Razonamiento automático (2017-18) - Contribuciones del usuario [es]

Relación 7

2018-02-07T09:59:08Z

Oscgonesc:

<source lang="isar">
chapter {* R7: Deducción natural proposicional en Isabelle/HOL *}

theory R7_Deduccion_natural_proposicional
imports Main
begin

text {*
Demostrar o refutar los siguientes lemas usando sólo las reglas
básicas de deducción natural de la lógica proposicional, de los
cuantificadores y de la igualdad:
· conjI: ⟦P; Q⟧ ⟹ P ∧ Q
· conjunct1: P ∧ Q ⟹ P
· conjunct2: P ∧ Q ⟹ Q
· notnotD: ¬¬ P ⟹ P
· mp: ⟦P ⟶ Q; P⟧ ⟹ Q
· impI: (P ⟹ Q) ⟹ P ⟶ Q
· disjI1: P ⟹ P ∨ Q
· disjI2: Q ⟹ P ∨ Q
· disjE: ⟦P ∨ Q; P ⟹ R; Q ⟹ R⟧ ⟹ R
· FalseE: False ⟹ P
· notE: ⟦¬P; P⟧ ⟹ R
· notI: (P ⟹ False) ⟹ ¬P
· iffI: ⟦P ⟹ Q; Q ⟹ P⟧ ⟹ P = Q
· iffD1: ⟦Q = P; Q⟧ ⟹ P
· iffD2: ⟦P = Q; Q⟧ ⟹ P
· ccontr: (¬P ⟹ False) ⟹ P

· allI: ⟦∀x. P x; P x ⟹ R⟧ ⟹ R
· allE: (⋀x. P x) ⟹ ∀x. P x
· exI: P x ⟹ ∃x. P x
· exE: ⟦∃x. P x; ⋀x. P x ⟹ Q⟧ ⟹ Q

· refl: t = t
· subst: ⟦s = t; P s⟧ ⟹ P t
· trans: ⟦r = s; s = t⟧ ⟹ r = t
· sym: s = t ⟹ t = s
· not_sym: t ≠ s ⟹ s ≠ t
· ssubst: ⟦t = s; P s⟧ ⟹ P t
· box_equals: ⟦a = b; a = c; b = d⟧ ⟹ a: = d
· arg_cong: x = y ⟹ f x = f y
· fun_cong: f = g ⟹ f x = g x
· cong: ⟦f = g; x = y⟧ ⟹ f x = g y
*}

text {*
Se usarán las reglas notnotI, mt y not_ex que demostramos a continuación.
*}

lemma notnotI: "P ⟹ ¬¬ P"
by auto

lemma mt: "⟦F ⟶ G; ¬G⟧ ⟹ ¬F"
by auto

lemma no_ex: "¬(∃x. P(x)) ⟹ ∀x. ¬P(x)"
by auto

text {* ---------------------------------------------------------------
Ejercicio 1. Demostrar
¬q ⟶ ¬p ⊢ p ⟶ q
------------------------------------------------------------------ *}

(*anddonram*)
lemma ej_1:
assumes 1: "¬q ⟶ ¬p"
shows "p ⟶ q"
proof (rule impI)
assume 2: "p"
show 3: "q"
proof (rule ccontr)
assume 4: "¬q"
have 5: "¬p" using 1 4 by (rule impE)
then show False using 2 by (rule notE)
qed
qed

(*luicedval cesgongut oscgonesc *)
lemma ej_1_2:
assumes 1: "¬q ⟶ ¬p"
shows "p ⟶ q"
proof -
{ assume 2: "p"
have 3: "¬¬p" using 2 by (rule notnotI)
have 4: "¬¬q" using 1 3 by (rule mt)
have 5: "q" using 4 by (rule notnotD) }
thus "p ⟶ q" by (rule impI)
qed

text {* ---------------------------------------------------------------
Ejercicio 2. Demostrar
¬(¬p ∧ ¬q) ⊢ p ∨ q
------------------------------------------------------------------ *}

(*anddonram luicedval oscgonesc *)
lemma l_e_m:
"F ∨ ¬F"
by auto

lemma ej_2:
assumes 1:"¬(¬p ∧ ¬q)"
shows "p ∨ q"
proof (rule disjE)
{
assume 2: "q"
show "p ∨ q" using 2 by (rule disjI2)
}
moreover{
assume 4:"¬q"
show "p ∨ q"
proof(rule disjI1)
show "p"
proof (rule ccontr)
assume 5:"¬p"
then have "¬p ∧ ¬q" using 4 by (rule conjI)
with 1 show False by (rule notE)
qed
qed
}
moreover
show 1: "q ∨ ¬q" by (rule l_e_m)
qed

(* cesgongut *)
lemma conjnotDM1:
assumes "¬(¬p ∧ ¬q)"
shows "p ∨ q"
proof (rule ccontr)
assume "¬(p ∨ q)"
have "p"
proof (rule ccontr)
assume "¬p"
have "q"
proof (rule ccontr)
assume "¬q"
have "¬p ∧ ¬q" using `¬p` `¬q` ..
then show False using assms by contradiction
qed
then have "p ∨ q" by (rule disjI2)
then show False using `¬(p ∨ q)` by contradiction
qed
then have "p ∨ q" by (rule disjI1)
then show False using `¬(p ∨ q)` by contradiction
qed

text {* ---------------------------------------------------------------
Ejercicio 3. Demostrar
¬(¬p ∨ ¬q) ⊢ p ∧ q
------------------------------------------------------------------ *}

(*anddonram luicedval cesgongut oscgonesc *)
lemma ej_3:
assumes 1:" ¬(¬p ∨ ¬q)"
shows "p ∧ q"
proof (rule conjI)
show "p"
proof (rule ccontr)
assume "¬p"
then have "(¬p ∨ ¬q)" by (rule disjI1)
with 1 show False by (rule notE)
qed
next
show "q"
proof (rule ccontr)
assume "¬q"
then have "(¬p ∨ ¬q)" by (rule disjI2)
with 1 show False by (rule notE)
qed
qed

text {* ---------------------------------------------------------------
Ejercicio 4. Demostrar
¬(p ∧ q) ⊢ ¬p ∨ ¬q
------------------------------------------------------------------ *}

(*anddonram luicedval*)
lemma ej_4:
assumes 1: "¬(p ∧ q)"
shows "¬p ∨ ¬q"
proof(rule disjE)
{
assume 2:"p"
show "¬p ∨ ¬q"
proof (rule disjI2)
show "¬q"
proof (rule notI)
assume "q"
with 2 have "p∧q" by (rule conjI)
with 1 show False by (rule notE)
qed
qed
}
moreover
{
assume "¬p"
then show "¬p ∨ ¬q" by (rule disjI1)
}
moreover
show 1: "p ∨ ¬p" by (rule l_e_m)
qed

(* cesgongut oscgonesc *)
lemma conjDM1:
assumes "¬(p ∧ q)"
shows "¬p ∨ ¬q"
proof (rule ccontr)
assume "¬(¬p ∨ ¬q)"
then have "p ∧ q" by (rule ej_3)
with assms show False by (rule notE)
qed

text {* ---------------------------------------------------------------
Ejercicio 5. Demostrar
⊢ (p ⟶ q) ∨ (q ⟶ p)
------------------------------------------------------------------ *}

(*anddonram luicedval oscgonesc *)
lemma ej_5:
shows "(p ⟶ q) ∨ (q ⟶ p)"
proof (rule disjE)

{
assume 2: "q"
show "(p ⟶ q) ∨ (q ⟶ p)"
proof(rule disjI1)
show "p ⟶ q" using 2 by (rule impI)
qed
}
moreover{
assume 4:" ¬q"
show "(p ⟶ q) ∨ (q ⟶ p)"
proof(rule disjI2)
show "q ⟶ p"
proof (rule impI)
assume 5:"q"
with 4 show "p" by (rule notE)
qed
qed
}
moreover
show 1: "q ∨ ¬q" by (rule l_e_m)
qed

(* cesgongut *)
lemma disjDM1:
assumes "¬(p ∨ q)"
shows "¬p ∧ ¬q"
proof
show "¬p" proof (rule notI)
assume "p"
then have "p ∨ q" ..
then show False using assms by contradiction
qed
show "¬q" proof (rule notI)
assume "q"
then have "p ∨ q" ..
then show False using assms by contradiction
qed
qed

lemma notimpD1:
assumes "¬(p ⟶ q)"
shows "p ∧ ¬q"
proof
show "¬q" proof (rule notI)
assume "q"
have "p ⟶ q" proof (rule impI)
assume "p"
show "q" proof (rule ccontr)
assume "¬q"
then show False using `q` by contradiction
qed
qed
then show False using assms by contradiction
qed
next
show "p" proof (rule ccontr)
assume "¬p"
have "p ⟶ q" proof (rule impI)
assume "p"
show "q" proof (rule ccontr)
assume "¬q"
show False using `p` `¬p` by contradiction
qed
qed
then show False using assms by contradiction
qed
qed

lemma ej5b: "(p ⟶ q) ∨ (q ⟶ p)"
proof (rule ccontr)
assume "¬?thesis"
then have "¬(p ⟶ q) ∧ ¬(q ⟶ p)" by (rule disjDM1)
then have "¬(p ⟶ q)" ..
then have "p ∧ ¬q" by (rule notimpD1)
then have "p" ..
have "¬(q ⟶ p)" using `¬(p ⟶ q) ∧ ¬(q ⟶ p)` ..
then have "q ∧ ¬p" by (rule notimpD1)
then have "¬p" ..
then show False using `p` by contradiction
qed

end
</source>

Relación 6

2018-01-16T09:41:07Z

Oscgonesc:

<source lang="isar">
chapter {* R7: Árboles binarios completos *}

theory R7_Arboles_binarios_completos
imports Main
begin

text {*
En esta relación se piden demostraciones automáticas (lo más cortas
posibles). Para ello, en algunos casos es necesario incluir lemas
auxiliares (que se demuestran automáticamente) y usar ejercicios
anteriores.

---------------------------------------------------------------------
Ejercicio 1. Definir el tipo de datos arbol para representar los
árboles binarios que no tienen información ni en los nodos y ni en las
hojas. Por ejemplo, el árbol
·
/ \
/ \
· ·
/ \ / \
· · · ·
se representa por "N (N H H) (N H H)".
---------------------------------------------------------------------
*}

datatype arbol = H | N arbol arbol

value "N (N H H) (N H H) = (N (N H H) (N H H) :: arbol)"

text {*
---------------------------------------------------------------------
Ejercicio 2. Definir la función
hojas :: "arbol => nat"
tal que (hojas a) es el número de hojas del árbol a. Por ejemplo,
hojas (N (N H H) (N H H)) = 4
---------------------------------------------------------------------
*}
(*anddonram luicedval oscgonesc*)
fun hojas :: "arbol => nat" where
"hojas H = 1"
| "hojas (N x y) =hojas x + hojas y"

value "hojas (N (N H H) (N H H)) = 4"

text {*
---------------------------------------------------------------------
Ejercicio 4. Definir la función
profundidad :: "arbol => nat"
tal que (profundidad a) es la profundidad del árbol a. Por ejemplo,
profundidad (N (N H H) (N H H)) = 2
---------------------------------------------------------------------
*}

(*anddonram luicedval oscgonesc*)
fun profundidad :: "arbol => nat" where
"profundidad H = 0"
| "profundidad (N x y) = Suc (max (profundidad x) (profundidad y))"

value "profundidad (N (N H H) (N H H)) = 2"

text {*
---------------------------------------------------------------------
Ejercicio 5. Definir la función
abc :: "nat ⇒ arbol"
tal que (abc n) es el árbol binario completo de profundidad n. Por
ejemplo,
abc 3 = N (N (N H H) (N H H)) (N (N H H) (N H H))
---------------------------------------------------------------------
*}

(*anddonram luicedval oscgonesc*)
fun abc :: "nat ⇒ arbol" where
"abc 0 = H"
| "abc (Suc n) = N (abc n) (abc n)"

value "abc 3 = N (N (N H H) (N H H)) (N (N H H) (N H H))"

text {*
---------------------------------------------------------------------
Ejercicio 6. Un árbol binario a es completo respecto de la medida f si
a es una hoja o bien a es de la forma (N i d) y se cumple que tanto i
como d son árboles binarios completos respecto de f y, además,
f(i) = f(r).

Definir la función
es_abc :: "(arbol => 'a) => arbol => bool
tal que (es_abc f a) se verifica si a es un árbol binario completo
respecto de f.
---------------------------------------------------------------------
*}

(*anddonram luicedval oscgonesc*)
fun es_abc :: "(arbol => 'a) => arbol => bool" where
"es_abc f H = True"
| "es_abc f (N i d) =((f i = f d) ∧ (es_abc f i) ∧ (es_abc f d))"

value "es_abc profundidad (abc 4) = True"

text {*
---------------------------------------------------------------------
Nota. (size a) es el número de nodos del árbol a. Por ejemplo,
size (N (N H H) (N H H)) = 3
---------------------------------------------------------------------
*}

value "size (N (N H H) (N H H)) = 3"
value "size (N (N (N H H) (N H H)) (N (N H H) (N H H))) = 7"

text {*
---------------------------------------------------------------------
Nota. Tenemos 3 funciones de medida sobre los árboles: número de
hojas, número de nodos y profundidad. A cada una le corresponde un
concepto de completitud. En los siguientes ejercicios demostraremos
que los tres conceptos de completitud son iguales.
---------------------------------------------------------------------
*}

text {*
---------------------------------------------------------------------
Ejercicio 7. Demostrar que un árbol binario a es completo respecto de
la profundidad syss es completo respecto del número de hojas.
---------------------------------------------------------------------
*}

(*anddonram luicedval oscgonesc*)
lemma hojas_prof:
"es_abc profundidad a ⟶ hojas a = 2^profundidad a"
by (induct a) auto
theorem comp_p_h: "es_abc profundidad a = es_abc hojas a"
by (induct a) (auto simp add: hojas_prof)

text {*
---------------------------------------------------------------------
Ejercicio 8. Demostrar que un árbol binario a es completo respecto del
número de hojas syss es completo respecto del número de nodos.
---------------------------------------------------------------------
*}

(*anddonram luicedval oscgonesc*)
lemma hojas_size:
" es_abc hojas a ⟶ hojas a = Suc (size a)"
by (induct a) auto

theorem comp_h_s: "es_abc hojas a = es_abc size a"
by (induct a) (auto simp add: hojas_size )

text {*
---------------------------------------------------------------------
Ejercicio 9. Demostrar que un árbol binario a es completo respecto de
la profundidad syss es completo respecto del número de nodos.
---------------------------------------------------------------------
*}

(*anddonram luicedval oscgonesc*)
theorem "es_abc profundidad a = es_abc size a"
by (simp_all add:comp_p_h comp_h_s)

text {*
---------------------------------------------------------------------
Ejercicio 10. Demostrar que (abc n) es un árbol binario completo.
---------------------------------------------------------------------
*}

(*anddonram luicedval oscgonesc*)
theorem "es_abc f (abc n) "
by (induct n) auto

text {*
---------------------------------------------------------------------
Ejercicio 11. Demostrar que si a es un árbolo binario completo
respecto de la profundidad, entonces a es igual a
(abc (profundidad a)).
---------------------------------------------------------------------
*}

(*anddonram luicedval oscgonesc*)
theorem "es_abc profundidad a =(a=abc (profundidad a)) "
by (induct a) auto

text {*
---------------------------------------------------------------------
Ejercicio 12. Encontrar una medida f tal que (es_abc f) es distinto de
(es_abc size).
---------------------------------------------------------------------
*}

(*anddonram(mas o menos llegué a la conclusión de que f(H)=f(N i d) ) luicedval*)
fun funcion :: "arbol => nat" where
"funcion H = 1"
| "funcion (N x y) = funcion x "

theorem "es_abc size a = es_abc funcion a"
quickcheck
oops

end
</source>

Relación 5

2017-12-18T10:01:59Z

Oscgonesc:

<source lang="isar">
chapter {* R5: Recorridos de árboles *}

theory R5_Recorridos_de_arboles
imports Main
begin

text {*
---------------------------------------------------------------------
Ejercicio 1. Definir el tipo de datos arbol para representar los
árboles binarios que tiene información en los nodos y en las hojas.
Por ejemplo, el árbol
e
/ \
/ \
c g
/ \ / \
a d f h
se representa por "N e (N c (H a) (H d)) (N g (H f) (H h))".
---------------------------------------------------------------------
*}

datatype 'a arbol = H "'a" | N "'a" "'a arbol" "'a arbol"

value "N e (N c (H a) (H d)) (N g (H f) (H h))"

text {*
---------------------------------------------------------------------
Ejercicio 2. Definir la función
preOrden :: "'a arbol ⇒ 'a list"
tal que (preOrden a) es el recorrido pre orden del árbol a. Por
ejemplo,
preOrden (N e (N c (H a) (H d)) (N g (H f) (H h)))
= [e,c,a,d,g,f,h]
---------------------------------------------------------------------
*}

(* luicedval anddonram oscgonesc *)
fun preOrden :: "'a arbol ⇒ 'a list" where
"preOrden (H x) = [x]"
| "preOrden (N x i d) = x # preOrden i @ preOrden d"

value "preOrden (N e (N c (H a) (H d)) (N g (H f) (H h)))
= [e,c,a,d,g,f,h]"

text {*
---------------------------------------------------------------------
Ejercicio 3. Definir la función
postOrden :: "'a arbol ⇒ 'a list"
tal que (postOrden a) es el recorrido post orden del árbol a. Por
ejemplo,
postOrden (N e (N c (H a) (H d)) (N g (H f) (H h)))
= [e,c,a,d,g,f,h]
---------------------------------------------------------------------
*}

(* luicedval anddonram oscgonesc*)
fun postOrden :: "'a arbol ⇒ 'a list" where
"postOrden (H x) = [x]"
| "postOrden (N x i d) = postOrden i @ postOrden d @ [x]"

value "postOrden (N e (N c (H a) (H d)) (N g (H f) (H h)))
= [a,d,c,f,h,g,e]"

text {*
---------------------------------------------------------------------
Ejercicio 4. Definir la función
inOrden :: "'a arbol ⇒ 'a list"
tal que (inOrden a) es el recorrido in orden del árbol a. Por
ejemplo,
inOrden (N e (N c (H a) (H d)) (N g (H f) (H h)))
= [a,c,d,e,f,g,h]
---------------------------------------------------------------------
*}

(* luicedval anddonram oscgonesc*)
fun inOrden :: "'a arbol ⇒ 'a list" where
"inOrden (H x) = [x]"
| "inOrden (N x i d) = inOrden i @ x # inOrden d"

value "inOrden (N e (N c (H a) (H d)) (N g (H f) (H h)))
= [a,c,d,e,f,g,h]"

text {*
---------------------------------------------------------------------
Ejercicio 5. Definir la función
espejo :: "'a arbol ⇒ 'a arbol"
tal que (espejo a) es la imagen especular del árbol a. Por ejemplo,
espejo (N e (N c (H a) (H d)) (N g (H f) (H h)))
= N e (N g (H h) (H f)) (N c (H d) (H a))
---------------------------------------------------------------------
*}

(* luicedval anddonram oscgonesc*)
fun espejo :: "'a arbol ⇒ 'a arbol" where
"espejo (H x) = (H x)"
| "espejo (N x i d) = (N x (espejo d) (espejo i))"

value "espejo (N e (N c (H a) (H d)) (N g (H f) (H h)))
= N e (N g (H h) (H f)) (N c (H d) (H a))"

text {*
---------------------------------------------------------------------
Ejercicio 6. Demostrar que
preOrden (espejo a) = rev (postOrden a)
---------------------------------------------------------------------
*}

(* luicedval oscgonesc*)
lemma "preOrden (espejo a) = rev (postOrden a)"
by (induct a) auto

lemma "preOrden (espejo a) = rev (postOrden a)" (is "?P a")
proof (induct a)
fix x::"'b arbol"
show "⋀x. preOrden (espejo (H x)) = rev (postOrden (H x))" by simp
next
fix x
fix i assume H1: "?P i"
fix d assume H2: "?P d"
have "preOrden (espejo (N x i d)) = preOrden (N x (espejo d) (espejo i))" by simp
also have "... = x # preOrden (espejo d) @ preOrden (espejo i)" by simp
also have "... = x # rev (postOrden d) @ rev (postOrden i)" using H1 H2 by simp
also have "... = rev((postOrden i) @ (postOrden d) @ [x])" by simp
finally show "preOrden (espejo (N x i d)) = rev (postOrden (N x i d))" by simp
qed

(*anddonram*)
lemma
fixes a :: "'b arbol"
shows "preOrden (espejo a) = rev (postOrden a)"
proof (induct a)
(*Si no pones el tipo da un warning. ¿Por qué?*)
fix x::'b
show " preOrden (espejo (H x)) = rev (postOrden (H x))" by simp
next
fix x1a
(*Si no pones el tipo da un error. ¿Por qué?*)
fix a1:: "'b arbol"
assume H1:" preOrden (espejo a1) = rev (postOrden a1)"
fix a2:: "'b arbol"
assume H2:" preOrden (espejo a2) = rev (postOrden a2)"
have "preOrden (espejo (N x1a a1 a2)) = preOrden (N x1a (espejo a2) (espejo a1)) "
by simp
also have "... = [x1a] @ (preOrden (espejo a2)) @ (preOrden (espejo a1)) " by simp
also have "... = [x1a] @ rev (postOrden a2) @ rev (postOrden a1) " using H1 H2 by simp
also have "... = rev( (postOrden a1) @ (postOrden a2) @ [x1a]) " by simp
finally show "preOrden (espejo (N x1a a1 a2)) = rev (postOrden (N x1a a1 a2)) "
by simp
qed

text {*
---------------------------------------------------------------------
Ejercicio 7. Demostrar que
postOrden (espejo a) = rev (preOrden a)
---------------------------------------------------------------------
*}

(* luicedval oscgonesc *)
lemma "postOrden (espejo a) = rev (preOrden a)"
by (induct a) auto

lemma "postOrden (espejo a) = rev (preOrden a)" (is "?P a")
proof (induct a)
fix x::"'b arbol"
show "⋀x. postOrden (espejo (H x)) = rev (preOrden (H x))" by simp
next
fix x
fix i assume H1: "?P i"
fix d assume H2: "?P d"
have "postOrden (espejo (N x i d)) = postOrden (N x (espejo d) (espejo i))" by simp
also have "... = postOrden (espejo d) @ postOrden (espejo i) @ [x]" by simp
also have "... = rev (preOrden d) @ rev (preOrden i) @ [x]" using H1 H2 by simp
also have "... = rev(x # (preOrden i) @ (preOrden d))" by simp
finally show "postOrden (espejo (N x i d)) = rev (preOrden (N x i d))" by simp
qed

(*anddonram*)
lemma
fixes a :: "'b arbol"
shows "postOrden (espejo a) = rev (preOrden a)"
proof (induct a)
fix x::'b
show "postOrden (espejo (H x)) = rev (preOrden (H x))" by simp
next
fix x1a
fix a1:: "'b arbol"
assume H1:"postOrden (espejo a1) = rev (preOrden a1)"
fix a2:: "'b arbol"
assume H2:"postOrden (espejo a2) = rev (preOrden a2)"
have "postOrden (espejo (N x1a a1 a2)) = postOrden (N x1a (espejo a2) (espejo a1)) "
by simp
also have "... = (postOrden (espejo a2)) @ (postOrden (espejo a1)) @ [x1a] " by simp
also have "... = rev (preOrden a2) @ rev (preOrden a1) @ [x1a]" using H1 H2 by simp
also have "... = rev([x1a] @ (preOrden a1) @ (preOrden a2) ) " by simp
finally show "postOrden (espejo (N x1a a1 a2)) = rev (preOrden (N x1a a1 a2))"
by simp
qed

text {*
---------------------------------------------------------------------
Ejercicio 8. Demostrar que
inOrden (espejo a) = rev (inOrden a)
---------------------------------------------------------------------
*}

(* luicedval oscgonesc *)
theorem "inOrden (espejo a) = rev (inOrden a)"
by (induct a) auto

theorem "inOrden (espejo a) = rev (inOrden a)" (is "?P a")
proof (induct a)
fix x::"'b arbol"
show "⋀x. inOrden (espejo (H x)) = rev (inOrden (H x))" by simp
next
fix x
fix i assume H1: "?P i"
fix d assume H2: "?P d"
have "inOrden (espejo (N x i d)) = inOrden (N x (espejo d) (espejo i))" by simp
also have "... = inOrden (espejo d) @ x # (inOrden (espejo i)) " by simp
also have "... = rev (inOrden d) @ x # rev (inOrden i) " using H1 H2 by simp
also have "... = rev( (inOrden i) @ x # (inOrden d)) " by simp
finally show "inOrden (espejo (N x i d)) = rev (inOrden (N x i d))" by simp
qed

(*anddonram*)
theorem
fixes a :: "'b arbol"
shows "inOrden (espejo a) = rev (inOrden a)"
proof (induct a)
fix x::'b
show "inOrden (espejo (H x)) = rev (inOrden (H x))" by simp
next
fix x1a
fix a1:: "'b arbol"
assume H1:" inOrden (espejo a1) = rev (inOrden a1)"
fix a2:: "'b arbol"
assume H2:" inOrden (espejo a2) = rev (inOrden a2)"
have "inOrden (espejo (N x1a a1 a2)) = inOrden (N x1a (espejo a2) (espejo a1)) "
by simp
also have "... = (inOrden (espejo a2)) @ [x1a]@ (inOrden (espejo a1)) " by simp
also have "... = rev (inOrden a2) @ [x1a] @ rev (inOrden a1) " using H1 H2 by simp
also have "... = rev((inOrden a1) @ [x1a] @ (inOrden a2) ) " by simp
finally show " inOrden (espejo (N x1a a1 a2)) = rev (inOrden (N x1a a1 a2))"
by simp
qed

text {*
---------------------------------------------------------------------
Ejercicio 9. Definir la función
raiz :: "'a arbol ⇒ 'a"
tal que (raiz a) es la raiz del árbol a. Por ejemplo,
raiz (N e (N c (H a) (H d)) (N g (H f) (H h))) = e
---------------------------------------------------------------------
*}

(* luicedval anddonram oscgonesc *)
fun raiz :: "'a arbol ⇒ 'a" where
"raiz (H x) = x"
| "raiz (N x i d) = x"

value "raiz (N e (N c (H a) (H d)) (N g (H f) (H h))) = e"

text {*
---------------------------------------------------------------------
Ejercicio 10. Definir la función
extremo_izquierda :: "'a arbol ⇒ 'a"
tal que (extremo_izquierda a) es el nodo más a la izquierda del árbol
a. Por ejemplo,
extremo_izquierda (N e (N c (H a) (H d)) (N g (H f) (H h))) = a
---------------------------------------------------------------------
*}

(* luicedval anddonram oscgonesc *)
fun extremo_izquierda :: "'a arbol ⇒ 'a" where
"extremo_izquierda (H x) = x"
| "extremo_izquierda (N x i d) = extremo_izquierda i"

value "extremo_izquierda (N e (N c (H a) (H d)) (N g (H f) (H h))) = a"

text {*
---------------------------------------------------------------------
Ejercicio 11. Definir la función
extremo_derecha :: "'a arbol ⇒ 'a"
tal que (extremo_derecha a) es el nodo más a la derecha del árbol
a. Por ejemplo,
extremo_derecha (N e (N c (H a) (H d)) (N g (H f) (H h))) = h
---------------------------------------------------------------------
*}

(* luicedval anddonram oscgonesc *)
fun extremo_derecha :: "'a arbol ⇒ 'a" where
"extremo_derecha (H x) = x"
| "extremo_derecha (N x i d) = extremo_derecha d"

value "extremo_derecha (N e (N c (H a) (H d)) (N g (H f) (H h))) = h"

text {*
---------------------------------------------------------------------
Ejercicio 12. Demostrar o refutar
last (inOrden a) = extremo_derecha a
---------------------------------------------------------------------
*}

(*anddonram*)
lemma inOrdenNoVacio: "inOrden a ≠ []" by (cases a) auto
theorem
fixes a :: "'b arbol"
shows "last (inOrden a) = extremo_derecha a"
proof (induct a)
fix x::'b
show "last (inOrden (H x)) = extremo_derecha (H x)" by simp
next
fix x1a
fix a1:: "'b arbol"
assume H1:"last (inOrden a1) = extremo_derecha a1"
fix a2:: "'b arbol"
assume H2:"last (inOrden a2) = extremo_derecha a2"
have "last (inOrden (N x1a a1 a2)) = last( (inOrden a1) @ [x1a] @ (inOrden a2)) "
by simp
also have "... = last( [x1a] @ inOrden a2) " by simp
also have "... = last (inOrden a2) " by (simp add:inOrdenNoVacio)
also have "... = extremo_derecha a2 " using H2 by simp
finally show "last (inOrden (N x1a a1 a2)) = extremo_derecha (N x1a a1 a2)"
by simp
qed

(* luicedval oscgonesc *)
lemma inOrdenNoVacio: "inOrden a ≠ []" by (cases a) auto
(* Créditos Andrés, no sabía como hacerlo *)

theorem "last (inOrden a) = extremo_derecha a" (is "?P a")
proof (induct a)
fix x::"'b arbol"
show "⋀x. ?P (H x)" by simp
next
fix x
fix i
fix d assume H1: "?P d"
have "last (inOrden (N x i d)) = last (inOrden i @ x # inOrden d)" by simp
also have "... = last (x # inOrden d)" by simp
also have "... = last (inOrden d)" by (simp add:inOrdenNoVacio)
also have "... = extremo_derecha d" using H1 by simp
finally show "last (inOrden (N x i d)) = extremo_derecha (N x i d)" by simp
qed

text {*
---------------------------------------------------------------------
Ejercicio 13. Demostrar o refutar
hd (inOrden a) = extremo_izquierda a
---------------------------------------------------------------------
*}

(*anddonram*)
theorem
fixes a :: "'b arbol"
shows "hd (inOrden a) = extremo_izquierda a"
proof (induct a)
fix x::'b
show "hd (inOrden (H x)) = extremo_izquierda (H x)" by simp
next
fix x1a
fix a1:: "'b arbol"
assume H1:" hd (inOrden a1) = extremo_izquierda a1"
fix a2:: "'b arbol"
assume H2:" hd (inOrden a2) = extremo_izquierda a2"
have "hd (inOrden (N x1a a1 a2)) = hd ( (inOrden a1) @ [x1a] @ (inOrden a2)) "
by simp
also have "... = hd (inOrden a1) " by (simp add:inOrdenNoVacio)
also have "... = extremo_izquierda a1 " using H1 by simp
finally show " hd (inOrden (N x1a a1 a2)) = extremo_izquierda (N x1a a1 a2)"
by simp
qed

(* luicedval oscgonesc *)
theorem "hd (inOrden a) = extremo_izquierda a" (is "?P a")
proof (induct a)
fix x::"'b arbol"
show "⋀x. ?P (H x)" by simp
next
fix x
fix i assume H1: "?P i"
fix d
have "hd (inOrden (N x i d)) = hd (inOrden i @ x # inOrden d)" by simp
also have "... = hd (inOrden i)" by (simp add:inOrdenNoVacio)
also have "... = extremo_izquierda i" using H1 by simp
finally show "hd (inOrden (N x i d)) = extremo_izquierda (N x i d)" by simp
qed

text {*
---------------------------------------------------------------------
Ejercicio 14. Demostrar o refutar
hd (preOrden a) = last (postOrden a)
---------------------------------------------------------------------
*}

(*anddonram*)
theorem
fixes a :: "'b arbol"
shows "hd (preOrden a) = last (postOrden a)"
proof (induct a)
fix x::'b
show "hd (preOrden (H x)) = last (postOrden (H x))" by simp
next
fix x1a
fix a1:: "'b arbol"
assume H1:"hd (preOrden a1) = last (postOrden a1)"
fix a2:: "'b arbol"
assume H2:"hd (preOrden a2) = last (postOrden a2)"
have "hd (preOrden (N x1a a1 a2)) = hd ( [x1a] @ (preOrden a1) @ (preOrden a2)) "
by simp
also have "... = x1a " by simp
also have "... = last ((preOrden a1) @ (preOrden a2)@ [x1a])" by simp
finally show "hd (preOrden (N x1a a1 a2)) = last (postOrden (N x1a a1 a2))"
by simp
qed

(* luicedval *)
theorem "hd (preOrden a) = last (postOrden a)" (is "?P a")
proof (induct a)
fix x::"'b arbol"
show "⋀x. ?P (H x)" by simp
next
fix x
fix i assume H1: "?P i"
fix d assume H2: "?P d"
have "hd (preOrden (N x i d)) = hd (x # preOrden i @ preOrden d)" by simp
also have "... = x" by simp
also have "... = last (postOrden d @ postOrden i @ [x])" by simp
finally show "hd (preOrden (N x i d)) = last (postOrden (N x i d))" by simp
qed

text {*
---------------------------------------------------------------------
Ejercicio 15. Demostrar o refutar
hd (preOrden a) = raiz a
---------------------------------------------------------------------
*}

(*anddonram*)
theorem "hd (preOrden a) = raiz a"
proof (induct a)
fix x::'b
show "hd (preOrden (H x)) = raiz (H x)" by simp
next
fix x1a
fix a1:: "'b arbol"
assume H1:" hd (preOrden a1) = raiz a1"
fix a2:: "'b arbol"
assume H2:" hd (preOrden a2) = raiz a2"
have "hd (preOrden (N x1a a1 a2)) = hd ( [x1a] @ (preOrden a1) @ (preOrden a2)) "
by simp
also have "... = x1a " by simp
finally show "hd (preOrden (N x1a a1 a2)) = raiz (N x1a a1 a2)"
by simp
qed

(* luicedval *)
theorem "hd (preOrden a) = raiz a" (is "?P a")
proof (induct a)
fix x::"'b arbol"
show "⋀x. ?P (H x)" by simp
next
fix x
fix i assume H1: "?P i"
fix d assume H2: "?P d"
have "hd (preOrden (N x i d)) = hd (x # preOrden i @ preOrden d)" by simp
also have "... = x" by simp
finally show "hd (preOrden (N x i d)) = raiz (N x i d)" by simp
qed

text {*
---------------------------------------------------------------------
Ejercicio 16. Demostrar o refutar
hd (inOrden a) = raiz a
---------------------------------------------------------------------
*}

(*anddonram luicedval oscgonesc *)
theorem "hd (inOrden a) = raiz a"
quickcheck
oops
(*
Quickcheck found a counterexample:
a = N a⇩1 (H a⇩2) (H a⇩1)
Evaluated terms:
hd (inOrden a) = a⇩2
raiz a = a⇩1
*)

text {*
---------------------------------------------------------------------
Ejercicio 17. Demostrar o refutar
last (postOrden a) = raiz a
---------------------------------------------------------------------
*}

(*anddonram*)
theorem "last (postOrden a) = raiz a"
proof (induct a)
fix x::'b
show "last (postOrden (H x)) = raiz (H x)" by simp
next
fix x1a
fix a1:: "'b arbol"
assume H1:"last (postOrden a1) = raiz a1 "
fix a2:: "'b arbol"
assume H2:"last (postOrden a2) = raiz a2"
have " last (postOrden (N x1a a1 a2)) = last ( (preOrden a1) @ (preOrden a2)@[x1a] ) "
by simp
also have "... = x1a " by simp
finally show " last (postOrden (N x1a a1 a2)) = raiz (N x1a a1 a2)"
by simp
qed

(* luicedval oscgonesc *)
theorem "last (postOrden a) = raiz a" (is "?P a")
proof (induct a)
fix x::"'b arbol"
show "⋀x. ?P (H x)" by simp
next
fix x
fix i assume H1: "?P i"
fix d assume H2: "?P d"
have "last (postOrden (N x i d)) = last (postOrden i @ postOrden d @ [x])" by simp
also have "... = last (postOrden d @ [x])" by simp
also have "... = x" by simp
finally show "last (postOrden (N x i d)) = raiz (N x i d)" by simp
qed

end
</source>