br />
have "estaEn x (a # xs) = ((a = x) ∨ (estaEn x xs))" by simp<br />
also have "... = (a = x ∨ algunos (op = x) xs)" using HI by simp<br />
finally show " estaEn x (a # xs) = algunos (op = x) (a # xs)" by auto<br />
qed<br />
<br />
{* danrodcha *}<br />
lemma "estaEn a xs = algunos (op = a) xs"<br />
by (induct xs) auto<br />
<br />
{* danrodcha *}<br />
lemma "estaEn a xs = algunos (op = a) xs"<br />
proof (induct xs)<br />
show "estaEn a [] = algunos (op = a) []" by simp<br />
next<br />
fix x xs <br />
assume HI: "estaEn a xs = algunos (op = a) xs"<br />
have "estaEn a (x # xs) = ((x = a) ∨ (estaEn a xs))" by simp<br />
also have "… = (x = a ∨ algunos (op = a) xs)" using HI by simp<br />
also have "… = ((op = a) x ∨ algunos (op = a) xs)" by (simp add:HOL.eq_commute)<br />
also have "… = algunos (op = a) (x # xs)" by simp<br />
finally show "estaEn a (x # xs) = algunos (op = a) (x # xs)" by simp<br />
qed<br />
<br />
end<br />
</source></div>Lucnovdoshttps://www.glc.us.es/~jalonso/RA2016/index.php?title=Relaci%C3%B3n_4&diff=581Relación 42016-11-22T17:44:40Z<p>Lucnovdos: </p>
<hr />
<div><source lang="isar"><br />
chapter {* R4: Cuantificadores sobre listas *}<br />
<br />
theory R4_Cuantificadores_sobre_listas<br />
imports Main <br />
begin<br />
<br />
text {* <br />
--------------------------------------------------------------------- <br />
Ejercicio 1. Definir la función <br />
todos :: ('a ⇒ bool) ⇒ 'a list ⇒ bool<br />
tal que (todos p xs) se verifica si todos los elementos de la lista <br />
xs cumplen la propiedad p. Por ejemplo, se verifica <br />
todos (λx. 1<length x) [[2,1,4],[1,3]]<br />
¬todos (λx. 1<length x) [[2,1,4],[3]]<br />
<br />
Nota: La función todos es equivalente a la predefinida list_all. <br />
--------------------------------------------------------------------- <br />
*}<br />
<br />
{*danrodcha*}<br />
fun todos :: "('a ⇒ bool) ⇒ 'a list ⇒ bool" where<br />
"todos p xs = "todos p [] = True"<br />
| "todos p (x#xs) = (p x ∧ todos p xs)""<br />
<br />
(* ivamenjim migtermor dancorgar wilmorort marpoldia1 ferrenseg paupeddeg pablucoto crigomgom anaprarod serrodcal juacabsou rubgonmar josgarsan fraortmoy lucnovdos*)<br />
fun todos :: "('a ⇒ bool) ⇒ 'a list ⇒ bool" where<br />
"todos p [] = True"<br />
| "todos p (x#xs) = (p x ∧ todos p xs)"<br />
<br />
text {* <br />
--------------------------------------------------------------------- <br />
Ejercicio 2. Definir la función <br />
algunos :: ('a ⇒ bool) ⇒ 'a list ⇒ bool<br />
tal que (algunos p xs) se verifica si algunos elementos de la lista <br />
xs cumplen la propiedad p. Por ejemplo, se verifica <br />
algunos (λx. 1<length x) [[2,1,4],[3]]<br />
¬algunos (λx. 1<length x) [[],[3]]"<br />
<br />
Nota: La función algunos es equivalente a la predefinida list_ex. <br />
--------------------------------------------------------------------- <br />
*}<br />
<br />
{*danrodcha ivamenjim migtermor dancorgar marpoldia1 ferrenseg wilmorort paupeddeg pablucoto crigomgom anaprarod serrodcal juacabsou rubgonmar josgarsan fraortmoy lucnovdos*}<br />
fun algunos :: "('a ⇒ bool) ⇒ 'a list ⇒ bool" where<br />
"algunos p [] = False"<br />
| "algunos p (x#xs) = (p x ∨ algunos p xs)"<br />
<br />
text {*<br />
--------------------------------------------------------------------- <br />
Ejercicio 3.1. Demostrar o refutar automáticamente <br />
todos (λx. P x ∧ Q x) xs = (todos P xs ∧ todos Q xs)<br />
--------------------------------------------------------------------- <br />
*}<br />
<br />
{*danrodcha ivamenjim migtermor dancorgar marpoldia1 ferrenseg wilmorort paupeddeg pablucoto anaprarod serrodcal juacabsou rubgonmar josgarsan fraortmoy lucnovdos*}<br />
lemma "todos (λx. P x ∧ Q x) xs = (todos P xs ∧ todos Q xs)"<br />
by (induct xs) auto<br />
<br />
text {*<br />
--------------------------------------------------------------------- <br />
Ejercicio 3.2. Demostrar o refutar detalladamente<br />
todos (λx. P x ∧ Q x) xs = (todos P xs ∧ todos Q xs)<br />
--------------------------------------------------------------------- <br />
*}<br />
<br />
{*danrodcha*}<br />
lemma "todos (λx. P x ∧ Q x) xs = (todos P xs ∧ todos Q xs)"<br />
proof (induct xs)<br />
show "?R []" by simp<br />
next<br />
fix a xs <br />
assume HI: "?R xs"<br />
have "todos (λx. P x ∧ Q x) (a#xs) = (P a ∧ Q a ∧ todos (λx. P x ∧ Q x) xs)" by simp<br />
also have "… = (P a ∧ Q a ∧ (todos P xs ∧ todos Q xs))" using HI by simp<br />
also have "… = ((P a ∧ todos P xs) ∧ (Q a ∧ todos Q xs))" by blast<br />
also have "… = (todos P (a#xs) ∧ todos Q (a#xs))" by simp<br />
finally show "?R (a#xs)" by simp<br />
qed<br />
<br />
(* ivamenjim wilmorort serrodcal josgarsan*)<br />
<br />
lemma "todos (λx. P x ∧ Q x) xs = (todos P xs ∧ todos Q xs)"<br />
proof (induct xs)<br />
show "todos (λx. P x ∧ Q x) [] = (todos P [] ∧ todos Q [])" by simp<br />
next <br />
fix a xs<br />
assume HI: "todos (λx. P x ∧ Q x) xs = (todos P xs ∧ todos Q xs)"<br />
have "todos (λx. P x ∧ Q x) (a # xs) = (P a ∧ Q a ∧ todos (λx. P x ∧ Q x) xs)" by simp<br />
also have "... = (P a ∧ Q a ∧ todos P xs ∧ todos Q xs)" using HI by simp<br />
finally show "todos (λx. P x ∧ Q x) (a # xs) = (todos P (a # xs) ∧ todos Q (a # xs))" by auto<br />
qed<br />
<br />
(* dancorgar paupeddeg *)<br />
lemma "todos (λx. P x ∧ Q x) xs = (todos P xs ∧ todos Q xs)"<br />
proof (induct xs)<br />
show "todos (λx. P x ∧ Q x) [] = (todos P [] ∧ todos Q [])" by simp<br />
next<br />
fix y xs<br />
assume HI: "todos (λx. P x ∧ Q x) xs = (todos P xs ∧ todos Q xs)"<br />
have "todos (λx. P x ∧ Q x) (y#xs) = ((P y ∧ Q y) ∧ (todos (λx. P x ∧ Q x) xs))" by simp<br />
also have "… = ((P y ∧ Q y) ∧ (todos P xs ∧ todos Q xs))" using HI by simp<br />
also have "… = ((P y ∧ todos P xs) ∧ (Q y ∧ todos Q xs))" by blast<br />
also have "… = ((todos P (y#xs)) ∧ (todos Q (y#xs)))" by simp<br />
finally show "todos (λx. P x ∧ Q x) (y#xs) = (todos P (y#xs) ∧ todos Q (y#xs))" by simp<br />
qed<br />
<br />
<br />
(* migtermor *)<br />
lemma "todos (λx. P x ∧ Q x) xs = (todos P xs ∧ todos Q xs)"<br />
proof (induct xs)<br />
show "todos (λx. P x ∧ Q x) [] = (todos P [] ∧ todos Q [])" by simp<br />
next <br />
fix x xs<br />
assume HI: "todos (λx. P x ∧ Q x) xs = (todos P xs ∧ todos Q xs)"<br />
have "(todos (λx. P x ∧ Q x) (x#xs)) = (( P x ∧ Q x) ∧ (todos (λx. P x ∧ Q x) xs))"<br />
by (simp only: todos.simps(2))<br />
also have "… = ((P x ∧ Q x) ∧ (todos P xs ∧ todos Q xs))" using HI by simp<br />
also have "… = (((P x)∧(todos P xs)) ∧ ((Q x) ∧ (todos Q xs)))" by arith<br />
also have "((P x ∧ Q x) ∧ (todos P xs ∧ todos Q xs)) = (((P x)∧(todos P xs)) ∧ ((Q x) ∧ (todos Q xs)))"<br />
by arith (* Este paso es exactamente el mismo que el anterior, pero sin cualquiera de los dos no funciona el "finally show" *)<br />
have "… = (((P x)∧(todos P xs))∧((Q x)∧(todos Q xs)))" by simp<br />
have "… = ((todos P (x#xs))∧(todos Q (x#xs)))" by simp<br />
finally show "(todos (λx. P x ∧ Q x) (x#xs)) = ((todos P (x#xs)) ∧ (todos Q (x#xs)))" by simp <br />
qed<br />
<br />
(* marpoldia1 *)<br />
lemma "todos (λx. P x ∧ Q x) xs = (todos P xs ∧ todos Q xs)"<br />
proof (induct xs)<br />
show "todos (λx. P x ∧ Q x) [] = (todos P [] ∧ todos Q [])" by simp<br />
next<br />
fix a xs<br />
assume HI: "todos (λx. P x ∧ Q x) xs = (todos P xs ∧ todos Q xs)" <br />
have "todos (λx. P x ∧ Q x) (a # xs) = ((P a ∧ Q a) ∧ (todos P xs ∧ todos Q xs))" using HI by simp<br />
also have "... = ((P a ∧ todos P xs) ∧ (Q a ∧ todos Q xs))" by blast<br />
also have "... = (todos P (a#xs) ∧ todos Q (a#xs)) " by simp<br />
finally show "todos (λx. P x ∧ Q x) (a#xs) = (todos P (a#xs) ∧ todos Q (a#xs))" by simp <br />
qed<br />
<br />
(* ferrenseg *)<br />
<br />
lemma "todos (λx. P x ∧ Q x) xs = (todos P xs ∧ todos Q xs)"<br />
proof (induct xs)<br />
show "todos (λx. P x ∧ Q x) [] = (todos P [] ∧ todos Q [])" by simp<br />
next <br />
fix n xs<br />
assume HI: "todos (λx. P x ∧ Q x) xs = (todos P xs ∧ todos Q xs)"<br />
have "todos (λx. P x ∧ Q x) (n # xs) = (P n ∧ Q n ∧ todos P xs ∧ todos Q xs)" using HI by simp<br />
also have "… = P n ∧ Q n ∧ todos P xs ∧ todos Q xs" by blast<br />
also have "⋯ = todos P (n # xs) ∧ todos Q (n # xs)" by simp <br />
finally show "todos (λx. P x ∧ Q x) (n # xs) = todos P (n # xs) ∧ todos Q (n # xs)" by simp<br />
qed<br />
<br />
{* wilmorort pablucoto crigomgom anaprarod juacabsou rubgonmar fraortmoy *}<br />
<br />
lemma "todos (λx. P x ∧ Q x) xs = (todos P xs ∧ todos Q xs)"<br />
proof (induct xs)<br />
show "todos (λx. P x ∧ Q x) [] = (todos P [] ∧ todos Q [])" by simp <br />
next<br />
fix a xs<br />
assume HI: "todos (λx. P x ∧ Q x) xs = (todos P xs ∧ todos Q xs)"<br />
have "todos (λx. P x ∧ Q x) (a # xs) = (P a ∧ Q a ∧ todos (λx. P x ∧ Q x) xs)" by simp<br />
also have "... = (P a ∧ Q a ∧ todos P xs ∧ todos Q xs)" using HI by simp<br />
also have "... = (P a ∧ todos P xs ∧ Q a ∧ todos Q xs)" by arith<br />
also have "... = (todos P (a # xs) ∧ todos Q (a # xs))" by simp (* Este paso se puede obviar*)<br />
finally show "todos (λx. P x ∧ Q x) (a # xs) = (todos P (a # xs) ∧ todos Q (a # xs))" by simp<br />
qed<br />
<br />
text {*<br />
--------------------------------------------------------------------- <br />
Ejercicio 4.1. Demostrar o refutar automáticamente<br />
todos P (x @ y) = (todos P x ∧ todos P y)<br />
--------------------------------------------------------------------- <br />
*}<br />
{*danrodcha ivamenjim marpoldia1 migtermor ferrenseg wilmorort paupeddeg crigomgom anaprarod serrodcal juacabsou rubgonmar pablucoto fraortmoy *}<br />
lemma "todos P (x @ y) = (todos P x ∧ todos P y)"<br />
by (induct x) auto<br />
<br />
{* anaprarod dancorgar *}<br />
lemma "todos P (x @ y) = (todos P x ∧ todos P y)"<br />
apply (induct x)<br />
apply auto<br />
done<br />
<br />
text {*<br />
--------------------------------------------------------------------- <br />
Ejercicio 4.2. Demostrar o refutar detalladamente<br />
todos P (x @ y) = (todos P x ∧ todos P y)<br />
--------------------------------------------------------------------- <br />
*}<br />
<br />
(* ivamenjim ferrenseg wilmorort dancorgar fraortmoy *)<br />
<br />
lemma todos_append:<br />
"todos P (x @ y) = (todos P x ∧ todos P y)"<br />
proof (induct x)<br />
show "todos P ([] @ y) = (todos P [] ∧ todos P y)" by simp<br />
next <br />
fix a x<br />
assume HI: "todos P (x @ y) = (todos P x ∧ todos P y)"<br />
have "todos P ((a#x) @ y) = (P a ∧ (todos P (x @ y)))" by simp<br />
also have "... = (P a ∧ (todos P x ∧ todos P y))" using HI by simp<br />
finally show "todos P ((a # x) @ y) = (todos P (a # x) ∧ todos P y)" by simp<br />
qed<br />
<br />
(* migtermor serrodcal *)<br />
<br />
lemma todos_append1:<br />
"todos P (x @ y) = (todos P x ∧ todos P y)" (is "?P x")<br />
proof (induct x)<br />
show "?P []" by simp<br />
next <br />
fix a x<br />
assume HI: "?P x"<br />
have "todos P ((a#x) @ y) = (P a ∧ (todos P (x @ y)))" by simp<br />
also have "… = (P a ∧ (todos P x ∧ todos P y))" using HI by simp<br />
finally show "?P (a#x)" by simp<br />
qed <br />
<br />
(* marpoldia1 *)<br />
lemma todos_append:<br />
"todos P (x @ y) = (todos P x ∧ todos P y)"<br />
proof (induct x)<br />
show "todos P ([] @ y) = (todos P [] ∧ todos P y)" by simp<br />
next<br />
fix a x<br />
assume HI: "todos P (x @ y) = (todos P x ∧ todos P y)"<br />
have "todos P ((a # x) @ y) = todos P (a#(x@y))" by simp<br />
also have "... = (P a ∧ todos P (x @ y))" by simp<br />
also have "... = (P a ∧ (todos P x ∧ todos P y))" using HI by simp<br />
finally show "todos P ((a # x) @ y) = (todos P (a # x) ∧ todos P y)" by simp <br />
qed<br />
<br />
(* paupeddeg *)<br />
lemma todos_append3:<br />
"todos P (x @ y) = (todos P x ∧ todos P y)"<br />
proof (induct x)<br />
show "todos P ([] @ y) = (todos P [] ∧ todos P y)" by simp<br />
next<br />
fix a x<br />
assume HI: "todos P (x @ y) = (todos P x ∧ todos P y)"<br />
have "todos P ((a # x) @ y) = todos P (a # (x @ y))" by simp<br />
also have "... = (todos P [a] ∧ todos P (x @ y)) " by simp<br />
also have "... = (todos P [a] ∧ (todos P x ∧ todos P y))" using HI by simp<br />
finally show "todos P ((a # x) @ y) = (todos P (a # x) ∧ todos P y)" by simp<br />
qed<br />
<br />
(* crigomgom juacabsou anaprarod*)<br />
<br />
lemma todos_append4:<br />
"todos P (x @ y) = (todos P x ∧ todos P y)"<br />
proof (induct x)<br />
show "todos P ([] @ y) = (todos P [] ∧ todos P y)" by simp<br />
next<br />
fix x xs<br />
assume HI:"todos P (xs @ y) = (todos P xs ∧ todos P y)"<br />
have "todos P ((x # xs) @ y) = (P x ∧ todos P (xs @ y ))" by simp<br />
also have "... = (P x ∧ (todos P xs ∧ todos P y)) " using HI by simp<br />
also have "... = ((P x ∧ todos P xs) ∧ todos P y)" by simp<br />
finally show "todos P ((x # xs) @ y) = (todos P (x # xs) ∧ todos P y)" by simp<br />
qed<br />
<br />
(* pablucoto *)<br />
lemma todos_append5:<br />
"todos P (x @ y) = (todos P x ∧ todos P y)"<br />
proof (induct x )<br />
show "todos P ([] @ y) = (todos P [] ∧ todos P y)" by simp<br />
next<br />
fix a x<br />
assume HI:"todos P (x @ y) = (todos P x ∧ todos P y) " <br />
have "todos P ((a # x) @ y) = ( P a ∧ todos P (x @ y))" by simp<br />
also have " ... =( P a ∧ (todos P x ∧ todos P y))" using HI by simp<br />
also have "... = (todos P (a#x) ∧ todos P y)" by simp<br />
finally show "todos P ((a # x) @ y) = (todos P (a # x) ∧ todos P y)" by auto<br />
qed<br />
<br />
text {*<br />
--------------------------------------------------------------------- <br />
Ejercicio 5.1. Demostrar o refutar automáticamente <br />
todos P (rev xs) = todos P xs<br />
--------------------------------------------------------------------- <br />
*}<br />
<br />
lemma "todos P (rev xs) = todos P xs"<br />
oops<br />
<br />
(* migtermor ivamenjim marpoldia1 serrodcal anaprarod *)<br />
lemma "todos P (rev xs) = todos P xs" <br />
apply (induct xs)<br />
apply simp<br />
apply (simp add: todos_append)<br />
apply auto<br />
done<br />
<br />
(* ferrenseg crigomgom rubgonmar fraortmoy *)<br />
<br />
lemma "todos P (rev xs) = todos P xs"<br />
by (induct xs) (auto simp add: todos_append)<br />
<br />
(* wilmorort *)<br />
lemma "todos P (rev xs) = todos P xs"<br />
by (induct xs,simp,simp add: todos_append,auto)<br />
<br />
(* juacabsou *)<br />
lemma "todos P (rev xs) = todos P xs"<br />
apply (induct xs,simp,simp add: todos_append,auto) done<br />
<br />
(* dancorgar *)<br />
lemma "todos P (rev xs) = todos P xs"<br />
apply (induct xs)<br />
apply (simp add: todos_append)<br />
apply (simp add: todos_append)<br />
apply auto<br />
done<br />
<br />
(* pablucoto * )<br />
lemma "todos P (rev xs) = todos P xs"<br />
by (induct xs) (auto, simp_all add: todos_append) <br />
lemma "todos P (rev xs) = todos P xs"<br />
by (induct xs) ( simp_all add: todos_append, auto)<br />
<br />
text {*<br />
--------------------------------------------------------------------- <br />
Ejercicio 5.2. Demostrar o refutar detalladamente<br />
todos P (rev xs) = todos P xs<br />
--------------------------------------------------------------------- <br />
*}<br />
<br />
lemma "todos P (rev xs) = todos P xs"<br />
oops<br />
<br />
(* migtermor *)<br />
<br />
lemma auxiliar:<br />
"rev (a#xs) = rev xs @ [a]"<br />
by auto<br />
<br />
lemma "todos P (rev xs) = todos P xs" (is "?Q xs")<br />
proof (induct xs)<br />
show "?Q []" by (simp only: rev.simps(1))<br />
next <br />
fix a xs<br />
assume HI: "?Q xs"<br />
have "todos P (rev (a#xs)) = (todos P (rev xs @ [a]))" by (simp add: auxiliar)<br />
have "… = ((todos P (rev xs)) ∧ (todos P [a]))" by (simp add: todos_append)<br />
have "… = (todos P (rev xs) ∧ P a)" by simp<br />
also have Aux: "… = (todos P xs ∧ P a)" using HI by simp<br />
have Aux1: "… = (P a ∧ todos P xs)" by arith<br />
have "(todos P (rev xs) ∧ P a) = (P a ∧ todos P xs)" using Aux Aux1 by simp<br />
finally show "todos P (rev (a#xs)) = todos P (a#xs)" by (simp add: todos_append)<br />
qed<br />
<br />
<br />
(* marpoldia1 ferrenseg crigomgom serrodcal juacabsou rubgonmar *)<br />
lemma "todos P (rev xs) = todos P xs"<br />
proof (induct xs)<br />
show "todos P (rev []) = todos P []" by simp<br />
next<br />
fix a xs<br />
assume HI: "todos P (rev xs) = todos P xs"<br />
have "todos P (rev (a # xs)) = (todos P ((rev xs)@[a]))" by simp<br />
also have "... = (todos P (rev xs) ∧ todos P [a])" by (simp add:todos_append)<br />
also have "... = (todos P xs ∧ P a)" using HI by simp<br />
also have "... = (P a ∧ todos P xs)" by arith<br />
also have "... = (todos P (a#xs))" by simp<br />
finally show "todos P (rev (a # xs)) = (todos P (a#xs))" by simp <br />
qed<br />
<br />
(* ivamenjim *)<br />
<br />
lemma "todos P (rev xs) = todos P xs" (is "?P xs")<br />
proof (induct xs)<br />
show "?P []" by simp<br />
next <br />
fix a xs<br />
assume HI: "?P xs"<br />
have "todos P (rev (a # xs)) = (todos P ((rev xs)@[a]))" by simp<br />
also have "... = ((todos P (rev xs)) ∧ todos P [a])" by (simp add: todos_append)<br />
also have "... = (todos P xs ∧ todos P [a])" using HI by simp<br />
also have "... = (todos P [a] ∧ todos P xs)" by arith<br />
finally show "todos P (rev (a # xs)) = todos P (a # xs)" by simp<br />
qed<br />
<br />
(* fraortmoy *)<br />
<br />
lemma "todos P (rev xs) = todos P xs"<br />
proof (induct xs)<br />
show "todos P (rev []) = todos P []" by simp<br />
next<br />
fix a xs <br />
assume H1: "todos P (rev xs) = todos P xs"<br />
have " todos P (rev (a # xs)) = todos P (rev xs @ [a])" by simp<br />
also have "… = (todos P (rev xs) ∧ todos P [a])" by (simp add:todos_append)<br />
also have "… = (todos P xs ∧ todos P [a])" using H1 by simp<br />
also have "… = (todos P [a] ∧ todos P xs)" by arith<br />
finally show "todos P (rev (a # xs)) = todos P (a#xs)" by simp<br />
qed<br />
<br />
(* paupeddeg dancorgar *)<br />
lemma "todos P (rev xs) = todos P xs"<br />
proof (induct xs)<br />
show " todos P (rev []) = todos P []" by simp<br />
next<br />
fix a xs<br />
assume HI: "todos P (rev xs) = todos P xs "<br />
have "todos P (rev (a # xs)) = todos P (rev(xs) @ [a])" by simp<br />
also have "... = (todos P (rev xs) ∧ todos P [a]) " by (simp add: todos_append)<br />
also have "... = (todos P xs ∧ todos P [a])" using HI by simp<br />
also have "... = (todos P [a] ∧ todos P xs)" by arith<br />
also have "... = todos P ([a] @ xs)" by (simp add: todos_append)<br />
finally show "todos P (rev (a # xs)) = todos P (a # xs)" by simp<br />
qed <br />
<br />
(* wilmorort *)<br />
lemma "todos P (rev xs) = todos P xs" (is "?P xs")<br />
proof (induct xs) <br />
show "?P []" by simp <br />
next<br />
fix a xs<br />
assume HI: "?P xs"<br />
have "todos P (rev (a#xs)) = todos P ((rev xs) @ [a])" by simp<br />
also have "... = (todos P(rev xs) ∧ todos P [a])" by (simp add: todos_append)<br />
also have "... = (todos P xs ∧ todos P [a])" using HI by simp<br />
also have "... = (todos P [a] ∧ todos P xs)" by (simp add: HOL.conj_commute)<br />
also have "... = todos P([a]@(xs))" by (simp)<br />
finally show "todos P (rev (a#xs))= todos P (a # xs)" by simp<br />
qed<br />
<br />
(* anaprarod *) <br />
(* es igual que las anteriores pero con el final también con patrones *)<br />
<br />
lemma "todos P (rev xs) = todos P xs" (is "?P xs")<br />
proof (induct xs)<br />
show "?P []" by auto<br />
next<br />
fix a xs<br />
assume HI: "?P xs"<br />
have "todos P (rev (a#xs)) = todos P (rev xs @ [a])" by simp<br />
also have "… = (todos P (rev xs) ∧ todos P [a])" by (simp add: todos_append)<br />
also have "… = (todos P xs ∧ todos P [a])" using HI by simp<br />
also have "… = (todos P [a] ∧ todos P xs)" by arith<br />
also have "… = todos P (a#xs)" by simp<br />
finally show "?P (a # xs)" by simp<br />
qed<br />
<br />
text {*<br />
--------------------------------------------------------------------- <br />
Ejercicio 6. Demostrar o refutar:<br />
algunos (λx. P x ∧ Q x) xs = (algunos P xs ∧ algunos Q xs)<br />
--------------------------------------------------------------------- <br />
*}<br />
<br />
lemma "algunos (λx. P x ∧ Q x) xs = (algunos P xs ∧ algunos Q xs)"<br />
oops<br />
<br />
(* migtermor ivamenjim ferrenseg paupeddeg crigomgom serrodcal wilmorort juacabsou rubgonmar anaprarod marpoldia1 fraortmoy *)<br />
lemma "algunos (λx. P x ∧ Q x) xs = (algunos P xs ∧ algunos Q xs)" <br />
quickcheck<br />
oops<br />
(* Quickcheck encuentra el siguiente contraejemplo: P={a1}, Q={a2}, xs={a1,a2}. En este ejemplo:<br />
· "algunos (λx. P x ∧ Q x) xs = False"<br />
· "(algunos P xs ∧ algunos Q xs) = True" *)<br />
<br />
text {*<br />
--------------------------------------------------------------------- <br />
Ejercicio 7.1. Demostrar o refutar automáticamente <br />
algunos P (map f xs) = algunos (P ∘ f) xs<br />
--------------------------------------------------------------------- <br />
*}<br />
<br />
(* ivamenjim migtermor marpoldia1 crigomgom rubgonmar wilmorort anaprarod fraortmoy *)<br />
<br />
lemma "algunos P (map f xs) = algunos (P o f) xs"<br />
by (induct xs) auto<br />
<br />
(* ferrenseg *)<br />
<br />
lemma "algunos P (map f xs) = algunos (P o f) xs"<br />
by (induct xs) simp_all<br />
<br />
(* anaprarod *)<br />
lemma "algunos P (map f xs) = algunos (P o f) xs"<br />
apply (induct xs)<br />
apply auto<br />
done<br />
<br />
text {*<br />
--------------------------------------------------------------------- <br />
Ejercicio 7.2. Demostrar o refutar datalladamente<br />
algunos P (map f xs) = algunos (P ∘ f) xs<br />
--------------------------------------------------------------------- <br />
*}<br />
<br />
lemma "algunos P (map f xs) = algunos (P o f) xs"<br />
oops<br />
<br />
(* migtermor *)<br />
<br />
lemma AUX: "algunos (λa. P (f a)) xs = algunos (P o f) xs"<br />
by (induct xs) auto<br />
<br />
lemma "algunos P (map f xs) = algunos (P o f) xs" (is "?Q xs")<br />
proof (induct xs)<br />
show "?Q []" by simp<br />
next<br />
fix x xs<br />
assume HI: "?Q xs"<br />
have "algunos P (map f (x#xs)) = (algunos P ((f x)#(map f xs)))" by simp<br />
also have "… = ((P (f x)) ∨ (algunos P (map f xs)))" by (simp only: algunos.simps(2))<br />
also have "… = ((P (f x)) ∨ (algunos (P o f) xs))" <br />
proof (cases)<br />
assume C1: "(P (f x))"<br />
have Aux: "((P (f x)) ∨ (algunos P (map f xs))) = True" using C1 by simp<br />
have Aux1: "… = ((P (f x)) ∨ (algunos (P o f) xs))" using C1 by simp<br />
then show "((P (f x)) ∨ (algunos P (map f xs))) = ((P (f x)) ∨ (algunos (P o f) xs))" <br />
using Aux Aux1 by simp<br />
next<br />
assume C2: "¬(P (f x))"<br />
have Aux2: "((P (f x)) ∨ (algunos P (map f xs))) = (algunos P (map f xs))" using C2 by simp<br />
have Aux3: "… = (algunos (P o f) xs)" using HI by (simp add: AUX)<br />
also have "… = ((P (f x)) ∨ (algunos (P o f) xs))" using C2 by simp<br />
then show "((P (f x)) ∨ (algunos P (map f xs))) = ((P (f x)) ∨ (algunos (P o f) xs))" <br />
using Aux2 Aux3 by simp<br />
qed<br />
also have "… = (((P o f) x) ∨ (algunos (P o f) xs))" by simp<br />
also have "… = (algunos (P o f) (x#xs))" by simp<br />
finally show "algunos P (map f (x#xs)) = (algunos (P o f) (x#xs))" by simp<br />
qed<br />
<br />
(* ferrenseg *)<br />
<br />
lemma "algunos P (map f xs) = algunos (P ∘ f) xs"<br />
proof (induct xs)<br />
show "algunos P (map f []) = algunos (P ∘ f) []" by simp<br />
next<br />
fix x xs<br />
assume HI: "algunos P (map f xs) = algunos (P ∘ f) xs"<br />
show "algunos P (map f (x # xs)) = algunos (P ∘ f) (x # xs)"<br />
proof -<br />
have "algunos P (map f (x # xs)) = algunos P ((f x) # (map f xs))" by simp<br />
also have "… = ((P (f x)) ∨ (algunos P (map f xs)))" by simp<br />
also have "… = (((P ∘ f) x) ∨ (algunos (P ∘ f) xs))" using HI by simp<br />
also have "… = algunos (P ∘ f) (x # xs)" by simp<br />
finally show ?thesis<br />
qed<br />
qed<br />
<br />
(* ivamenjim *)<br />
<br />
lemma "algunos P (map f xs) = algunos (P o f) xs"<br />
proof (induct xs)<br />
show "algunos P (map f []) = algunos (P ∘ f) []" by simp<br />
next <br />
fix a xs<br />
assume HI: "algunos P (map f xs) = algunos (P ∘ f) xs"<br />
have "algunos P (map f (a # xs)) = algunos P ((f a)#map f xs)" by simp<br />
also have "... = (algunos P (map f [a]) ∨ algunos P (map f xs))" by simp <br />
also have "... = (algunos P (map f [a]) ∨ algunos (P ∘ f) xs)" using HI by simp <br />
finally show "algunos P (map f (a # xs)) = algunos (P ∘ f) (a # xs)" by simp<br />
qed<br />
<br />
(* crigomgom rubgonmar anaprarod marpoldia1 *)<br />
lemma "algunos P (map f xs) = algunos (P o f) xs"<br />
proof (induct xs)<br />
show "algunos P (map f []) = algunos (P ∘ f) []" by simp<br />
next<br />
fix x xs<br />
assume HI: "algunos P (map f xs) = algunos (P ∘ f) xs"<br />
have "algunos P (map f (x # xs)) = algunos P ((f x) # (map f xs))" by simp<br />
also have "... = (P (f x) ∨ algunos P (map f xs))" by simp<br />
also have "... = (P (f x) ∨ algunos (P ∘ f) xs)" using HI by simp<br />
also have "... = ((P ∘ f) x ∨ algunos (P ∘ f) xs)" by simp<br />
finally show "algunos P (map f (x # xs)) = algunos (P ∘ f) (x # xs)" by simp<br />
qed<br />
<br />
(* wilmorort *)<br />
lemma "algunos P (map f xs) = algunos (P o f) xs" (is "?P xs" )<br />
proof (induct xs)<br />
show "?P []" by simp<br />
next<br />
fix a xs<br />
assume HI: "algunos P (map f xs) = algunos (P ∘ f) xs"<br />
have "algunos P (map f (a # xs)) = (P (f a) ∨ algunos P (map f xs))" by simp<br />
also have "... = (P (f a) ∨ algunos (P ∘ f) xs )" using HI by simp<br />
also have "... = ((P ∘ f) a ∨ algunos (P ∘ f)xs)" by simp<br />
finally show " algunos P (map f (a # xs)) = algunos (P ∘ f) (a # xs)" by simp<br />
qed<br />
<br />
text {*<br />
--------------------------------------------------------------------- <br />
Ejercicio 8.1. Demostrar o refutar automáticamente <br />
algunos P (xs @ ys) = (algunos P xs ∨ algunos P ys)<br />
--------------------------------------------------------------------- <br />
*}<br />
<br />
(* ivamenjim marpoldia1 paupeddeg crigomgom rubgonmar wilmorort*)<br />
<br />
lemma "algunos P (xs @ ys) = (algunos P xs ∨ algunos P ys)"<br />
by (induct xs) auto<br />
<br />
(* ferrenseg *)<br />
lemma "algunos P (xs @ ys) = (algunos P xs ∨ algunos P ys)"<br />
by (induct xs) simp_all<br />
<br />
text {*<br />
--------------------------------------------------------------------- <br />
Ejercicio 8.2. Demostrar o refutar detalladamente<br />
algunos P (xs @ ys) = (algunos P xs ∨ algunos P ys)<br />
--------------------------------------------------------------------- <br />
*}<br />
<br />
(* ivamenjim ferrenseg marpoldia1 wilmorort*)<br />
<br />
lemma algunos_append:<br />
"algunos P (xs @ ys) = (algunos P xs ∨ algunos P ys)"<br />
proof (induct xs)<br />
show "algunos P ([] @ ys) = (algunos P [] ∨ algunos P ys)" by simp<br />
next <br />
fix a xs<br />
assume HI: "algunos P (xs @ ys) = (algunos P xs ∨ algunos P ys)"<br />
have "algunos P ((a # xs) @ ys) = (P a ∨ (algunos P (xs @ ys)))" by simp<br />
also have "... = (P a ∨ (algunos P xs ∨ algunos P ys))" using HI by simp<br />
finally show "algunos P ((a # xs) @ ys) = (algunos P (a # xs) ∨ algunos P ys)" by simp<br />
qed<br />
<br />
(* migtermor crigomgom *)<br />
<br />
lemma algunos_append:<br />
"algunos P (xs @ ys) = (algunos P xs ∨ algunos P ys)" (is "?Q xs")<br />
proof (induct xs)<br />
show "?Q []" by simp<br />
next <br />
fix x xs<br />
assume HI: "?Q xs"<br />
have "algunos P ((x#xs) @ ys) = algunos P (x#(xs @ ys))" by simp<br />
also have "… = ((P x) ∨ (algunos P (xs @ ys)))" by simp<br />
also have "… = ((P x) ∨ (algunos P xs) ∨ (algunos P ys))" using HI by simp<br />
also have "… = ((algunos P (x#xs)) ∨ (algunos P ys))" by simp<br />
finally show "algunos P ((x#xs) @ ys) = (algunos P (x#xs) ∨ algunos P ys)" by simp<br />
qed<br />
<br />
(* paupeddeg*)<br />
lemma algunos_append:<br />
"algunos P (xs @ ys) = (algunos P xs ∨ algunos P ys)"<br />
proof (induct xs)<br />
show "algunos P ([] @ ys) = (algunos P [] ∨ algunos P ys)" by simp<br />
next<br />
fix a xs<br />
assume HI: "algunos P (xs @ ys) = (algunos P xs ∨ algunos P ys)"<br />
have "algunos P ((a # xs) @ ys) = (algunos P [a] ∨ ( algunos P (xs @ ys)))" by simp<br />
also have "... = (algunos P [a] ∨ (algunos P xs ∨ algunos P ys))" using HI by simp<br />
finally show "algunos P ((a # xs) @ ys) = (algunos P (a # xs) ∨ algunos P ys)" by simp<br />
qed<br />
<br />
text {*<br />
--------------------------------------------------------------------- <br />
Ejercicio 9.1. Demostrar o refutar automáticamente<br />
algunos P (rev xs) = algunos P xs<br />
--------------------------------------------------------------------- <br />
*}<br />
<br />
(* ivamenjim migtermor marpoldia1 rubgonmar *)<br />
<br />
lemma "algunos P (rev xs) = algunos P xs"<br />
apply (induct xs)<br />
apply simp<br />
apply (simp add: algunos_append)<br />
apply auto<br />
done<br />
<br />
(* ferrenseg crigomgom*)<br />
lemma "algunos P (rev xs) = algunos P xs"<br />
by (induct xs) (auto simp add: algunos_append)<br />
<br />
(* wilmorort *)<br />
lemma "algunos P (rev xs) = algunos P xs"<br />
by (induct xs,simp,simp add: algunos_append,auto) <br />
<br />
text {*<br />
--------------------------------------------------------------------- <br />
Ejercicio 9.2. Demostrar o refutar detalladamente<br />
algunos P (rev xs) = algunos P xs<br />
--------------------------------------------------------------------- <br />
*}<br />
<br />
lemma "algunos P (rev xs) = algunos P xs"<br />
oops<br />
<br />
(* migtermor *)<br />
lemma auxiliar1:<br />
"rev (a#xs) = rev xs @ [a]"<br />
by auto <br />
<br />
lemma "algunos P (rev xs) = algunos P xs" (is "?Q xs")<br />
proof (induct xs)<br />
show "?Q []" by simp<br />
next<br />
fix x xs<br />
assume HI: "?Q xs"<br />
have "algunos P (rev (x#xs)) = (algunos P (rev xs @ [x]))" using auxiliar1 by simp<br />
also have "… = ((algunos P (rev xs)) ∨ (algunos P [x]))" by (simp add: algunos_append)<br />
also have "… = ((P x) ∨ (algunos P (rev xs)))" by simp arith<br />
also have "… = ((P x) ∨ (algunos P xs))" using HI by simp<br />
finally show "algunos P (rev (x#xs)) = (algunos P (x#xs))" by simp<br />
qed<br />
<br />
(* ferrenseg *)<br />
<br />
lemma "algunos P (rev xs) = algunos P xs"<br />
proof (induct xs)<br />
show "algunos P (rev []) = algunos P []" by simp<br />
next<br />
fix x xs<br />
assume HI: "algunos P (rev xs) = algunos P xs"<br />
show "algunos P (rev (x # xs)) = algunos P (x # xs)"<br />
proof -<br />
have "algunos P (rev (x # xs)) = algunos P ((rev xs) @ [x])" by simp<br />
also have "… = (algunos P (rev xs) ∨ algunos P [x])" <br />
by (simp add: algunos_append)<br />
also have "… = (algunos P xs ∨ algunos P [x])" using HI by simp<br />
also have "… = (algunos P xs ∨ P x)" by simp<br />
also have "… = (P x ∨ algunos P xs)" by auto<br />
also have "… = algunos P (x # xs)" by simp<br />
finally show ?thesis<br />
qed<br />
qed<br />
<br />
(* ivamenjim marpoldia1*)<br />
<br />
lemma "algunos P (rev xs) = algunos P xs" (is "?P xs")<br />
proof (induct xs)<br />
show "?P []" by simp<br />
next <br />
fix a xs<br />
assume HI: "?P xs"<br />
have "algunos P (rev (a # xs)) = (algunos P ((rev xs)@[a]))" by simp<br />
also have "... = ((algunos P (rev xs)) ∨ algunos P [a])" by (simp add: algunos_append)<br />
also have "... = (algunos P xs ∨ algunos P [a])" using HI by simp<br />
also have "... = (algunos P [a] ∨ algunos P xs)" by arith<br />
finally show "algunos P (rev (a # xs)) = algunos P (a # xs)" by simp<br />
qed<br />
<br />
(*paupeddeg*)<br />
lemma "algunos P (rev xs) = algunos P xs"<br />
proof (induct xs)<br />
show "algunos P (rev []) = algunos P []" by simp<br />
next<br />
fix a xs<br />
assume HI:"algunos P (rev xs) = algunos P xs"<br />
have "algunos P (rev (a # xs)) = algunos P ((rev xs) @ [a])" by simp<br />
also have "... = ((algunos P (rev xs)) ∨ (algunos P [a]))" by (simp add: algunos_append)<br />
also have "... = ((algunos P xs) ∨ (algunos P [a]))" using HI by simp<br />
also have "... = ((algunos P [a]) ∨ (algunos P xs))" by arith<br />
finally show "algunos P (rev (a # xs)) = algunos P (a # xs)" by simp<br />
qed<br />
<br />
(*crigomgom *)<br />
lemma "algunos P (rev xs) = algunos P xs"<br />
proof (induct xs)<br />
show "algunos P (rev []) = algunos P []" by simp<br />
next<br />
fix x xs<br />
assume HI: " algunos P (rev xs) = algunos P xs" <br />
have "algunos P (rev (x # xs)) = algunos P ((rev xs) @ [x])" by simp<br />
also have "... = (algunos P (rev xs) ∨ algunos P [x])" by (simp add: algunos_append)<br />
also have "... = (algunos P xs ∨ algunos P [x])" using HI by simp<br />
also have "... = (algunos P xs ∨ P x)" by simp<br />
also have "... = (P x ∨ algunos P xs)" by arith<br />
also have "... = algunos P (x#xs)" by simp<br />
finally show "algunos P (rev (x # xs)) = algunos P (x # xs)" by simp<br />
qed<br />
<br />
<br />
(*wilmorort*)<br />
lemma "algunos P (rev xs) = algunos P xs" (is "?P xs")<br />
proof (induct xs) <br />
show "?P []" by simp <br />
next<br />
fix a xs<br />
assume HI: "?P xs"<br />
have "algunos P (rev (a#xs)) = algunos P ((rev xs) @ [a])" by simp<br />
also have "... = (algunos P(rev xs) ∨ algunos P [a])" by (simp add: algunos_append)<br />
also have "... = (algunos P xs ∨ algunos P [a])" using HI by simp<br />
also have "... = (algunos P [a] ∨ algunos P xs)" by (simp add: HOL.disj_commute)<br />
also have "... = algunos P([a]@(xs))" by (simp)<br />
finally show "algunos P (rev (a#xs))= algunos P (a # xs)" by simp<br />
qed<br />
<br />
text {*<br />
--------------------------------------------------------------------- <br />
Ejercicio 10. Encontrar un término no trivial Z tal que sea cierta la <br />
siguiente ecuación:<br />
algunos (λx. P x ∨ Q x) xs = Z<br />
y demostrar la equivalencia de forma automática y detallada.<br />
--------------------------------------------------------------------- <br />
*}<br />
<br />
(* migtermor *)<br />
<br />
lemma "algunos (λx. P x ∨ Q x) xs = ((algunos (λx. P x) xs) ∨ (algunos (λx. Q x) xs))"<br />
by (induct xs) auto<br />
<br />
lemma "algunos (λx. P x ∨ Q x) xs = ((algunos (λx. P x) xs) ∨ (algunos (λx. Q x) xs))" (is "?R xs")<br />
proof (induct xs)<br />
show "?R []" by simp<br />
next<br />
fix x xs<br />
assume HI: "?R xs"<br />
have "algunos (λx. P x ∨ Q x) (x#xs) = (((λx. P x ∨ Q x) x) ∨ (algunos (λx. P x ∨ Q x) xs))"<br />
by simp<br />
also have "… = ((P x ∨ Q x) ∨ (algunos (λx. P x ∨ Q x) xs))" by simp<br />
also have H1: "… = ((((λx. P x) x) ∨ (algunos (λx. P x) xs)) ∨ (((λx. Q x) x) ∨ (algunos (λx. Q x) xs)))"<br />
using HI by simp arith<br />
have H2: "… = ((algunos (λx. P x) (x#xs)) ∨ (algunos (λx. Q x) (x#xs)))" <br />
by simp<br />
have C: "(algunos (λx. P x ∨ Q x) (x#xs)) = <br />
((algunos (λx. P x) (x#xs)) ∨ (algunos (λx. Q x) (x#xs)))" <br />
using H1 H2 by simp<br />
finally show "(algunos (λx. P x ∨ Q x) (x#xs)) = <br />
((algunos (λx. P x) (x#xs)) ∨ (algunos (λx. Q x) (x#xs)))" <br />
using C by simp<br />
qed<br />
<br />
<br />
(* ferrenseg *)<br />
<br />
lemma "algunos (λx. P x ∨ Q x) xs = (algunos P xs ∨ algunos Q xs)"<br />
proof (induct xs)<br />
show "algunos (λx. P x ∨ Q x) [] = (algunos P [] ∨ algunos Q [])" by simp<br />
next<br />
fix x xs<br />
assume HI: "algunos (λx. (P x ∨ Q x)) xs = (algunos P xs ∨ algunos Q xs)"<br />
show "algunos (λx. P x ∨ Q x) (x # xs) = (algunos P (x # xs) ∨ algunos Q (x # xs))"<br />
proof -<br />
have "algunos (λx. P x ∨ Q x) (x # xs) = <br />
((P x) ∨ (Q x) ∨ algunos (λx. P x ∨ Q x) xs)" by simp<br />
also have "… = ((P x) ∨ (Q x) ∨ algunos P xs ∨ algunos Q xs)" <br />
using HI by simp<br />
also have "… = (((P x) ∨ algunos P xs) ∨ ((Q x) ∨ algunos Q xs))" by auto<br />
also have "… = (algunos P (x # xs) ∨ algunos Q (x # xs))" by simp<br />
finally show ?thesis <br />
qed<br />
qed<br />
<br />
(* ivamenjim marpoldia1 *)<br />
<br />
lemma "algunos (λx. P x ∨ Q x) xs = (algunos P xs ∨ algunos Q xs)" <br />
by (induct xs) auto<br />
<br />
lemma "algunos (λx. P x ∨ Q x) xs = (algunos P xs ∨ algunos Q xs)"<br />
proof (induct xs)<br />
show "algunos (λx. P x ∨ Q x) [] = (algunos P [] ∨ algunos Q [])" by simp<br />
next <br />
fix a xs<br />
assume HI: "algunos (λx. P x ∨ Q x) xs = (algunos P xs ∨ algunos Q xs)"<br />
have "algunos (λx. P x ∨ Q x) (a # xs) = (P a ∨ Q a ∨ algunos (λx. P x ∨ Q x) xs)" by simp<br />
also have "... = (P a ∨ Q a ∨ algunos P xs ∨ algunos Q xs)" using HI by simp<br />
finally show "algunos (λx. P x ∨ Q x) (a # xs) = (algunos P (a # xs) ∨ algunos Q (a # xs))" by auto<br />
qed<br />
<br />
<br />
(* wilmorort *)<br />
lemma "algunos (λx. P x ∨ Q x) xs = (¬todos(λx. ¬P x)xs ∨ ¬todos(λx. ¬Q x)xs)"<br />
by (induct xs) auto<br />
<br />
<br />
lemma "algunos (λx. P x ∨ Q x) xs = (¬todos(λx. ¬P x)xs ∨ ¬todos(λx. ¬Q x)xs)"<br />
(is "?P xs")<br />
proof (induct xs)<br />
show "?P []" by simp<br />
next<br />
fix a xs<br />
assume HI: "?P xs"<br />
have "algunos (λx. P x ∨ Q x) (a # xs) = (P a ∨ Q a ∨ algunos (λx. P x ∨ Q x) xs)" by simp<br />
also have "... = (P a ∨ Q a ∨ (¬ todos (λx. ¬ P x) xs ∨ ¬ todos (λx. ¬ Q x) xs))" using HI by simp<br />
also have "... = (P a ∨ ¬ todos (λx. ¬ P x) xs ∨ Q a ∨ ¬ todos (λx. ¬ Q x) xs )" by arith<br />
finally show "algunos (λx. P x ∨ Q x) (a # xs) = (~ todos (λx. ¬ P x) (a # xs) ∨ ~ todos (λx. ¬ Q x) (a # xs))" by simp<br />
qed<br />
<br />
<br />
<br />
text {*<br />
--------------------------------------------------------------------- <br />
Ejercicio 11.1. Demostrar o refutar automáticamente<br />
algunos P xs = (¬ todos (λx. (¬ P x)) xs)<br />
--------------------------------------------------------------------- <br />
*}<br />
<br />
(* ivamenjim wilmorort migtermor marpoldia1 crigomgom rubgonmar *)<br />
<br />
lemma "algunos P xs = (¬ todos (λx. (¬ P x)) xs)"<br />
by (induct xs) auto<br />
<br />
(* ferrenseg *)<br />
<br />
lemma "algunos P xs = (¬ todos (λx. (¬ P x)) xs)"<br />
by (induct xs) simp_all<br />
<br />
text {*<br />
--------------------------------------------------------------------- <br />
Ejercicio 11.2. Demostrar o refutar datalladamente<br />
algunos P xs = (¬ todos (λx. (¬ P x)) xs)<br />
--------------------------------------------------------------------- <br />
*}<br />
<br />
(* ivamenjim migtermor marpoldia1 crigomgom*)<br />
<br />
lemma "algunos P xs = (¬ todos (λx. (¬ P x)) xs)"<br />
proof (induct xs)<br />
show "algunos P [] = (¬ todos (λx. ¬ P x) [])" by simp<br />
next <br />
fix a xs<br />
assume HI: "algunos P xs = (¬ todos (λx. ¬ P x) xs)"<br />
have "algunos P (a # xs) = (P a ∨ (algunos P xs))" by simp<br />
also have "... = (P a ∨ (¬ todos (λx. ¬ P x) xs))" using HI by simp<br />
finally show "algunos P (a # xs) = (¬ todos (λx. ¬ P x) (a # xs))" by simp<br />
qed<br />
<br />
(* ferrenseg *)<br />
<br />
lemma "algunos P xs = (¬ todos (λx. (¬ P x)) xs)"<br />
proof (induct xs)<br />
show "algunos P [] = (¬ todos (λx. (¬ P x)) [])" by simp<br />
next<br />
fix x xs<br />
assume HI: "algunos P xs = (¬ todos (λx. (¬ P x)) xs)"<br />
show "algunos P (x # xs) = (¬ todos (λx. (¬ P x)) (x # xs))"<br />
proof - <br />
have "algunos P (x # xs) = ((P x) ∨ algunos P xs)" by simp<br />
also have "… = ((P x) ∨ ¬ todos (λx. (¬ P x)) xs)" using HI by simp<br />
also have "… = (¬ (¬ (P x) ∧ todos (λx. (¬ P x)) xs))" by simp<br />
also have "… = (¬ todos (λx. (¬ P x)) (x # xs))" by simp<br />
finally show ?thesis<br />
qed<br />
qed<br />
<br />
text {*<br />
--------------------------------------------------------------------- <br />
Ejercicio 12. Definir la funcion primitiva recursiva <br />
estaEn :: 'a ⇒ 'a list ⇒ bool<br />
tal que (estaEn x xs) se verifica si el elemento x está en la lista<br />
xs. Por ejemplo, <br />
estaEn (2::nat) [3,2,4] = True<br />
estaEn (1::nat) [3,2,4] = False<br />
--------------------------------------------------------------------- <br />
*}<br />
<br />
(* ivamenjim ferrenseg migtermor serrodcal crigomgom rubgonmar marpoldia1*)<br />
<br />
fun estaEn :: "'a ⇒ 'a list ⇒ bool" where<br />
"estaEn x [] = False"<br />
| "estaEn x (a#xs) = ((a = x) ∨ (estaEn x xs))"<br />
<br />
value "estaEn (2::nat) [3,2,4] = True"<br />
value "estaEn (1::nat) [3,2,4] = False"<br />
<br />
<br />
text {*<br />
--------------------------------------------------------------------- <br />
Ejercicio 13. Expresar la relación existente entre estaEn y algunos. <br />
Demostrar dicha relación de forma automática y detallada.<br />
--------------------------------------------------------------------- <br />
*}<br />
<br />
(* migtermor crigomgom*)<br />
<br />
lemma "estaEn x xs = (algunos (λa. a=x) xs)"<br />
by (induct xs) auto<br />
<br />
lemma auxiliar13:<br />
"(x=a) = ((λa. a=x) a)"<br />
by auto<br />
<br />
lemma "estaEn x xs = (algunos (λa. a=x) xs)" (is "?P xs")<br />
proof (induct xs)<br />
show "?P []" by simp<br />
next<br />
fix a xs<br />
assume HI: "?P xs"<br />
have "estaEn x (a#xs) = ((a=x) ∨ (estaEn x xs))" by simp<br />
also have H: "… = ((a=x) ∨ (algunos (λa. a=x) xs))" using HI by simp<br />
also have "… = (((λa. a=x) a) ∨ (algunos (λa. a=x) xs))" using auxiliar13 by simp<br />
also have "… = (algunos (λa. a=x) (a#xs))" by simp<br />
have C: "estaEn x (a#xs) = (algunos (λa. a=x) (a#xs))" using H by simp<br />
finally show "estaEn x (a#xs) = (algunos (λa. a=x) (a#xs))" using C by simp<br />
qed<br />
<br />
<br />
(* ferrenseg *)<br />
<br />
lemma estaEn_algunos:<br />
"estaEn y xs = algunos (λx. x=y) xs"<br />
by (induct xs) auto<br />
<br />
lemma estaEn_algunos_2:<br />
"estaEn y xs = algunos (λx. x=y) xs"<br />
proof (induct xs)<br />
show "estaEn y [] = algunos (λx. x=y) []" by simp<br />
next<br />
fix x xs<br />
assume HI: "estaEn y xs = algunos (λx. x=y) xs"<br />
show "estaEn y (x # xs) = algunos (λx. x=y) (x # xs)"<br />
proof -<br />
have "estaEn y (x # xs) = (y=x ∨ estaEn y xs)" by simp<br />
also have "… = (y=x ∨ algunos (λx. x=y) xs)" using HI by simp<br />
also have "… = (x=y ∨ algunos (λx. x=y) xs)" by auto<br />
also have "… = algunos (λx. x=y) (x # xs)" by simp<br />
finally show ?thesis<br />
qed<br />
qed<br />
<br />
(*crigomgom*)<br />
lemma "estaEn x xs = (algunos (λa. a=x) xs)" <br />
proof (induct xs)<br />
show " estaEn x [] = algunos (λa. a = x) []" by simp<br />
next<br />
fix a xs<br />
assume HI: "estaEn x xs = algunos (λa. a = x) xs"<br />
have "estaEn x (a # xs) = (a = x ∨ estaEn x xs)" by simp<br />
also have "... = (a = x ∨ algunos (λa. a = x) xs)" using HI by simp<br />
finally show " estaEn x (a # xs) = algunos (λa. a = x) (a # xs)" by simp<br />
qed<br />
<br />
(* ivamenjim marpoldia1 *)<br />
<br />
lemma "estaEn x xs = algunos (λy. x=y) xs"<br />
by (induct xs) auto<br />
<br />
lemma "estaEn x xs = algunos (λy. x=y) xs"<br />
proof (induct xs)<br />
show "estaEn x [] = algunos (op = x) []" by simp<br />
next <br />
fix a xs<br />
assume HI: "estaEn x xs = algunos (op = x) xs"<br />
have "estaEn x (a # xs) = ((a = x) ∨ (estaEn x xs))" by simp<br />
also have "... = (a = x ∨ algunos (op = x) xs)" using HI by simp<br />
finally show " estaEn x (a # xs) = algunos (op = x) (a # xs)" by auto<br />
qed<br />
<br />
end<br />
</source></div>Lucnovdoshttps://www.glc.us.es/~jalonso/RA2016/index.php?title=Relaci%C3%B3n_4&diff=580Relación 42016-11-22T17:43:15Z<p>Lucnovdos: </p>
<hr />
<div><source lang="isar"><br />
chapter {* R4: Cuantificadores sobre listas *}<br />
<br />
theory R4_Cuantificadores_sobre_listas<br />
imports Main <br />
begin<br />
<br />
text {* <br />
--------------------------------------------------------------------- <br />
Ejercicio 1. Definir la función <br />
todos :: ('a ⇒ bool) ⇒ 'a list ⇒ bool<br />
tal que (todos p xs) se verifica si todos los elementos de la lista <br />
xs cumplen la propiedad p. Por ejemplo, se verifica <br />
todos (λx. 1<length x) [[2,1,4],[1,3]]<br />
¬todos (λx. 1<length x) [[2,1,4],[3]]<br />
<br />
Nota: La función todos es equivalente a la predefinida list_all. <br />
--------------------------------------------------------------------- <br />
*}<br />
<br />
{*danrodcha*}<br />
fun todos :: "('a ⇒ bool) ⇒ 'a list ⇒ bool" where<br />
"todos p xs = "todos p [] = True"<br />
| "todos p (x#xs) = (p x ∧ todos p xs)""<br />
<br />
(* ivamenjim migtermor dancorgar wilmorort marpoldia1 ferrenseg paupeddeg pablucoto crigomgom anaprarod serrodcal juacabsou rubgonmar josgarsan fraortmoy lucnovdos*)<br />
fun todos :: "('a ⇒ bool) ⇒ 'a list ⇒ bool" where<br />
"todos p [] = True"<br />
| "todos p (x#xs) = (p x ∧ todos p xs)"<br />
<br />
text {* <br />
--------------------------------------------------------------------- <br />
Ejercicio 2. Definir la función <br />
algunos :: ('a ⇒ bool) ⇒ 'a list ⇒ bool<br />
tal que (algunos p xs) se verifica si algunos elementos de la lista <br />
xs cumplen la propiedad p. Por ejemplo, se verifica <br />
algunos (λx. 1<length x) [[2,1,4],[3]]<br />
¬algunos (λx. 1<length x) [[],[3]]"<br />
<br />
Nota: La función algunos es equivalente a la predefinida list_ex. <br />
--------------------------------------------------------------------- <br />
*}<br />
<br />
{*danrodcha ivamenjim migtermor dancorgar marpoldia1 ferrenseg wilmorort paupeddeg pablucoto crigomgom anaprarod serrodcal juacabsou rubgonmar josgarsan fraortmoy lucnovdos*}<br />
fun algunos :: "('a ⇒ bool) ⇒ 'a list ⇒ bool" where<br />
"algunos p [] = False"<br />
| "algunos p (x#xs) = (p x ∨ algunos p xs)"<br />
<br />
text {*<br />
--------------------------------------------------------------------- <br />
Ejercicio 3.1. Demostrar o refutar automáticamente <br />
todos (λx. P x ∧ Q x) xs = (todos P xs ∧ todos Q xs)<br />
--------------------------------------------------------------------- <br />
*}<br />
<br />
{*danrodcha ivamenjim migtermor dancorgar marpoldia1 ferrenseg wilmorort paupeddeg pablucoto anaprarod serrodcal juacabsou rubgonmar josgarsan fraortmoy*}<br />
lemma "todos (λx. P x ∧ Q x) xs = (todos P xs ∧ todos Q xs)"<br />
by (induct xs) auto<br />
<br />
text {*<br />
--------------------------------------------------------------------- <br />
Ejercicio 3.2. Demostrar o refutar detalladamente<br />
todos (λx. P x ∧ Q x) xs = (todos P xs ∧ todos Q xs)<br />
--------------------------------------------------------------------- <br />
*}<br />
<br />
{*danrodcha*}<br />
lemma "todos (λx. P x ∧ Q x) xs = (todos P xs ∧ todos Q xs)"<br />
proof (induct xs)<br />
show "?R []" by simp<br />
next<br />
fix a xs <br />
assume HI: "?R xs"<br />
have "todos (λx. P x ∧ Q x) (a#xs) = (P a ∧ Q a ∧ todos (λx. P x ∧ Q x) xs)" by simp<br />
also have "… = (P a ∧ Q a ∧ (todos P xs ∧ todos Q xs))" using HI by simp<br />
also have "… = ((P a ∧ todos P xs) ∧ (Q a ∧ todos Q xs))" by blast<br />
also have "… = (todos P (a#xs) ∧ todos Q (a#xs))" by simp<br />
finally show "?R (a#xs)" by simp<br />
qed<br />
<br />
(* ivamenjim wilmorort serrodcal josgarsan*)<br />
<br />
lemma "todos (λx. P x ∧ Q x) xs = (todos P xs ∧ todos Q xs)"<br />
proof (induct xs)<br />
show "todos (λx. P x ∧ Q x) [] = (todos P [] ∧ todos Q [])" by simp<br />
next <br />
fix a xs<br />
assume HI: "todos (λx. P x ∧ Q x) xs = (todos P xs ∧ todos Q xs)"<br />
have "todos (λx. P x ∧ Q x) (a # xs) = (P a ∧ Q a ∧ todos (λx. P x ∧ Q x) xs)" by simp<br />
also have "... = (P a ∧ Q a ∧ todos P xs ∧ todos Q xs)" using HI by simp<br />
finally show "todos (λx. P x ∧ Q x) (a # xs) = (todos P (a # xs) ∧ todos Q (a # xs))" by auto<br />
qed<br />
<br />
(* dancorgar paupeddeg *)<br />
lemma "todos (λx. P x ∧ Q x) xs = (todos P xs ∧ todos Q xs)"<br />
proof (induct xs)<br />
show "todos (λx. P x ∧ Q x) [] = (todos P [] ∧ todos Q [])" by simp<br />
next<br />
fix y xs<br />
assume HI: "todos (λx. P x ∧ Q x) xs = (todos P xs ∧ todos Q xs)"<br />
have "todos (λx. P x ∧ Q x) (y#xs) = ((P y ∧ Q y) ∧ (todos (λx. P x ∧ Q x) xs))" by simp<br />
also have "… = ((P y ∧ Q y) ∧ (todos P xs ∧ todos Q xs))" using HI by simp<br />
also have "… = ((P y ∧ todos P xs) ∧ (Q y ∧ todos Q xs))" by blast<br />
also have "… = ((todos P (y#xs)) ∧ (todos Q (y#xs)))" by simp<br />
finally show "todos (λx. P x ∧ Q x) (y#xs) = (todos P (y#xs) ∧ todos Q (y#xs))" by simp<br />
qed<br />
<br />
<br />
(* migtermor *)<br />
lemma "todos (λx. P x ∧ Q x) xs = (todos P xs ∧ todos Q xs)"<br />
proof (induct xs)<br />
show "todos (λx. P x ∧ Q x) [] = (todos P [] ∧ todos Q [])" by simp<br />
next <br />
fix x xs<br />
assume HI: "todos (λx. P x ∧ Q x) xs = (todos P xs ∧ todos Q xs)"<br />
have "(todos (λx. P x ∧ Q x) (x#xs)) = (( P x ∧ Q x) ∧ (todos (λx. P x ∧ Q x) xs))"<br />
by (simp only: todos.simps(2))<br />
also have "… = ((P x ∧ Q x) ∧ (todos P xs ∧ todos Q xs))" using HI by simp<br />
also have "… = (((P x)∧(todos P xs)) ∧ ((Q x) ∧ (todos Q xs)))" by arith<br />
also have "((P x ∧ Q x) ∧ (todos P xs ∧ todos Q xs)) = (((P x)∧(todos P xs)) ∧ ((Q x) ∧ (todos Q xs)))"<br />
by arith (* Este paso es exactamente el mismo que el anterior, pero sin cualquiera de los dos no funciona el "finally show" *)<br />
have "… = (((P x)∧(todos P xs))∧((Q x)∧(todos Q xs)))" by simp<br />
have "… = ((todos P (x#xs))∧(todos Q (x#xs)))" by simp<br />
finally show "(todos (λx. P x ∧ Q x) (x#xs)) = ((todos P (x#xs)) ∧ (todos Q (x#xs)))" by simp <br />
qed<br />
<br />
(* marpoldia1 *)<br />
lemma "todos (λx. P x ∧ Q x) xs = (todos P xs ∧ todos Q xs)"<br />
proof (induct xs)<br />
show "todos (λx. P x ∧ Q x) [] = (todos P [] ∧ todos Q [])" by simp<br />
next<br />
fix a xs<br />
assume HI: "todos (λx. P x ∧ Q x) xs = (todos P xs ∧ todos Q xs)" <br />
have "todos (λx. P x ∧ Q x) (a # xs) = ((P a ∧ Q a) ∧ (todos P xs ∧ todos Q xs))" using HI by simp<br />
also have "... = ((P a ∧ todos P xs) ∧ (Q a ∧ todos Q xs))" by blast<br />
also have "... = (todos P (a#xs) ∧ todos Q (a#xs)) " by simp<br />
finally show "todos (λx. P x ∧ Q x) (a#xs) = (todos P (a#xs) ∧ todos Q (a#xs))" by simp <br />
qed<br />
<br />
(* ferrenseg *)<br />
<br />
lemma "todos (λx. P x ∧ Q x) xs = (todos P xs ∧ todos Q xs)"<br />
proof (induct xs)<br />
show "todos (λx. P x ∧ Q x) [] = (todos P [] ∧ todos Q [])" by simp<br />
next <br />
fix n xs<br />
assume HI: "todos (λx. P x ∧ Q x) xs = (todos P xs ∧ todos Q xs)"<br />
have "todos (λx. P x ∧ Q x) (n # xs) = (P n ∧ Q n ∧ todos P xs ∧ todos Q xs)" using HI by simp<br />
also have "… = P n ∧ Q n ∧ todos P xs ∧ todos Q xs" by blast<br />
also have "⋯ = todos P (n # xs) ∧ todos Q (n # xs)" by simp <br />
finally show "todos (λx. P x ∧ Q x) (n # xs) = todos P (n # xs) ∧ todos Q (n # xs)" by simp<br />
qed<br />
<br />
{* wilmorort pablucoto crigomgom anaprarod juacabsou rubgonmar fraortmoy *}<br />
<br />
lemma "todos (λx. P x ∧ Q x) xs = (todos P xs ∧ todos Q xs)"<br />
proof (induct xs)<br />
show "todos (λx. P x ∧ Q x) [] = (todos P [] ∧ todos Q [])" by simp <br />
next<br />
fix a xs<br />
assume HI: "todos (λx. P x ∧ Q x) xs = (todos P xs ∧ todos Q xs)"<br />
have "todos (λx. P x ∧ Q x) (a # xs) = (P a ∧ Q a ∧ todos (λx. P x ∧ Q x) xs)" by simp<br />
also have "... = (P a ∧ Q a ∧ todos P xs ∧ todos Q xs)" using HI by simp<br />
also have "... = (P a ∧ todos P xs ∧ Q a ∧ todos Q xs)" by arith<br />
also have "... = (todos P (a # xs) ∧ todos Q (a # xs))" by simp (* Este paso se puede obviar*)<br />
finally show "todos (λx. P x ∧ Q x) (a # xs) = (todos P (a # xs) ∧ todos Q (a # xs))" by simp<br />
qed<br />
<br />
text {*<br />
--------------------------------------------------------------------- <br />
Ejercicio 4.1. Demostrar o refutar automáticamente<br />
todos P (x @ y) = (todos P x ∧ todos P y)<br />
--------------------------------------------------------------------- <br />
*}<br />
{*danrodcha ivamenjim marpoldia1 migtermor ferrenseg wilmorort paupeddeg crigomgom anaprarod serrodcal juacabsou rubgonmar pablucoto fraortmoy *}<br />
lemma "todos P (x @ y) = (todos P x ∧ todos P y)"<br />
by (induct x) auto<br />
<br />
{* anaprarod dancorgar *}<br />
lemma "todos P (x @ y) = (todos P x ∧ todos P y)"<br />
apply (induct x)<br />
apply auto<br />
done<br />
<br />
text {*<br />
--------------------------------------------------------------------- <br />
Ejercicio 4.2. Demostrar o refutar detalladamente<br />
todos P (x @ y) = (todos P x ∧ todos P y)<br />
--------------------------------------------------------------------- <br />
*}<br />
<br />
(* ivamenjim ferrenseg wilmorort dancorgar fraortmoy *)<br />
<br />
lemma todos_append:<br />
"todos P (x @ y) = (todos P x ∧ todos P y)"<br />
proof (induct x)<br />
show "todos P ([] @ y) = (todos P [] ∧ todos P y)" by simp<br />
next <br />
fix a x<br />
assume HI: "todos P (x @ y) = (todos P x ∧ todos P y)"<br />
have "todos P ((a#x) @ y) = (P a ∧ (todos P (x @ y)))" by simp<br />
also have "... = (P a ∧ (todos P x ∧ todos P y))" using HI by simp<br />
finally show "todos P ((a # x) @ y) = (todos P (a # x) ∧ todos P y)" by simp<br />
qed<br />
<br />
(* migtermor serrodcal *)<br />
<br />
lemma todos_append1:<br />
"todos P (x @ y) = (todos P x ∧ todos P y)" (is "?P x")<br />
proof (induct x)<br />
show "?P []" by simp<br />
next <br />
fix a x<br />
assume HI: "?P x"<br />
have "todos P ((a#x) @ y) = (P a ∧ (todos P (x @ y)))" by simp<br />
also have "… = (P a ∧ (todos P x ∧ todos P y))" using HI by simp<br />
finally show "?P (a#x)" by simp<br />
qed <br />
<br />
(* marpoldia1 *)<br />
lemma todos_append:<br />
"todos P (x @ y) = (todos P x ∧ todos P y)"<br />
proof (induct x)<br />
show "todos P ([] @ y) = (todos P [] ∧ todos P y)" by simp<br />
next<br />
fix a x<br />
assume HI: "todos P (x @ y) = (todos P x ∧ todos P y)"<br />
have "todos P ((a # x) @ y) = todos P (a#(x@y))" by simp<br />
also have "... = (P a ∧ todos P (x @ y))" by simp<br />
also have "... = (P a ∧ (todos P x ∧ todos P y))" using HI by simp<br />
finally show "todos P ((a # x) @ y) = (todos P (a # x) ∧ todos P y)" by simp <br />
qed<br />
<br />
(* paupeddeg *)<br />
lemma todos_append3:<br />
"todos P (x @ y) = (todos P x ∧ todos P y)"<br />
proof (induct x)<br />
show "todos P ([] @ y) = (todos P [] ∧ todos P y)" by simp<br />
next<br />
fix a x<br />
assume HI: "todos P (x @ y) = (todos P x ∧ todos P y)"<br />
have "todos P ((a # x) @ y) = todos P (a # (x @ y))" by simp<br />
also have "... = (todos P [a] ∧ todos P (x @ y)) " by simp<br />
also have "... = (todos P [a] ∧ (todos P x ∧ todos P y))" using HI by simp<br />
finally show "todos P ((a # x) @ y) = (todos P (a # x) ∧ todos P y)" by simp<br />
qed<br />
<br />
(* crigomgom juacabsou anaprarod*)<br />
<br />
lemma todos_append4:<br />
"todos P (x @ y) = (todos P x ∧ todos P y)"<br />
proof (induct x)<br />
show "todos P ([] @ y) = (todos P [] ∧ todos P y)" by simp<br />
next<br />
fix x xs<br />
assume HI:"todos P (xs @ y) = (todos P xs ∧ todos P y)"<br />
have "todos P ((x # xs) @ y) = (P x ∧ todos P (xs @ y ))" by simp<br />
also have "... = (P x ∧ (todos P xs ∧ todos P y)) " using HI by simp<br />
also have "... = ((P x ∧ todos P xs) ∧ todos P y)" by simp<br />
finally show "todos P ((x # xs) @ y) = (todos P (x # xs) ∧ todos P y)" by simp<br />
qed<br />
<br />
(* pablucoto *)<br />
lemma todos_append5:<br />
"todos P (x @ y) = (todos P x ∧ todos P y)"<br />
proof (induct x )<br />
show "todos P ([] @ y) = (todos P [] ∧ todos P y)" by simp<br />
next<br />
fix a x<br />
assume HI:"todos P (x @ y) = (todos P x ∧ todos P y) " <br />
have "todos P ((a # x) @ y) = ( P a ∧ todos P (x @ y))" by simp<br />
also have " ... =( P a ∧ (todos P x ∧ todos P y))" using HI by simp<br />
also have "... = (todos P (a#x) ∧ todos P y)" by simp<br />
finally show "todos P ((a # x) @ y) = (todos P (a # x) ∧ todos P y)" by auto<br />
qed<br />
<br />
text {*<br />
--------------------------------------------------------------------- <br />
Ejercicio 5.1. Demostrar o refutar automáticamente <br />
todos P (rev xs) = todos P xs<br />
--------------------------------------------------------------------- <br />
*}<br />
<br />
lemma "todos P (rev xs) = todos P xs"<br />
oops<br />
<br />
(* migtermor ivamenjim marpoldia1 serrodcal anaprarod *)<br />
lemma "todos P (rev xs) = todos P xs" <br />
apply (induct xs)<br />
apply simp<br />
apply (simp add: todos_append)<br />
apply auto<br />
done<br />
<br />
(* ferrenseg crigomgom rubgonmar fraortmoy *)<br />
<br />
lemma "todos P (rev xs) = todos P xs"<br />
by (induct xs) (auto simp add: todos_append)<br />
<br />
(* wilmorort *)<br />
lemma "todos P (rev xs) = todos P xs"<br />
by (induct xs,simp,simp add: todos_append,auto)<br />
<br />
(* juacabsou *)<br />
lemma "todos P (rev xs) = todos P xs"<br />
apply (induct xs,simp,simp add: todos_append,auto) done<br />
<br />
(* dancorgar *)<br />
lemma "todos P (rev xs) = todos P xs"<br />
apply (induct xs)<br />
apply (simp add: todos_append)<br />
apply (simp add: todos_append)<br />
apply auto<br />
done<br />
<br />
(* pablucoto * )<br />
lemma "todos P (rev xs) = todos P xs"<br />
by (induct xs) (auto, simp_all add: todos_append) <br />
lemma "todos P (rev xs) = todos P xs"<br />
by (induct xs) ( simp_all add: todos_append, auto)<br />
<br />
text {*<br />
--------------------------------------------------------------------- <br />
Ejercicio 5.2. Demostrar o refutar detalladamente<br />
todos P (rev xs) = todos P xs<br />
--------------------------------------------------------------------- <br />
*}<br />
<br />
lemma "todos P (rev xs) = todos P xs"<br />
oops<br />
<br />
(* migtermor *)<br />
<br />
lemma auxiliar:<br />
"rev (a#xs) = rev xs @ [a]"<br />
by auto<br />
<br />
lemma "todos P (rev xs) = todos P xs" (is "?Q xs")<br />
proof (induct xs)<br />
show "?Q []" by (simp only: rev.simps(1))<br />
next <br />
fix a xs<br />
assume HI: "?Q xs"<br />
have "todos P (rev (a#xs)) = (todos P (rev xs @ [a]))" by (simp add: auxiliar)<br />
have "… = ((todos P (rev xs)) ∧ (todos P [a]))" by (simp add: todos_append)<br />
have "… = (todos P (rev xs) ∧ P a)" by simp<br />
also have Aux: "… = (todos P xs ∧ P a)" using HI by simp<br />
have Aux1: "… = (P a ∧ todos P xs)" by arith<br />
have "(todos P (rev xs) ∧ P a) = (P a ∧ todos P xs)" using Aux Aux1 by simp<br />
finally show "todos P (rev (a#xs)) = todos P (a#xs)" by (simp add: todos_append)<br />
qed<br />
<br />
<br />
(* marpoldia1 ferrenseg crigomgom serrodcal juacabsou rubgonmar *)<br />
lemma "todos P (rev xs) = todos P xs"<br />
proof (induct xs)<br />
show "todos P (rev []) = todos P []" by simp<br />
next<br />
fix a xs<br />
assume HI: "todos P (rev xs) = todos P xs"<br />
have "todos P (rev (a # xs)) = (todos P ((rev xs)@[a]))" by simp<br />
also have "... = (todos P (rev xs) ∧ todos P [a])" by (simp add:todos_append)<br />
also have "... = (todos P xs ∧ P a)" using HI by simp<br />
also have "... = (P a ∧ todos P xs)" by arith<br />
also have "... = (todos P (a#xs))" by simp<br />
finally show "todos P (rev (a # xs)) = (todos P (a#xs))" by simp <br />
qed<br />
<br />
(* ivamenjim *)<br />
<br />
lemma "todos P (rev xs) = todos P xs" (is "?P xs")<br />
proof (induct xs)<br />
show "?P []" by simp<br />
next <br />
fix a xs<br />
assume HI: "?P xs"<br />
have "todos P (rev (a # xs)) = (todos P ((rev xs)@[a]))" by simp<br />
also have "... = ((todos P (rev xs)) ∧ todos P [a])" by (simp add: todos_append)<br />
also have "... = (todos P xs ∧ todos P [a])" using HI by simp<br />
also have "... = (todos P [a] ∧ todos P xs)" by arith<br />
finally show "todos P (rev (a # xs)) = todos P (a # xs)" by simp<br />
qed<br />
<br />
(* fraortmoy *)<br />
<br />
lemma "todos P (rev xs) = todos P xs"<br />
proof (induct xs)<br />
show "todos P (rev []) = todos P []" by simp<br />
next<br />
fix a xs <br />
assume H1: "todos P (rev xs) = todos P xs"<br />
have " todos P (rev (a # xs)) = todos P (rev xs @ [a])" by simp<br />
also have "… = (todos P (rev xs) ∧ todos P [a])" by (simp add:todos_append)<br />
also have "… = (todos P xs ∧ todos P [a])" using H1 by simp<br />
also have "… = (todos P [a] ∧ todos P xs)" by arith<br />
finally show "todos P (rev (a # xs)) = todos P (a#xs)" by simp<br />
qed<br />
<br />
(* paupeddeg dancorgar *)<br />
lemma "todos P (rev xs) = todos P xs"<br />
proof (induct xs)<br />
show " todos P (rev []) = todos P []" by simp<br />
next<br />
fix a xs<br />
assume HI: "todos P (rev xs) = todos P xs "<br />
have "todos P (rev (a # xs)) = todos P (rev(xs) @ [a])" by simp<br />
also have "... = (todos P (rev xs) ∧ todos P [a]) " by (simp add: todos_append)<br />
also have "... = (todos P xs ∧ todos P [a])" using HI by simp<br />
also have "... = (todos P [a] ∧ todos P xs)" by arith<br />
also have "... = todos P ([a] @ xs)" by (simp add: todos_append)<br />
finally show "todos P (rev (a # xs)) = todos P (a # xs)" by simp<br />
qed <br />
<br />
(* wilmorort *)<br />
lemma "todos P (rev xs) = todos P xs" (is "?P xs")<br />
proof (induct xs) <br />
show "?P []" by simp <br />
next<br />
fix a xs<br />
assume HI: "?P xs"<br />
have "todos P (rev (a#xs)) = todos P ((rev xs) @ [a])" by simp<br />
also have "... = (todos P(rev xs) ∧ todos P [a])" by (simp add: todos_append)<br />
also have "... = (todos P xs ∧ todos P [a])" using HI by simp<br />
also have "... = (todos P [a] ∧ todos P xs)" by (simp add: HOL.conj_commute)<br />
also have "... = todos P([a]@(xs))" by (simp)<br />
finally show "todos P (rev (a#xs))= todos P (a # xs)" by simp<br />
qed<br />
<br />
(* anaprarod *) <br />
(* es igual que las anteriores pero con el final también con patrones *)<br />
<br />
lemma "todos P (rev xs) = todos P xs" (is "?P xs")<br />
proof (induct xs)<br />
show "?P []" by auto<br />
next<br />
fix a xs<br />
assume HI: "?P xs"<br />
have "todos P (rev (a#xs)) = todos P (rev xs @ [a])" by simp<br />
also have "… = (todos P (rev xs) ∧ todos P [a])" by (simp add: todos_append)<br />
also have "… = (todos P xs ∧ todos P [a])" using HI by simp<br />
also have "… = (todos P [a] ∧ todos P xs)" by arith<br />
also have "… = todos P (a#xs)" by simp<br />
finally show "?P (a # xs)" by simp<br />
qed<br />
<br />
text {*<br />
--------------------------------------------------------------------- <br />
Ejercicio 6. Demostrar o refutar:<br />
algunos (λx. P x ∧ Q x) xs = (algunos P xs ∧ algunos Q xs)<br />
--------------------------------------------------------------------- <br />
*}<br />
<br />
lemma "algunos (λx. P x ∧ Q x) xs = (algunos P xs ∧ algunos Q xs)"<br />
oops<br />
<br />
(* migtermor ivamenjim ferrenseg paupeddeg crigomgom serrodcal wilmorort juacabsou rubgonmar anaprarod marpoldia1 fraortmoy *)<br />
lemma "algunos (λx. P x ∧ Q x) xs = (algunos P xs ∧ algunos Q xs)" <br />
quickcheck<br />
oops<br />
(* Quickcheck encuentra el siguiente contraejemplo: P={a1}, Q={a2}, xs={a1,a2}. En este ejemplo:<br />
· "algunos (λx. P x ∧ Q x) xs = False"<br />
· "(algunos P xs ∧ algunos Q xs) = True" *)<br />
<br />
text {*<br />
--------------------------------------------------------------------- <br />
Ejercicio 7.1. Demostrar o refutar automáticamente <br />
algunos P (map f xs) = algunos (P ∘ f) xs<br />
--------------------------------------------------------------------- <br />
*}<br />
<br />
(* ivamenjim migtermor marpoldia1 crigomgom rubgonmar wilmorort anaprarod fraortmoy *)<br />
<br />
lemma "algunos P (map f xs) = algunos (P o f) xs"<br />
by (induct xs) auto<br />
<br />
(* ferrenseg *)<br />
<br />
lemma "algunos P (map f xs) = algunos (P o f) xs"<br />
by (induct xs) simp_all<br />
<br />
(* anaprarod *)<br />
lemma "algunos P (map f xs) = algunos (P o f) xs"<br />
apply (induct xs)<br />
apply auto<br />
done<br />
<br />
text {*<br />
--------------------------------------------------------------------- <br />
Ejercicio 7.2. Demostrar o refutar datalladamente<br />
algunos P (map f xs) = algunos (P ∘ f) xs<br />
--------------------------------------------------------------------- <br />
*}<br />
<br />
lemma "algunos P (map f xs) = algunos (P o f) xs"<br />
oops<br />
<br />
(* migtermor *)<br />
<br />
lemma AUX: "algunos (λa. P (f a)) xs = algunos (P o f) xs"<br />
by (induct xs) auto<br />
<br />
lemma "algunos P (map f xs) = algunos (P o f) xs" (is "?Q xs")<br />
proof (induct xs)<br />
show "?Q []" by simp<br />
next<br />
fix x xs<br />
assume HI: "?Q xs"<br />
have "algunos P (map f (x#xs)) = (algunos P ((f x)#(map f xs)))" by simp<br />
also have "… = ((P (f x)) ∨ (algunos P (map f xs)))" by (simp only: algunos.simps(2))<br />
also have "… = ((P (f x)) ∨ (algunos (P o f) xs))" <br />
proof (cases)<br />
assume C1: "(P (f x))"<br />
have Aux: "((P (f x)) ∨ (algunos P (map f xs))) = True" using C1 by simp<br />
have Aux1: "… = ((P (f x)) ∨ (algunos (P o f) xs))" using C1 by simp<br />
then show "((P (f x)) ∨ (algunos P (map f xs))) = ((P (f x)) ∨ (algunos (P o f) xs))" <br />
using Aux Aux1 by simp<br />
next<br />
assume C2: "¬(P (f x))"<br />
have Aux2: "((P (f x)) ∨ (algunos P (map f xs))) = (algunos P (map f xs))" using C2 by simp<br />
have Aux3: "… = (algunos (P o f) xs)" using HI by (simp add: AUX)<br />
also have "… = ((P (f x)) ∨ (algunos (P o f) xs))" using C2 by simp<br />
then show "((P (f x)) ∨ (algunos P (map f xs))) = ((P (f x)) ∨ (algunos (P o f) xs))" <br />
using Aux2 Aux3 by simp<br />
qed<br />
also have "… = (((P o f) x) ∨ (algunos (P o f) xs))" by simp<br />
also have "… = (algunos (P o f) (x#xs))" by simp<br />
finally show "algunos P (map f (x#xs)) = (algunos (P o f) (x#xs))" by simp<br />
qed<br />
<br />
(* ferrenseg *)<br />
<br />
lemma "algunos P (map f xs) = algunos (P ∘ f) xs"<br />
proof (induct xs)<br />
show "algunos P (map f []) = algunos (P ∘ f) []" by simp<br />
next<br />
fix x xs<br />
assume HI: "algunos P (map f xs) = algunos (P ∘ f) xs"<br />
show "algunos P (map f (x # xs)) = algunos (P ∘ f) (x # xs)"<br />
proof -<br />
have "algunos P (map f (x # xs)) = algunos P ((f x) # (map f xs))" by simp<br />
also have "… = ((P (f x)) ∨ (algunos P (map f xs)))" by simp<br />
also have "… = (((P ∘ f) x) ∨ (algunos (P ∘ f) xs))" using HI by simp<br />
also have "… = algunos (P ∘ f) (x # xs)" by simp<br />
finally show ?thesis<br />
qed<br />
qed<br />
<br />
(* ivamenjim *)<br />
<br />
lemma "algunos P (map f xs) = algunos (P o f) xs"<br />
proof (induct xs)<br />
show "algunos P (map f []) = algunos (P ∘ f) []" by simp<br />
next <br />
fix a xs<br />
assume HI: "algunos P (map f xs) = algunos (P ∘ f) xs"<br />
have "algunos P (map f (a # xs)) = algunos P ((f a)#map f xs)" by simp<br />
also have "... = (algunos P (map f [a]) ∨ algunos P (map f xs))" by simp <br />
also have "... = (algunos P (map f [a]) ∨ algunos (P ∘ f) xs)" using HI by simp <br />
finally show "algunos P (map f (a # xs)) = algunos (P ∘ f) (a # xs)" by simp<br />
qed<br />
<br />
(* crigomgom rubgonmar anaprarod marpoldia1 *)<br />
lemma "algunos P (map f xs) = algunos (P o f) xs"<br />
proof (induct xs)<br />
show "algunos P (map f []) = algunos (P ∘ f) []" by simp<br />
next<br />
fix x xs<br />
assume HI: "algunos P (map f xs) = algunos (P ∘ f) xs"<br />
have "algunos P (map f (x # xs)) = algunos P ((f x) # (map f xs))" by simp<br />
also have "... = (P (f x) ∨ algunos P (map f xs))" by simp<br />
also have "... = (P (f x) ∨ algunos (P ∘ f) xs)" using HI by simp<br />
also have "... = ((P ∘ f) x ∨ algunos (P ∘ f) xs)" by simp<br />
finally show "algunos P (map f (x # xs)) = algunos (P ∘ f) (x # xs)" by simp<br />
qed<br />
<br />
(* wilmorort *)<br />
lemma "algunos P (map f xs) = algunos (P o f) xs" (is "?P xs" )<br />
proof (induct xs)<br />
show "?P []" by simp<br />
next<br />
fix a xs<br />
assume HI: "algunos P (map f xs) = algunos (P ∘ f) xs"<br />
have "algunos P (map f (a # xs)) = (P (f a) ∨ algunos P (map f xs))" by simp<br />
also have "... = (P (f a) ∨ algunos (P ∘ f) xs )" using HI by simp<br />
also have "... = ((P ∘ f) a ∨ algunos (P ∘ f)xs)" by simp<br />
finally show " algunos P (map f (a # xs)) = algunos (P ∘ f) (a # xs)" by simp<br />
qed<br />
<br />
text {*<br />
--------------------------------------------------------------------- <br />
Ejercicio 8.1. Demostrar o refutar automáticamente <br />
algunos P (xs @ ys) = (algunos P xs ∨ algunos P ys)<br />
--------------------------------------------------------------------- <br />
*}<br />
<br />
(* ivamenjim marpoldia1 paupeddeg crigomgom rubgonmar wilmorort*)<br />
<br />
lemma "algunos P (xs @ ys) = (algunos P xs ∨ algunos P ys)"<br />
by (induct xs) auto<br />
<br />
(* ferrenseg *)<br />
lemma "algunos P (xs @ ys) = (algunos P xs ∨ algunos P ys)"<br />
by (induct xs) simp_all<br />
<br />
text {*<br />
--------------------------------------------------------------------- <br />
Ejercicio 8.2. Demostrar o refutar detalladamente<br />
algunos P (xs @ ys) = (algunos P xs ∨ algunos P ys)<br />
--------------------------------------------------------------------- <br />
*}<br />
<br />
(* ivamenjim ferrenseg marpoldia1 wilmorort*)<br />
<br />
lemma algunos_append:<br />
"algunos P (xs @ ys) = (algunos P xs ∨ algunos P ys)"<br />
proof (induct xs)<br />
show "algunos P ([] @ ys) = (algunos P [] ∨ algunos P ys)" by simp<br />
next <br />
fix a xs<br />
assume HI: "algunos P (xs @ ys) = (algunos P xs ∨ algunos P ys)"<br />
have "algunos P ((a # xs) @ ys) = (P a ∨ (algunos P (xs @ ys)))" by simp<br />
also have "... = (P a ∨ (algunos P xs ∨ algunos P ys))" using HI by simp<br />
finally show "algunos P ((a # xs) @ ys) = (algunos P (a # xs) ∨ algunos P ys)" by simp<br />
qed<br />
<br />
(* migtermor crigomgom *)<br />
<br />
lemma algunos_append:<br />
"algunos P (xs @ ys) = (algunos P xs ∨ algunos P ys)" (is "?Q xs")<br />
proof (induct xs)<br />
show "?Q []" by simp<br />
next <br />
fix x xs<br />
assume HI: "?Q xs"<br />
have "algunos P ((x#xs) @ ys) = algunos P (x#(xs @ ys))" by simp<br />
also have "… = ((P x) ∨ (algunos P (xs @ ys)))" by simp<br />
also have "… = ((P x) ∨ (algunos P xs) ∨ (algunos P ys))" using HI by simp<br />
also have "… = ((algunos P (x#xs)) ∨ (algunos P ys))" by simp<br />
finally show "algunos P ((x#xs) @ ys) = (algunos P (x#xs) ∨ algunos P ys)" by simp<br />
qed<br />
<br />
(* paupeddeg*)<br />
lemma algunos_append:<br />
"algunos P (xs @ ys) = (algunos P xs ∨ algunos P ys)"<br />
proof (induct xs)<br />
show "algunos P ([] @ ys) = (algunos P [] ∨ algunos P ys)" by simp<br />
next<br />
fix a xs<br />
assume HI: "algunos P (xs @ ys) = (algunos P xs ∨ algunos P ys)"<br />
have "algunos P ((a # xs) @ ys) = (algunos P [a] ∨ ( algunos P (xs @ ys)))" by simp<br />
also have "... = (algunos P [a] ∨ (algunos P xs ∨ algunos P ys))" using HI by simp<br />
finally show "algunos P ((a # xs) @ ys) = (algunos P (a # xs) ∨ algunos P ys)" by simp<br />
qed<br />
<br />
text {*<br />
--------------------------------------------------------------------- <br />
Ejercicio 9.1. Demostrar o refutar automáticamente<br />
algunos P (rev xs) = algunos P xs<br />
--------------------------------------------------------------------- <br />
*}<br />
<br />
(* ivamenjim migtermor marpoldia1 rubgonmar *)<br />
<br />
lemma "algunos P (rev xs) = algunos P xs"<br />
apply (induct xs)<br />
apply simp<br />
apply (simp add: algunos_append)<br />
apply auto<br />
done<br />
<br />
(* ferrenseg crigomgom*)<br />
lemma "algunos P (rev xs) = algunos P xs"<br />
by (induct xs) (auto simp add: algunos_append)<br />
<br />
(* wilmorort *)<br />
lemma "algunos P (rev xs) = algunos P xs"<br />
by (induct xs,simp,simp add: algunos_append,auto) <br />
<br />
text {*<br />
--------------------------------------------------------------------- <br />
Ejercicio 9.2. Demostrar o refutar detalladamente<br />
algunos P (rev xs) = algunos P xs<br />
--------------------------------------------------------------------- <br />
*}<br />
<br />
lemma "algunos P (rev xs) = algunos P xs"<br />
oops<br />
<br />
(* migtermor *)<br />
lemma auxiliar1:<br />
"rev (a#xs) = rev xs @ [a]"<br />
by auto <br />
<br />
lemma "algunos P (rev xs) = algunos P xs" (is "?Q xs")<br />
proof (induct xs)<br />
show "?Q []" by simp<br />
next<br />
fix x xs<br />
assume HI: "?Q xs"<br />
have "algunos P (rev (x#xs)) = (algunos P (rev xs @ [x]))" using auxiliar1 by simp<br />
also have "… = ((algunos P (rev xs)) ∨ (algunos P [x]))" by (simp add: algunos_append)<br />
also have "… = ((P x) ∨ (algunos P (rev xs)))" by simp arith<br />
also have "… = ((P x) ∨ (algunos P xs))" using HI by simp<br />
finally show "algunos P (rev (x#xs)) = (algunos P (x#xs))" by simp<br />
qed<br />
<br />
(* ferrenseg *)<br />
<br />
lemma "algunos P (rev xs) = algunos P xs"<br />
proof (induct xs)<br />
show "algunos P (rev []) = algunos P []" by simp<br />
next<br />
fix x xs<br />
assume HI: "algunos P (rev xs) = algunos P xs"<br />
show "algunos P (rev (x # xs)) = algunos P (x # xs)"<br />
proof -<br />
have "algunos P (rev (x # xs)) = algunos P ((rev xs) @ [x])" by simp<br />
also have "… = (algunos P (rev xs) ∨ algunos P [x])" <br />
by (simp add: algunos_append)<br />
also have "… = (algunos P xs ∨ algunos P [x])" using HI by simp<br />
also have "… = (algunos P xs ∨ P x)" by simp<br />
also have "… = (P x ∨ algunos P xs)" by auto<br />
also have "… = algunos P (x # xs)" by simp<br />
finally show ?thesis<br />
qed<br />
qed<br />
<br />
(* ivamenjim marpoldia1*)<br />
<br />
lemma "algunos P (rev xs) = algunos P xs" (is "?P xs")<br />
proof (induct xs)<br />
show "?P []" by simp<br />
next <br />
fix a xs<br />
assume HI: "?P xs"<br />
have "algunos P (rev (a # xs)) = (algunos P ((rev xs)@[a]))" by simp<br />
also have "... = ((algunos P (rev xs)) ∨ algunos P [a])" by (simp add: algunos_append)<br />
also have "... = (algunos P xs ∨ algunos P [a])" using HI by simp<br />
also have "... = (algunos P [a] ∨ algunos P xs)" by arith<br />
finally show "algunos P (rev (a # xs)) = algunos P (a # xs)" by simp<br />
qed<br />
<br />
(*paupeddeg*)<br />
lemma "algunos P (rev xs) = algunos P xs"<br />
proof (induct xs)<br />
show "algunos P (rev []) = algunos P []" by simp<br />
next<br />
fix a xs<br />
assume HI:"algunos P (rev xs) = algunos P xs"<br />
have "algunos P (rev (a # xs)) = algunos P ((rev xs) @ [a])" by simp<br />
also have "... = ((algunos P (rev xs)) ∨ (algunos P [a]))" by (simp add: algunos_append)<br />
also have "... = ((algunos P xs) ∨ (algunos P [a]))" using HI by simp<br />
also have "... = ((algunos P [a]) ∨ (algunos P xs))" by arith<br />
finally show "algunos P (rev (a # xs)) = algunos P (a # xs)" by simp<br />
qed<br />
<br />
(*crigomgom *)<br />
lemma "algunos P (rev xs) = algunos P xs"<br />
proof (induct xs)<br />
show "algunos P (rev []) = algunos P []" by simp<br />
next<br />
fix x xs<br />
assume HI: " algunos P (rev xs) = algunos P xs" <br />
have "algunos P (rev (x # xs)) = algunos P ((rev xs) @ [x])" by simp<br />
also have "... = (algunos P (rev xs) ∨ algunos P [x])" by (simp add: algunos_append)<br />
also have "... = (algunos P xs ∨ algunos P [x])" using HI by simp<br />
also have "... = (algunos P xs ∨ P x)" by simp<br />
also have "... = (P x ∨ algunos P xs)" by arith<br />
also have "... = algunos P (x#xs)" by simp<br />
finally show "algunos P (rev (x # xs)) = algunos P (x # xs)" by simp<br />
qed<br />
<br />
<br />
(*wilmorort*)<br />
lemma "algunos P (rev xs) = algunos P xs" (is "?P xs")<br />
proof (induct xs) <br />
show "?P []" by simp <br />
next<br />
fix a xs<br />
assume HI: "?P xs"<br />
have "algunos P (rev (a#xs)) = algunos P ((rev xs) @ [a])" by simp<br />
also have "... = (algunos P(rev xs) ∨ algunos P [a])" by (simp add: algunos_append)<br />
also have "... = (algunos P xs ∨ algunos P [a])" using HI by simp<br />
also have "... = (algunos P [a] ∨ algunos P xs)" by (simp add: HOL.disj_commute)<br />
also have "... = algunos P([a]@(xs))" by (simp)<br />
finally show "algunos P (rev (a#xs))= algunos P (a # xs)" by simp<br />
qed<br />
<br />
text {*<br />
--------------------------------------------------------------------- <br />
Ejercicio 10. Encontrar un término no trivial Z tal que sea cierta la <br />
siguiente ecuación:<br />
algunos (λx. P x ∨ Q x) xs = Z<br />
y demostrar la equivalencia de forma automática y detallada.<br />
--------------------------------------------------------------------- <br />
*}<br />
<br />
(* migtermor *)<br />
<br />
lemma "algunos (λx. P x ∨ Q x) xs = ((algunos (λx. P x) xs) ∨ (algunos (λx. Q x) xs))"<br />
by (induct xs) auto<br />
<br />
lemma "algunos (λx. P x ∨ Q x) xs = ((algunos (λx. P x) xs) ∨ (algunos (λx. Q x) xs))" (is "?R xs")<br />
proof (induct xs)<br />
show "?R []" by simp<br />
next<br />
fix x xs<br />
assume HI: "?R xs"<br />
have "algunos (λx. P x ∨ Q x) (x#xs) = (((λx. P x ∨ Q x) x) ∨ (algunos (λx. P x ∨ Q x) xs))"<br />
by simp<br />
also have "… = ((P x ∨ Q x) ∨ (algunos (λx. P x ∨ Q x) xs))" by simp<br />
also have H1: "… = ((((λx. P x) x) ∨ (algunos (λx. P x) xs)) ∨ (((λx. Q x) x) ∨ (algunos (λx. Q x) xs)))"<br />
using HI by simp arith<br />
have H2: "… = ((algunos (λx. P x) (x#xs)) ∨ (algunos (λx. Q x) (x#xs)))" <br />
by simp<br />
have C: "(algunos (λx. P x ∨ Q x) (x#xs)) = <br />
((algunos (λx. P x) (x#xs)) ∨ (algunos (λx. Q x) (x#xs)))" <br />
using H1 H2 by simp<br />
finally show "(algunos (λx. P x ∨ Q x) (x#xs)) = <br />
((algunos (λx. P x) (x#xs)) ∨ (algunos (λx. Q x) (x#xs)))" <br />
using C by simp<br />
qed<br />
<br />
<br />
(* ferrenseg *)<br />
<br />
lemma "algunos (λx. P x ∨ Q x) xs = (algunos P xs ∨ algunos Q xs)"<br />
proof (induct xs)<br />
show "algunos (λx. P x ∨ Q x) [] = (algunos P [] ∨ algunos Q [])" by simp<br />
next<br />
fix x xs<br />
assume HI: "algunos (λx. (P x ∨ Q x)) xs = (algunos P xs ∨ algunos Q xs)"<br />
show "algunos (λx. P x ∨ Q x) (x # xs) = (algunos P (x # xs) ∨ algunos Q (x # xs))"<br />
proof -<br />
have "algunos (λx. P x ∨ Q x) (x # xs) = <br />
((P x) ∨ (Q x) ∨ algunos (λx. P x ∨ Q x) xs)" by simp<br />
also have "… = ((P x) ∨ (Q x) ∨ algunos P xs ∨ algunos Q xs)" <br />
using HI by simp<br />
also have "… = (((P x) ∨ algunos P xs) ∨ ((Q x) ∨ algunos Q xs))" by auto<br />
also have "… = (algunos P (x # xs) ∨ algunos Q (x # xs))" by simp<br />
finally show ?thesis <br />
qed<br />
qed<br />
<br />
(* ivamenjim marpoldia1 *)<br />
<br />
lemma "algunos (λx. P x ∨ Q x) xs = (algunos P xs ∨ algunos Q xs)" <br />
by (induct xs) auto<br />
<br />
lemma "algunos (λx. P x ∨ Q x) xs = (algunos P xs ∨ algunos Q xs)"<br />
proof (induct xs)<br />
show "algunos (λx. P x ∨ Q x) [] = (algunos P [] ∨ algunos Q [])" by simp<br />
next <br />
fix a xs<br />
assume HI: "algunos (λx. P x ∨ Q x) xs = (algunos P xs ∨ algunos Q xs)"<br />
have "algunos (λx. P x ∨ Q x) (a # xs) = (P a ∨ Q a ∨ algunos (λx. P x ∨ Q x) xs)" by simp<br />
also have "... = (P a ∨ Q a ∨ algunos P xs ∨ algunos Q xs)" using HI by simp<br />
finally show "algunos (λx. P x ∨ Q x) (a # xs) = (algunos P (a # xs) ∨ algunos Q (a # xs))" by auto<br />
qed<br />
<br />
<br />
(* wilmorort *)<br />
lemma "algunos (λx. P x ∨ Q x) xs = (¬todos(λx. ¬P x)xs ∨ ¬todos(λx. ¬Q x)xs)"<br />
by (induct xs) auto<br />
<br />
<br />
lemma "algunos (λx. P x ∨ Q x) xs = (¬todos(λx. ¬P x)xs ∨ ¬todos(λx. ¬Q x)xs)"<br />
(is "?P xs")<br />
proof (induct xs)<br />
show "?P []" by simp<br />
next<br />
fix a xs<br />
assume HI: "?P xs"<br />
have "algunos (λx. P x ∨ Q x) (a # xs) = (P a ∨ Q a ∨ algunos (λx. P x ∨ Q x) xs)" by simp<br />
also have "... = (P a ∨ Q a ∨ (¬ todos (λx. ¬ P x) xs ∨ ¬ todos (λx. ¬ Q x) xs))" using HI by simp<br />
also have "... = (P a ∨ ¬ todos (λx. ¬ P x) xs ∨ Q a ∨ ¬ todos (λx. ¬ Q x) xs )" by arith<br />
finally show "algunos (λx. P x ∨ Q x) (a # xs) = (~ todos (λx. ¬ P x) (a # xs) ∨ ~ todos (λx. ¬ Q x) (a # xs))" by simp<br />
qed<br />
<br />
<br />
<br />
text {*<br />
--------------------------------------------------------------------- <br />
Ejercicio 11.1. Demostrar o refutar automáticamente<br />
algunos P xs = (¬ todos (λx. (¬ P x)) xs)<br />
--------------------------------------------------------------------- <br />
*}<br />
<br />
(* ivamenjim wilmorort migtermor marpoldia1 crigomgom rubgonmar *)<br />
<br />
lemma "algunos P xs = (¬ todos (λx. (¬ P x)) xs)"<br />
by (induct xs) auto<br />
<br />
(* ferrenseg *)<br />
<br />
lemma "algunos P xs = (¬ todos (λx. (¬ P x)) xs)"<br />
by (induct xs) simp_all<br />
<br />
text {*<br />
--------------------------------------------------------------------- <br />
Ejercicio 11.2. Demostrar o refutar datalladamente<br />
algunos P xs = (¬ todos (λx. (¬ P x)) xs)<br />
--------------------------------------------------------------------- <br />
*}<br />
<br />
(* ivamenjim migtermor marpoldia1 crigomgom*)<br />
<br />
lemma "algunos P xs = (¬ todos (λx. (¬ P x)) xs)"<br />
proof (induct xs)<br />
show "algunos P [] = (¬ todos (λx. ¬ P x) [])" by simp<br />
next <br />
fix a xs<br />
assume HI: "algunos P xs = (¬ todos (λx. ¬ P x) xs)"<br />
have "algunos P (a # xs) = (P a ∨ (algunos P xs))" by simp<br />
also have "... = (P a ∨ (¬ todos (λx. ¬ P x) xs))" using HI by simp<br />
finally show "algunos P (a # xs) = (¬ todos (λx. ¬ P x) (a # xs))" by simp<br />
qed<br />
<br />
(* ferrenseg *)<br />
<br />
lemma "algunos P xs = (¬ todos (λx. (¬ P x)) xs)"<br />
proof (induct xs)<br />
show "algunos P [] = (¬ todos (λx. (¬ P x)) [])" by simp<br />
next<br />
fix x xs<br />
assume HI: "algunos P xs = (¬ todos (λx. (¬ P x)) xs)"<br />
show "algunos P (x # xs) = (¬ todos (λx. (¬ P x)) (x # xs))"<br />
proof - <br />
have "algunos P (x # xs) = ((P x) ∨ algunos P xs)" by simp<br />
also have "… = ((P x) ∨ ¬ todos (λx. (¬ P x)) xs)" using HI by simp<br />
also have "… = (¬ (¬ (P x) ∧ todos (λx. (¬ P x)) xs))" by simp<br />
also have "… = (¬ todos (λx. (¬ P x)) (x # xs))" by simp<br />
finally show ?thesis<br />
qed<br />
qed<br />
<br />
text {*<br />
--------------------------------------------------------------------- <br />
Ejercicio 12. Definir la funcion primitiva recursiva <br />
estaEn :: 'a ⇒ 'a list ⇒ bool<br />
tal que (estaEn x xs) se verifica si el elemento x está en la lista<br />
xs. Por ejemplo, <br />
estaEn (2::nat) [3,2,4] = True<br />
estaEn (1::nat) [3,2,4] = False<br />
--------------------------------------------------------------------- <br />
*}<br />
<br />
(* ivamenjim ferrenseg migtermor serrodcal crigomgom rubgonmar marpoldia1*)<br />
<br />
fun estaEn :: "'a ⇒ 'a list ⇒ bool" where<br />
"estaEn x [] = False"<br />
| "estaEn x (a#xs) = ((a = x) ∨ (estaEn x xs))"<br />
<br />
value "estaEn (2::nat) [3,2,4] = True"<br />
value "estaEn (1::nat) [3,2,4] = False"<br />
<br />
<br />
text {*<br />
--------------------------------------------------------------------- <br />
Ejercicio 13. Expresar la relación existente entre estaEn y algunos. <br />
Demostrar dicha relación de forma automática y detallada.<br />
--------------------------------------------------------------------- <br />
*}<br />
<br />
(* migtermor crigomgom*)<br />
<br />
lemma "estaEn x xs = (algunos (λa. a=x) xs)"<br />
by (induct xs) auto<br />
<br />
lemma auxiliar13:<br />
"(x=a) = ((λa. a=x) a)"<br />
by auto<br />
<br />
lemma "estaEn x xs = (algunos (λa. a=x) xs)" (is "?P xs")<br />
proof (induct xs)<br />
show "?P []" by simp<br />
next<br />
fix a xs<br />
assume HI: "?P xs"<br />
have "estaEn x (a#xs) = ((a=x) ∨ (estaEn x xs))" by simp<br />
also have H: "… = ((a=x) ∨ (algunos (λa. a=x) xs))" using HI by simp<br />
also have "… = (((λa. a=x) a) ∨ (algunos (λa. a=x) xs))" using auxiliar13 by simp<br />
also have "… = (algunos (λa. a=x) (a#xs))" by simp<br />
have C: "estaEn x (a#xs) = (algunos (λa. a=x) (a#xs))" using H by simp<br />
finally show "estaEn x (a#xs) = (algunos (λa. a=x) (a#xs))" using C by simp<br />
qed<br />
<br />
<br />
(* ferrenseg *)<br />
<br />
lemma estaEn_algunos:<br />
"estaEn y xs = algunos (λx. x=y) xs"<br />
by (induct xs) auto<br />
<br />
lemma estaEn_algunos_2:<br />
"estaEn y xs = algunos (λx. x=y) xs"<br />
proof (induct xs)<br />
show "estaEn y [] = algunos (λx. x=y) []" by simp<br />
next<br />
fix x xs<br />
assume HI: "estaEn y xs = algunos (λx. x=y) xs"<br />
show "estaEn y (x # xs) = algunos (λx. x=y) (x # xs)"<br />
proof -<br />
have "estaEn y (x # xs) = (y=x ∨ estaEn y xs)" by simp<br />
also have "… = (y=x ∨ algunos (λx. x=y) xs)" using HI by simp<br />
also have "… = (x=y ∨ algunos (λx. x=y) xs)" by auto<br />
also have "… = algunos (λx. x=y) (x # xs)" by simp<br />
finally show ?thesis<br />
qed<br />
qed<br />
<br />
(*crigomgom*)<br />
lemma "estaEn x xs = (algunos (λa. a=x) xs)" <br />
proof (induct xs)<br />
show " estaEn x [] = algunos (λa. a = x) []" by simp<br />
next<br />
fix a xs<br />
assume HI: "estaEn x xs = algunos (λa. a = x) xs"<br />
have "estaEn x (a # xs) = (a = x ∨ estaEn x xs)" by simp<br />
also have "... = (a = x ∨ algunos (λa. a = x) xs)" using HI by simp<br />
finally show " estaEn x (a # xs) = algunos (λa. a = x) (a # xs)" by simp<br />
qed<br />
<br />
(* ivamenjim marpoldia1 *)<br />
<br />
lemma "estaEn x xs = algunos (λy. x=y) xs"<br />
by (induct xs) auto<br />
<br />
lemma "estaEn x xs = algunos (λy. x=y) xs"<br />
proof (induct xs)<br />
show "estaEn x [] = algunos (op = x) []" by simp<br />
next <br />
fix a xs<br />
assume HI: "estaEn x xs = algunos (op = x) xs"<br />
have "estaEn x (a # xs) = ((a = x) ∨ (estaEn x xs))" by simp<br />
also have "... = (a = x ∨ algunos (op = x) xs)" using HI by simp<br />
finally show " estaEn x (a # xs) = algunos (op = x) (a # xs)" by auto<br />
qed<br />
<br />
end<br />
</source></div>Lucnovdoshttps://www.glc.us.es/~jalonso/RA2016/index.php?title=Relaci%C3%B3n_4&diff=579Relación 42016-11-22T17:42:09Z<p>Lucnovdos: </p>
<hr />
<div><source lang="isar"><br />
chapter {* R4: Cuantificadores sobre listas *}<br />
<br />
theory R4_Cuantificadores_sobre_listas<br />
imports Main <br />
begin<br />
<br />
text {* <br />
--------------------------------------------------------------------- <br />
Ejercicio 1. Definir la función <br />
todos :: ('a ⇒ bool) ⇒ 'a list ⇒ bool<br />
tal que (todos p xs) se verifica si todos los elementos de la lista <br />
xs cumplen la propiedad p. Por ejemplo, se verifica <br />
todos (λx. 1<length x) [[2,1,4],[1,3]]<br />
¬todos (λx. 1<length x) [[2,1,4],[3]]<br />
<br />
Nota: La función todos es equivalente a la predefinida list_all. <br />
--------------------------------------------------------------------- <br />
*}<br />
<br />
{*danrodcha*}<br />
fun todos :: "('a ⇒ bool) ⇒ 'a list ⇒ bool" where<br />
"todos p xs = "todos p [] = True"<br />
| "todos p (x#xs) = (p x ∧ todos p xs)""<br />
<br />
(* ivamenjim migtermor dancorgar wilmorort marpoldia1 ferrenseg paupeddeg pablucoto crigomgom anaprarod serrodcal juacabsou rubgonmar josgarsan fraortmoy *)<br />
fun todos :: "('a ⇒ bool) ⇒ 'a list ⇒ bool" where<br />
"todos p [] = True"<br />
| "todos p (x#xs) = (p x ∧ todos p xs)"<br />
<br />
text {* <br />
--------------------------------------------------------------------- <br />
Ejercicio 2. Definir la función <br />
algunos :: ('a ⇒ bool) ⇒ 'a list ⇒ bool<br />
tal que (algunos p xs) se verifica si algunos elementos de la lista <br />
xs cumplen la propiedad p. Por ejemplo, se verifica <br />
algunos (λx. 1<length x) [[2,1,4],[3]]<br />
¬algunos (λx. 1<length x) [[],[3]]"<br />
<br />
Nota: La función algunos es equivalente a la predefinida list_ex. <br />
--------------------------------------------------------------------- <br />
*}<br />
<br />
{*danrodcha ivamenjim migtermor dancorgar marpoldia1 ferrenseg wilmorort paupeddeg pablucoto crigomgom anaprarod serrodcal juacabsou rubgonmar josgarsan fraortmoy lucnovdos*}<br />
fun algunos :: "('a ⇒ bool) ⇒ 'a list ⇒ bool" where<br />
"algunos p [] = False"<br />
| "algunos p (x#xs) = (p x ∨ algunos p xs)"<br />
<br />
text {*<br />
--------------------------------------------------------------------- <br />
Ejercicio 3.1. Demostrar o refutar automáticamente <br />
todos (λx. P x ∧ Q x) xs = (todos P xs ∧ todos Q xs)<br />
--------------------------------------------------------------------- <br />
*}<br />
<br />
{*danrodcha ivamenjim migtermor dancorgar marpoldia1 ferrenseg wilmorort paupeddeg pablucoto anaprarod serrodcal juacabsou rubgonmar josgarsan fraortmoy*}<br />
lemma "todos (λx. P x ∧ Q x) xs = (todos P xs ∧ todos Q xs)"<br />
by (induct xs) auto<br />
<br />
text {*<br />
--------------------------------------------------------------------- <br />
Ejercicio 3.2. Demostrar o refutar detalladamente<br />
todos (λx. P x ∧ Q x) xs = (todos P xs ∧ todos Q xs)<br />
--------------------------------------------------------------------- <br />
*}<br />
<br />
{*danrodcha*}<br />
lemma "todos (λx. P x ∧ Q x) xs = (todos P xs ∧ todos Q xs)"<br />
proof (induct xs)<br />
show "?R []" by simp<br />
next<br />
fix a xs <br />
assume HI: "?R xs"<br />
have "todos (λx. P x ∧ Q x) (a#xs) = (P a ∧ Q a ∧ todos (λx. P x ∧ Q x) xs)" by simp<br />
also have "… = (P a ∧ Q a ∧ (todos P xs ∧ todos Q xs))" using HI by simp<br />
also have "… = ((P a ∧ todos P xs) ∧ (Q a ∧ todos Q xs))" by blast<br />
also have "… = (todos P (a#xs) ∧ todos Q (a#xs))" by simp<br />
finally show "?R (a#xs)" by simp<br />
qed<br />
<br />
(* ivamenjim wilmorort serrodcal josgarsan*)<br />
<br />
lemma "todos (λx. P x ∧ Q x) xs = (todos P xs ∧ todos Q xs)"<br />
proof (induct xs)<br />
show "todos (λx. P x ∧ Q x) [] = (todos P [] ∧ todos Q [])" by simp<br />
next <br />
fix a xs<br />
assume HI: "todos (λx. P x ∧ Q x) xs = (todos P xs ∧ todos Q xs)"<br />
have "todos (λx. P x ∧ Q x) (a # xs) = (P a ∧ Q a ∧ todos (λx. P x ∧ Q x) xs)" by simp<br />
also have "... = (P a ∧ Q a ∧ todos P xs ∧ todos Q xs)" using HI by simp<br />
finally show "todos (λx. P x ∧ Q x) (a # xs) = (todos P (a # xs) ∧ todos Q (a # xs))" by auto<br />
qed<br />
<br />
(* dancorgar paupeddeg *)<br />
lemma "todos (λx. P x ∧ Q x) xs = (todos P xs ∧ todos Q xs)"<br />
proof (induct xs)<br />
show "todos (λx. P x ∧ Q x) [] = (todos P [] ∧ todos Q [])" by simp<br />
next<br />
fix y xs<br />
assume HI: "todos (λx. P x ∧ Q x) xs = (todos P xs ∧ todos Q xs)"<br />
have "todos (λx. P x ∧ Q x) (y#xs) = ((P y ∧ Q y) ∧ (todos (λx. P x ∧ Q x) xs))" by simp<br />
also have "… = ((P y ∧ Q y) ∧ (todos P xs ∧ todos Q xs))" using HI by simp<br />
also have "… = ((P y ∧ todos P xs) ∧ (Q y ∧ todos Q xs))" by blast<br />
also have "… = ((todos P (y#xs)) ∧ (todos Q (y#xs)))" by simp<br />
finally show "todos (λx. P x ∧ Q x) (y#xs) = (todos P (y#xs) ∧ todos Q (y#xs))" by simp<br />
qed<br />
<br />
<br />
(* migtermor *)<br />
lemma "todos (λx. P x ∧ Q x) xs = (todos P xs ∧ todos Q xs)"<br />
proof (induct xs)<br />
show "todos (λx. P x ∧ Q x) [] = (todos P [] ∧ todos Q [])" by simp<br />
next <br />
fix x xs<br />
assume HI: "todos (λx. P x ∧ Q x) xs = (todos P xs ∧ todos Q xs)"<br />
have "(todos (λx. P x ∧ Q x) (x#xs)) = (( P x ∧ Q x) ∧ (todos (λx. P x ∧ Q x) xs))"<br />
by (simp only: todos.simps(2))<br />
also have "… = ((P x ∧ Q x) ∧ (todos P xs ∧ todos Q xs))" using HI by simp<br />
also have "… = (((P x)∧(todos P xs)) ∧ ((Q x) ∧ (todos Q xs)))" by arith<br />
also have "((P x ∧ Q x) ∧ (todos P xs ∧ todos Q xs)) = (((P x)∧(todos P xs)) ∧ ((Q x) ∧ (todos Q xs)))"<br />
by arith (* Este paso es exactamente el mismo que el anterior, pero sin cualquiera de los dos no funciona el "finally show" *)<br />
have "… = (((P x)∧(todos P xs))∧((Q x)∧(todos Q xs)))" by simp<br />
have "… = ((todos P (x#xs))∧(todos Q (x#xs)))" by simp<br />
finally show "(todos (λx. P x ∧ Q x) (x#xs)) = ((todos P (x#xs)) ∧ (todos Q (x#xs)))" by simp <br />
qed<br />
<br />
(* marpoldia1 *)<br />
lemma "todos (λx. P x ∧ Q x) xs = (todos P xs ∧ todos Q xs)"<br />
proof (induct xs)<br />
show "todos (λx. P x ∧ Q x) [] = (todos P [] ∧ todos Q [])" by simp<br />
next<br />
fix a xs<br />
assume HI: "todos (λx. P x ∧ Q x) xs = (todos P xs ∧ todos Q xs)" <br />
have "todos (λx. P x ∧ Q x) (a # xs) = ((P a ∧ Q a) ∧ (todos P xs ∧ todos Q xs))" using HI by simp<br />
also have "... = ((P a ∧ todos P xs) ∧ (Q a ∧ todos Q xs))" by blast<br />
also have "... = (todos P (a#xs) ∧ todos Q (a#xs)) " by simp<br />
finally show "todos (λx. P x ∧ Q x) (a#xs) = (todos P (a#xs) ∧ todos Q (a#xs))" by simp <br />
qed<br />
<br />
(* ferrenseg *)<br />
<br />
lemma "todos (λx. P x ∧ Q x) xs = (todos P xs ∧ todos Q xs)"<br />
proof (induct xs)<br />
show "todos (λx. P x ∧ Q x) [] = (todos P [] ∧ todos Q [])" by simp<br />
next <br />
fix n xs<br />
assume HI: "todos (λx. P x ∧ Q x) xs = (todos P xs ∧ todos Q xs)"<br />
have "todos (λx. P x ∧ Q x) (n # xs) = (P n ∧ Q n ∧ todos P xs ∧ todos Q xs)" using HI by simp<br />
also have "… = P n ∧ Q n ∧ todos P xs ∧ todos Q xs" by blast<br />
also have "⋯ = todos P (n # xs) ∧ todos Q (n # xs)" by simp <br />
finally show "todos (λx. P x ∧ Q x) (n # xs) = todos P (n # xs) ∧ todos Q (n # xs)" by simp<br />
qed<br />
<br />
{* wilmorort pablucoto crigomgom anaprarod juacabsou rubgonmar fraortmoy *}<br />
<br />
lemma "todos (λx. P x ∧ Q x) xs = (todos P xs ∧ todos Q xs)"<br />
proof (induct xs)<br />
show "todos (λx. P x ∧ Q x) [] = (todos P [] ∧ todos Q [])" by simp <br />
next<br />
fix a xs<br />
assume HI: "todos (λx. P x ∧ Q x) xs = (todos P xs ∧ todos Q xs)"<br />
have "todos (λx. P x ∧ Q x) (a # xs) = (P a ∧ Q a ∧ todos (λx. P x ∧ Q x) xs)" by simp<br />
also have "... = (P a ∧ Q a ∧ todos P xs ∧ todos Q xs)" using HI by simp<br />
also have "... = (P a ∧ todos P xs ∧ Q a ∧ todos Q xs)" by arith<br />
also have "... = (todos P (a # xs) ∧ todos Q (a # xs))" by simp (* Este paso se puede obviar*)<br />
finally show "todos (λx. P x ∧ Q x) (a # xs) = (todos P (a # xs) ∧ todos Q (a # xs))" by simp<br />
qed<br />
<br />
text {*<br />
--------------------------------------------------------------------- <br />
Ejercicio 4.1. Demostrar o refutar automáticamente<br />
todos P (x @ y) = (todos P x ∧ todos P y)<br />
--------------------------------------------------------------------- <br />
*}<br />
{*danrodcha ivamenjim marpoldia1 migtermor ferrenseg wilmorort paupeddeg crigomgom anaprarod serrodcal juacabsou rubgonmar pablucoto fraortmoy *}<br />
lemma "todos P (x @ y) = (todos P x ∧ todos P y)"<br />
by (induct x) auto<br />
<br />
{* anaprarod dancorgar *}<br />
lemma "todos P (x @ y) = (todos P x ∧ todos P y)"<br />
apply (induct x)<br />
apply auto<br />
done<br />
<br />
text {*<br />
--------------------------------------------------------------------- <br />
Ejercicio 4.2. Demostrar o refutar detalladamente<br />
todos P (x @ y) = (todos P x ∧ todos P y)<br />
--------------------------------------------------------------------- <br />
*}<br />
<br />
(* ivamenjim ferrenseg wilmorort dancorgar fraortmoy *)<br />
<br />
lemma todos_append:<br />
"todos P (x @ y) = (todos P x ∧ todos P y)"<br />
proof (induct x)<br />
show "todos P ([] @ y) = (todos P [] ∧ todos P y)" by simp<br />
next <br />
fix a x<br />
assume HI: "todos P (x @ y) = (todos P x ∧ todos P y)"<br />
have "todos P ((a#x) @ y) = (P a ∧ (todos P (x @ y)))" by simp<br />
also have "... = (P a ∧ (todos P x ∧ todos P y))" using HI by simp<br />
finally show "todos P ((a # x) @ y) = (todos P (a # x) ∧ todos P y)" by simp<br />
qed<br />
<br />
(* migtermor serrodcal *)<br />
<br />
lemma todos_append1:<br />
"todos P (x @ y) = (todos P x ∧ todos P y)" (is "?P x")<br />
proof (induct x)<br />
show "?P []" by simp<br />
next <br />
fix a x<br />
assume HI: "?P x"<br />
have "todos P ((a#x) @ y) = (P a ∧ (todos P (x @ y)))" by simp<br />
also have "… = (P a ∧ (todos P x ∧ todos P y))" using HI by simp<br />
finally show "?P (a#x)" by simp<br />
qed <br />
<br />
(* marpoldia1 *)<br />
lemma todos_append:<br />
"todos P (x @ y) = (todos P x ∧ todos P y)"<br />
proof (induct x)<br />
show "todos P ([] @ y) = (todos P [] ∧ todos P y)" by simp<br />
next<br />
fix a x<br />
assume HI: "todos P (x @ y) = (todos P x ∧ todos P y)"<br />
have "todos P ((a # x) @ y) = todos P (a#(x@y))" by simp<br />
also have "... = (P a ∧ todos P (x @ y))" by simp<br />
also have "... = (P a ∧ (todos P x ∧ todos P y))" using HI by simp<br />
finally show "todos P ((a # x) @ y) = (todos P (a # x) ∧ todos P y)" by simp <br />
qed<br />
<br />
(* paupeddeg *)<br />
lemma todos_append3:<br />
"todos P (x @ y) = (todos P x ∧ todos P y)"<br />
proof (induct x)<br />
show "todos P ([] @ y) = (todos P [] ∧ todos P y)" by simp<br />
next<br />
fix a x<br />
assume HI: "todos P (x @ y) = (todos P x ∧ todos P y)"<br />
have "todos P ((a # x) @ y) = todos P (a # (x @ y))" by simp<br />
also have "... = (todos P [a] ∧ todos P (x @ y)) " by simp<br />
also have "... = (todos P [a] ∧ (todos P x ∧ todos P y))" using HI by simp<br />
finally show "todos P ((a # x) @ y) = (todos P (a # x) ∧ todos P y)" by simp<br />
qed<br />
<br />
(* crigomgom juacabsou anaprarod*)<br />
<br />
lemma todos_append4:<br />
"todos P (x @ y) = (todos P x ∧ todos P y)"<br />
proof (induct x)<br />
show "todos P ([] @ y) = (todos P [] ∧ todos P y)" by simp<br />
next<br />
fix x xs<br />
assume HI:"todos P (xs @ y) = (todos P xs ∧ todos P y)"<br />
have "todos P ((x # xs) @ y) = (P x ∧ todos P (xs @ y ))" by simp<br />
also have "... = (P x ∧ (todos P xs ∧ todos P y)) " using HI by simp<br />
also have "... = ((P x ∧ todos P xs) ∧ todos P y)" by simp<br />
finally show "todos P ((x # xs) @ y) = (todos P (x # xs) ∧ todos P y)" by simp<br />
qed<br />
<br />
(* pablucoto *)<br />
lemma todos_append5:<br />
"todos P (x @ y) = (todos P x ∧ todos P y)"<br />
proof (induct x )<br />
show "todos P ([] @ y) = (todos P [] ∧ todos P y)" by simp<br />
next<br />
fix a x<br />
assume HI:"todos P (x @ y) = (todos P x ∧ todos P y) " <br />
have "todos P ((a # x) @ y) = ( P a ∧ todos P (x @ y))" by simp<br />
also have " ... =( P a ∧ (todos P x ∧ todos P y))" using HI by simp<br />
also have "... = (todos P (a#x) ∧ todos P y)" by simp<br />
finally show "todos P ((a # x) @ y) = (todos P (a # x) ∧ todos P y)" by auto<br />
qed<br />
<br />
text {*<br />
--------------------------------------------------------------------- <br />
Ejercicio 5.1. Demostrar o refutar automáticamente <br />
todos P (rev xs) = todos P xs<br />
--------------------------------------------------------------------- <br />
*}<br />
<br />
lemma "todos P (rev xs) = todos P xs"<br />
oops<br />
<br />
(* migtermor ivamenjim marpoldia1 serrodcal anaprarod *)<br />
lemma "todos P (rev xs) = todos P xs" <br />
apply (induct xs)<br />
apply simp<br />
apply (simp add: todos_append)<br />
apply auto<br />
done<br />
<br />
(* ferrenseg crigomgom rubgonmar fraortmoy *)<br />
<br />
lemma "todos P (rev xs) = todos P xs"<br />
by (induct xs) (auto simp add: todos_append)<br />
<br />
(* wilmorort *)<br />
lemma "todos P (rev xs) = todos P xs"<br />
by (induct xs,simp,simp add: todos_append,auto)<br />
<br />
(* juacabsou *)<br />
lemma "todos P (rev xs) = todos P xs"<br />
apply (induct xs,simp,simp add: todos_append,auto) done<br />
<br />
(* dancorgar *)<br />
lemma "todos P (rev xs) = todos P xs"<br />
apply (induct xs)<br />
apply (simp add: todos_append)<br />
apply (simp add: todos_append)<br />
apply auto<br />
done<br />
<br />
(* pablucoto * )<br />
lemma "todos P (rev xs) = todos P xs"<br />
by (induct xs) (auto, simp_all add: todos_append) <br />
lemma "todos P (rev xs) = todos P xs"<br />
by (induct xs) ( simp_all add: todos_append, auto)<br />
<br />
text {*<br />
--------------------------------------------------------------------- <br />
Ejercicio 5.2. Demostrar o refutar detalladamente<br />
todos P (rev xs) = todos P xs<br />
--------------------------------------------------------------------- <br />
*}<br />
<br />
lemma "todos P (rev xs) = todos P xs"<br />
oops<br />
<br />
(* migtermor *)<br />
<br />
lemma auxiliar:<br />
"rev (a#xs) = rev xs @ [a]"<br />
by auto<br />
<br />
lemma "todos P (rev xs) = todos P xs" (is "?Q xs")<br />
proof (induct xs)<br />
show "?Q []" by (simp only: rev.simps(1))<br />
next <br />
fix a xs<br />
assume HI: "?Q xs"<br />
have "todos P (rev (a#xs)) = (todos P (rev xs @ [a]))" by (simp add: auxiliar)<br />
have "… = ((todos P (rev xs)) ∧ (todos P [a]))" by (simp add: todos_append)<br />
have "… = (todos P (rev xs) ∧ P a)" by simp<br />
also have Aux: "… = (todos P xs ∧ P a)" using HI by simp<br />
have Aux1: "… = (P a ∧ todos P xs)" by arith<br />
have "(todos P (rev xs) ∧ P a) = (P a ∧ todos P xs)" using Aux Aux1 by simp<br />
finally show "todos P (rev (a#xs)) = todos P (a#xs)" by (simp add: todos_append)<br />
qed<br />
<br />
<br />
(* marpoldia1 ferrenseg crigomgom serrodcal juacabsou rubgonmar *)<br />
lemma "todos P (rev xs) = todos P xs"<br />
proof (induct xs)<br />
show "todos P (rev []) = todos P []" by simp<br />
next<br />
fix a xs<br />
assume HI: "todos P (rev xs) = todos P xs"<br />
have "todos P (rev (a # xs)) = (todos P ((rev xs)@[a]))" by simp<br />
also have "... = (todos P (rev xs) ∧ todos P [a])" by (simp add:todos_append)<br />
also have "... = (todos P xs ∧ P a)" using HI by simp<br />
also have "... = (P a ∧ todos P xs)" by arith<br />
also have "... = (todos P (a#xs))" by simp<br />
finally show "todos P (rev (a # xs)) = (todos P (a#xs))" by simp <br />
qed<br />
<br />
(* ivamenjim *)<br />
<br />
lemma "todos P (rev xs) = todos P xs" (is "?P xs")<br />
proof (induct xs)<br />
show "?P []" by simp<br />
next <br />
fix a xs<br />
assume HI: "?P xs"<br />
have "todos P (rev (a # xs)) = (todos P ((rev xs)@[a]))" by simp<br />
also have "... = ((todos P (rev xs)) ∧ todos P [a])" by (simp add: todos_append)<br />
also have "... = (todos P xs ∧ todos P [a])" using HI by simp<br />
also have "... = (todos P [a] ∧ todos P xs)" by arith<br />
finally show "todos P (rev (a # xs)) = todos P (a # xs)" by simp<br />
qed<br />
<br />
(* fraortmoy *)<br />
<br />
lemma "todos P (rev xs) = todos P xs"<br />
proof (induct xs)<br />
show "todos P (rev []) = todos P []" by simp<br />
next<br />
fix a xs <br />
assume H1: "todos P (rev xs) = todos P xs"<br />
have " todos P (rev (a # xs)) = todos P (rev xs @ [a])" by simp<br />
also have "… = (todos P (rev xs) ∧ todos P [a])" by (simp add:todos_append)<br />
also have "… = (todos P xs ∧ todos P [a])" using H1 by simp<br />
also have "… = (todos P [a] ∧ todos P xs)" by arith<br />
finally show "todos P (rev (a # xs)) = todos P (a#xs)" by simp<br />
qed<br />
<br />
(* paupeddeg dancorgar *)<br />
lemma "todos P (rev xs) = todos P xs"<br />
proof (induct xs)<br />
show " todos P (rev []) = todos P []" by simp<br />
next<br />
fix a xs<br />
assume HI: "todos P (rev xs) = todos P xs "<br />
have "todos P (rev (a # xs)) = todos P (rev(xs) @ [a])" by simp<br />
also have "... = (todos P (rev xs) ∧ todos P [a]) " by (simp add: todos_append)<br />
also have "... = (todos P xs ∧ todos P [a])" using HI by simp<br />
also have "... = (todos P [a] ∧ todos P xs)" by arith<br />
also have "... = todos P ([a] @ xs)" by (simp add: todos_append)<br />
finally show "todos P (rev (a # xs)) = todos P (a # xs)" by simp<br />
qed <br />
<br />
(* wilmorort *)<br />
lemma "todos P (rev xs) = todos P xs" (is "?P xs")<br />
proof (induct xs) <br />
show "?P []" by simp <br />
next<br />
fix a xs<br />
assume HI: "?P xs"<br />
have "todos P (rev (a#xs)) = todos P ((rev xs) @ [a])" by simp<br />
also have "... = (todos P(rev xs) ∧ todos P [a])" by (simp add: todos_append)<br />
also have "... = (todos P xs ∧ todos P [a])" using HI by simp<br />
also have "... = (todos P [a] ∧ todos P xs)" by (simp add: HOL.conj_commute)<br />
also have "... = todos P([a]@(xs))" by (simp)<br />
finally show "todos P (rev (a#xs))= todos P (a # xs)" by simp<br />
qed<br />
<br />
(* anaprarod *) <br />
(* es igual que las anteriores pero con el final también con patrones *)<br />
<br />
lemma "todos P (rev xs) = todos P xs" (is "?P xs")<br />
proof (induct xs)<br />
show "?P []" by auto<br />
next<br />
fix a xs<br />
assume HI: "?P xs"<br />
have "todos P (rev (a#xs)) = todos P (rev xs @ [a])" by simp<br />
also have "… = (todos P (rev xs) ∧ todos P [a])" by (simp add: todos_append)<br />
also have "… = (todos P xs ∧ todos P [a])" using HI by simp<br />
also have "… = (todos P [a] ∧ todos P xs)" by arith<br />
also have "… = todos P (a#xs)" by simp<br />
finally show "?P (a # xs)" by simp<br />
qed<br />
<br />
text {*<br />
--------------------------------------------------------------------- <br />
Ejercicio 6. Demostrar o refutar:<br />
algunos (λx. P x ∧ Q x) xs = (algunos P xs ∧ algunos Q xs)<br />
--------------------------------------------------------------------- <br />
*}<br />
<br />
lemma "algunos (λx. P x ∧ Q x) xs = (algunos P xs ∧ algunos Q xs)"<br />
oops<br />
<br />
(* migtermor ivamenjim ferrenseg paupeddeg crigomgom serrodcal wilmorort juacabsou rubgonmar anaprarod marpoldia1 fraortmoy *)<br />
lemma "algunos (λx. P x ∧ Q x) xs = (algunos P xs ∧ algunos Q xs)" <br />
quickcheck<br />
oops<br />
(* Quickcheck encuentra el siguiente contraejemplo: P={a1}, Q={a2}, xs={a1,a2}. En este ejemplo:<br />
· "algunos (λx. P x ∧ Q x) xs = False"<br />
· "(algunos P xs ∧ algunos Q xs) = True" *)<br />
<br />
text {*<br />
--------------------------------------------------------------------- <br />
Ejercicio 7.1. Demostrar o refutar automáticamente <br />
algunos P (map f xs) = algunos (P ∘ f) xs<br />
--------------------------------------------------------------------- <br />
*}<br />
<br />
(* ivamenjim migtermor marpoldia1 crigomgom rubgonmar wilmorort anaprarod fraortmoy *)<br />
<br />
lemma "algunos P (map f xs) = algunos (P o f) xs"<br />
by (induct xs) auto<br />
<br />
(* ferrenseg *)<br />
<br />
lemma "algunos P (map f xs) = algunos (P o f) xs"<br />
by (induct xs) simp_all<br />
<br />
(* anaprarod *)<br />
lemma "algunos P (map f xs) = algunos (P o f) xs"<br />
apply (induct xs)<br />
apply auto<br />
done<br />
<br />
text {*<br />
--------------------------------------------------------------------- <br />
Ejercicio 7.2. Demostrar o refutar datalladamente<br />
algunos P (map f xs) = algunos (P ∘ f) xs<br />
--------------------------------------------------------------------- <br />
*}<br />
<br />
lemma "algunos P (map f xs) = algunos (P o f) xs"<br />
oops<br />
<br />
(* migtermor *)<br />
<br />
lemma AUX: "algunos (λa. P (f a)) xs = algunos (P o f) xs"<br />
by (induct xs) auto<br />
<br />
lemma "algunos P (map f xs) = algunos (P o f) xs" (is "?Q xs")<br />
proof (induct xs)<br />
show "?Q []" by simp<br />
next<br />
fix x xs<br />
assume HI: "?Q xs"<br />
have "algunos P (map f (x#xs)) = (algunos P ((f x)#(map f xs)))" by simp<br />
also have "… = ((P (f x)) ∨ (algunos P (map f xs)))" by (simp only: algunos.simps(2))<br />
also have "… = ((P (f x)) ∨ (algunos (P o f) xs))" <br />
proof (cases)<br />
assume C1: "(P (f x))"<br />
have Aux: "((P (f x)) ∨ (algunos P (map f xs))) = True" using C1 by simp<br />
have Aux1: "… = ((P (f x)) ∨ (algunos (P o f) xs))" using C1 by simp<br />
then show "((P (f x)) ∨ (algunos P (map f xs))) = ((P (f x)) ∨ (algunos (P o f) xs))" <br />
using Aux Aux1 by simp<br />
next<br />
assume C2: "¬(P (f x))"<br />
have Aux2: "((P (f x)) ∨ (algunos P (map f xs))) = (algunos P (map f xs))" using C2 by simp<br />
have Aux3: "… = (algunos (P o f) xs)" using HI by (simp add: AUX)<br />
also have "… = ((P (f x)) ∨ (algunos (P o f) xs))" using C2 by simp<br />
then show "((P (f x)) ∨ (algunos P (map f xs))) = ((P (f x)) ∨ (algunos (P o f) xs))" <br />
using Aux2 Aux3 by simp<br />
qed<br />
also have "… = (((P o f) x) ∨ (algunos (P o f) xs))" by simp<br />
also have "… = (algunos (P o f) (x#xs))" by simp<br />
finally show "algunos P (map f (x#xs)) = (algunos (P o f) (x#xs))" by simp<br />
qed<br />
<br />
(* ferrenseg *)<br />
<br />
lemma "algunos P (map f xs) = algunos (P ∘ f) xs"<br />
proof (induct xs)<br />
show "algunos P (map f []) = algunos (P ∘ f) []" by simp<br />
next<br />
fix x xs<br />
assume HI: "algunos P (map f xs) = algunos (P ∘ f) xs"<br />
show "algunos P (map f (x # xs)) = algunos (P ∘ f) (x # xs)"<br />
proof -<br />
have "algunos P (map f (x # xs)) = algunos P ((f x) # (map f xs))" by simp<br />
also have "… = ((P (f x)) ∨ (algunos P (map f xs)))" by simp<br />
also have "… = (((P ∘ f) x) ∨ (algunos (P ∘ f) xs))" using HI by simp<br />
also have "… = algunos (P ∘ f) (x # xs)" by simp<br />
finally show ?thesis<br />
qed<br />
qed<br />
<br />
(* ivamenjim *)<br />
<br />
lemma "algunos P (map f xs) = algunos (P o f) xs"<br />
proof (induct xs)<br />
show "algunos P (map f []) = algunos (P ∘ f) []" by simp<br />
next <br />
fix a xs<br />
assume HI: "algunos P (map f xs) = algunos (P ∘ f) xs"<br />
have "algunos P (map f (a # xs)) = algunos P ((f a)#map f xs)" by simp<br />
also have "... = (algunos P (map f [a]) ∨ algunos P (map f xs))" by simp <br />
also have "... = (algunos P (map f [a]) ∨ algunos (P ∘ f) xs)" using HI by simp <br />
finally show "algunos P (map f (a # xs)) = algunos (P ∘ f) (a # xs)" by simp<br />
qed<br />
<br />
(* crigomgom rubgonmar anaprarod marpoldia1 *)<br />
lemma "algunos P (map f xs) = algunos (P o f) xs"<br />
proof (induct xs)<br />
show "algunos P (map f []) = algunos (P ∘ f) []" by simp<br />
next<br />
fix x xs<br />
assume HI: "algunos P (map f xs) = algunos (P ∘ f) xs"<br />
have "algunos P (map f (x # xs)) = algunos P ((f x) # (map f xs))" by simp<br />
also have "... = (P (f x) ∨ algunos P (map f xs))" by simp<br />
also have "... = (P (f x) ∨ algunos (P ∘ f) xs)" using HI by simp<br />
also have "... = ((P ∘ f) x ∨ algunos (P ∘ f) xs)" by simp<br />
finally show "algunos P (map f (x # xs)) = algunos (P ∘ f) (x # xs)" by simp<br />
qed<br />
<br />
(* wilmorort *)<br />
lemma "algunos P (map f xs) = algunos (P o f) xs" (is "?P xs" )<br />
proof (induct xs)<br />
show "?P []" by simp<br />
next<br />
fix a xs<br />
assume HI: "algunos P (map f xs) = algunos (P ∘ f) xs"<br />
have "algunos P (map f (a # xs)) = (P (f a) ∨ algunos P (map f xs))" by simp<br />
also have "... = (P (f a) ∨ algunos (P ∘ f) xs )" using HI by simp<br />
also have "... = ((P ∘ f) a ∨ algunos (P ∘ f)xs)" by simp<br />
finally show " algunos P (map f (a # xs)) = algunos (P ∘ f) (a # xs)" by simp<br />
qed<br />
<br />
text {*<br />
--------------------------------------------------------------------- <br />
Ejercicio 8.1. Demostrar o refutar automáticamente <br />
algunos P (xs @ ys) = (algunos P xs ∨ algunos P ys)<br />
--------------------------------------------------------------------- <br />
*}<br />
<br />
(* ivamenjim marpoldia1 paupeddeg crigomgom rubgonmar wilmorort*)<br />
<br />
lemma "algunos P (xs @ ys) = (algunos P xs ∨ algunos P ys)"<br />
by (induct xs) auto<br />
<br />
(* ferrenseg *)<br />
lemma "algunos P (xs @ ys) = (algunos P xs ∨ algunos P ys)"<br />
by (induct xs) simp_all<br />
<br />
text {*<br />
--------------------------------------------------------------------- <br />
Ejercicio 8.2. Demostrar o refutar detalladamente<br />
algunos P (xs @ ys) = (algunos P xs ∨ algunos P ys)<br />
--------------------------------------------------------------------- <br />
*}<br />
<br />
(* ivamenjim ferrenseg marpoldia1 wilmorort*)<br />
<br />
lemma algunos_append:<br />
"algunos P (xs @ ys) = (algunos P xs ∨ algunos P ys)"<br />
proof (induct xs)<br />
show "algunos P ([] @ ys) = (algunos P [] ∨ algunos P ys)" by simp<br />
next <br />
fix a xs<br />
assume HI: "algunos P (xs @ ys) = (algunos P xs ∨ algunos P ys)"<br />
have "algunos P ((a # xs) @ ys) = (P a ∨ (algunos P (xs @ ys)))" by simp<br />
also have "... = (P a ∨ (algunos P xs ∨ algunos P ys))" using HI by simp<br />
finally show "algunos P ((a # xs) @ ys) = (algunos P (a # xs) ∨ algunos P ys)" by simp<br />
qed<br />
<br />
(* migtermor crigomgom *)<br />
<br />
lemma algunos_append:<br />
"algunos P (xs @ ys) = (algunos P xs ∨ algunos P ys)" (is "?Q xs")<br />
proof (induct xs)<br />
show "?Q []" by simp<br />
next <br />
fix x xs<br />
assume HI: "?Q xs"<br />
have "algunos P ((x#xs) @ ys) = algunos P (x#(xs @ ys))" by simp<br />
also have "… = ((P x) ∨ (algunos P (xs @ ys)))" by simp<br />
also have "… = ((P x) ∨ (algunos P xs) ∨ (algunos P ys))" using HI by simp<br />
also have "… = ((algunos P (x#xs)) ∨ (algunos P ys))" by simp<br />
finally show "algunos P ((x#xs) @ ys) = (algunos P (x#xs) ∨ algunos P ys)" by simp<br />
qed<br />
<br />
(* paupeddeg*)<br />
lemma algunos_append:<br />
"algunos P (xs @ ys) = (algunos P xs ∨ algunos P ys)"<br />
proof (induct xs)<br />
show "algunos P ([] @ ys) = (algunos P [] ∨ algunos P ys)" by simp<br />
next<br />
fix a xs<br />
assume HI: "algunos P (xs @ ys) = (algunos P xs ∨ algunos P ys)"<br />
have "algunos P ((a # xs) @ ys) = (algunos P [a] ∨ ( algunos P (xs @ ys)))" by simp<br />
also have "... = (algunos P [a] ∨ (algunos P xs ∨ algunos P ys))" using HI by simp<br />
finally show "algunos P ((a # xs) @ ys) = (algunos P (a # xs) ∨ algunos P ys)" by simp<br />
qed<br />
<br />
text {*<br />
--------------------------------------------------------------------- <br />
Ejercicio 9.1. Demostrar o refutar automáticamente<br />
algunos P (rev xs) = algunos P xs<br />
--------------------------------------------------------------------- <br />
*}<br />
<br />
(* ivamenjim migtermor marpoldia1 rubgonmar *)<br />
<br />
lemma "algunos P (rev xs) = algunos P xs"<br />
apply (induct xs)<br />
apply simp<br />
apply (simp add: algunos_append)<br />
apply auto<br />
done<br />
<br />
(* ferrenseg crigomgom*)<br />
lemma "algunos P (rev xs) = algunos P xs"<br />
by (induct xs) (auto simp add: algunos_append)<br />
<br />
(* wilmorort *)<br />
lemma "algunos P (rev xs) = algunos P xs"<br />
by (induct xs,simp,simp add: algunos_append,auto) <br />
<br />
text {*<br />
--------------------------------------------------------------------- <br />
Ejercicio 9.2. Demostrar o refutar detalladamente<br />
algunos P (rev xs) = algunos P xs<br />
--------------------------------------------------------------------- <br />
*}<br />
<br />
lemma "algunos P (rev xs) = algunos P xs"<br />
oops<br />
<br />
(* migtermor *)<br />
lemma auxiliar1:<br />
"rev (a#xs) = rev xs @ [a]"<br />
by auto <br />
<br />
lemma "algunos P (rev xs) = algunos P xs" (is "?Q xs")<br />
proof (induct xs)<br />
show "?Q []" by simp<br />
next<br />
fix x xs<br />
assume HI: "?Q xs"<br />
have "algunos P (rev (x#xs)) = (algunos P (rev xs @ [x]))" using auxiliar1 by simp<br />
also have "… = ((algunos P (rev xs)) ∨ (algunos P [x]))" by (simp add: algunos_append)<br />
also have "… = ((P x) ∨ (algunos P (rev xs)))" by simp arith<br />
also have "… = ((P x) ∨ (algunos P xs))" using HI by simp<br />
finally show "algunos P (rev (x#xs)) = (algunos P (x#xs))" by simp<br />
qed<br />
<br />
(* ferrenseg *)<br />
<br />
lemma "algunos P (rev xs) = algunos P xs"<br />
proof (induct xs)<br />
show "algunos P (rev []) = algunos P []" by simp<br />
next<br />
fix x xs<br />
assume HI: "algunos P (rev xs) = algunos P xs"<br />
show "algunos P (rev (x # xs)) = algunos P (x # xs)"<br />
proof -<br />
have "algunos P (rev (x # xs)) = algunos P ((rev xs) @ [x])" by simp<br />
also have "… = (algunos P (rev xs) ∨ algunos P [x])" <br />
by (simp add: algunos_append)<br />
also have "… = (algunos P xs ∨ algunos P [x])" using HI by simp<br />
also have "… = (algunos P xs ∨ P x)" by simp<br />
also have "… = (P x ∨ algunos P xs)" by auto<br />
also have "… = algunos P (x # xs)" by simp<br />
finally show ?thesis<br />
qed<br />
qed<br />
<br />
(* ivamenjim marpoldia1*)<br />
<br />
lemma "algunos P (rev xs) = algunos P xs" (is "?P xs")<br />
proof (induct xs)<br />
show "?P []" by simp<br />
next <br />
fix a xs<br />
assume HI: "?P xs"<br />
have "algunos P (rev (a # xs)) = (algunos P ((rev xs)@[a]))" by simp<br />
also have "... = ((algunos P (rev xs)) ∨ algunos P [a])" by (simp add: algunos_append)<br />
also have "... = (algunos P xs ∨ algunos P [a])" using HI by simp<br />
also have "... = (algunos P [a] ∨ algunos P xs)" by arith<br />
finally show "algunos P (rev (a # xs)) = algunos P (a # xs)" by simp<br />
qed<br />
<br />
(*paupeddeg*)<br />
lemma "algunos P (rev xs) = algunos P xs"<br />
proof (induct xs)<br />
show "algunos P (rev []) = algunos P []" by simp<br />
next<br />
fix a xs<br />
assume HI:"algunos P (rev xs) = algunos P xs"<br />
have "algunos P (rev (a # xs)) = algunos P ((rev xs) @ [a])" by simp<br />
also have "... = ((algunos P (rev xs)) ∨ (algunos P [a]))" by (simp add: algunos_append)<br />
also have "... = ((algunos P xs) ∨ (algunos P [a]))" using HI by simp<br />
also have "... = ((algunos P [a]) ∨ (algunos P xs))" by arith<br />
finally show "algunos P (rev (a # xs)) = algunos P (a # xs)" by simp<br />
qed<br />
<br />
(*crigomgom *)<br />
lemma "algunos P (rev xs) = algunos P xs"<br />
proof (induct xs)<br />
show "algunos P (rev []) = algunos P []" by simp<br />
next<br />
fix x xs<br />
assume HI: " algunos P (rev xs) = algunos P xs" <br />
have "algunos P (rev (x # xs)) = algunos P ((rev xs) @ [x])" by simp<br />
also have "... = (algunos P (rev xs) ∨ algunos P [x])" by (simp add: algunos_append)<br />
also have "... = (algunos P xs ∨ algunos P [x])" using HI by simp<br />
also have "... = (algunos P xs ∨ P x)" by simp<br />
also have "... = (P x ∨ algunos P xs)" by arith<br />
also have "... = algunos P (x#xs)" by simp<br />
finally show "algunos P (rev (x # xs)) = algunos P (x # xs)" by simp<br />
qed<br />
<br />
<br />
(*wilmorort*)<br />
lemma "algunos P (rev xs) = algunos P xs" (is "?P xs")<br />
proof (induct xs) <br />
show "?P []" by simp <br />
next<br />
fix a xs<br />
assume HI: "?P xs"<br />
have "algunos P (rev (a#xs)) = algunos P ((rev xs) @ [a])" by simp<br />
also have "... = (algunos P(rev xs) ∨ algunos P [a])" by (simp add: algunos_append)<br />
also have "... = (algunos P xs ∨ algunos P [a])" using HI by simp<br />
also have "... = (algunos P [a] ∨ algunos P xs)" by (simp add: HOL.disj_commute)<br />
also have "... = algunos P([a]@(xs))" by (simp)<br />
finally show "algunos P (rev (a#xs))= algunos P (a # xs)" by simp<br />
qed<br />
<br />
text {*<br />
--------------------------------------------------------------------- <br />
Ejercicio 10. Encontrar un término no trivial Z tal que sea cierta la <br />
siguiente ecuación:<br />
algunos (λx. P x ∨ Q x) xs = Z<br />
y demostrar la equivalencia de forma automática y detallada.<br />
--------------------------------------------------------------------- <br />
*}<br />
<br />
(* migtermor *)<br />
<br />
lemma "algunos (λx. P x ∨ Q x) xs = ((algunos (λx. P x) xs) ∨ (algunos (λx. Q x) xs))"<br />
by (induct xs) auto<br />
<br />
lemma "algunos (λx. P x ∨ Q x) xs = ((algunos (λx. P x) xs) ∨ (algunos (λx. Q x) xs))" (is "?R xs")<br />
proof (induct xs)<br />
show "?R []" by simp<br />
next<br />
fix x xs<br />
assume HI: "?R xs"<br />
have "algunos (λx. P x ∨ Q x) (x#xs) = (((λx. P x ∨ Q x) x) ∨ (algunos (λx. P x ∨ Q x) xs))"<br />
by simp<br />
also have "… = ((P x ∨ Q x) ∨ (algunos (λx. P x ∨ Q x) xs))" by simp<br />
also have H1: "… = ((((λx. P x) x) ∨ (algunos (λx. P x) xs)) ∨ (((λx. Q x) x) ∨ (algunos (λx. Q x) xs)))"<br />
using HI by simp arith<br />
have H2: "… = ((algunos (λx. P x) (x#xs)) ∨ (algunos (λx. Q x) (x#xs)))" <br />
by simp<br />
have C: "(algunos (λx. P x ∨ Q x) (x#xs)) = <br />
((algunos (λx. P x) (x#xs)) ∨ (algunos (λx. Q x) (x#xs)))" <br />
using H1 H2 by simp<br />
finally show "(algunos (λx. P x ∨ Q x) (x#xs)) = <br />
((algunos (λx. P x) (x#xs)) ∨ (algunos (λx. Q x) (x#xs)))" <br />
using C by simp<br />
qed<br />
<br />
<br />
(* ferrenseg *)<br />
<br />
lemma "algunos (λx. P x ∨ Q x) xs = (algunos P xs ∨ algunos Q xs)"<br />
proof (induct xs)<br />
show "algunos (λx. P x ∨ Q x) [] = (algunos P [] ∨ algunos Q [])" by simp<br />
next<br />
fix x xs<br />
assume HI: "algunos (λx. (P x ∨ Q x)) xs = (algunos P xs ∨ algunos Q xs)"<br />
show "algunos (λx. P x ∨ Q x) (x # xs) = (algunos P (x # xs) ∨ algunos Q (x # xs))"<br />
proof -<br />
have "algunos (λx. P x ∨ Q x) (x # xs) = <br />
((P x) ∨ (Q x) ∨ algunos (λx. P x ∨ Q x) xs)" by simp<br />
also have "… = ((P x) ∨ (Q x) ∨ algunos P xs ∨ algunos Q xs)" <br />
using HI by simp<br />
also have "… = (((P x) ∨ algunos P xs) ∨ ((Q x) ∨ algunos Q xs))" by auto<br />
also have "… = (algunos P (x # xs) ∨ algunos Q (x # xs))" by simp<br />
finally show ?thesis <br />
qed<br />
qed<br />
<br />
(* ivamenjim marpoldia1 *)<br />
<br />
lemma "algunos (λx. P x ∨ Q x) xs = (algunos P xs ∨ algunos Q xs)" <br />
by (induct xs) auto<br />
<br />
lemma "algunos (λx. P x ∨ Q x) xs = (algunos P xs ∨ algunos Q xs)"<br />
proof (induct xs)<br />
show "algunos (λx. P x ∨ Q x) [] = (algunos P [] ∨ algunos Q [])" by simp<br />
next <br />
fix a xs<br />
assume HI: "algunos (λx. P x ∨ Q x) xs = (algunos P xs ∨ algunos Q xs)"<br />
have "algunos (λx. P x ∨ Q x) (a # xs) = (P a ∨ Q a ∨ algunos (λx. P x ∨ Q x) xs)" by simp<br />
also have "... = (P a ∨ Q a ∨ algunos P xs ∨ algunos Q xs)" using HI by simp<br />
finally show "algunos (λx. P x ∨ Q x) (a # xs) = (algunos P (a # xs) ∨ algunos Q (a # xs))" by auto<br />
qed<br />
<br />
<br />
(* wilmorort *)<br />
lemma "algunos (λx. P x ∨ Q x) xs = (¬todos(λx. ¬P x)xs ∨ ¬todos(λx. ¬Q x)xs)"<br />
by (induct xs) auto<br />
<br />
<br />
lemma "algunos (λx. P x ∨ Q x) xs = (¬todos(λx. ¬P x)xs ∨ ¬todos(λx. ¬Q x)xs)"<br />
(is "?P xs")<br />
proof (induct xs)<br />
show "?P []" by simp<br />
next<br />
fix a xs<br />
assume HI: "?P xs"<br />
have "algunos (λx. P x ∨ Q x) (a # xs) = (P a ∨ Q a ∨ algunos (λx. P x ∨ Q x) xs)" by simp<br />
also have "... = (P a ∨ Q a ∨ (¬ todos (λx. ¬ P x) xs ∨ ¬ todos (λx. ¬ Q x) xs))" using HI by simp<br />
also have "... = (P a ∨ ¬ todos (λx. ¬ P x) xs ∨ Q a ∨ ¬ todos (λx. ¬ Q x) xs )" by arith<br />
finally show "algunos (λx. P x ∨ Q x) (a # xs) = (~ todos (λx. ¬ P x) (a # xs) ∨ ~ todos (λx. ¬ Q x) (a # xs))" by simp<br />
qed<br />
<br />
<br />
<br />
text {*<br />
--------------------------------------------------------------------- <br />
Ejercicio 11.1. Demostrar o refutar automáticamente<br />
algunos P xs = (¬ todos (λx. (¬ P x)) xs)<br />
--------------------------------------------------------------------- <br />
*}<br />
<br />
(* ivamenjim wilmorort migtermor marpoldia1 crigomgom rubgonmar *)<br />
<br />
lemma "algunos P xs = (¬ todos (λx. (¬ P x)) xs)"<br />
by (induct xs) auto<br />
<br />
(* ferrenseg *)<br />
<br />
lemma "algunos P xs = (¬ todos (λx. (¬ P x)) xs)"<br />
by (induct xs) simp_all<br />
<br />
text {*<br />
--------------------------------------------------------------------- <br />
Ejercicio 11.2. Demostrar o refutar datalladamente<br />
algunos P xs = (¬ todos (λx. (¬ P x)) xs)<br />
--------------------------------------------------------------------- <br />
*}<br />
<br />
(* ivamenjim migtermor marpoldia1 crigomgom*)<br />
<br />
lemma "algunos P xs = (¬ todos (λx. (¬ P x)) xs)"<br />
proof (induct xs)<br />
show "algunos P [] = (¬ todos (λx. ¬ P x) [])" by simp<br />
next <br />
fix a xs<br />
assume HI: "algunos P xs = (¬ todos (λx. ¬ P x) xs)"<br />
have "algunos P (a # xs) = (P a ∨ (algunos P xs))" by simp<br />
also have "... = (P a ∨ (¬ todos (λx. ¬ P x) xs))" using HI by simp<br />
finally show "algunos P (a # xs) = (¬ todos (λx. ¬ P x) (a # xs))" by simp<br />
qed<br />
<br />
(* ferrenseg *)<br />
<br />
lemma "algunos P xs = (¬ todos (λx. (¬ P x)) xs)"<br />
proof (induct xs)<br />
show "algunos P [] = (¬ todos (λx. (¬ P x)) [])" by simp<br />
next<br />
fix x xs<br />
assume HI: "algunos P xs = (¬ todos (λx. (¬ P x)) xs)"<br />
show "algunos P (x # xs) = (¬ todos (λx. (¬ P x)) (x # xs))"<br />
proof - <br />
have "algunos P (x # xs) = ((P x) ∨ algunos P xs)" by simp<br />
also have "… = ((P x) ∨ ¬ todos (λx. (¬ P x)) xs)" using HI by simp<br />
also have "… = (¬ (¬ (P x) ∧ todos (λx. (¬ P x)) xs))" by simp<br />
also have "… = (¬ todos (λx. (¬ P x)) (x # xs))" by simp<br />
finally show ?thesis<br />
qed<br />
qed<br />
<br />
text {*<br />
--------------------------------------------------------------------- <br />
Ejercicio 12. Definir la funcion primitiva recursiva <br />
estaEn :: 'a ⇒ 'a list ⇒ bool<br />
tal que (estaEn x xs) se verifica si el elemento x está en la lista<br />
xs. Por ejemplo, <br />
estaEn (2::nat) [3,2,4] = True<br />
estaEn (1::nat) [3,2,4] = False<br />
--------------------------------------------------------------------- <br />
*}<br />
<br />
(* ivamenjim ferrenseg migtermor serrodcal crigomgom rubgonmar marpoldia1*)<br />
<br />
fun estaEn :: "'a ⇒ 'a list ⇒ bool" where<br />
"estaEn x [] = False"<br />
| "estaEn x (a#xs) = ((a = x) ∨ (estaEn x xs))"<br />
<br />
value "estaEn (2::nat) [3,2,4] = True"<br />
value "estaEn (1::nat) [3,2,4] = False"<br />
<br />
<br />
text {*<br />
--------------------------------------------------------------------- <br />
Ejercicio 13. Expresar la relación existente entre estaEn y algunos. <br />
Demostrar dicha relación de forma automática y detallada.<br />
--------------------------------------------------------------------- <br />
*}<br />
<br />
(* migtermor crigomgom*)<br />
<br />
lemma "estaEn x xs = (algunos (λa. a=x) xs)"<br />
by (induct xs) auto<br />
<br />
lemma auxiliar13:<br />
"(x=a) = ((λa. a=x) a)"<br />
by auto<br />
<br />
lemma "estaEn x xs = (algunos (λa. a=x) xs)" (is "?P xs")<br />
proof (induct xs)<br />
show "?P []" by simp<br />
next<br />
fix a xs<br />
assume HI: "?P xs"<br />
have "estaEn x (a#xs) = ((a=x) ∨ (estaEn x xs))" by simp<br />
also have H: "… = ((a=x) ∨ (algunos (λa. a=x) xs))" using HI by simp<br />
also have "… = (((λa. a=x) a) ∨ (algunos (λa. a=x) xs))" using auxiliar13 by simp<br />
also have "… = (algunos (λa. a=x) (a#xs))" by simp<br />
have C: "estaEn x (a#xs) = (algunos (λa. a=x) (a#xs))" using H by simp<br />
finally show "estaEn x (a#xs) = (algunos (λa. a=x) (a#xs))" using C by simp<br />
qed<br />
<br />
<br />
(* ferrenseg *)<br />
<br />
lemma estaEn_algunos:<br />
"estaEn y xs = algunos (λx. x=y) xs"<br />
by (induct xs) auto<br />
<br />
lemma estaEn_algunos_2:<br />
"estaEn y xs = algunos (λx. x=y) xs"<br />
proof (induct xs)<br />
show "estaEn y [] = algunos (λx. x=y) []" by simp<br />
next<br />
fix x xs<br />
assume HI: "estaEn y xs = algunos (λx. x=y) xs"<br />
show "estaEn y (x # xs) = algunos (λx. x=y) (x # xs)"<br />
proof -<br />
have "estaEn y (x # xs) = (y=x ∨ estaEn y xs)" by simp<br />
also have "… = (y=x ∨ algunos (λx. x=y) xs)" using HI by simp<br />
also have "… = (x=y ∨ algunos (λx. x=y) xs)" by auto<br />
also have "… = algunos (λx. x=y) (x # xs)" by simp<br />
finally show ?thesis<br />
qed<br />
qed<br />
<br />
(*crigomgom*)<br />
lemma "estaEn x xs = (algunos (λa. a=x) xs)" <br />
proof (induct xs)<br />
show " estaEn x [] = algunos (λa. a = x) []" by simp<br />
next<br />
fix a xs<br />
assume HI: "estaEn x xs = algunos (λa. a = x) xs"<br />
have "estaEn x (a # xs) = (a = x ∨ estaEn x xs)" by simp<br />
also have "... = (a = x ∨ algunos (λa. a = x) xs)" using HI by simp<br />
finally show " estaEn x (a # xs) = algunos (λa. a = x) (a # xs)" by simp<br />
qed<br />
<br />
(* ivamenjim marpoldia1 *)<br />
<br />
lemma "estaEn x xs = algunos (λy. x=y) xs"<br />
by (induct xs) auto<br />
<br />
lemma "estaEn x xs = algunos (λy. x=y) xs"<br />
proof (induct xs)<br />
show "estaEn x [] = algunos (op = x) []" by simp<br />
next <br />
fix a xs<br />
assume HI: "estaEn x xs = algunos (op = x) xs"<br />
have "estaEn x (a # xs) = ((a = x) ∨ (estaEn x xs))" by simp<br />
also have "... = (a = x ∨ algunos (op = x) xs)" using HI by simp<br />
finally show " estaEn x (a # xs) = algunos (op = x) (a # xs)" by auto<br />
qed<br />
<br />
end<br />
</source></div>Lucnovdoshttps://www.glc.us.es/~jalonso/RA2016/index.php?title=Relaci%C3%B3n_3&diff=394Relación 32016-11-16T18:39:29Z<p>Lucnovdos: </p>
<hr />
<div><source lang="isar"><br />
chapter {* R3: Razonamiento sobre programas *}<br />
<br />
theory R3_Razonamiento_sobre_programas<br />
imports Main <br />
begin<br />
<br />
text {* --------------------------------------------------------------- <br />
Ejercicio 1.1. Definir la función<br />
sumaImpares :: nat ⇒ nat<br />
tal que (sumaImpares n) es la suma de los n primeros números<br />
impares. Por ejemplo,<br />
sumaImpares 5 = 25<br />
------------------------------------------------------------------ *}<br />
<br />
fun sumaImpares :: "nat ⇒ nat" where<br />
"sumaImpares 0 = 0"<br />
| "sumaImpares (Suc n) = sumaImpares n + (2*n+1)"<br />
<br />
text {* --------------------------------------------------------------- <br />
Ejercicio 1.2. Escribir la demostración detallada de <br />
sumaImpares n = n*n<br />
------------------------------------------------------------------- *}<br />
<br />
-- "La demostración detallada es"<br />
(* crigomgom fraortmoy marpoldia1 ivamenjim serrodcal rubgonmar ferrenseg juacabsou wilmorort josgarsan lucnovdos paupeddeg bowma pabrodmac*)<br />
lemma "sumaImpares n = n*n"<br />
proof (induct n)<br />
show "sumaImpares 0 = 0 * 0" by simp<br />
next<br />
fix n<br />
assume HI: "sumaImpares n = n * n"<br />
have "sumaImpares (Suc n) = sumaImpares n + 2*n + 1" by simp<br />
also have "... = n*n + 2*n + 1" using HI by simp<br />
also have "... = Suc n * Suc n" by simp<br />
finally show "sumaImpares (Suc n) = Suc n * Suc n" by simp<br />
qed<br />
<br />
(*danrodcha ; es la misma demostración que la anterior pero uso ?P para sustituir la propiedad.*)<br />
lemma "sumaImpares n = n*n" (is "?P n")<br />
proof (induct n)<br />
show "?P 0" by simp<br />
next<br />
fix n<br />
assume HI: "?P n"<br />
have "sumaImpares (Suc n) = sumaImpares n + 2*n + 1" by simp<br />
also have "... = n*n + 2*n + 1" using HI by simp<br />
also have "... = Suc n * Suc n" by simp <br />
finally show "?P (Suc n)" by simp<br />
qed<br />
<br />
(* migtermor *)<br />
lemma "sumaImpares n = n*n"<br />
proof (induct n)<br />
show "sumaImpares 0 = 0*0" by simp<br />
next<br />
fix n<br />
assume HI: "sumaImpares n = n*n"<br />
have "sumaImpares (Suc n) = sumaImpares n + 2*n+1" <br />
by (simp only: sumaImpares.simps(2))<br />
also have "sumaImpares (Suc n) = n*n + 2*n+1" using HI by simp<br />
finally show "sumaImpares (Suc n) = (Suc n)*(Suc n)" by simp<br />
qed<br />
<br />
(* fracorjim1 - Especifico la regla de simplificación y el paso del desarrollo al cuadrado para hacerlo más legible*)<br />
lemma "sumaImpares n = n*n"<br />
proof (induct n)<br />
show "sumaImpares 0 = 0 * 0" by simp<br />
next<br />
fix n<br />
assume HI : "sumaImpares n = n * n"<br />
have "sumaImpares (Suc n) = sumaImpares n + (2 * n + 1)" by (simp only: sumaImpares.simps(2))<br />
also have "... = n * n + 2 * n + 1" using HI by simp<br />
also have "... = Suc n * Suc n" by simp<br />
finally show "sumaImpares (Suc n) = Suc n * Suc n" by simp<br />
qed<br />
<br />
(*palucoto anaprarod marcarmor13*)<br />
lemma "sumaImpares n = n*n"<br />
proof (induct n)<br />
show "sumaImpares 0 = 0 * 0" by simp<br />
next <br />
fix n <br />
assume HI : "sumaImpares n = n * n"<br />
have " sumaImpares (Suc n) = sumaImpares n + 2*n+1" by simp<br />
also have "... = n*n + 2*n+1" using HI by simp<br />
finally show "sumaImpares (Suc n) = Suc n * Suc n" by simp<br />
qed<br />
<br />
<br />
text {* --------------------------------------------------------------- <br />
Ejercicio 2.1. Definir la función<br />
sumaPotenciasDeDosMasUno :: nat ⇒ nat<br />
tal que <br />
(sumaPotenciasDeDosMasUno n) = 1 + 2^0 + 2^1 + 2^2 + ... + 2^n. <br />
Por ejemplo, <br />
sumaPotenciasDeDosMasUno 3 = 16<br />
------------------------------------------------------------------ *}<br />
<br />
fun sumaPotenciasDeDosMasUno :: "nat ⇒ nat" where<br />
"sumaPotenciasDeDosMasUno 0 = 2"<br />
| "sumaPotenciasDeDosMasUno (Suc n) = <br />
sumaPotenciasDeDosMasUno n + 2^(n+1)"<br />
<br />
text {* --------------------------------------------------------------- <br />
Ejercicio 2.2. Escribir la demostración detallada de <br />
sumaPotenciasDeDosMasUno n = 2^(n+1)<br />
------------------------------------------------------------------- *}<br />
<br />
(*crigomgom ivamenjim danrodcha serrodcal rubgonmar ferrenseg juacabsou wilmorort anaprarod marpoldia1 josgarsan paupeddeg pabrodmac bowma*)<br />
lemma "sumaPotenciasDeDosMasUno n = 2^(n+1)"<br />
proof (induct n)<br />
show "sumaPotenciasDeDosMasUno 0 = 2 ^ (0 + 1)" by simp<br />
next<br />
fix n<br />
assume HI: "sumaPotenciasDeDosMasUno n = 2 ^ (n + 1)"<br />
have "sumaPotenciasDeDosMasUno (Suc n) = sumaPotenciasDeDosMasUno n + 2^(n+1)" by simp<br />
also have "... = 2 ^ (n + 1) + 2 ^ (n + 1)" using HI by simp<br />
also have "... = 2 ^ ((Suc n) + 1)" by simp<br />
finally show "sumaPotenciasDeDosMasUno (Suc n) = 2 ^ (Suc n + 1)" by simp<br />
qed<br />
<br />
(* fraortmoy *)<br />
(* es la misma demostración, pero quise probar a delimitar lo que se usa en el "by simp" *)<br />
<br />
lemma "sumaPotenciasDeDosMasUno n = 2^(n+1)"<br />
proof (induct n)<br />
show "sumaPotenciasDeDosMasUno 0 = 2 ^ (0 + 1)" by simp<br />
next<br />
fix n<br />
assume H1:" sumaPotenciasDeDosMasUno n = 2 ^ (n + 1)"<br />
have "sumaPotenciasDeDosMasUno (Suc n) = sumaPotenciasDeDosMasUno n + 2^(n + 1)" by (simp only : sumaPotenciasDeDosMasUno.simps(2))<br />
also have "... = 2 ^ (n + 1) + 2 ^ (n + 1)" using H1 by simp<br />
also have "... = 2 ^ (Suc n + 1)" by simp<br />
finally show "sumaPotenciasDeDosMasUno (Suc n) = 2 ^ (Suc n + 1)" by simp<br />
qed<br />
<br />
(* migtermor *)<br />
lemma "sumaPotenciasDeDosMasUno n = 2^(n+1)"<br />
proof (induct n)<br />
show "sumaPotenciasDeDosMasUno 0 = 2^(0+1)" by simp<br />
next<br />
fix n<br />
assume HI: "sumaPotenciasDeDosMasUno n = 2^(n+1)"<br />
have "sumaPotenciasDeDosMasUno (Suc n) = sumaPotenciasDeDosMasUno n + 2^(n+1)"<br />
by (simp only: sumaPotenciasDeDosMasUno.simps(2))<br />
also have "sumaPotenciasDeDosMasUno (Suc n) = 2^(n+1)+2^(n+1)" using HI by simp<br />
finally show "sumaPotenciasDeDosMasUno (Suc n) = 2^((Suc n)+1)" by simp<br />
qed<br />
<br />
(*pablucoto marcarmor13 lucnovdos*)<br />
lemma "sumaPotenciasDeDosMasUno n = 2^(n+1)"<br />
proof (induct n)<br />
show "sumaPotenciasDeDosMasUno 0 = 2 ^ (0 + 1)" by simp<br />
next <br />
fix n <br />
assume HI : "sumaPotenciasDeDosMasUno n = 2 ^ (n + 1)"<br />
have "sumaPotenciasDeDosMasUno (Suc n) = sumaPotenciasDeDosMasUno n + 2^(n+1) " by simp<br />
also have "... = 2^(n + 1) + 2^( n + 1) " using HI by simp <br />
finally show " sumaPotenciasDeDosMasUno (Suc n) = 2 ^ (Suc n + 1)" by simp<br />
qed<br />
<br />
(* fracorjim1 - Hago explícita toda la manipulación algebraica. Quizás excesivo.*)<br />
lemma "sumaPotenciasDeDosMasUno n = 2^(n+1)"<br />
proof (induction n)<br />
show "sumaPotenciasDeDosMasUno 0 = 2 ^ (0 + 1)" by simp<br />
next<br />
fix n<br />
assume HI : "sumaPotenciasDeDosMasUno n = 2 ^ (n + 1)"<br />
have "sumaPotenciasDeDosMasUno (Suc n) = sumaPotenciasDeDosMasUno n + 2 ^ (n + 1)"<br />
by (simp only:sumaPotenciasDeDosMasUno.simps(2))<br />
also have "... = 2 ^ (n + 1) + 2 ^ (n + 1)" using HI by simp<br />
also have "... = 2 ^ (n + 1) * 2" by simp<br />
also have "... = 2 ^ ((n + 1) + 1)" by simp<br />
also have "... = 2 ^ (Suc n + 1)" by simp<br />
finally show "sumaPotenciasDeDosMasUno (Suc n) = 2 ^ (Suc n + 1)" by simp<br />
qed<br />
<br />
<br />
text {* --------------------------------------------------------------- <br />
Ejercicio 3.1. Definir la función<br />
copia :: nat ⇒ 'a ⇒ 'a list<br />
tal que (copia n x) es la lista formado por n copias del elemento<br />
x. Por ejemplo, <br />
copia 3 x = [x,x,x]<br />
------------------------------------------------------------------ *}<br />
<br />
fun copia :: "nat ⇒ 'a ⇒ 'a list" where<br />
"copia 0 x = []"<br />
| "copia (Suc n) x = x # copia n x"<br />
<br />
text {* --------------------------------------------------------------- <br />
Ejercicio 3.2. Definir la función<br />
todos :: ('a ⇒ bool) ⇒ 'a list ⇒ bool<br />
tal que (todos p xs) se verifica si todos los elementos de xs cumplen<br />
la propiedad p. Por ejemplo,<br />
todos (λx. x>(1::nat)) [2,6,4] = True<br />
todos (λx. x>(2::nat)) [2,6,4] = False<br />
------------------------------------------------------------------ *}<br />
<br />
fun todos :: "('a ⇒ bool) ⇒ 'a list ⇒ bool" where<br />
"todos p [] = True"<br />
| "todos p (x#xs) = (p x ∧ todos p xs)"<br />
<br />
text {* --------------------------------------------------------------- <br />
Ejercicio 3.2. Demostrar detalladamente que todos los elementos de<br />
(copia n x) son iguales a x. <br />
------------------------------------------------------------------- *}<br />
<br />
(* crigomgom ivamenjim serrodcal ferrenseg wilmorort juacabsou josgarsan lucnovdos*)<br />
lemma "todos (λy. y=x) (copia n x)"<br />
proof (induct n)<br />
show "todos (λy. y = x) (copia 0 x)" by simp<br />
next<br />
fix n<br />
assume HI: "todos (λy. y = x) (copia n x)" <br />
have "todos (λy. y = x) (copia (Suc n) x) = todos (λy. y = x) (x # (copia n x))" by simp<br />
also have "... = ((x = x) ∧ (todos (λy. y = x) (copia n x)))" by simp<br />
also have "... = True" using HI by simp<br />
finally show "todos (λy. y = x) (copia (Suc n) x)" by simp<br />
qed<br />
<br />
<br />
(* fraortmoy *)<br />
lemma "todos (λy. y=x) (copia n x)"<br />
proof (induct n)<br />
have "todos (λy. y = x) (copia 0 x) = todos (λy. y = x) []" by (simp only: copia.simps(1))<br />
also have "... = True" by (simp only: todos.simps(1))<br />
show "todos (λy. y = x) (copia 0 x)" by simp<br />
next<br />
fix n <br />
assume H1 : " todos (λy. y = x) (copia n x) "<br />
have "todos (λy. y = x) (copia (Suc n) x) = ((todos (λy. y = x) (x#[])) ∧ (todos (λy. y = x) (copia n x) )) " by simp<br />
also have " ... = (todos (λy. y = x) (x#[]))" using H1 by simp<br />
also have " ... = ((λy. y = x) x ∧ todos (λy. y = x) [])" by simp<br />
also have " ... = True" by simp<br />
finally show " todos (λy. y = x) (copia (Suc n) x)" by simp<br />
qed<br />
<br />
(*danrodcha rubgonmar*)<br />
lemma "todos (λy. y=x) (copia n x)" (is "?P n")<br />
proof (induct n)<br />
show "?P 0" by simp<br />
next<br />
fix n<br />
assume HI: "?P n"<br />
have "todos (λy. y=x) (copia (Suc n) x) = todos (λy. y=x) (x # copia n x)" by simp<br />
also have "... = ((λy. y=x) x ∧ todos (λy. y=x) (copia n x))" by simp<br />
also have "... = todos (λy. y=x) (copia n x)" by simp<br />
finally show "?P (Suc n)" using HI by simp<br />
qed<br />
<br />
(* migtermor *)<br />
lemma "todos (λy. y=x) (copia n x)"<br />
proof (induct n)<br />
have "todos (λy. y=x) (copia 0 x) = todos (λy. y=x) []" by (simp only: copia.simps(1))<br />
also have "todos (λy. y=x) []" by (simp only: todos.simps(1))<br />
show "todos (λy. y=x) (copia 0 x)" by simp<br />
next <br />
fix n<br />
assume HI: "todos (λy. y=x) (copia n x)"<br />
have "todos (λy. y=x) (copia (Suc n) x) = todos (λy. y=x) (x # (copia n x))"<br />
by (simp only: copia.simps(2))<br />
also have "todos (λy. y=x) (x#(copia n x)) = ((λy. y=x) x ∧ todos (λy. y=x) (copia n x))"<br />
by (simp only: todos.simps(2))<br />
also have "todos (λy. y=x) (copia (Suc n) x) = ((λy. y=x) x)" using HI by simp<br />
also have "((λy. y=x) x) = True" by simp <br />
finally show "(todos (λy. y=x) (copia (Suc n) x))" by simp<br />
qed<br />
<br />
(*pablucoto pabrodmac marcarmor13*)<br />
<br />
lemma "todos (λy. y=x) (copia n x)"<br />
proof (induct n)<br />
show "todos (λy. y = x) (copia 0 x)" by simp<br />
next <br />
fix n<br />
assume HI: "todos (λy. y=x) (copia n x)"<br />
have "todos (λy. y = x) (copia (Suc n) x) = todos (λy. y=x) (x # copia n x)" by simp<br />
also have "... = ( (λy. y=x) x ∧ todos (λy. y=x) (copia (Suc n) x) ) " by simp<br />
also have "... = True" using HI by simp<br />
finally show " todos (λy. y = x) (copia (Suc n) x) " by simp<br />
qed<br />
<br />
(* anaprarod marpoldia1 *)<br />
lemma "todos (λy. y=x) (copia n x)"<br />
proof (induct n)<br />
show "todos (λy. y = x) (copia 0 x)" by simp<br />
next<br />
fix n<br />
assume HI: "todos (λy. y = x) (copia n x)"<br />
have "todos (λy. y = x) (copia (Suc n) x) =todos (λy. y = x) (x # copia n x)" by simp<br />
also have "... = ((λy. y = x) x ∧ todos (λy. y = x) ( copia n x))" by simp<br />
finally show "todos (λy. y = x) (copia (Suc n) x)" using HI by simp<br />
qed<br />
<br />
(*paupeddeg*)<br />
lemma "todos (λy. y=x) (copia n x)"<br />
proof (induct n)<br />
show "todos (λy. y = x) (copia 0 x)" by simp<br />
next <br />
fix n<br />
assume HI: "todos (λy. y = x) (copia n x)" <br />
have "todos (λy. y = x) (copia (Suc n) x) = (todos (λy. y = x) (x # copia n x))" by simp<br />
also have "... = ((todos (λy. y = x) [x]) ∧ (todos (λy. y = x) (copia n x)))" by simp<br />
also have "... = True" using HI by simp<br />
finally show "todos(λy. y = x) (copia (Suc n) x)" by simp<br />
qed<br />
<br />
text {* --------------------------------------------------------------- <br />
Ejercicio 4.1. Definir la función<br />
factR :: nat ⇒ nat<br />
tal que (factR n) es el factorial de n. Por ejemplo,<br />
factR 4 = 24<br />
------------------------------------------------------------------ *}<br />
<br />
fun factR :: "nat ⇒ nat" where<br />
"factR 0 = 1"<br />
| "factR (Suc n) = Suc n * factR n"<br />
<br />
text {* --------------------------------------------------------------- <br />
Ejercicio 4.2. Se considera la siguiente definición iterativa de la<br />
función factorial <br />
factI :: "nat ⇒ nat" where<br />
factI n = factI' n 1<br />
<br />
factI' :: nat ⇒ nat ⇒ nat" where<br />
factI' 0 x = x<br />
factI' (Suc n) x = factI' n (Suc n)*x<br />
Demostrar que, para todo n y todo x, se tiene <br />
factI' n x = x * factR n<br />
Indicación: La propiedad mult_Suc es <br />
(Suc m) * n = n + m * n<br />
Puede que se necesite desactivarla en un paso con <br />
(simp del: mult_Suc)<br />
------------------------------------------------------------------- *}<br />
<br />
fun factI' :: "nat ⇒ nat ⇒ nat" where<br />
"factI' 0 x = x"<br />
| "factI' (Suc n) x = factI' n (x * Suc n)"<br />
<br />
fun factI :: "nat ⇒ nat" where<br />
"factI n = factI' n 1"<br />
<br />
(* fraortmoy , danrodcha serrodcal pablucoto wilmorort anaprarod marpoldia1 juacabsou josgarsan rubgonmar paupeddeg pabrodmac bowma*)<br />
lemma fact: "factI' n x = x * factR n"<br />
proof (induct n)<br />
show "factI' 0 x = x * factR 0" by simp<br />
next<br />
fix n<br />
assume H1 : "factI' n x = x * factR n"<br />
have "factI' (Suc n) x = factI' n (x * Suc n)" by (simp only:factI'.simps(2))<br />
also have "... = (x * Suc n) * factR n" using H1 by simp (* no entiendo por qué no hace esto bien y luego todo funciona *) (*Creo que es porque no le estas dando la hipótesis de inducción sino otra que se le parece*)<br />
also have "... = x * factR (Suc n)" by (simp del: mult_Suc)<br />
finally show "factI' (Suc n) x = x * factR (Suc n)" by simp<br />
qed<br />
<br />
(* crigomgom ivamenjim ferrenseg marcarmor13 *)<br />
lemma fact: "factI' n x = x * factR n"<br />
proof (induct n arbitrary: x)<br />
show "⋀x. factI' 0 x = x * factR 0" by simp<br />
next<br />
fix n<br />
assume HI: "⋀x. factI' n x = x * factR n"<br />
show "⋀x. factI' (Suc n) x = x * factR (Suc n)"<br />
proof -<br />
fix x<br />
have "factI' (Suc n) x = factI' n (x * Suc n)" by simp<br />
also have "... = (x * Suc n) * factR n" using HI by simp<br />
also have "... = x * (Suc n * factR n)" by (simp del: mult_Suc)<br />
also have "... = x * factR (Suc n)" by simp<br />
finally show "factI' (Suc n) x = x * factR (Suc n)" by simp<br />
qed<br />
qed<br />
<br />
(* migtermor *)<br />
<br />
lemma fact: "factI' n x = x * factR n"<br />
proof (induct n)<br />
have "x * factR 0 = x" by (simp only: factR.simps(1))<br />
also have "factI' 0 x = x" by (simp only: factI'.simps(1))<br />
show "factI' 0 x = x * factR 0" by simp<br />
next<br />
fix n<br />
assume HI: "∀x. factI' n x = x * factR n"<br />
fix x<br />
have "factI' (Suc n) x = factI' n (x * Suc n)" by (simp only: factI'.simps(2))<br />
also have "factI' (Suc n) x = factI' n (x * Suc n)" by simp<br />
have "factI' (Suc n) x = x * Suc n * factR n" using HI by simp<br />
also have "... = x * factR (Suc n)" by (simp del: mult_Suc)<br />
finally show "factI' (Suc n) x = x * factR (Suc n)" by simp (* No entiendo por qué no acepta esto como demostrado *)<br />
qed<br />
<br />
<br />
text {* --------------------------------------------------------------- <br />
Ejercicio 4.3. Escribir la demostración detallada de<br />
factI n = factR n<br />
------------------------------------------------------------------- *}<br />
<br />
(* fraortmoy danrodcha crigomgom ivamenjim serrodcal ferrenseg pablucoto wilmorort rubgonmar marpoldia1 juacabsou josgarsan paupeddeg*)<br />
corollary "factI n = factR n"<br />
proof -<br />
have "factI n = factI' n 1" by simp<br />
also have "... = 1 * factR n" by (simp add: fact)<br />
finally show "factI n = factR n" by simp<br />
qed<br />
<br />
(* migtermor *)<br />
corollary "factI n = factR n"<br />
proof -<br />
have "factI n = factI' n 1" by simp<br />
also have "... = 1 * factI' n 1" by simp<br />
also have "... = 1 * factR n" using fact by simp<br />
finally show "factI n = factR n" by simp<br />
qed<br />
<br />
(*pabrodmac*)<br />
corollary "factI n = factR n"<br />
proof -<br />
have "factI n = factI' n 1" by simp<br />
also have "... = 1 * factR n" using fact by simp<br />
finally show "factI n = factR n" by simp<br />
qed<br />
<br />
text {* --------------------------------------------------------------- <br />
Ejercicio 5.1. Definir, recursivamente y sin usar (@), la función<br />
amplia :: 'a list ⇒ 'a ⇒ 'a list<br />
tal que (amplia xs y) es la lista obtenida añadiendo el elemento y al<br />
final de la lista xs. Por ejemplo,<br />
amplia [d,a] t = [d,a,t]<br />
------------------------------------------------------------------ *}<br />
<br />
fun amplia :: "'a list ⇒ 'a ⇒ 'a list" where<br />
"amplia [] y = [y]"<br />
| "amplia (x#xs) y = x # amplia xs y"<br />
<br />
text {* --------------------------------------------------------------- <br />
Ejercicio 5.2. Escribir la demostración detallada de<br />
amplia xs y = xs @ [y]<br />
------------------------------------------------------------------- *}<br />
<br />
(* crigomgom fraortmoy rubgonmar ivamenjim serrodcal pablucoto wilmorort anaprarod marpoldia1 juacabsou paupeddeg pabrodmac lucnovdos*)<br />
lemma "amplia xs y = xs @ [y]"<br />
proof (induct xs)<br />
show "amplia [] y = [] @ [y]" by simp<br />
next<br />
fix x xs<br />
assume HI: "amplia xs y = xs @ [y]"<br />
have "amplia (x # xs) y = x # (amplia xs y)" by simp<br />
also have "... = x # (xs @ [y])" using HI by simp<br />
also have "... = (x # xs) @ [y]" by simp<br />
finally show "amplia (x # xs) y = (x # xs) @ [y]" by simp<br />
qed<br />
<br />
(*danrodcha ferrenseg *)<br />
lemma "amplia xs y = xs @ [y]" (is "?P xs")<br />
proof (induct xs)<br />
show "?P []" by simp<br />
next<br />
fix x xs<br />
assume HI: "?P xs"<br />
have "amplia (x # xs) y = x # amplia xs y" by simp<br />
also have "... = (x # xs) @ [y]" using HI by simp<br />
finally show "?P (x#xs)" by simp<br />
qed<br />
<br />
(* migtermor *)<br />
lemma "amplia xs y = xs @ [y]"<br />
proof (induct xs)<br />
have "amplia [] y = [y]" by (simp only: amplia.simps(1)) <br />
have "[] @ [y] = [y]" by simp<br />
show "amplia [] y = [] @ [y]" by simp<br />
next <br />
fix x xs<br />
assume HI: "amplia xs y = xs @ [y]"<br />
have "amplia (x # xs) y = x # amplia xs y" by (simp only: amplia.simps(2))<br />
also have "... = x # (xs @ [y])" using HI by simp<br />
also have "... = (x # xs) @ [y]" by simp<br />
finally show "amplia (x # xs) y = (x # xs) @ [y]" by simp<br />
qed<br />
<br />
end<br />
</source></div>Lucnovdoshttps://www.glc.us.es/~jalonso/RA2016/index.php?title=Relaci%C3%B3n_3&diff=372Relación 32016-11-15T17:46:58Z<p>Lucnovdos: </p>
<hr />
<div><source lang="isar"><br />
chapter {* R3: Razonamiento sobre programas *}<br />
<br />
theory R3_Razonamiento_sobre_programas<br />
imports Main <br />
begin<br />
<br />
text {* --------------------------------------------------------------- <br />
Ejercicio 1.1. Definir la función<br />
sumaImpares :: nat ⇒ nat<br />
tal que (sumaImpares n) es la suma de los n primeros números<br />
impares. Por ejemplo,<br />
sumaImpares 5 = 25<br />
------------------------------------------------------------------ *}<br />
<br />
fun sumaImpares :: "nat ⇒ nat" where<br />
"sumaImpares 0 = 0"<br />
| "sumaImpares (Suc n) = sumaImpares n + (2*n+1)"<br />
<br />
text {* --------------------------------------------------------------- <br />
Ejercicio 1.2. Escribir la demostración detallada de <br />
sumaImpares n = n*n<br />
------------------------------------------------------------------- *}<br />
<br />
-- "La demostración detallada es"<br />
(* crigomgom fraortmoy marpoldia1 ivamenjim serrodcal rubgonmar ferrenseg juacabsou wilmorort josgarsan lucnovdos*)<br />
lemma "sumaImpares n = n*n"<br />
proof (induct n)<br />
show "sumaImpares 0 = 0 * 0" by simp<br />
next<br />
fix n<br />
assume HI: "sumaImpares n = n * n"<br />
have "sumaImpares (Suc n) = sumaImpares n + 2*n + 1" by simp<br />
also have "... = n*n + 2*n + 1" using HI by simp<br />
also have "... = Suc n * Suc n" by simp<br />
finally show "sumaImpares (Suc n) = Suc n * Suc n" by simp<br />
qed<br />
<br />
(*danrodcha ; es la misma demostración que la anterior pero uso ?P para sustituir la propiedad.*)<br />
lemma "sumaImpares n = n*n" (is "?P n")<br />
proof (induct n)<br />
show "?P 0" by simp<br />
next<br />
fix n<br />
assume HI: "?P n"<br />
have "sumaImpares (Suc n) = sumaImpares n + 2*n + 1" by simp<br />
also have "... = n*n + 2*n + 1" using HI by simp<br />
also have "... = Suc n * Suc n" by simp <br />
finally show "?P (Suc n)" by simp<br />
qed<br />
<br />
(* migtermor *)<br />
lemma "sumaImpares n = n*n"<br />
proof (induct n)<br />
show "sumaImpares 0 = 0*0" by simp<br />
next<br />
fix n<br />
assume HI: "sumaImpares n = n*n"<br />
have "sumaImpares (Suc n) = sumaImpares n + 2*n+1" <br />
by (simp only: sumaImpares.simps(2))<br />
also have "sumaImpares (Suc n) = n*n + 2*n+1" using HI by simp<br />
finally show "sumaImpares (Suc n) = (Suc n)*(Suc n)" by simp<br />
qed<br />
<br />
(* fracorjim1 - Especifico la regla de simplificación y el paso del desarrollo al cuadrado para hacerlo más legible*)<br />
lemma "sumaImpares n = n*n"<br />
proof (induct n)<br />
show "sumaImpares 0 = 0 * 0" by simp<br />
next<br />
fix n<br />
assume HI : "sumaImpares n = n * n"<br />
have "sumaImpares (Suc n) = sumaImpares n + (2 * n + 1)" by (simp only: sumaImpares.simps(2))<br />
also have "... = n * n + 2 * n + 1" using HI by simp<br />
also have "... = Suc n * Suc n" by simp<br />
finally show "sumaImpares (Suc n) = Suc n * Suc n" by simp<br />
qed<br />
<br />
(*palucoto anaprarod*)<br />
lemma "sumaImpares n = n*n"<br />
proof (induct n)<br />
show "sumaImpares 0 = 0 * 0" by simp<br />
next <br />
fix n <br />
assume HI : "sumaImpares n = n * n"<br />
have " sumaImpares (Suc n) = sumaImpares n + 2*n+1" by simp<br />
also have "... = n*n + 2*n+1" using HI by simp<br />
finally show "sumaImpares (Suc n) = Suc n * Suc n" by simp<br />
qed<br />
<br />
<br />
text {* --------------------------------------------------------------- <br />
Ejercicio 2.1. Definir la función<br />
sumaPotenciasDeDosMasUno :: nat ⇒ nat<br />
tal que <br />
(sumaPotenciasDeDosMasUno n) = 1 + 2^0 + 2^1 + 2^2 + ... + 2^n. <br />
Por ejemplo, <br />
sumaPotenciasDeDosMasUno 3 = 16<br />
------------------------------------------------------------------ *}<br />
<br />
fun sumaPotenciasDeDosMasUno :: "nat ⇒ nat" where<br />
"sumaPotenciasDeDosMasUno 0 = 2"<br />
| "sumaPotenciasDeDosMasUno (Suc n) = <br />
sumaPotenciasDeDosMasUno n + 2^(n+1)"<br />
<br />
text {* --------------------------------------------------------------- <br />
Ejercicio 2.2. Escribir la demostración detallada de <br />
sumaPotenciasDeDosMasUno n = 2^(n+1)<br />
------------------------------------------------------------------- *}<br />
<br />
(*crigomgom ivamenjim danrodcha serrodcal rubgonmar ferrenseg juacabsou wilmorort anaprarod marpoldia1 josgarsan*)<br />
lemma "sumaPotenciasDeDosMasUno n = 2^(n+1)"<br />
proof (induct n)<br />
show "sumaPotenciasDeDosMasUno 0 = 2 ^ (0 + 1)" by simp<br />
next<br />
fix n<br />
assume HI: "sumaPotenciasDeDosMasUno n = 2 ^ (n + 1)"<br />
have "sumaPotenciasDeDosMasUno (Suc n) = sumaPotenciasDeDosMasUno n + 2^(n+1)" by simp<br />
also have "... = 2 ^ (n + 1) + 2 ^ (n + 1)" using HI by simp<br />
also have "... = 2 ^ ((Suc n) + 1)" by simp<br />
finally show "sumaPotenciasDeDosMasUno (Suc n) = 2 ^ (Suc n + 1)" by simp<br />
qed<br />
<br />
(* fraortmoy *)<br />
(* es la misma demostración, pero quise probar a delimitar lo que se usa en el "by simp" *)<br />
<br />
lemma "sumaPotenciasDeDosMasUno n = 2^(n+1)"<br />
proof (induct n)<br />
show "sumaPotenciasDeDosMasUno 0 = 2 ^ (0 + 1)" by simp<br />
next<br />
fix n<br />
assume H1:" sumaPotenciasDeDosMasUno n = 2 ^ (n + 1)"<br />
have "sumaPotenciasDeDosMasUno (Suc n) = sumaPotenciasDeDosMasUno n + 2^(n + 1)" by (simp only : sumaPotenciasDeDosMasUno.simps(2))<br />
also have "... = 2 ^ (n + 1) + 2 ^ (n + 1)" using H1 by simp<br />
also have "... = 2 ^ (Suc n + 1)" by simp<br />
finally show "sumaPotenciasDeDosMasUno (Suc n) = 2 ^ (Suc n + 1)" by simp<br />
qed<br />
<br />
(* migtermor *)<br />
lemma "sumaPotenciasDeDosMasUno n = 2^(n+1)"<br />
proof (induct n)<br />
show "sumaPotenciasDeDosMasUno 0 = 2^(0+1)" by simp<br />
next<br />
fix n<br />
assume HI: "sumaPotenciasDeDosMasUno n = 2^(n+1)"<br />
have "sumaPotenciasDeDosMasUno (Suc n) = sumaPotenciasDeDosMasUno n + 2^(n+1)"<br />
by (simp only: sumaPotenciasDeDosMasUno.simps(2))<br />
also have "sumaPotenciasDeDosMasUno (Suc n) = 2^(n+1)+2^(n+1)" using HI by simp<br />
finally show "sumaPotenciasDeDosMasUno (Suc n) = 2^((Suc n)+1)" by simp<br />
qed<br />
<br />
(*pablucoto*)<br />
lemma "sumaPotenciasDeDosMasUno n = 2^(n+1)"<br />
proof (induct n)<br />
show "sumaPotenciasDeDosMasUno 0 = 2 ^ (0 + 1)" by simp<br />
next <br />
fix n <br />
assume HI : "sumaPotenciasDeDosMasUno n = 2 ^ (n + 1)"<br />
have "sumaPotenciasDeDosMasUno (Suc n) = sumaPotenciasDeDosMasUno n + 2^(n+1) " by simp<br />
also have "... = 2^(n + 1) + 2^( n + 1) " using HI by simp <br />
finally show " sumaPotenciasDeDosMasUno (Suc n) = 2 ^ (Suc n + 1)" by simp<br />
qed<br />
<br />
(* fracorjim1 - Hago explícita toda la manipulación algebraica. Quizás excesivo.*)<br />
lemma "sumaPotenciasDeDosMasUno n = 2^(n+1)"<br />
proof (induction n)<br />
show "sumaPotenciasDeDosMasUno 0 = 2 ^ (0 + 1)" by simp<br />
next<br />
fix n<br />
assume HI : "sumaPotenciasDeDosMasUno n = 2 ^ (n + 1)"<br />
have "sumaPotenciasDeDosMasUno (Suc n) = sumaPotenciasDeDosMasUno n + 2 ^ (n + 1)"<br />
by (simp only:sumaPotenciasDeDosMasUno.simps(2))<br />
also have "... = 2 ^ (n + 1) + 2 ^ (n + 1)" using HI by simp<br />
also have "... = 2 ^ (n + 1) * 2" by simp<br />
also have "... = 2 ^ ((n + 1) + 1)" by simp<br />
also have "... = 2 ^ (Suc n + 1)" by simp<br />
finally show "sumaPotenciasDeDosMasUno (Suc n) = 2 ^ (Suc n + 1)" by simp<br />
qed<br />
<br />
(*lucnovdos*)<br />
lemma "sumaPotenciasDeDosMasUno n = 2^(n+1)"<br />
proof (induct n)<br />
show "sumaPotenciasDeDosMasUno 0 = 2 ^ (0 + 1)" by simp<br />
next<br />
fix n<br />
assume HI: "sumaPotenciasDeDosMasUno n = 2 ^ (n + 1)"<br />
have "sumaPotenciasDeDosMasUno (Suc n) = sumaPotenciasDeDosMasUno n + 2 ^ (n + 1)" by simp<br />
also have "... = 2 ^ (n + 1) + 2 ^ (n + 1)" using HI by simp<br />
finally show "sumaPotenciasDeDosMasUno (Suc n) = 2 ^ (Suc n + 1)" by simp<br />
qed<br />
<br />
text {* --------------------------------------------------------------- <br />
Ejercicio 3.1. Definir la función<br />
copia :: nat ⇒ 'a ⇒ 'a list<br />
tal que (copia n x) es la lista formado por n copias del elemento<br />
x. Por ejemplo, <br />
copia 3 x = [x,x,x]<br />
------------------------------------------------------------------ *}<br />
<br />
fun copia :: "nat ⇒ 'a ⇒ 'a list" where<br />
"copia 0 x = []"<br />
| "copia (Suc n) x = x # copia n x"<br />
<br />
text {* --------------------------------------------------------------- <br />
Ejercicio 3.2. Definir la función<br />
todos :: ('a ⇒ bool) ⇒ 'a list ⇒ bool<br />
tal que (todos p xs) se verifica si todos los elementos de xs cumplen<br />
la propiedad p. Por ejemplo,<br />
todos (λx. x>(1::nat)) [2,6,4] = True<br />
todos (λx. x>(2::nat)) [2,6,4] = False<br />
------------------------------------------------------------------ *}<br />
<br />
fun todos :: "('a ⇒ bool) ⇒ 'a list ⇒ bool" where<br />
"todos p [] = True"<br />
| "todos p (x#xs) = (p x ∧ todos p xs)"<br />
<br />
text {* --------------------------------------------------------------- <br />
Ejercicio 3.2. Demostrar detalladamente que todos los elementos de<br />
(copia n x) son iguales a x. <br />
------------------------------------------------------------------- *}<br />
<br />
(* crigomgom ivamenjim serrodcal ferrenseg wilmorort *)<br />
lemma "todos (λy. y=x) (copia n x)"<br />
proof (induct n)<br />
show "todos (λy. y = x) (copia 0 x)" by simp<br />
next<br />
fix n<br />
assume HI: "todos (λy. y = x) (copia n x)" <br />
have "todos (λy. y = x) (copia (Suc n) x) = todos (λy. y = x) (x # (copia n x))" by simp<br />
also have "... = ((x = x) ∧ (todos (λy. y = x) (copia n x)))" by simp<br />
also have "... = True" using HI by simp<br />
finally show "todos (λy. y = x) (copia (Suc n) x)" by simp<br />
qed<br />
<br />
<br />
(* fraortmoy *)<br />
lemma "todos (λy. y=x) (copia n x)"<br />
proof (induct n)<br />
have "todos (λy. y = x) (copia 0 x) = todos (λy. y = x) []" by (simp only: copia.simps(1))<br />
also have "... = True" by (simp only: todos.simps(1))<br />
show "todos (λy. y = x) (copia 0 x)" by simp<br />
next<br />
fix n <br />
assume H1 : " todos (λy. y = x) (copia n x) "<br />
have "todos (λy. y = x) (copia (Suc n) x) = ((todos (λy. y = x) (x#[])) ∧ (todos (λy. y = x) (copia n x) )) " by simp<br />
also have " ... = (todos (λy. y = x) (x#[]))" using H1 by simp<br />
also have " ... = ((λy. y = x) x ∧ todos (λy. y = x) [])" by simp<br />
also have " ... = True" by simp<br />
finally show " todos (λy. y = x) (copia (Suc n) x)" by simp<br />
qed<br />
<br />
(*danrodcha rubgonmar*)<br />
lemma "todos (λy. y=x) (copia n x)" (is "?P n")<br />
proof (induct n)<br />
show "?P 0" by simp<br />
next<br />
fix n<br />
assume HI: "?P n"<br />
have "todos (λy. y=x) (copia (Suc n) x) = todos (λy. y=x) (x # copia n x)" by simp<br />
also have "... = ((λy. y=x) x ∧ todos (λy. y=x) (copia n x))" by simp<br />
also have "... = todos (λy. y=x) (copia n x)" by simp<br />
finally show "?P (Suc n)" using HI by simp<br />
qed<br />
<br />
(* migtermor *)<br />
lemma "todos (λy. y=x) (copia n x)"<br />
proof (induct n)<br />
have "todos (λy. y=x) (copia 0 x) = todos (λy. y=x) []" by (simp only: copia.simps(1))<br />
also have "todos (λy. y=x) []" by (simp only: todos.simps(1))<br />
show "todos (λy. y=x) (copia 0 x)" by simp<br />
next <br />
fix n<br />
assume HI: "todos (λy. y=x) (copia n x)"<br />
have "todos (λy. y=x) (copia (Suc n) x) = todos (λy. y=x) (x # (copia n x))"<br />
by (simp only: copia.simps(2))<br />
also have "todos (λy. y=x) (x#(copia n x)) = ((λy. y=x) x ∧ todos (λy. y=x) (copia n x))"<br />
by (simp only: todos.simps(2))<br />
also have "todos (λy. y=x) (copia (Suc n) x) = ((λy. y=x) x)" using HI by simp<br />
also have "((λy. y=x) x) = True" by simp <br />
finally show "(todos (λy. y=x) (copia (Suc n) x))" by simp<br />
qed<br />
<br />
(*pablucoto*)<br />
<br />
lemma "todos (λy. y=x) (copia n x)"<br />
proof (induct n)<br />
show "todos (λy. y = x) (copia 0 x)" by simp<br />
next <br />
fix n<br />
assume HI: "todos (λy. y=x) (copia n x)"<br />
have "todos (λy. y = x) (copia (Suc n) x) = todos (λy. y=x) (x # copia n x)" by simp<br />
also have "... = ( (λy. y=x) x ∧ todos (λy. y=x) (copia (Suc n) x) ) " by simp<br />
also have "... = True" using HI by simp<br />
finally show " todos (λy. y = x) (copia (Suc n) x) " by simp<br />
qed<br />
<br />
(* anaprarod marpoldia1 *)<br />
lemma "todos (λy. y=x) (copia n x)"<br />
proof (induct n)<br />
show "todos (λy. y = x) (copia 0 x)" by simp<br />
next<br />
fix n<br />
assume HI: "todos (λy. y = x) (copia n x)"<br />
have "todos (λy. y = x) (copia (Suc n) x) =todos (λy. y = x) (x # copia n x)" by simp<br />
also have "... = ((λy. y = x) x ∧ todos (λy. y = x) ( copia n x))" by simp<br />
finally show "todos (λy. y = x) (copia (Suc n) x)" using HI by simp<br />
qed<br />
<br />
text {* --------------------------------------------------------------- <br />
Ejercicio 4.1. Definir la función<br />
factR :: nat ⇒ nat<br />
tal que (factR n) es el factorial de n. Por ejemplo,<br />
factR 4 = 24<br />
------------------------------------------------------------------ *}<br />
<br />
fun factR :: "nat ⇒ nat" where<br />
"factR 0 = 1"<br />
| "factR (Suc n) = Suc n * factR n"<br />
<br />
text {* --------------------------------------------------------------- <br />
Ejercicio 4.2. Se considera la siguiente definición iterativa de la<br />
función factorial <br />
factI :: "nat ⇒ nat" where<br />
factI n = factI' n 1<br />
<br />
factI' :: nat ⇒ nat ⇒ nat" where<br />
factI' 0 x = x<br />
factI' (Suc n) x = factI' n (Suc n)*x<br />
Demostrar que, para todo n y todo x, se tiene <br />
factI' n x = x * factR n<br />
Indicación: La propiedad mult_Suc es <br />
(Suc m) * n = n + m * n<br />
Puede que se necesite desactivarla en un paso con <br />
(simp del: mult_Suc)<br />
------------------------------------------------------------------- *}<br />
<br />
fun factI' :: "nat ⇒ nat ⇒ nat" where<br />
"factI' 0 x = x"<br />
| "factI' (Suc n) x = factI' n (x * Suc n)"<br />
<br />
fun factI :: "nat ⇒ nat" where<br />
"factI n = factI' n 1"<br />
<br />
(* fraortmoy , danrodcha serrodcal pablucoto wilmorort anaprarod marpoldia1*)<br />
lemma fact: "factI' n x = x * factR n"<br />
proof (induct n)<br />
show "factI' 0 x = x * factR 0" by simp<br />
next<br />
fix n<br />
assume H1 : "factI' n x = x * factR n"<br />
have "factI' (Suc n) x = factI' n (x * Suc n)" by (simp only:factI'.simps(2))<br />
also have "... = (x * Suc n) * factR n" using H1 by simp (* no entiendo por qué no hace esto bien y luego todo funciona *)<br />
also have "... = x * factR (Suc n)" by (simp del: mult_Suc)<br />
finally show "factI' (Suc n) x = x * factR (Suc n)" by simp<br />
qed<br />
<br />
(* crigomgom ivamenjim ferrenseg *)<br />
lemma fact: "factI' n x = x * factR n"<br />
proof (induct n arbitrary: x)<br />
show "⋀x. factI' 0 x = x * factR 0" by simp<br />
next<br />
fix n<br />
assume HI: "⋀x. factI' n x = x * factR n"<br />
show "⋀x. factI' (Suc n) x = x * factR (Suc n)"<br />
proof -<br />
fix x<br />
have "factI' (Suc n) x = factI' n (x * Suc n)" by simp<br />
also have "... = (x * Suc n) * factR n" using HI by simp<br />
also have "... = x * (Suc n * factR n)" by (simp del: mult_Suc)<br />
also have "... = x * factR (Suc n)" by simp<br />
finally show "factI' (Suc n) x = x * factR (Suc n)" by simp<br />
qed<br />
qed<br />
<br />
(* migtermor *)<br />
<br />
lemma fact: "factI' n x = x * factR n"<br />
proof (induct n)<br />
have "x * factR 0 = x" by (simp only: factR.simps(1))<br />
also have "factI' 0 x = x" by (simp only: factI'.simps(1))<br />
show "factI' 0 x = x * factR 0" by simp<br />
next<br />
fix n<br />
assume HI: "∀x. factI' n x = x * factR n"<br />
fix x<br />
have "factI' (Suc n) x = factI' n (x * Suc n)" by (simp only: factI'.simps(2))<br />
also have "factI' (Suc n) x = factI' n (x * Suc n)" by simp<br />
have "factI' (Suc n) x = x * Suc n * factR n" using HI by simp<br />
also have "... = x * factR (Suc n)" by (simp del: mult_Suc)<br />
finally show "factI' (Suc n) x = x * factR (Suc n)" by simp (* No entiendo por qué no acepta esto como demostrado *)<br />
qed<br />
<br />
<br />
text {* --------------------------------------------------------------- <br />
Ejercicio 4.3. Escribir la demostración detallada de<br />
factI n = factR n<br />
------------------------------------------------------------------- *}<br />
<br />
(* fraortmoy danrodcha crigomgom ivamenjim serrodcal ferrenseg pablucoto wilmorort rubgonmar marpoldia1*)<br />
corollary "factI n = factR n"<br />
proof -<br />
have "factI n = factI' n 1" by simp<br />
also have "... = 1 * factR n" by (simp add: fact)<br />
finally show "factI n = factR n" by simp<br />
qed<br />
<br />
(* migtermor *)<br />
corollary "factI n = factR n"<br />
proof -<br />
have "factI n = factI' n 1" by simp<br />
also have "... = 1 * factI' n 1" by simp<br />
also have "... = 1 * factR n" using fact by simp<br />
finally show "factI n = factR n" by simp<br />
qed<br />
<br />
<br />
text {* --------------------------------------------------------------- <br />
Ejercicio 5.1. Definir, recursivamente y sin usar (@), la función<br />
amplia :: 'a list ⇒ 'a ⇒ 'a list<br />
tal que (amplia xs y) es la lista obtenida añadiendo el elemento y al<br />
final de la lista xs. Por ejemplo,<br />
amplia [d,a] t = [d,a,t]<br />
------------------------------------------------------------------ *}<br />
<br />
fun amplia :: "'a list ⇒ 'a ⇒ 'a list" where<br />
"amplia [] y = [y]"<br />
| "amplia (x#xs) y = x # amplia xs y"<br />
<br />
text {* --------------------------------------------------------------- <br />
Ejercicio 5.2. Escribir la demostración detallada de<br />
amplia xs y = xs @ [y]<br />
------------------------------------------------------------------- *}<br />
<br />
(* crigomgom fraortmoy rubgonmar ivamenjim serrodcal pablucoto wilmorort anaprarod marpoldia1 *)<br />
lemma "amplia xs y = xs @ [y]"<br />
proof (induct xs)<br />
show "amplia [] y = [] @ [y]" by simp<br />
next<br />
fix x xs<br />
assume HI: "amplia xs y = xs @ [y]"<br />
have "amplia (x # xs) y = x # (amplia xs y)" by simp<br />
also have "... = x # (xs @ [y])" using HI by simp<br />
also have "... = (x # xs) @ [y]" by simp<br />
finally show "amplia (x # xs) y = (x # xs) @ [y]" by simp<br />
qed<br />
<br />
(*danrodcha ferrenseg *)<br />
lemma "amplia xs y = xs @ [y]" (is "?P xs")<br />
proof (induct xs)<br />
show "?P []" by simp<br />
next<br />
fix x xs<br />
assume HI: "?P xs"<br />
have "amplia (x # xs) y = x # amplia xs y" by simp<br />
also have "... = (x # xs) @ [y]" using HI by simp<br />
finally show "?P (x#xs)" by simp<br />
qed<br />
<br />
(* migtermor *)<br />
lemma "amplia xs y = xs @ [y]"<br />
proof (induct xs)<br />
have "amplia [] y = [y]" by (simp only: amplia.simps(1)) <br />
have "[] @ [y] = [y]" by simp<br />
show "amplia [] y = [] @ [y]" by simp<br />
next <br />
fix x xs<br />
assume HI: "amplia xs y = xs @ [y]"<br />
have "amplia (x # xs) y = x # amplia xs y" by (simp only: amplia.simps(2))<br />
also have "... = x # (xs @ [y])" using HI by simp<br />
also have "... = (x # xs) @ [y]" by simp<br />
finally show "amplia (x # xs) y = (x # xs) @ [y]" by simp<br />
qed<br />
<br />
end<br />
</source></div>Lucnovdoshttps://www.glc.us.es/~jalonso/RA2016/index.php?title=Relaci%C3%B3n_3&diff=371Relación 32016-11-15T17:40:20Z<p>Lucnovdos: </p>
<hr />
<div><source lang="isar"><br />
chapter {* R3: Razonamiento sobre programas *}<br />
<br />
theory R3_Razonamiento_sobre_programas<br />
imports Main <br />
begin<br />
<br />
text {* --------------------------------------------------------------- <br />
Ejercicio 1.1. Definir la función<br />
sumaImpares :: nat ⇒ nat<br />
tal que (sumaImpares n) es la suma de los n primeros números<br />
impares. Por ejemplo,<br />
sumaImpares 5 = 25<br />
------------------------------------------------------------------ *}<br />
<br />
fun sumaImpares :: "nat ⇒ nat" where<br />
"sumaImpares 0 = 0"<br />
| "sumaImpares (Suc n) = sumaImpares n + (2*n+1)"<br />
<br />
text {* --------------------------------------------------------------- <br />
Ejercicio 1.2. Escribir la demostración detallada de <br />
sumaImpares n = n*n<br />
------------------------------------------------------------------- *}<br />
<br />
-- "La demostración detallada es"<br />
(* crigomgom fraortmoy marpoldia1 ivamenjim serrodcal rubgonmar ferrenseg juacabsou wilmorort josgarsan lucnovdos*)<br />
lemma "sumaImpares n = n*n"<br />
proof (induct n)<br />
show "sumaImpares 0 = 0 * 0" by simp<br />
next<br />
fix n<br />
assume HI: "sumaImpares n = n * n"<br />
have "sumaImpares (Suc n) = sumaImpares n + 2*n + 1" by simp<br />
also have "... = n*n + 2*n + 1" using HI by simp<br />
also have "... = Suc n * Suc n" by simp<br />
finally show "sumaImpares (Suc n) = Suc n * Suc n" by simp<br />
qed<br />
<br />
(*danrodcha ; es la misma demostración que la anterior pero uso ?P para sustituir la propiedad.*)<br />
lemma "sumaImpares n = n*n" (is "?P n")<br />
proof (induct n)<br />
show "?P 0" by simp<br />
next<br />
fix n<br />
assume HI: "?P n"<br />
have "sumaImpares (Suc n) = sumaImpares n + 2*n + 1" by simp<br />
also have "... = n*n + 2*n + 1" using HI by simp<br />
also have "... = Suc n * Suc n" by simp <br />
finally show "?P (Suc n)" by simp<br />
qed<br />
<br />
(* migtermor *)<br />
lemma "sumaImpares n = n*n"<br />
proof (induct n)<br />
show "sumaImpares 0 = 0*0" by simp<br />
next<br />
fix n<br />
assume HI: "sumaImpares n = n*n"<br />
have "sumaImpares (Suc n) = sumaImpares n + 2*n+1" <br />
by (simp only: sumaImpares.simps(2))<br />
also have "sumaImpares (Suc n) = n*n + 2*n+1" using HI by simp<br />
finally show "sumaImpares (Suc n) = (Suc n)*(Suc n)" by simp<br />
qed<br />
<br />
(* fracorjim1 - Especifico la regla de simplificación y el paso del desarrollo al cuadrado para hacerlo más legible*)<br />
lemma "sumaImpares n = n*n"<br />
proof (induct n)<br />
show "sumaImpares 0 = 0 * 0" by simp<br />
next<br />
fix n<br />
assume HI : "sumaImpares n = n * n"<br />
have "sumaImpares (Suc n) = sumaImpares n + (2 * n + 1)" by (simp only: sumaImpares.simps(2))<br />
also have "... = n * n + 2 * n + 1" using HI by simp<br />
also have "... = Suc n * Suc n" by simp<br />
finally show "sumaImpares (Suc n) = Suc n * Suc n" by simp<br />
qed<br />
<br />
(*palucoto anaprarod*)<br />
lemma "sumaImpares n = n*n"<br />
proof (induct n)<br />
show "sumaImpares 0 = 0 * 0" by simp<br />
next <br />
fix n <br />
assume HI : "sumaImpares n = n * n"<br />
have " sumaImpares (Suc n) = sumaImpares n + 2*n+1" by simp<br />
also have "... = n*n + 2*n+1" using HI by simp<br />
finally show "sumaImpares (Suc n) = Suc n * Suc n" by simp<br />
qed<br />
<br />
<br />
text {* --------------------------------------------------------------- <br />
Ejercicio 2.1. Definir la función<br />
sumaPotenciasDeDosMasUno :: nat ⇒ nat<br />
tal que <br />
(sumaPotenciasDeDosMasUno n) = 1 + 2^0 + 2^1 + 2^2 + ... + 2^n. <br />
Por ejemplo, <br />
sumaPotenciasDeDosMasUno 3 = 16<br />
------------------------------------------------------------------ *}<br />
<br />
fun sumaPotenciasDeDosMasUno :: "nat ⇒ nat" where<br />
"sumaPotenciasDeDosMasUno 0 = 2"<br />
| "sumaPotenciasDeDosMasUno (Suc n) = <br />
sumaPotenciasDeDosMasUno n + 2^(n+1)"<br />
<br />
text {* --------------------------------------------------------------- <br />
Ejercicio 2.2. Escribir la demostración detallada de <br />
sumaPotenciasDeDosMasUno n = 2^(n+1)<br />
------------------------------------------------------------------- *}<br />
<br />
(*crigomgom ivamenjim danrodcha serrodcal rubgonmar ferrenseg juacabsou wilmorort anaprarod marpoldia1 josgarsan*)<br />
lemma "sumaPotenciasDeDosMasUno n = 2^(n+1)"<br />
proof (induct n)<br />
show "sumaPotenciasDeDosMasUno 0 = 2 ^ (0 + 1)" by simp<br />
next<br />
fix n<br />
assume HI: "sumaPotenciasDeDosMasUno n = 2 ^ (n + 1)"<br />
have "sumaPotenciasDeDosMasUno (Suc n) = sumaPotenciasDeDosMasUno n + 2^(n+1)" by simp<br />
also have "... = 2 ^ (n + 1) + 2 ^ (n + 1)" using HI by simp<br />
also have "... = 2 ^ ((Suc n) + 1)" by simp<br />
finally show "sumaPotenciasDeDosMasUno (Suc n) = 2 ^ (Suc n + 1)" by simp<br />
qed<br />
<br />
(* fraortmoy *)<br />
(* es la misma demostración, pero quise probar a delimitar lo que se usa en el "by simp" *)<br />
<br />
lemma "sumaPotenciasDeDosMasUno n = 2^(n+1)"<br />
proof (induct n)<br />
show "sumaPotenciasDeDosMasUno 0 = 2 ^ (0 + 1)" by simp<br />
next<br />
fix n<br />
assume H1:" sumaPotenciasDeDosMasUno n = 2 ^ (n + 1)"<br />
have "sumaPotenciasDeDosMasUno (Suc n) = sumaPotenciasDeDosMasUno n + 2^(n + 1)" by (simp only : sumaPotenciasDeDosMasUno.simps(2))<br />
also have "... = 2 ^ (n + 1) + 2 ^ (n + 1)" using H1 by simp<br />
also have "... = 2 ^ (Suc n + 1)" by simp<br />
finally show "sumaPotenciasDeDosMasUno (Suc n) = 2 ^ (Suc n + 1)" by simp<br />
qed<br />
<br />
(* migtermor *)<br />
lemma "sumaPotenciasDeDosMasUno n = 2^(n+1)"<br />
proof (induct n)<br />
show "sumaPotenciasDeDosMasUno 0 = 2^(0+1)" by simp<br />
next<br />
fix n<br />
assume HI: "sumaPotenciasDeDosMasUno n = 2^(n+1)"<br />
have "sumaPotenciasDeDosMasUno (Suc n) = sumaPotenciasDeDosMasUno n + 2^(n+1)"<br />
by (simp only: sumaPotenciasDeDosMasUno.simps(2))<br />
also have "sumaPotenciasDeDosMasUno (Suc n) = 2^(n+1)+2^(n+1)" using HI by simp<br />
finally show "sumaPotenciasDeDosMasUno (Suc n) = 2^((Suc n)+1)" by simp<br />
qed<br />
<br />
(*pablucoto*)<br />
lemma "sumaPotenciasDeDosMasUno n = 2^(n+1)"<br />
proof (induct n)<br />
show "sumaPotenciasDeDosMasUno 0 = 2 ^ (0 + 1)" by simp<br />
next <br />
fix n <br />
assume HI : "sumaPotenciasDeDosMasUno n = 2 ^ (n + 1)"<br />
have "sumaPotenciasDeDosMasUno (Suc n) = sumaPotenciasDeDosMasUno n + 2^(n+1) " by simp<br />
also have "... = 2^(n + 1) + 2^( n + 1) " using HI by simp <br />
finally show " sumaPotenciasDeDosMasUno (Suc n) = 2 ^ (Suc n + 1)" by simp<br />
qed<br />
<br />
(* fracorjim1 - Hago explícita toda la manipulación algebraica. Quizás excesivo.*)<br />
lemma "sumaPotenciasDeDosMasUno n = 2^(n+1)"<br />
proof (induction n)<br />
show "sumaPotenciasDeDosMasUno 0 = 2 ^ (0 + 1)" by simp<br />
next<br />
fix n<br />
assume HI : "sumaPotenciasDeDosMasUno n = 2 ^ (n + 1)"<br />
have "sumaPotenciasDeDosMasUno (Suc n) = sumaPotenciasDeDosMasUno n + 2 ^ (n + 1)"<br />
by (simp only:sumaPotenciasDeDosMasUno.simps(2))<br />
also have "... = 2 ^ (n + 1) + 2 ^ (n + 1)" using HI by simp<br />
also have "... = 2 ^ (n + 1) * 2" by simp<br />
also have "... = 2 ^ ((n + 1) + 1)" by simp<br />
also have "... = 2 ^ (Suc n + 1)" by simp<br />
finally show "sumaPotenciasDeDosMasUno (Suc n) = 2 ^ (Suc n + 1)" by simp<br />
qed<br />
<br />
text {* --------------------------------------------------------------- <br />
Ejercicio 3.1. Definir la función<br />
copia :: nat ⇒ 'a ⇒ 'a list<br />
tal que (copia n x) es la lista formado por n copias del elemento<br />
x. Por ejemplo, <br />
copia 3 x = [x,x,x]<br />
------------------------------------------------------------------ *}<br />
<br />
fun copia :: "nat ⇒ 'a ⇒ 'a list" where<br />
"copia 0 x = []"<br />
| "copia (Suc n) x = x # copia n x"<br />
<br />
text {* --------------------------------------------------------------- <br />
Ejercicio 3.2. Definir la función<br />
todos :: ('a ⇒ bool) ⇒ 'a list ⇒ bool<br />
tal que (todos p xs) se verifica si todos los elementos de xs cumplen<br />
la propiedad p. Por ejemplo,<br />
todos (λx. x>(1::nat)) [2,6,4] = True<br />
todos (λx. x>(2::nat)) [2,6,4] = False<br />
------------------------------------------------------------------ *}<br />
<br />
fun todos :: "('a ⇒ bool) ⇒ 'a list ⇒ bool" where<br />
"todos p [] = True"<br />
| "todos p (x#xs) = (p x ∧ todos p xs)"<br />
<br />
text {* --------------------------------------------------------------- <br />
Ejercicio 3.2. Demostrar detalladamente que todos los elementos de<br />
(copia n x) son iguales a x. <br />
------------------------------------------------------------------- *}<br />
<br />
(* crigomgom ivamenjim serrodcal ferrenseg wilmorort *)<br />
lemma "todos (λy. y=x) (copia n x)"<br />
proof (induct n)<br />
show "todos (λy. y = x) (copia 0 x)" by simp<br />
next<br />
fix n<br />
assume HI: "todos (λy. y = x) (copia n x)" <br />
have "todos (λy. y = x) (copia (Suc n) x) = todos (λy. y = x) (x # (copia n x))" by simp<br />
also have "... = ((x = x) ∧ (todos (λy. y = x) (copia n x)))" by simp<br />
also have "... = True" using HI by simp<br />
finally show "todos (λy. y = x) (copia (Suc n) x)" by simp<br />
qed<br />
<br />
<br />
(* fraortmoy *)<br />
lemma "todos (λy. y=x) (copia n x)"<br />
proof (induct n)<br />
have "todos (λy. y = x) (copia 0 x) = todos (λy. y = x) []" by (simp only: copia.simps(1))<br />
also have "... = True" by (simp only: todos.simps(1))<br />
show "todos (λy. y = x) (copia 0 x)" by simp<br />
next<br />
fix n <br />
assume H1 : " todos (λy. y = x) (copia n x) "<br />
have "todos (λy. y = x) (copia (Suc n) x) = ((todos (λy. y = x) (x#[])) ∧ (todos (λy. y = x) (copia n x) )) " by simp<br />
also have " ... = (todos (λy. y = x) (x#[]))" using H1 by simp<br />
also have " ... = ((λy. y = x) x ∧ todos (λy. y = x) [])" by simp<br />
also have " ... = True" by simp<br />
finally show " todos (λy. y = x) (copia (Suc n) x)" by simp<br />
qed<br />
<br />
(*danrodcha rubgonmar*)<br />
lemma "todos (λy. y=x) (copia n x)" (is "?P n")<br />
proof (induct n)<br />
show "?P 0" by simp<br />
next<br />
fix n<br />
assume HI: "?P n"<br />
have "todos (λy. y=x) (copia (Suc n) x) = todos (λy. y=x) (x # copia n x)" by simp<br />
also have "... = ((λy. y=x) x ∧ todos (λy. y=x) (copia n x))" by simp<br />
also have "... = todos (λy. y=x) (copia n x)" by simp<br />
finally show "?P (Suc n)" using HI by simp<br />
qed<br />
<br />
(* migtermor *)<br />
lemma "todos (λy. y=x) (copia n x)"<br />
proof (induct n)<br />
have "todos (λy. y=x) (copia 0 x) = todos (λy. y=x) []" by (simp only: copia.simps(1))<br />
also have "todos (λy. y=x) []" by (simp only: todos.simps(1))<br />
show "todos (λy. y=x) (copia 0 x)" by simp<br />
next <br />
fix n<br />
assume HI: "todos (λy. y=x) (copia n x)"<br />
have "todos (λy. y=x) (copia (Suc n) x) = todos (λy. y=x) (x # (copia n x))"<br />
by (simp only: copia.simps(2))<br />
also have "todos (λy. y=x) (x#(copia n x)) = ((λy. y=x) x ∧ todos (λy. y=x) (copia n x))"<br />
by (simp only: todos.simps(2))<br />
also have "todos (λy. y=x) (copia (Suc n) x) = ((λy. y=x) x)" using HI by simp<br />
also have "((λy. y=x) x) = True" by simp <br />
finally show "(todos (λy. y=x) (copia (Suc n) x))" by simp<br />
qed<br />
<br />
(*pablucoto*)<br />
<br />
lemma "todos (λy. y=x) (copia n x)"<br />
proof (induct n)<br />
show "todos (λy. y = x) (copia 0 x)" by simp<br />
next <br />
fix n<br />
assume HI: "todos (λy. y=x) (copia n x)"<br />
have "todos (λy. y = x) (copia (Suc n) x) = todos (λy. y=x) (x # copia n x)" by simp<br />
also have "... = ( (λy. y=x) x ∧ todos (λy. y=x) (copia (Suc n) x) ) " by simp<br />
also have "... = True" using HI by simp<br />
finally show " todos (λy. y = x) (copia (Suc n) x) " by simp<br />
qed<br />
<br />
(* anaprarod marpoldia1 *)<br />
lemma "todos (λy. y=x) (copia n x)"<br />
proof (induct n)<br />
show "todos (λy. y = x) (copia 0 x)" by simp<br />
next<br />
fix n<br />
assume HI: "todos (λy. y = x) (copia n x)"<br />
have "todos (λy. y = x) (copia (Suc n) x) =todos (λy. y = x) (x # copia n x)" by simp<br />
also have "... = ((λy. y = x) x ∧ todos (λy. y = x) ( copia n x))" by simp<br />
finally show "todos (λy. y = x) (copia (Suc n) x)" using HI by simp<br />
qed<br />
<br />
text {* --------------------------------------------------------------- <br />
Ejercicio 4.1. Definir la función<br />
factR :: nat ⇒ nat<br />
tal que (factR n) es el factorial de n. Por ejemplo,<br />
factR 4 = 24<br />
------------------------------------------------------------------ *}<br />
<br />
fun factR :: "nat ⇒ nat" where<br />
"factR 0 = 1"<br />
| "factR (Suc n) = Suc n * factR n"<br />
<br />
text {* --------------------------------------------------------------- <br />
Ejercicio 4.2. Se considera la siguiente definición iterativa de la<br />
función factorial <br />
factI :: "nat ⇒ nat" where<br />
factI n = factI' n 1<br />
<br />
factI' :: nat ⇒ nat ⇒ nat" where<br />
factI' 0 x = x<br />
factI' (Suc n) x = factI' n (Suc n)*x<br />
Demostrar que, para todo n y todo x, se tiene <br />
factI' n x = x * factR n<br />
Indicación: La propiedad mult_Suc es <br />
(Suc m) * n = n + m * n<br />
Puede que se necesite desactivarla en un paso con <br />
(simp del: mult_Suc)<br />
------------------------------------------------------------------- *}<br />
<br />
fun factI' :: "nat ⇒ nat ⇒ nat" where<br />
"factI' 0 x = x"<br />
| "factI' (Suc n) x = factI' n (x * Suc n)"<br />
<br />
fun factI :: "nat ⇒ nat" where<br />
"factI n = factI' n 1"<br />
<br />
(* fraortmoy , danrodcha serrodcal pablucoto wilmorort anaprarod marpoldia1*)<br />
lemma fact: "factI' n x = x * factR n"<br />
proof (induct n)<br />
show "factI' 0 x = x * factR 0" by simp<br />
next<br />
fix n<br />
assume H1 : "factI' n x = x * factR n"<br />
have "factI' (Suc n) x = factI' n (x * Suc n)" by (simp only:factI'.simps(2))<br />
also have "... = (x * Suc n) * factR n" using H1 by simp (* no entiendo por qué no hace esto bien y luego todo funciona *)<br />
also have "... = x * factR (Suc n)" by (simp del: mult_Suc)<br />
finally show "factI' (Suc n) x = x * factR (Suc n)" by simp<br />
qed<br />
<br />
(* crigomgom ivamenjim ferrenseg *)<br />
lemma fact: "factI' n x = x * factR n"<br />
proof (induct n arbitrary: x)<br />
show "⋀x. factI' 0 x = x * factR 0" by simp<br />
next<br />
fix n<br />
assume HI: "⋀x. factI' n x = x * factR n"<br />
show "⋀x. factI' (Suc n) x = x * factR (Suc n)"<br />
proof -<br />
fix x<br />
have "factI' (Suc n) x = factI' n (x * Suc n)" by simp<br />
also have "... = (x * Suc n) * factR n" using HI by simp<br />
also have "... = x * (Suc n * factR n)" by (simp del: mult_Suc)<br />
also have "... = x * factR (Suc n)" by simp<br />
finally show "factI' (Suc n) x = x * factR (Suc n)" by simp<br />
qed<br />
qed<br />
<br />
(* migtermor *)<br />
<br />
lemma fact: "factI' n x = x * factR n"<br />
proof (induct n)<br />
have "x * factR 0 = x" by (simp only: factR.simps(1))<br />
also have "factI' 0 x = x" by (simp only: factI'.simps(1))<br />
show "factI' 0 x = x * factR 0" by simp<br />
next<br />
fix n<br />
assume HI: "∀x. factI' n x = x * factR n"<br />
fix x<br />
have "factI' (Suc n) x = factI' n (x * Suc n)" by (simp only: factI'.simps(2))<br />
also have "factI' (Suc n) x = factI' n (x * Suc n)" by simp<br />
have "factI' (Suc n) x = x * Suc n * factR n" using HI by simp<br />
also have "... = x * factR (Suc n)" by (simp del: mult_Suc)<br />
finally show "factI' (Suc n) x = x * factR (Suc n)" by simp (* No entiendo por qué no acepta esto como demostrado *)<br />
qed<br />
<br />
<br />
text {* --------------------------------------------------------------- <br />
Ejercicio 4.3. Escribir la demostración detallada de<br />
factI n = factR n<br />
------------------------------------------------------------------- *}<br />
<br />
(* fraortmoy danrodcha crigomgom ivamenjim serrodcal ferrenseg pablucoto wilmorort rubgonmar marpoldia1*)<br />
corollary "factI n = factR n"<br />
proof -<br />
have "factI n = factI' n 1" by simp<br />
also have "... = 1 * factR n" by (simp add: fact)<br />
finally show "factI n = factR n" by simp<br />
qed<br />
<br />
(* migtermor *)<br />
corollary "factI n = factR n"<br />
proof -<br />
have "factI n = factI' n 1" by simp<br />
also have "... = 1 * factI' n 1" by simp<br />
also have "... = 1 * factR n" using fact by simp<br />
finally show "factI n = factR n" by simp<br />
qed<br />
<br />
<br />
text {* --------------------------------------------------------------- <br />
Ejercicio 5.1. Definir, recursivamente y sin usar (@), la función<br />
amplia :: 'a list ⇒ 'a ⇒ 'a list<br />
tal que (amplia xs y) es la lista obtenida añadiendo el elemento y al<br />
final de la lista xs. Por ejemplo,<br />
amplia [d,a] t = [d,a,t]<br />
------------------------------------------------------------------ *}<br />
<br />
fun amplia :: "'a list ⇒ 'a ⇒ 'a list" where<br />
"amplia [] y = [y]"<br />
| "amplia (x#xs) y = x # amplia xs y"<br />
<br />
text {* --------------------------------------------------------------- <br />
Ejercicio 5.2. Escribir la demostración detallada de<br />
amplia xs y = xs @ [y]<br />
------------------------------------------------------------------- *}<br />
<br />
(* crigomgom fraortmoy rubgonmar ivamenjim serrodcal pablucoto wilmorort anaprarod marpoldia1 *)<br />
lemma "amplia xs y = xs @ [y]"<br />
proof (induct xs)<br />
show "amplia [] y = [] @ [y]" by simp<br />
next<br />
fix x xs<br />
assume HI: "amplia xs y = xs @ [y]"<br />
have "amplia (x # xs) y = x # (amplia xs y)" by simp<br />
also have "... = x # (xs @ [y])" using HI by simp<br />
also have "... = (x # xs) @ [y]" by simp<br />
finally show "amplia (x # xs) y = (x # xs) @ [y]" by simp<br />
qed<br />
<br />
(*danrodcha ferrenseg *)<br />
lemma "amplia xs y = xs @ [y]" (is "?P xs")<br />
proof (induct xs)<br />
show "?P []" by simp<br />
next<br />
fix x xs<br />
assume HI: "?P xs"<br />
have "amplia (x # xs) y = x # amplia xs y" by simp<br />
also have "... = (x # xs) @ [y]" using HI by simp<br />
finally show "?P (x#xs)" by simp<br />
qed<br />
<br />
(* migtermor *)<br />
lemma "amplia xs y = xs @ [y]"<br />
proof (induct xs)<br />
have "amplia [] y = [y]" by (simp only: amplia.simps(1)) <br />
have "[] @ [y] = [y]" by simp<br />
show "amplia [] y = [] @ [y]" by simp<br />
next <br />
fix x xs<br />
assume HI: "amplia xs y = xs @ [y]"<br />
have "amplia (x # xs) y = x # amplia xs y" by (simp only: amplia.simps(2))<br />
also have "... = x # (xs @ [y])" using HI by simp<br />
also have "... = (x # xs) @ [y]" by simp<br />
finally show "amplia (x # xs) y = (x # xs) @ [y]" by simp<br />
qed<br />
<br />
end<br />
</source></div>Lucnovdos