Razonamiento automático (2016-17) - Contribuciones del usuario [es]

Relación 10

2017-01-31T23:10:58Z

Pablucoto:

<source lang="isar">
chapter {* R10: Formalización y argumentación con Isabelle/HOL *}

theory R10_Formalizacion_y_argmentacion
imports Main
begin

text {*
---------------------------------------------------------------------
El objetivo de esta es relación formalizar y demostrar la corrección
de los argumentos automáticamente y detalladamente usando sólo las reglas
básicas de deducción natural.

· 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, no_ex y no_para_todo 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

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

text {* ---------------------------------------------------------------
Ejercicio 1. Formalizar, y demostrar la corrección, del siguiente
argumento
Si la válvula está abierta o la monitorización está preparada,
entonces se envía una señal de reconocimiento y un mensaje de
funcionamiento al controlador del ordenador. Si se envía un mensaje
de funcionamiento al controlador del ordenador o el sistema está en
estado normal, entonces se aceptan las órdenes del operador. Por lo
tanto, si la válvula está abierta, entonces se aceptan las órdenes
del operador.
Usar A : La válvula está abierta.
P : La monitorización está preparada.
R : Envía una señal de reconocimiento.
F : Envía un mensaje de funcionamiento.
N : El sistema está en estado normal.
O, AO, OK : Se aceptan órdenes del operador.
------------------------------------------------------------------ *}

(* ivamenjim manmorjim1 crigomgom migtermor serrodcal *)
(* Buscando, he detectado que 'O' es un carácter especial en Isabelle y que forma parte de su
sintaxis pre-definida, por lo que da problemas a la hora de formalizar y demostrar
el argumento planteado. Por lo tanto, en su lugar he usado "AO: Se aceptan órdenes del operador" *)

lemma ejercicio_1:
assumes 1: "(A ∨ P) ⟶ (R ∧ F)"
assumes 2: "(F ∨ N) ⟶ AO"
shows "A ⟶ AO"
proof -
{assume 3: "A"
have 4: "A ∨ P" using 3 by (rule disjI1)
have 5: "R ∧ F" using 1 4 by (rule mp)
have 6: "F" using 5 by (rule conjunct2)
have 7: "F ∨ N" using 6 by (rule disjI1)
have 8: "AO" using 2 7 by (rule mp)}
then show "A ⟶ AO" by (rule impI)
qed

(*danrodcha ferrenseg anaprarod marcarmor13 pablucoto*)

lemma ej_1:
assumes "A ∨ P ⟶ R ∧ F" and
"F ∨ N ⟶ OK"
shows "A ⟶ OK"
proof (rule impI)
{assume "A"
hence "A ∨ P" by (rule disjI1)
with assms(1) have "R ∧ F" by (rule mp)
hence "F" by (rule conjE)
hence "F ∨ N" by (rule disjI1)
with assms(2) show "OK" by (rule mp)}
qed

(*pabrodmac*)
lemma ejercicio_1_1:
assumes "A ∨ P ⟶ R ∧ F"
"F ∨ N ⟶ Op"
shows "A ⟶ Op"
proof
assume "A"
hence "A ∨ P" by (rule disjI1)
have "R ∧ F" using `A ∨ P ⟶ R ∧ F` `A ∨ P` by (rule mp)
hence "F" by (rule conjE)
hence "F ∨ N" by (rule disjI1)
show "Op" using `F ∨ N ⟶ Op` `F ∨ N` by (rule mp)
qed

(*josgarsan*)
lemma ejericio_1_2:
assumes 1: "(A ∨ P) ⟶ (R ∧ F)" and
2: "(F ∨ N) ⟶ Op"
shows "A ⟶ Op"
proof
assume 3: A
have 4:"(A ∨ P)" using 3 by (rule disjI1)
have "(R ∧ F)" using 1 4 by (rule mp)
then have "F" by (rule conjunct2)
then have 5: "(F ∨ N)" by (rule disjI1)
show "Op" using 2 5 by (rule mp)
qed

text {* ---------------------------------------------------------------
Ejercicio 2. Formalizar, y decidir la corrección, del siguiente
argumento
Hay estudiantes inteligentes y hay estudiantes trabajadores. Por
tanto, hay estudiantes inteligentes y trabajadores.
Usar I(x) para x es inteligente
T(x) para x es trabajador
------------------------------------------------------------------ *}

(* ivamenjim ferrenseg danrodcha anaprarod crigomgom marcarmor13 pabrodmac migtermor josgarsan serrodcal pablucoto*)

lemma ejercicio_2:
assumes "(∃x. I(x)) ∧ (∃x. T(x))"
shows "∃x. (I(x) ∧ T(x))"
quickcheck
oops

(* Encontrando el contraejemplo:
I = {a1}
x = a1
T = {a2}
xa = a2
*)

text {* ---------------------------------------------------------------
Ejercicio 3. Formalizar, y decidir la corrección, del siguiente
argumento
Los hermanos tienen el mismo padre. Juan es hermano de Luis. Carlos
es padre de Luis. Por tanto, Carlos es padre de Juan.
Usar H(x,y) para x es hermano de y
P(x,y) para x es padre de y
j para Juan
l para Luis
c para Carlos
------------------------------------------------------------------ *}

(* ivamenjim crigomgom serrodcal*)

lemma ejercicio_3:
assumes 1: "∀x y. P(x,y) ⟶ (∀z. (H(z,y) ⟶ P(x,z)))"
assumes 2: "H(j,l)"
assumes 3: "P(c,l)"
shows "P(c,j)"
proof -
have 4 : "∀y. P(c,y) ⟶ (∀z. (H(z,y) ⟶ P(c,z)))" using 1 by (rule allE)
have 5 : "P(c,l) ⟶ (∀z. (H(z,l) ⟶ P(c,z)))" using 4 by (rule allE)
then have 6 : "(∀z. (H(z,l) ⟶ P(c,z)))" using 3 by (rule mp)
have 7 : "H(j,l) ⟶ P(c,j)" using 6 by (rule allE)
then show "P(c,j)" using 2 by (rule mp)
qed

(* danrodcha anaprarod ferrenseg marcarmor13 pablucoto*)
(* es casi igual que la anterior *)
lemma ej_3:
assumes "∀x y. P(x,y) ⟶ (∀z. (H(z,y) ⟶ P(x,z)))"
assumes "H(j,l)"
assumes "P(c,l)"
shows "P(c,j)"
proof (rule mp)
have 4 : "∀y. P(c,y) ⟶ (∀z. (H(z,y) ⟶ P(c,z)))" using assms(1) by (rule allE)
hence "P(c,l) ⟶ (∀z. (H(z,l) ⟶ P(c,z)))" by (rule allE)
hence "(∀z. (H(z,l) ⟶ P(c,z)))" using assms(3) by (rule mp)
thus "H(j,l) ⟶ P(c,j)" by (rule allE)
next
show "H(j,l)" using assms(2) by this
qed

(*pabrodmac*)
lemma ejercicio_3_1:
assumes "∀x y. H(x,y) ⟶ (∀ z. (P(z,y) ⟶ P(z,x)))"
"H(j,l)"
"P(c,l)"
shows "P(c,j)"
proof-
have "∀y. H(j,y) ⟶ (∀ z. (P(z,y) ⟶ P(z,j)))" using assms(1) by (rule allE)
hence "H(j,l) ⟶ (∀ z. (P(z,l) ⟶ P(z,j)))" by (rule allE)
hence "∀z. (P(z,l) ⟶ P(z,j))" using assms(2) by (rule mp)
hence "P(c,l) ⟶ P(c,j)" by (rule allE)
thus "P(c,j)" using assms(3) by (rule mp)
qed

(* migtermor josgarsan*)

lemma ejercicio_3_2:
assumes "∀x y z. (P(x,y) ∧ H(y,z)) ⟶ P(x,z)"
assumes "H(l,j)"
assumes "P(c,l)"
shows "P(c,j)"
proof -
have 1: "P(c,l) ∧ H(l,j)" using assms(3) assms(2) by (rule conjI)
have 2: "∀y z. (P(c,y) ∧ H(y,z)) ⟶ P(c,z)" using assms(1) by (rule allE)
then have 3: "∀z. (P(c,l) ∧ H(l,z)) ⟶ P(c,z)" by (rule allE)
then have 4: "(P(c,l) ∧ H(l,j)) ⟶ P(c,j)" by (rule allE)
then show "P(c,j)" using 1 by (rule mp)
qed

text {* ---------------------------------------------------------------
Ejercicio 4. Formalizar, y decidir la corrección, del siguiente
argumento
Los aficionados al fútbol aplauden a cualquier futbolista
extranjero. Juanito no aplaude a futbolistas extranjeros. Por
tanto, si hay algún futbolista extranjero nacionalizado español,
Juanito no es aficionado al fútbol.
Usar Af(x) para x es aficicionado al fútbol
Ap(x,y) para x aplaude a y
E(x) para x es un futbolista extranjero
N(x) para x es un futbolista nacionalizado español
j para Juanito
------------------------------------------------------------------ *}
(* danrodcha ferrenseg pablucoto*)
(* danrodcha: Me gusta más la formalización de ana prado, aunque
son equivalentes *)
lemma ej_4:
assumes "∀x y. Af(x) ∧ E(y) ⟶ Ap(x,y)"
"∀x. Ap(j,x) ⟶ ¬ E(x)"
shows "(∃x. E(x) ∧ N(x)) ⟶ ¬Af(j)"
proof (rule impI)
assume "∃x. E(x) ∧ N(x)"
then obtain a where "E(a) ∧ N(a)" by (rule exE)
hence "E(a)" by (rule conjE)
show "¬ Af(j)"
proof (rule notI)
assume "Af(j)"
hence "Af(j) ∧ E(a)" using `E(a)` by (rule conjI)
have "∀y. Af(j) ∧ E(y) ⟶ Ap(j,y)" using assms(1) by (rule allE)
hence "Af(j) ∧ E(a) ⟶ Ap(j,a)" by (rule allE)
hence "Ap(j,a)" using `Af(j) ∧ E(a)` by (rule mp)
have "Ap(j,a) ⟶ ¬ E(a)" using assms(2) by (rule allE)
hence "¬ E(a)" using `Ap(j,a)` by (rule mp)
thus False using `E(a)` by (rule notE)
qed
qed

(* ivamenjim crigomgom marcarmor13 serrodcal*)

lemma ejercicio_4:
assumes 1: "∀x y. Af(x) ∧ E(y) ⟶ Ap(x,y)"
assumes 2: "¬(∃x. E(x) ∧ Ap(j,x))"
shows "(∃x. E(x) ∧ N(x)) ⟶ ¬Af(j)"
proof (rule impI)
assume 3: "∃x. E(x) ∧ N(x)"
then obtain a where 4: "E(a) ∧ N(a)" by (rule exE)
then have 5: "E(a)" by (rule conjunct1)
show 6: "¬Af(j)"
proof (rule notI)
assume 7: "Af(j)"
then have 8: "Af(j) ∧ E(a)" using 5 by (rule conjI)
have 9: "∀y. Af(j) ∧ E(y) ⟶ Ap(j,y)" using 1 by (rule allE)
have 10: "Af(j) ∧ E(a) ⟶ Ap(j,a)" using 9 by (rule allE)
have 11: "Ap(j,a)" using 10 8 by (rule mp)
have 12: "E(a) ∧ Ap(j,a)" using 5 11 by (rule conjI)
have 13: "∃x. E(x) ∧ Ap(j,x)" using 12 by (rule exI)
show "False" using 2 13 by (rule notE)
qed
qed

(* anaprarod migtermor *)

(* Este auxiliar ya se probó en el ejercicio 4 de la rel 8 *)
lemma aux4: "¬(p∧q) ⟹¬p ∨ ¬q"
by (auto)

lemma ej_4:
assumes "∀x y. Af(x) ∧ E(y) ⟶ Ap(x,y)"
"∀ x.(E(x) ⟶ ¬ Ap(j,x))"
shows "(∃x. E(x) ∧ N(x)) ⟶ ¬Af(j)"
proof (rule impI)
{ assume "∃x. E(x) ∧ N(x)"
then obtain a where 1: "E(a) ∧ N(a)" by (rule exE)
have "∀ y. Af(j) ∧ E(y) ⟶ Ap(j,y)" using assms(1) by (rule allE)
hence 2: "Af(j) ∧ E(a) ⟶ Ap(j,a)" by (rule allE)
have 3: "E(a)" using 1 by (rule conjunct1)
have 4: "E(a) ⟶ ¬ Ap(j,a)" using assms(2) by (rule allE)
have 5: "¬ Ap(j,a)" using 4 3 by (rule mp)
have "¬(Af(j) ∧ E(a))" using 2 5 by (rule mt)
hence "¬ Af(j) ∨ ¬ E(a)" by (rule aux4)
thus " ¬Af(j)"
proof (rule disjE)
{assume "¬Af(j)"
thus "¬Af(j)" by this}
next
{assume 6: "¬ E(a)"
show "¬ Af(j)" using 6 3 by (rule notE)}
qed}
qed

(*pabrodmac*)

lemma ejercicio_4_1:
assumes "∀x y. Af(x) ∧ E(y) ⟶ Ap(x,y)"
"∀x. E(x)⟶ ¬Ap(j,x)"
shows "(∃x. N(x) ∧ E(x)) ⟶ ¬Af(j)"
proof
assume "∃x. N(x) ∧ E(x)"
then obtain b where "N(b) ∧ E(b)" by (rule exE)
hence "E(b)" by (rule conjunct2)
have "E(b)⟶ ¬Ap(j,b)" using assms(2) by (rule allE)
hence "¬Ap(j,b)" using `E(b)` by (rule mp)
have "∀y. Af(j) ∧ E(y) ⟶ Ap(j,y)" using assms(1) by (rule allE)
hence "Af(j) ∧ E(b) ⟶ Ap(j,b)" by (rule allE)
hence "¬Ap(j,b) ⟶ ¬(Af(j) ∧ E(b))" by (rule Set.not_mono)
hence "¬(Af(j) ∧ E(b))" using `¬Ap(j,b)` by (rule mp)
show "¬Af(j)"
proof
assume "Af(j)"
hence "Af(j) ∧ E(b)" using `E(b)` by (rule conjI)
show False using `¬(Af(j) ∧ E(b))` `Af(j) ∧ E(b)` by (rule notE)
qed
qed

text {* ---------------------------------------------------------------
Ejercicio 5. Formalizar, y decidir la corrección, del siguiente
argumento
El esposo de la hermana de Toni es Roberto. La hermana de Toni es
María. Por tanto, el esposo de María es Roberto.
Usar e(x) para el esposo de x
h para la hermana de Toni
m para María
r para Roberto
------------------------------------------------------------------ *}

(* ivamenjim *)

lemma ejercicio_5:
assumes 1: "e(h) = r"
assumes 2: "h = m"
shows "e(m) = r"
proof -
have 3: "e(h) = e(m)" using 2 by (rule arg_cong)
have 4: "e(m) = e(h)" using 3 by (rule sym)
then show "e(m) = r" using 1 by (rule trans)
qed

(* danrodcha anaprarod crigomgom ferrenseg marcarmor13 migtermor serrodcal pablucoto*)
lemma ej_5:
assumes "e(h) = r" and
"h = m"
shows "e(m) = r"
proof -
show "e(m) = r" using assms(2) assms(1) by (rule subst)
qed

(*pabrodmac*)
lemma ejercicio_5_1:
assumes "e(h)=r"
"h=m"
shows "e(m)=r"
proof-
show "e(m)=r" using assms(1) assms(2) by (rule HOL.back_subst)
qed

(*pabrodmac*)
lemma ejercicio_5_2:
assumes "e(h)=r"
"h=m"
shows "e(m)=r"
proof-
show "e(m)=r" using assms(2) assms(1) by (rule HOL.subst)
qed

(*pabrodmac*)
lemma ejercicio_5_3:
assumes "e(h)=r"
"h=m"
shows "e(m)=r"
proof-
have "e(h) = e(m)" using assms(2) by (rule arg_cong)
show "e(m)=r" using `e(h)=r` `e(h) = e(m)` by simp
qed

end
</source>

Relación 9

2017-01-26T19:33:44Z

Pablucoto:

<source lang="isar">
chapter {* R9: Deducción natural LPO en Isabelle/HOL *}

theory R9_Deduccion_natural_LPO
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
P a ⟶ (∃x. Q x) ⊢ ∃x. P a ⟶ Q x
------------------------------------------------------------------ *}

lemma ejercicio_1:
fixes P Q :: "'b ⇒ bool"
assumes "P a ⟶ (∃x. Q x)"
shows "∃x. P a ⟶ Q x"
oops

(* migtermor ferrenseg juacabsou josgarsan pablucoto*)
lemma ejercicio_1:
fixes P Q :: "'b ⇒ bool"
assumes "P a ⟶ (∃x. Q x)"
shows "∃x. P a ⟶ Q x"
proof -
{ assume 1: "P a"
have 2: "∃x. Q x" using assms 1 by (rule mp)}
then obtain b where 3: "Q b" by (rule exE)
(* No sé por qué salta un aviso aquí. Aún así, sin esto no se
finaliza correctamente la demostración, y con ello sí. *)
then have 4: "(P a) ⟶ (Q b)" by (rule impI)
then show 5: "∃x. P a ⟶ Q x" by (rule exI)
qed

(* ivamenjim bowma rubgonmar *)

lemma ejercicio_1_1:
assumes 1: "P a ⟶ (∃x. Q x)"
shows "∃x. P a ⟶ Q x"
proof -
{assume 2: "P a"
have 3: "(∃x. Q x)" using 1 2 by (rule mp)
obtain b where 4: "Q b" using 3 by (rule exE)
then have 5: "P a ⟶ Q b" by (rule impI)
then have 6: "∃x. P a ⟶ Q x" by (rule exI)}
then show "∃x. P a ⟶ Q x" by simp
qed

(* crigomgom *)
lemma ejercicio_1_2:
assumes 0: "P a ⟶ (∃x. Q x)"
shows "∃x. P a ⟶ Q x"
proof -
have "¬(P a) ∨ P a" by (rule excluded_middle)
then show "∃x. P a ⟶ Q x"
proof (rule disjE)
assume 1 : "¬(P a)"
{assume 2 : "P a"
have "Q b" using 1 2 by (rule notE)}
then have "P a ⟶ Q b" by (rule impI)
then show "∃x. P a ⟶ Q x" by (rule exI)
next
assume 3 : "P a"
have "∃x. Q x" using 0 3 by (rule mp)
then obtain b where "Q b" by (rule exE)
then have "P a ⟶ Q b" by (rule impI)
then show "∃x. P a ⟶ Q x" by (rule exI)
qed
qed

(* fracorjim1 *)
(* Entre corchetes se demuestra que efectivamente existe un caso *)
lemma ejercicio_1_3:
fixes P Q :: "'b ⇒ bool"
assumes "P a ⟶ (∃x. Q x)"
shows "∃x. P a ⟶ Q x"
proof -
{
assume "P a"
with assms have "∃x. Q x" by (rule mp)
then obtain x where "Q x" by (rule exE)
hence "P a ⟶ Q x" by (rule impI)
hence "∃x. P a ⟶ Q x" by (rule exI)
}
thus ?thesis by simp
qed

(* Comentario: Usa simp *)

text {* ---------------------------------------------------------------
Ejercicio 2. Demostrar
{∀x y z. R x y ∧ R y z ⟶ R x z,
∀x. ¬(R x x)}
⊢ ∀x y. R x y ⟶ ¬(R y x)
------------------------------------------------------------------ *}

(* migtermor ferrenseg ivamenjim marcarmor13 serrodcal juacabsou
marpoldia1 crigomgom bowma josgarsan rubgonmar pablucoto*)
lemma ejercicio_2:
fixes R :: "'b ⇒ 'b ⇒ bool"
assumes 1: "∀x y z. R x y ∧ R y z ⟶ R x z"
assumes 2: "∀x. ¬(R x x)"
shows "∀x y. R x y ⟶ ¬(R y x)"
proof (rule allI)
fix x
show "∀y. R x y ⟶ ¬(R y x)"
proof (rule allI)
fix y
{assume 3: "R x y"
{assume 4: "R y x"
have 5: "R x y ∧ R y x" using 3 4 by (rule conjI)
also have 6: "∀ z1 z2. R x z1 ∧ R z1 z2 ⟶ R x z2" using 1 by (rule allE)
then have 7: "∀ z. R x y ∧ R y z ⟶ R x z" by (rule allE)
then have 8: "R x y ∧ R y x ⟶ R x x" by (rule allE)
then have 9: "R x x" using 5 by (rule mp)
have 10: "¬(R x x)" using 2 by (rule allE)
then have 11: "False" using 9 by (rule notE)}
then have 12: "¬ (R y x)" by (rule notI)}
thus "R x y ⟶ ¬(R y x)" by (rule impI)
qed
qed

(* ivamenjim *)
(* Semejante al anterior, pero indicando que se pruebe por la regla
correspondiente *)
lemma ejercicio_2_1:
assumes 1: "∀x y z. R x y ∧ R y z ⟶ R x z"
assumes 2: "∀x. ¬(R x x)"
shows "∀x y. R x y ⟶ ¬(R y x)"
proof (rule allI)
fix x
show "∀y. R x y ⟶ ¬(R y x)"
proof (rule allI)
fix y
{assume 3: "R x y"
{assume 4: "R y x"
have 5: "R x y ∧ R y x" using 3 4 ..
have 6: "∀ y z. R x y ∧ R y z ⟶ R x z" using 1 ..
then have 7: "∀ z. R x y ∧ R y z ⟶ R x z" ..
then have 8: "R x y ∧ R y x ⟶ R x x" ..
then have 9: "R x x" using 5 ..
have 10: "¬(R x x)" using 2 ..
then have 11: "False" using 9 ..}
then have 12: "¬ (R y x)" ..}
thus "R x y ⟶ ¬(R y x)" ..
qed
qed

text {* ---------------------------------------------------------------
Ejercicio 3. Demostrar o refutar
(∀x. ∃y. P x y) ⟶ (∃y. ∀x. P x y)
------------------------------------------------------------------ *}

(* ivamenjim ferrenseg paupeddeg serrodcal juacabsou marpoldia1
crigomgom bowma josgarsan rubgonmar pablucoto*)
lemma ejercicio_3:
assumes "(∀x. ∃y. P x y)"
shows "(∃y. ∀x. P x y)"
quickcheck
oops

(* Y se encuentra el contraejemplo: P = (λx. undefined)(a1 := {a2}, a2 := {a1}) *)

(* migtermor *)
fun P :: "'a ⇒ 'a ⇒ bool" where
"P x y = (x=y)"

lemma ejercicio3:
"(∀x. ∃y. P x y) ⟶ (∃y. ∀x. P x y)"
quickcheck
oops

(* dancorgar *)
lemma ejercicio_3_3:
shows "(∀x. ∃y. P x y) ⟶ (∃y. ∀x. P x y)"
quickcheck
oops

text {* ---------------------------------------------------------------
Ejercicio 4. Demostrar o refutar
{∀x. P a x x,
∀x y z. P x y z ⟶ P (f x) y (f z)⟧
⊢ ∃z. P (f a) z (f (f a))
------------------------------------------------------------------ *}

(* ferrenseg ivamenjim *)
lemma ejercicio_4:
fixes P :: "'b ⇒ 'b ⇒ 'b ⇒ bool"
assumes 1: "∀x. P a x x" and
2: "∀x y z. P x y z ⟶ P (f x) y (f z)"
shows "∃z. P (f a) z (f (f a))"
proof -
have 4:"P a (f a) (f a)" using 1 ..
also have 5:"∀y z. P a y z ⟶ P (f a) y (f z)" using 2 ..
then have 6:"∀z. P a (f a) z ⟶ P (f a) (f a) (f z)" ..
then have 7:"P a (f a) (f a) ⟶ P (f a) (f a) (f (f a))" ..
also have 8:"P (f a) (f a) (f (f a))" using 7 4 by (rule mp)
then show "∃z. P (f a) z (f (f a))" ..
qed

(* migtermor serrodcal crigomgom pablucoto*)
lemma ejercicio_4_2:
fixes P :: "'b ⇒ 'b ⇒ 'b ⇒ bool"
assumes 1: "∀x. P a x x"
assumes 2: "∀x y z. P x y z ⟶ P (f x) y (f z)"
shows " ∃z. P (f a) z (f (f a))"
proof (rule exI)
have 3: "P a (f a) (f a)" using 1 by (rule allE)
have 4: "∀y z. P a y z ⟶ P (f a) y (f z)" using 2 by (rule allE)
then have 5: "∀z. P a (f a) z ⟶ P (f a) (f a) (f z)" by (rule allE)
then have 6: "P a (f a) (f a) ⟶ P (f a) (f a) (f (f a))" by (rule allE)
then show "P (f a) (f a) (f (f a))" using 3 by (rule mp)
qed

(* paupeddeg juacabsou marpoldia1 bowma rubgonmar josgarsan*)
lemma ejercicio_4_3:
fixes P :: "'b ⇒ 'b ⇒ 'b ⇒ bool"
assumes " ∀x. P a x x "
" ∀x y z. P x y z ⟶ P (f x) y (f z)"
shows "∃z. P (f a) z (f (f a))"
proof
have "∀ y z. P a y z ⟶ P (f a) y (f z)" using assms(2) by (rule allE)
hence "∀z. P a (f a) z ⟶ P (f a) (f a) (f z)" by (rule allE)
hence "P a (f a) (f a) ⟶ P (f a) (f a) (f (f a))" by (rule allE)
have "P a (f a) (f a)" using assms(1) by (rule allE)
show "P (f a) (f a) (f (f a))"
using `P a (f a) (f a) ⟶ P (f a) (f a) (f (f a))` `P a (f a) (f a)`
by (rule mp)
qed

text {* ---------------------------------------------------------------
Ejercicio 5. Demostrar o refutar
{∀y. Q a y,
∀x y. Q x y ⟶ Q (s x) (s y)}
⊢ ∃z. Qa z ∧ Q z (s (s a))
------------------------------------------------------------------ *}

(* ferrenseg ivamenjim *)
lemma ejercicio_5:
assumes 1: "∀y. Q a y" and
2: "∀x y. Q x y ⟶ Q (s x) (s y)"
shows "∃z. Q a z ∧ Q z (s (s a))"
proof
have 3:"Q a (s a)" using 1 ..
also have 4:"∀y. Q a y ⟶ Q (s a) (s y)" using 2 ..
then have 5:"Q a (s a) ⟶ Q (s a) (s (s a))" ..
then have 6:"Q (s a) (s (s a))" using 3 by (rule mp)
show "Q a (s a) ∧ Q (s a) (s (s a))" using 3 6 by (rule conjI)
qed

(* migtermor *)
lemma ejercicio_5_2:
fixes P :: "'b ⇒ 'b ⇒ bool"
assumes 1: "∀y. Q a y"
assumes 2: "∀x y. Q x y ⟶ Q (s x) (s y)"
shows " ∃z. Q a z ∧ Q z (s (s a))"
proof -
have 3: "Q a (s a)" using 1 by (rule allE)
have 4: "∀y. Q a y ⟶ Q (s a) (s y)" using 2 by (rule allE)
then have 5: "Q a (s a) ⟶ Q (s a) (s (s a))" by (rule allE)
then have 6: "Q (s a) (s (s a))" using 3 by (rule mp)
have "Q a (s a) ∧ Q (s a) (s (s a))" using 3 6 by (rule conjI)
then show "∃z. Q a z ∧ Q z (s (s a))" by (rule exI)
qed

(* paupeddeg serrodcal juacabsou marpoldia1 crigomgom bowma rubgonmar pablucoto*)
lemma ejercicio_5_3:
assumes "∀y. Q a y"
"∀x y. Q x y ⟶ Q (s x) (s y)"
shows "∃z. Q a z ∧ Q z (s (s a))"
proof
have "∀y. Q a y ⟶ Q (s a) (s y)" using assms(2) by (rule allE)
hence "Q a (s a) ⟶ Q (s a) (s (s a))" by (rule allE)
have "Q a (s a)" using assms(1) by (rule allE)
have "Q (s a) (s (s a))" using `Q a (s a) ⟶ Q (s a) (s (s a))` `Q a (s a)` by (rule mp)
show "Q a (s a) ∧ Q (s a) (s (s a))"
using `Q a (s a)` `Q (s a) (s (s a))` by (rule conjI)
qed

end
</source>

Relación 8

2017-01-16T16:00:40Z

Pablucoto:

<source lang="isar">
chapter {* R8: Deducción natural proposicional en Isabelle/HOL *}

theory R8_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
------------------------------------------------------------------ *}
(* marcarmor13 jeamacpov *)
--"usando un supuesto ¬¬p"
lemma ejercicio_1:
assumes 1: "¬q ⟶ ¬p" and
2: "¬¬p"
shows "p ⟶ q"
proof -
have 3: "¬¬q" using 1 2 by (rule mt)
have 4: "q" using 3 by (rule notnotD)
show "p ⟶ q" using 4 by (rule impI)
qed

(*pablucoto*)
lemma ejercicio_1_2:
assumes "¬q ⟶ ¬p"
shows "p ⟶ q"
proof -
{assume "p"
hence "¬¬p" by (rule notnotI)
with `¬q ⟶ ¬p` have "¬¬q" by (rule mt)
hence "q" by (rule notnotD)}
then show "p ⟶ q" by (rule impI)
qed

(* ivamenjim serrodcal anaprarod*)
lemma ejercicio_1_3:
assumes 1: "¬q ⟶ ¬p"
shows "p ⟶ q"
proof -
{assume 2:"p"
then have 3: "¬¬p" by (rule notnotI)
have 4: "¬¬q" using 1 3 by (rule mt)
then have 5: "q" by (rule notnotD)}
thus "p ⟶ q" by (rule impI)
qed

(* bowma *)
lemma ejercicio_1_4:
assumes "¬q ⟶ ¬p"
shows "p ⟶ q"
proof -
{assume "p"
hence "¬¬p" by (rule notnotI)
with assms have "¬¬q" by (rule mt)
then have "q" by (rule notnotD)}
thus "p ⟶ q" by (rule impI)
qed

（* bowma danrodcha *)
"quita la limitación de -"
lemma ejercicio_1_5:
assumes "¬q ⟶ ¬p"
shows "p ⟶ q"
proof
assume "p"
hence "¬¬p" by (rule notnotI)
with assms have "¬¬q" by (rule mt)
thus "q" by (rule notnotD)
qed

text {* ---------------------------------------------------------------
Ejercicio 2. Demostrar
¬(¬p ∧ ¬q) ⊢ p ∨ q
------------------------------------------------------------------ *}
(* marcarmor13 jeamacpov *)
--"usando un supuesto ¬p ∧ ¬q"
lemma ejercicio_2:
assumes 1: "¬(¬p ∧ ¬q)" and
2:"¬p ∧ ¬q"
shows "p ∨ q"
proof-
have 3: "p"using 1 2 by (rule notE)
show "p ∨ q" using 3 by (rule disjI1)
qed

(* ivamenjim serrodcal *)
lemma ejercicio_2_2:
assumes 1: "¬(¬p ∧ ¬q)"
shows "p ∨ q"
proof -
{assume 2:"(¬p ∧ ¬q)"
have "p" using 1 2 by (rule notE)
then have "p ∨ q" by (rule disjI1)}
thus "p ∨ q" by auto
qed

(* pablucoto *)

lemma aux_ejercicio2:
assumes "¬(p ∨ q)"
shows "¬p ∧ ¬q"
proof
{assume "p"
hence "p ∨ q" by (rule disjI1)
with `¬(p ∨ q)` have "False" by (rule notE)}
then show "¬p" ..
next
{assume "q"
hence "p ∨ q" by (rule disjI2)
with `¬(p ∨ q)` have "False" by (rule notE)}
then show "¬q" ..
qed

lemma ejercicio_2_3:
assumes "¬(¬p ∧ ¬q)"
shows "p ∨ q"
proof -
{assume 2:"¬(p ∨ q)"
hence "¬p ∧ ¬q" by (rule aux_ejercicio2)
with `¬(¬p ∧ ¬q)` have "False" ..}
then show "p ∨ q" by (rule ccontr)
qed

(* danrodcha *)
lemma ej_2:
assumes "¬(¬p ∧ ¬q)"
shows "p ∨ q"
proof -
have "¬p ∨ p" by (rule excluded_middle)
thus "p ∨ q"
proof (rule disjE)
{ assume "¬p"
have "¬q ∨ q" by (rule excluded_middle)
thus "p ∨ q"
proof (rule disjE)
{ assume "¬q"
with `¬p` have "¬p ∧ ¬q" by (rule conjI)
with assms show "p ∨ q" by (rule notE)}
next
{ assume "q"
then show "p ∨ q" by (rule disjI2)}
qed}
next
{ assume "p"
thus "p ∨ q" by (rule disjI1)}
qed
qed

(* anaprarod *)
(* Igual que el anterior pero con etiquetas *)
lemma ejercicio_2:
assumes 0: "¬(¬p ∧ ¬q)"
shows "p ∨ q"
proof-
have "¬p ∨ p" by (rule excluded_middle)
thus "p ∨ q"
proof (rule disjE)
{assume 1: "¬p"
have "¬q ∨ q" by (rule excluded_middle)
thus "p ∨ q"
proof (rule disjE)
{assume 2: "¬q"
have 3: "(¬p ∧ ¬q)" using 1 2 by (rule conjI)
have "p ∨ q" using 0 3 by (rule notE)
thus "p ∨ q" by this}
next
{assume 4: "q"
have "p ∨ q" using 4 by (rule disjI2)
thus "p ∨ q" by this}
qed}
next
{assume 5: "p"
have "p ∨ q" using 5 by (rule disjI1)
thus "p ∨ q" by this}
qed
qed

text {* ---------------------------------------------------------------
Ejercicio 3. Demostrar
¬(¬p ∨ ¬q) ⊢ p ∧ q
------------------------------------------------------------------ *}
(* marcarmor13 jeamacpov serrodcal *)
--"usando un supuesto ¬p ∨ ¬q"
lemma ejercicio_3:
assumes 1: "¬(¬p ∨ ¬q)" and
2:"¬p ∨ ¬q"
shows "p ∧ q"
proof-
have 3: "p"using 1 2 by (rule notE)
have 4: "q"using 1 2 by (rule notE)
show "p ∧ q" using 3 4 by (rule conjI)
qed

(* pablucoto *)

lemma ejercicio_3_2:
assumes "¬(¬p ∨ ¬q)"
shows "p ∧ q"
proof
have "¬¬p ∧ ¬¬q" using assms(1) by (rule aux_ejercicio2)
hence "¬¬p" by (rule conjunct1)
show "p" using `¬¬p` by (rule notnotD)
next
have "¬¬p ∧ ¬¬q" using assms(1) by (rule aux_ejercicio2)
have "¬¬q" using `¬¬p ∧ ¬¬q` by (rule conjunct2)
show "q" using `¬¬q` by (rule notnotD)
qed

(* ivamenjim *)
lemma aux: "¬(p ∨ q) ⟹ ¬p ∧ ¬q"
by auto

lemma ejercicio_3_3:
assumes 1: "¬(¬p ∨ ¬q)"
shows "p ∧ q"
proof
have 2: "¬¬p ∧ ¬¬q" using 1 by (rule aux)
have 3: "¬¬p" using 2 ..
have 4: "¬¬q" using 2 ..
show "p" using 3 by (rule notnotD)
show "q" using 4 by (rule notnotD)
qed

(* danrodcha *)
lemma ej_3:
assumes "¬(¬p ∨ ¬q)"
shows "p ∧ q"
proof (rule conjI)
{ assume "¬p"
hence "¬p ∨ ¬q" by (rule disjI1)
with assms have False by (rule notE)}
then show "p" by (rule ccontr)
{ assume "¬q"
hence "¬p ∨ ¬q" by (rule disjI2)
with assms have False by (rule notE)}
then show "q" by (rule ccontr)
qed

(* anaprarod *)
(* Igual que el anterior pero con etiquetas *)
lemma ejercicio_3:
assumes "¬(¬p ∨ ¬q)"
shows "p ∧ q"
proof (rule conjI)
{assume 1: "¬p"
hence 2: "¬p ∨ ¬q" by (rule disjI1)
have "False" using assms 2 by (rule notE)}
thus 3: "p" by (rule ccontr)
{assume 4: "¬q"
hence 5: "¬p ∨ ¬q" by (rule disjI2)
have "False" using assms 5 by (rule notE)}
thus 6: "q" by (rule ccontr)
qed

text {* ---------------------------------------------------------------
Ejercicio 4. Demostrar
¬(p ∧ q) ⊢ ¬p ∨ ¬q
------------------------------------------------------------------ *}
(* marcarmor13 jeamacpov serrodcal *)
--"usando un supuesto p ∧ q"
lemma ejercicio_4:
assumes 1: " ¬(p ∧ q)" and
2:"p ∧ q"
shows "¬p ∨ ¬q"
proof-
have 3: "¬p"using 1 2 by (rule notE)
show "¬p ∨ ¬q" using 3 by (rule disjI1)
qed

(* ivamenjim *)
lemma ejercicio_4:
assumes 1: "¬(p ∧ q)"
shows "¬p ∨ ¬q"
proof -
{assume 2:"(p ∧ q)"
have "¬p" using 1 2 by (rule notE)
then have "¬p ∨ ¬q" by (rule disjI1)}
thus "¬p ∨ ¬q" by auto
qed

( * pablucoto *)
lemma ejercicio_4:
assumes " ¬(p ∧ q)"
shows " ¬p ∨ ¬q"
proof -
{ assume 2: "¬(¬p ∨ ¬q)"
hence "p ∧ q" by (rule ejercicio_3_2)
with assms(1) have "False" .. }
then show " ¬p ∨ ¬q" by (rule ccontr)
qed

(*danrodcha anaprarod*)
lemma ej_4:
assumes "¬(p ∧ q)"
shows "¬p ∨ ¬q"
proof (rule ccontr)
assume "¬ (¬ p ∨ ¬ q)"
hence "p ∧ q" by (rule ej_3)
with assms show False by (rule notE)
qed

(* anaprarod *)
(* sin usar el ejercicio anterior *)
lemma ejercicio_4:
assumes "¬(p ∧ q)"
shows "¬p ∨ ¬q"
proof -
have "¬p ∨ p" by (rule excluded_middle)
thus "¬p ∨ ¬q"
proof (rule disjE)
{assume "¬p"
thus "¬p ∨ ¬q" by (rule disjI1)}
next
{assume 1:"p"
have "¬q ∨ q" by (rule excluded_middle)
thus "¬p ∨ ¬q"
proof (rule disjE)
{assume "¬q"
thus "¬p ∨ ¬q" by (rule disjI2)}
next
{assume 2:"q"
have 3:"p ∧ q" using 1 2 by (rule conjI)
have "¬p ∨ ¬q" using assms 3 by (rule notE)
thus "¬p ∨ ¬q" by this}
qed}
qed
qed

text {* ---------------------------------------------------------------
Ejercicio 5. Demostrar
⊢ (p ⟶ q) ∨ (q ⟶ p)
------------------------------------------------------------------ *}
(* marcarmor13 jeamacpov serrodcal *)
--"usando un supuesto q"
lemma ejercicio_5:
assumes 1: "q"

shows "(p ⟶ q) ∨ (q ⟶ p)"
proof-
have 2: "p ⟶ q" using 1 by (rule impI)
show "(p ⟶ q) ∨ (q ⟶ p)" using 2 by (rule disjI1)
qed

(* ivamenjim *)
lemma ejercicio_5:
shows "(p ⟶ q) ∨ (q ⟶ p)"
proof -
{assume 1:"q"
have "(p ⟶ q)" using 1 by (rule impI)
then have "(p ⟶ q) ∨ (q ⟶ p)" by (rule disjI1)}
thus "(p ⟶ q) ∨ (q ⟶ p)" by auto
qed

(* danrodcha pablucoto *)
lemma ej_5:
shows "(p ⟶ q) ∨ (q ⟶ p)"
proof -
have "¬p ∨ p" by (rule excluded_middle)
thus "(p ⟶ q) ∨ (q ⟶ p)"
proof (rule disjE)
{ assume "¬p"
have "¬q ∨ q" by (rule excluded_middle)
thus "(p ⟶ q) ∨ (q ⟶ p)"
proof (rule disjE)
{assume "¬q"
hence "¬p ⟶ ¬q" by (rule impI)
{ assume "q"
hence "¬¬q" by (rule notnotI)
with `¬p ⟶ ¬q` have "¬¬p" by (rule mt)
hence "p" by (rule notnotD)}
hence "q ⟶ p" by (rule impI)
thus "(p ⟶ q) ∨ (q ⟶ p)" by (rule disjI2)}
next
{assume "q"
hence "(p ⟶ q)" by (rule impI)
thus "(p ⟶ q) ∨ (q ⟶ p)" by (rule disjI1)}
qed}
next
{assume "p"
hence "q ⟶ p" by (rule impI)
thus "(p ⟶ q) ∨ (q ⟶ p)" by (rule disjI2)}
qed
qed

end
</source>

Relación 8

2017-01-14T17:33:45Z

Pablucoto:

Relación 8

2017-01-14T16:31:57Z

Pablucoto:

Relación 8

2017-01-14T15:58:52Z

Pablucoto:

Relación 8

2017-01-13T07:01:47Z

Pablucoto:

Relación 8

2017-01-13T06:56:56Z

Pablucoto:

Relación 7

2016-12-19T00:38:32Z

Pablucoto:

<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
---------------------------------------------------------------------
*}

(* fraortmoy *)
fun hojas :: "arbol => nat" where
"hojas H = Suc 0"
| "hojas (N a b) = hojas a + hojas b"

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

(* marpoldia1 anaprarod paupeddeg migtermor wilmorort pablucoto ivamenjim *)
(* Es muy parecida a la definición anterior *)
fun hojas2 :: "arbol => nat" where
"hojas2 H = 1" |
"hojas2 (N i d) = hojas2 i + hojas2 d"

value "hojas2 (N (N H H) (N H H)) = 4"

(* anaprarod *)
(* Equivalencia de las definiciones *)
lemma "hojas a = hojas2 a"
by (induct a) simp_all

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
---------------------------------------------------------------------
*}

(* fraortmoy marpoldia1 anaprarod migtermor wilmorort*)
fun profundidad :: "arbol => nat" where
"profundidad H = 0"
| "profundidad (N a b) = (if profundidad a > profundidad b
then 1 + profundidad a
else 1 + profundidad b)"

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

(* anaprarod wilmorort pablucoto ivamenjim *)
fun profundidad2 :: "arbol => nat" where
"profundidad2 H = 0"
|"profundidad2 (N i d) = 1 + (max (profundidad2 i)(profundidad2 d))"

value "profundidad2 (N (N H H) (N H H)) = 2"

(* anaprarod *)
(* Equivalencia de las definiciones *)
lemma "profundidad a= profundidad2 a"
by (induct a) auto

(* paupeddeg *)
fun maximo :: "nat × nat => nat" where
"maximo (a,b) = (if a > b
then a else b)"

fun profundidad :: "arbol => nat" where
"profundidad H = 0"
| "profundidad (N i d) = 1 + maximo(profundidad i, profundidad d)"

(* ivamenjim: llamando a la función anterior profundidad3 *)

lemma "profundidad a = profundidad3 a"
by (induct a) auto

(* ivamenjim *)
fun profundidad4 :: "arbol => nat" where
"profundidad4 H = 0"
| "profundidad4 (N i d) = Suc (max (profundidad4 i)(profundidad4 d))"

lemma "profundidad a = profundidad4 a"
by (induct a) auto

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))
---------------------------------------------------------------------
*}
(* fraortmoy marpoldia1 anaprarod paupeddeg migtermor wilmorort*)
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))"

(* ivamenjim pablucoto*)
fun abc2 :: "nat ⇒ arbol" where
"abc2 0 = H"
| "abc2 t = N (abc2 (t-1)) (abc2 (t-1))"

value "abc2 3 = N (N (N H H) (N H H)) (N (N H H) (N H H))"

(* ivamenjim: Metaejercicio de demostración *)
lemma "abc t = abc2 t"
by (induct t) auto

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.
---------------------------------------------------------------------
*}

(* fraortmoy anaprarod migtermor *)
fun es_abc :: "(arbol => 'a) => arbol => bool" where
"es_abc _ H = True"
| "es_abc f (N a b) = (es_abc f a ∧ es_abc f b ∧ (f a = f b))"

(* marpoldia1 paupeddeg ivamenjim pablucoto*)
fun es_abc2 :: "(arbol => 'a) => arbol => bool" where
"es_abc2 f H = True" |
"es_abc2 f (N i d) = ((f i = f d) ∧ (es_abc2 f i) ∧ (es_abc2 f d))"

(* anaprarod *)
(* Equivalencia de las definiciones *)
lemma "es_abc f a = es_abc2 f a"
by (induct a) auto

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.
---------------------------------------------------------------------
*}

(* fraortmoy marpoldia1 migtermor wilmorort pablucoto*)

lemma abc_prof_num_hojas:
assumes "es_abc profundidad a"
shows "hojas a = 2^(profundidad a)"
using assms
by (induct a) auto

(* Otra forma de escribir lo mismo *)
(* anaprarod *)
lemma AUX7: "es_abc profundidad a ⟶ (hojas a = 2^(profundidad a))"
by (induct a) simp_all

(* fraortmoy marpoldia1 migtermor anaprarod wilmorort *)
(* También funciona con AUX7 *)

lemma lej7: "es_abc profundidad a = es_abc hojas a"
by (induct a) (auto simp add: abc_prof_num_hojas)

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

(* fraortmoy marpoldia1 migtermor wilmorort pablucoto*)

lemma abc_hojas_num_nodos:
assumes "es_abc hojas a"
shows "Suc(size a) = hojas a"
using assms
by (induct a) auto

(* Otra forma de escribir lo mismo *)
(* anaprarod *)
lemma AUX8: "es_abc hojas a ⟶ (hojas a = (Suc (size a)))"
by (induct a) auto

(* fraortmoy marpoldia1 migtermor anaprarod wilmorort pablucoto*)

lemma lej8: "es_abc hojas a = es_abc size a"
by (induct a) (auto simp add:abc_hojas_num_nodos [symmetric])

(* anaprarod *)
(* Usando AUX8 *)
lemma L8: "es_abc hojas a= es_abc size a"
by (induct a) (auto simp add: AUX8)

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

(* fraortmoy marpoldia1 migtermor anaprarod wilmorort pablucoto*)

lemma lej9: "es_abc profundidad a = es_abc size a"
by (simp add: lej7 lej8)

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

(* fraortmoy marpoldia1 migtermor wilmorort pablucoto*)

lemma lej10: "es_abc profundidad (abc n)"
by (induct n) auto

(* anaprarod *)
(* con un demostrador más débil *)
lemma L10: "es_abc f (abc a)"
by (induct a) simp_all

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

(* fraortmoy marpoldia1 migtermor wilmorort pablucoto*)

lemma lej11:
assumes " es_abc profundidad a"
shows "a = (abc (profundidad a))"
using assms
by (induct a) auto

(* Otra forma de escribir lo mismo *)
(* anaprarod *)
lemma "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).
---------------------------------------------------------------------
*}

(* migtermor *)
fun medida_nula :: "arbol => nat" where
"medida_nula H = 0"
| "medida_nula (N i d) = 0"

lemma "es_abc medida_nula a = es_abc size a"
quickcheck
oops
(* Quickcheck encuentra el siguiente contraejemplo:
a= N H (N H H)
Tras evaluar:
es_abc medida_nula a = True
es_abc size a = False*)

(* anaprarod wilmorort pablucoto*)
lemma "es_abc f a = es_abc size a"
quickcheck
(* Quickcheck found a counterexample:
f = λx. a⇩1
a = N H (N H H)
Evaluated terms:
es_abc f a = True
es_abc size a = False *)
oops

end
</source>

Relación 7

2016-12-19T00:19:12Z

Pablucoto:

<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
---------------------------------------------------------------------
*}

(* fraortmoy *)
fun hojas :: "arbol => nat" where
"hojas H = Suc 0"
| "hojas (N a b) = hojas a + hojas b"

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

(* marpoldia1 anaprarod paupeddeg migtermor wilmorort pablucoto ivamenjim *)
(* Es muy parecida a la definición anterior *)
fun hojas2 :: "arbol => nat" where
"hojas2 H = 1" |
"hojas2 (N i d) = hojas2 i + hojas2 d"

value "hojas2 (N (N H H) (N H H)) = 4"

(* anaprarod *)
(* Equivalencia de las definiciones *)
lemma "hojas a = hojas2 a"
by (induct a) simp_all

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
---------------------------------------------------------------------
*}

(* fraortmoy marpoldia1 anaprarod migtermor wilmorort*)
fun profundidad :: "arbol => nat" where
"profundidad H = 0"
| "profundidad (N a b) = (if profundidad a > profundidad b
then 1 + profundidad a
else 1 + profundidad b)"

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

(* anaprarod wilmorort pablucoto ivamenjim *)
fun profundidad2 :: "arbol => nat" where
"profundidad2 H = 0"
|"profundidad2 (N i d) = 1 + (max (profundidad2 i)(profundidad2 d))"

value "profundidad2 (N (N H H) (N H H)) = 2"

(* anaprarod *)
(* Equivalencia de las definiciones *)
lemma "profundidad a= profundidad2 a"
by (induct a) auto

(* paupeddeg *)
fun maximo :: "nat × nat => nat" where
"maximo (a,b) = (if a > b
then a else b)"

fun profundidad :: "arbol => nat" where
"profundidad H = 0"
| "profundidad (N i d) = 1 + maximo(profundidad i, profundidad d)"

(* ivamenjim: llamando a la función anterior profundidad3 *)

lemma "profundidad a = profundidad3 a"
by (induct a) auto

(* ivamenjim *)
fun profundidad4 :: "arbol => nat" where
"profundidad4 H = 0"
| "profundidad4 (N i d) = Suc (max (profundidad4 i)(profundidad4 d))"

lemma "profundidad a = profundidad4 a"
by (induct a) auto

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))
---------------------------------------------------------------------
*}
(* fraortmoy marpoldia1 anaprarod paupeddeg migtermor wilmorort*)
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))"

(* ivamenjim pablucoto*)
fun abc2 :: "nat ⇒ arbol" where
"abc2 0 = H"
| "abc2 t = N (abc2 (t-1)) (abc2 (t-1))"

value "abc2 3 = N (N (N H H) (N H H)) (N (N H H) (N H H))"

(* ivamenjim: Metaejercicio de demostración *)
lemma "abc t = abc2 t"
by (induct t) auto

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.
---------------------------------------------------------------------
*}

(* fraortmoy anaprarod migtermor *)
fun es_abc :: "(arbol => 'a) => arbol => bool" where
"es_abc _ H = True"
| "es_abc f (N a b) = (es_abc f a ∧ es_abc f b ∧ (f a = f b))"

(* marpoldia1 paupeddeg ivamenjim pablucoto*)
fun es_abc2 :: "(arbol => 'a) => arbol => bool" where
"es_abc2 f H = True" |
"es_abc2 f (N i d) = ((f i = f d) ∧ (es_abc2 f i) ∧ (es_abc2 f d))"

(* anaprarod *)
(* Equivalencia de las definiciones *)
lemma "es_abc f a = es_abc2 f a"
by (induct a) auto

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.
---------------------------------------------------------------------
*}

(* fraortmoy marpoldia1 migtermor wilmorort pablucoto*)

lemma abc_prof_num_hojas:
assumes "es_abc profundidad a"
shows "hojas a = 2^(profundidad a)"
using assms
by (induct a) auto

(* Otra forma de escribir lo mismo *)
(* anaprarod *)
lemma AUX7: "es_abc profundidad a ⟶ (hojas a = 2^(profundidad a))"
by (induct a) simp_all

(* fraortmoy marpoldia1 migtermor anaprarod wilmorort *)
(* También funciona con AUX7 *)

lemma lej7: "es_abc profundidad a = es_abc hojas a"
by (induct a) (auto simp add: abc_prof_num_hojas)

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

(* fraortmoy marpoldia1 migtermor wilmorort*)

lemma abc_hojas_num_nodos:
assumes "es_abc hojas a"
shows "Suc(size a) = hojas a"
using assms
by (induct a) auto

(* Otra forma de escribir lo mismo *)
(* anaprarod *)
lemma AUX8: "es_abc hojas a ⟶ (hojas a = (Suc (size a)))"
by (induct a) auto

(* fraortmoy marpoldia1 migtermor anaprarod wilmorort*)

lemma lej8: "es_abc hojas a = es_abc size a"
by (induct a) (auto simp add:abc_hojas_num_nodos [symmetric])

(* anaprarod *)
(* Usando AUX8 *)
lemma L8: "es_abc hojas a= es_abc size a"
by (induct a) (auto simp add: AUX8)

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

(* fraortmoy marpoldia1 migtermor anaprarod wilmorort*)

lemma lej9: "es_abc profundidad a = es_abc size a"
by (simp add: lej7 lej8)

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

(* fraortmoy marpoldia1 migtermor wilmorort *)

lemma lej10: "es_abc profundidad (abc n)"
by (induct n) auto

(* anaprarod *)
(* con un demostrador más débil *)
lemma L10: "es_abc f (abc a)"
by (induct a) simp_all

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

(* fraortmoy marpoldia1 migtermor wilmorort *)

lemma lej11:
assumes " es_abc profundidad a"
shows "a = (abc (profundidad a))"
using assms
by (induct a) auto

(* Otra forma de escribir lo mismo *)
(* anaprarod *)
lemma "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).
---------------------------------------------------------------------
*}

(* migtermor *)
fun medida_nula :: "arbol => nat" where
"medida_nula H = 0"
| "medida_nula (N i d) = 0"

lemma "es_abc medida_nula a = es_abc size a"
quickcheck
oops
(* Quickcheck encuentra el siguiente contraejemplo:
a= N H (N H H)
Tras evaluar:
es_abc medida_nula a = True
es_abc size a = False*)

(* anaprarod wilmorort*)
lemma "es_abc f a = es_abc size a"
quickcheck
(* Quickcheck found a counterexample:
f = λx. a⇩1
a = N H (N H H)
Evaluated terms:
es_abc f a = True
es_abc size a = False *)
oops

end
</source>

Relación 7

2016-12-18T23:13:59Z

Pablucoto:

<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
---------------------------------------------------------------------
*}

(* fraortmoy *)
fun hojas :: "arbol => nat" where
"hojas H = Suc 0"
| "hojas (N a b) = hojas a + hojas b"

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

(* marpoldia1 anaprarod paupeddeg migtermor wilmorort pablucoto ivamenjim *)
(* Es muy parecida a la definición anterior *)
fun hojas2 :: "arbol => nat" where
"hojas2 H = 1" |
"hojas2 (N i d) = hojas2 i + hojas2 d"

value "hojas2 (N (N H H) (N H H)) = 4"

(* anaprarod *)
(* Equivalencia de las definiciones *)
lemma "hojas a = hojas2 a"
by (induct a) simp_all

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
---------------------------------------------------------------------
*}

(* fraortmoy marpoldia1 anaprarod migtermor wilmorort*)
fun profundidad :: "arbol => nat" where
"profundidad H = 0"
| "profundidad (N a b) = (if profundidad a > profundidad b
then 1 + profundidad a
else 1 + profundidad b)"

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

(* anaprarod wilmorort pablucoto ivamenjim *)
fun profundidad2 :: "arbol => nat" where
"profundidad2 H = 0"
|"profundidad2 (N i d) = 1 + (max (profundidad2 i)(profundidad2 d))"

value "profundidad2 (N (N H H) (N H H)) = 2"

(* anaprarod *)
(* Equivalencia de las definiciones *)
lemma "profundidad a= profundidad2 a"
by (induct a) auto

(* paupeddeg *)
fun maximo :: "nat × nat => nat" where
"maximo (a,b) = (if a > b
then a else b)"

fun profundidad :: "arbol => nat" where
"profundidad H = 0"
| "profundidad (N i d) = 1 + maximo(profundidad i, profundidad d)"

(* ivamenjim: llamando a la función anterior profundidad3 *)

lemma "profundidad a = profundidad3 a"
by (induct a) auto

(* ivamenjim *)
fun profundidad4 :: "arbol => nat" where
"profundidad4 H = 0"
| "profundidad4 (N i d) = Suc (max (profundidad4 i)(profundidad4 d))"

lemma "profundidad a = profundidad4 a"
by (induct a) auto

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))
---------------------------------------------------------------------
*}
(* fraortmoy marpoldia1 anaprarod paupeddeg migtermor wilmorort*)
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))"

(* ivamenjim pablucoto*)
fun abc2 :: "nat ⇒ arbol" where
"abc2 0 = H"
| "abc2 t = N (abc2 (t-1)) (abc2 (t-1))"

value "abc2 3 = N (N (N H H) (N H H)) (N (N H H) (N H H))"

(* ivamenjim: Metaejercicio de demostración *)
lemma "abc t = abc2 t"
by (induct t) auto

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.
---------------------------------------------------------------------
*}

(* fraortmoy anaprarod migtermor *)
fun es_abc :: "(arbol => 'a) => arbol => bool" where
"es_abc _ H = True"
| "es_abc f (N a b) = (es_abc f a ∧ es_abc f b ∧ (f a = f b))"

(* marpoldia1 paupeddeg ivamenjim pablucoto*)
fun es_abc2 :: "(arbol => 'a) => arbol => bool" where
"es_abc2 f H = True" |
"es_abc2 f (N i d) = ((f i = f d) ∧ (es_abc2 f i) ∧ (es_abc2 f d))"

(* anaprarod *)
(* Equivalencia de las definiciones *)
lemma "es_abc f a = es_abc2 f a"
by (induct a) auto

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.
---------------------------------------------------------------------
*}

(* fraortmoy marpoldia1 migtermor wilmorort *)

lemma abc_prof_num_hojas:
assumes "es_abc profundidad a"
shows "hojas a = 2^(profundidad a)"
using assms
by (induct a) auto

(* Otra forma de escribir lo mismo *)
(* anaprarod *)
lemma AUX7: "es_abc profundidad a ⟶ (hojas a = 2^(profundidad a))"
by (induct a) simp_all

(* fraortmoy marpoldia1 migtermor anaprarod wilmorort *)
(* También funciona con AUX7 *)

lemma lej7: "es_abc profundidad a = es_abc hojas a"
by (induct a) (auto simp add: abc_prof_num_hojas)

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

(* fraortmoy marpoldia1 migtermor wilmorort*)

lemma abc_hojas_num_nodos:
assumes "es_abc hojas a"
shows "Suc(size a) = hojas a"
using assms
by (induct a) auto

(* Otra forma de escribir lo mismo *)
(* anaprarod *)
lemma AUX8: "es_abc hojas a ⟶ (hojas a = (Suc (size a)))"
by (induct a) auto

(* fraortmoy marpoldia1 migtermor anaprarod wilmorort*)

lemma lej8: "es_abc hojas a = es_abc size a"
by (induct a) (auto simp add:abc_hojas_num_nodos [symmetric])

(* anaprarod *)
(* Usando AUX8 *)
lemma L8: "es_abc hojas a= es_abc size a"
by (induct a) (auto simp add: AUX8)

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

(* fraortmoy marpoldia1 migtermor anaprarod wilmorort*)

lemma lej9: "es_abc profundidad a = es_abc size a"
by (simp add: lej7 lej8)

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

(* fraortmoy marpoldia1 migtermor wilmorort *)

lemma lej10: "es_abc profundidad (abc n)"
by (induct n) auto

(* anaprarod *)
(* con un demostrador más débil *)
lemma L10: "es_abc f (abc a)"
by (induct a) simp_all

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

(* fraortmoy marpoldia1 migtermor wilmorort *)

lemma lej11:
assumes " es_abc profundidad a"
shows "a = (abc (profundidad a))"
using assms
by (induct a) auto

(* Otra forma de escribir lo mismo *)
(* anaprarod *)
lemma "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).
---------------------------------------------------------------------
*}

(* migtermor *)
fun medida_nula :: "arbol => nat" where
"medida_nula H = 0"
| "medida_nula (N i d) = 0"

lemma "es_abc medida_nula a = es_abc size a"
quickcheck
oops
(* Quickcheck encuentra el siguiente contraejemplo:
a= N H (N H H)
Tras evaluar:
es_abc medida_nula a = True
es_abc size a = False*)

(* anaprarod wilmorort*)
lemma "es_abc f a = es_abc size a"
quickcheck
(* Quickcheck found a counterexample:
f = λx. a⇩1
a = N H (N H H)
Evaluated terms:
es_abc f a = True
es_abc size a = False *)
oops

end
</source>

Relación 7

2016-12-18T23:02:06Z

Pablucoto:

<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
---------------------------------------------------------------------
*}

(* fraortmoy *)
fun hojas :: "arbol => nat" where
"hojas H = Suc 0"
| "hojas (N a b) = hojas a + hojas b"

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

(* marpoldia1 anaprarod paupeddeg migtermor wilmorort pablucoto ivamenjim *)
(* Es muy parecida a la definición anterior *)
fun hojas2 :: "arbol => nat" where
"hojas2 H = 1" |
"hojas2 (N i d) = hojas2 i + hojas2 d"

value "hojas2 (N (N H H) (N H H)) = 4"

(* anaprarod *)
(* Equivalencia de las definiciones *)
lemma "hojas a = hojas2 a"
by (induct a) simp_all

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
---------------------------------------------------------------------
*}

(* fraortmoy marpoldia1 anaprarod migtermor wilmorort*)
fun profundidad :: "arbol => nat" where
"profundidad H = 0"
| "profundidad (N a b) = (if profundidad a > profundidad b
then 1 + profundidad a
else 1 + profundidad b)"

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

(* anaprarod wilmorort pablucoto ivamenjim *)
fun profundidad2 :: "arbol => nat" where
"profundidad2 H = 0"
|"profundidad2 (N i d) = 1 + (max (profundidad2 i)(profundidad2 d))"

value "profundidad2 (N (N H H) (N H H)) = 2"

(* anaprarod *)
(* Equivalencia de las definiciones *)
lemma "profundidad a= profundidad2 a"
by (induct a) auto

(* paupeddeg *)
fun maximo :: "nat × nat => nat" where
"maximo (a,b) = (if a > b
then a else b)"

fun profundidad :: "arbol => nat" where
"profundidad H = 0"
| "profundidad (N i d) = 1 + maximo(profundidad i, profundidad d)"

(* ivamenjim: llamando a la función anterior profundidad3 *)

lemma "profundidad a = profundidad3 a"
by (induct a) auto

(* ivamenjim *)
fun profundidad4 :: "arbol => nat" where
"profundidad4 H = 0"
| "profundidad4 (N i d) = Suc (max (profundidad4 i)(profundidad4 d))"

lemma "profundidad a = profundidad4 a"
by (induct a) auto

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))
---------------------------------------------------------------------
*}
(* fraortmoy marpoldia1 anaprarod paupeddeg migtermor wilmorort*)
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))"

(* ivamenjim pablucoto*)
fun abc2 :: "nat ⇒ arbol" where
"abc2 0 = H"
| "abc2 t = N (abc2 (t-1)) (abc2 (t-1))"

value "abc2 3 = N (N (N H H) (N H H)) (N (N H H) (N H H))"

(* ivamenjim: Metaejercicio de demostración *)
lemma "abc t = abc2 t"
by (induct t) auto

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.
---------------------------------------------------------------------
*}

(* fraortmoy anaprarod migtermor *)
fun es_abc :: "(arbol => 'a) => arbol => bool" where
"es_abc _ H = True"
| "es_abc f (N a b) = (es_abc f a ∧ es_abc f b ∧ (f a = f b))"

(* marpoldia1 paupeddeg *)
fun es_abc2 :: "(arbol => 'a) => arbol => bool" where
"es_abc2 f H = True" |
"es_abc2 f (N i d) = ((f i = f d) ∧ (es_abc2 f i) ∧ (es_abc2 f d))"

(* anaprarod *)
(* Equivalencia de las definiciones *)
lemma "es_abc f a = es_abc2 f a"
by (induct a) auto

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.
---------------------------------------------------------------------
*}

(* fraortmoy marpoldia1 migtermor wilmorort *)

lemma abc_prof_num_hojas:
assumes "es_abc profundidad a"
shows "hojas a = 2^(profundidad a)"
using assms
by (induct a) auto

(* Otra forma de escribir lo mismo *)
(* anaprarod *)
lemma AUX7: "es_abc profundidad a ⟶ (hojas a = 2^(profundidad a))"
by (induct a) simp_all

(* fraortmoy marpoldia1 migtermor anaprarod wilmorort *)
(* También funciona con AUX7 *)

lemma lej7: "es_abc profundidad a = es_abc hojas a"
by (induct a) (auto simp add: abc_prof_num_hojas)

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

(* fraortmoy marpoldia1 migtermor wilmorort*)

lemma abc_hojas_num_nodos:
assumes "es_abc hojas a"
shows "Suc(size a) = hojas a"
using assms
by (induct a) auto

(* Otra forma de escribir lo mismo *)
(* anaprarod *)
lemma AUX8: "es_abc hojas a ⟶ (hojas a = (Suc (size a)))"
by (induct a) auto

(* fraortmoy marpoldia1 migtermor anaprarod wilmorort*)

lemma lej8: "es_abc hojas a = es_abc size a"
by (induct a) (auto simp add:abc_hojas_num_nodos [symmetric])

(* anaprarod *)
(* Usando AUX8 *)
lemma L8: "es_abc hojas a= es_abc size a"
by (induct a) (auto simp add: AUX8)

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

(* fraortmoy marpoldia1 migtermor anaprarod wilmorort*)

lemma lej9: "es_abc profundidad a = es_abc size a"
by (simp add: lej7 lej8)

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

(* fraortmoy marpoldia1 migtermor wilmorort *)

lemma lej10: "es_abc profundidad (abc n)"
by (induct n) auto

(* anaprarod *)
(* con un demostrador más débil *)
lemma L10: "es_abc f (abc a)"
by (induct a) simp_all

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

(* fraortmoy marpoldia1 migtermor wilmorort *)

lemma lej11:
assumes " es_abc profundidad a"
shows "a = (abc (profundidad a))"
using assms
by (induct a) auto

(* Otra forma de escribir lo mismo *)
(* anaprarod *)
lemma "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).
---------------------------------------------------------------------
*}

(* migtermor *)
fun medida_nula :: "arbol => nat" where
"medida_nula H = 0"
| "medida_nula (N i d) = 0"

lemma "es_abc medida_nula a = es_abc size a"
quickcheck
oops
(* Quickcheck encuentra el siguiente contraejemplo:
a= N H (N H H)
Tras evaluar:
es_abc medida_nula a = True
es_abc size a = False*)

(* anaprarod wilmorort*)
lemma "es_abc f a = es_abc size a"
quickcheck
(* Quickcheck found a counterexample:
f = λx. a⇩1
a = N H (N H H)
Evaluated terms:
es_abc f a = True
es_abc size a = False *)
oops

end
</source>

Relación 7

2016-12-18T22:25:11Z

Pablucoto:

<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
---------------------------------------------------------------------
*}

(* fraortmoy *)
fun hojas :: "arbol => nat" where
"hojas H = Suc 0"
| "hojas (N a b) = hojas a + hojas b"

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

(* marpoldia1 anaprarod paupeddeg migtermor wilmorort pablucoto ivamenjim *)
(* Es muy parecida a la definición anterior *)
fun hojas2 :: "arbol => nat" where
"hojas2 H = 1" |
"hojas2 (N i d) = hojas2 i + hojas2 d"

value "hojas2 (N (N H H) (N H H)) = 4"

(* anaprarod *)
(* Equivalencia de las definiciones *)
lemma "hojas a = hojas2 a"
by (induct a) simp_all

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
---------------------------------------------------------------------
*}

(* fraortmoy marpoldia1 anaprarod migtermor wilmorort*)
fun profundidad :: "arbol => nat" where
"profundidad H = 0"
| "profundidad (N a b) = (if profundidad a > profundidad b
then 1 + profundidad a
else 1 + profundidad b)"

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

(* anaprarod wilmorort pablucoto*)
fun profundidad2 :: "arbol => nat" where
"profundidad2 H = 0"
|"profundidad2 (N i d) = 1 + (max (profundidad2 i)(profundidad2 d))"

value "profundidad2 (N (N H H) (N H H)) = 2"

(* anaprarod *)
(* Equivalencia de las definiciones *)
lemma "profundidad a= profundidad2 a"
by (induct a) auto

(* paupeddeg *)
fun maximo :: "nat × nat => nat" where
"maximo (a,b) = (if a > b
then a else b)"

fun profundidad :: "arbol => nat" where
"profundidad H = 0"
| "profundidad (N i d) = 1 + maximo(profundidad i, profundidad d)"

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))
---------------------------------------------------------------------
*}
(* fraortmoy marpoldia1 anaprarod paupeddeg migtermor wilmorort*)
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.
---------------------------------------------------------------------
*}

(* fraortmoy anaprarod migtermor *)
fun es_abc :: "(arbol => 'a) => arbol => bool" where
"es_abc _ H = True"
| "es_abc f (N a b) = (es_abc f a ∧ es_abc f b ∧ (f a = f b))"

(* marpoldia1 paupeddeg *)
fun es_abc2 :: "(arbol => 'a) => arbol => bool" where
"es_abc2 f H = True" |
"es_abc2 f (N i d) = ((f i = f d) ∧ (es_abc2 f i) ∧ (es_abc2 f d))"

(* anaprarod *)
(* Equivalencia de las definiciones *)
lemma "es_abc f a = es_abc2 f a"
by (induct a) auto

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.
---------------------------------------------------------------------
*}

(* fraortmoy marpoldia1 migtermor wilmorort *)

lemma abc_prof_num_hojas:
assumes "es_abc profundidad a"
shows "hojas a = 2^(profundidad a)"
using assms
by (induct a) auto

(* Otra forma de escribir lo mismo *)
(* anaprarod *)
lemma AUX7: "es_abc profundidad a ⟶ (hojas a = 2^(profundidad a))"
by (induct a) simp_all

(* fraortmoy marpoldia1 migtermor anaprarod wilmorort *)
(* También funciona con AUX7 *)

lemma lej7: "es_abc profundidad a = es_abc hojas a"
by (induct a) (auto simp add: abc_prof_num_hojas)

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

(* fraortmoy marpoldia1 migtermor wilmorort*)

lemma abc_hojas_num_nodos:
assumes "es_abc hojas a"
shows "Suc(size a) = hojas a"
using assms
by (induct a) auto

(* Otra forma de escribir lo mismo *)
(* anaprarod *)
lemma AUX8: "es_abc hojas a ⟶ (hojas a = (Suc (size a)))"
by (induct a) auto

(* fraortmoy marpoldia1 migtermor anaprarod wilmorort*)

lemma lej8: "es_abc hojas a = es_abc size a"
by (induct a) (auto simp add:abc_hojas_num_nodos [symmetric])

(* anaprarod *)
(* Usando AUX8 *)
lemma L8: "es_abc hojas a= es_abc size a"
by (induct a) (auto simp add: AUX8)

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

(* fraortmoy marpoldia1 migtermor anaprarod wilmorort*)

lemma lej9: "es_abc profundidad a = es_abc size a"
by (simp add: lej7 lej8)

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

(* fraortmoy marpoldia1 migtermor *)

lemma lej10: "es_abc profundidad (abc n)"
by (induct n) auto

(* anaprarod *)
(* con un demostrador más débil *)
lemma L10: "es_abc f (abc a)"
by (induct a) simp_all

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

(* fraortmoy marpoldia1 migtermor *)

lemma lej11:
assumes " es_abc profundidad a"
shows "a = (abc (profundidad a))"
using assms
by (induct a) auto

(* Otra forma de escribir lo mismo *)
(* anaprarod *)
lemma "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).
---------------------------------------------------------------------
*}

(* migtermor *)
fun medida_nula :: "arbol => nat" where
"medida_nula H = 0"
| "medida_nula (N i d) = 0"

lemma "es_abc medida_nula a = es_abc size a"
quickcheck
oops
(* Quickcheck encuentra el siguiente contraejemplo:
a= N H (N H H)
Tras evaluar:
es_abc medida_nula a = True
es_abc size a = False*)

(* anaprarod wilmorort*)
lemma "es_abc f a = es_abc size a"
quickcheck
(* Quickcheck found a counterexample:
f = λx. a⇩1
a = N H (N H H)
Evaluated terms:
es_abc f a = True
es_abc size a = False *)
oops

end
</source>

Relación 7

2016-12-18T22:00:24Z

Pablucoto:

<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
---------------------------------------------------------------------
*}

(* fraortmoy *)
fun hojas :: "arbol => nat" where
"hojas H = Suc 0"
| "hojas (N a b) = hojas a + hojas b"

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

(* marpoldia1 anaprarod paupeddeg migtermor wilmorort pablucoto*)
(* Es muy parecida a la definición anterior *)
fun hojas2 :: "arbol => nat" where
"hojas2 H = 1" |
"hojas2 (N i d) = hojas2 i + hojas2 d"

value "hojas2 (N (N H H) (N H H)) = 4"

(* anaprarod *)
(* Equivalencia de las definiciones *)
lemma "hojas a = hojas2 a"
by (induct a) simp_all

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
---------------------------------------------------------------------
*}

(* fraortmoy marpoldia1 anaprarod migtermor wilmorort*)
fun profundidad :: "arbol => nat" where
"profundidad H = 0"
| "profundidad (N a b) = (if profundidad a > profundidad b
then 1 + profundidad a
else 1 + profundidad b)"

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

(* anaprarod wilmorort*)
fun profundidad2 :: "arbol => nat" where
"profundidad2 H = 0"
|"profundidad2 (N i d) = 1 + (max (profundidad2 i)(profundidad2 d))"

value "profundidad2 (N (N H H) (N H H)) = 2"

(* anaprarod *)
(* Equivalencia de las definiciones *)
lemma "profundidad a= profundidad2 a"
by (induct a) auto

(* paupeddeg *)
fun maximo :: "nat × nat => nat" where
"maximo (a,b) = (if a > b
then a else b)"

fun profundidad :: "arbol => nat" where
"profundidad H = 0"
| "profundidad (N i d) = 1 + maximo(profundidad i, profundidad d)"

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))
---------------------------------------------------------------------
*}
(* fraortmoy marpoldia1 anaprarod paupeddeg migtermor wilmorort*)
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.
---------------------------------------------------------------------
*}

(* fraortmoy anaprarod migtermor *)
fun es_abc :: "(arbol => 'a) => arbol => bool" where
"es_abc _ H = True"
| "es_abc f (N a b) = (es_abc f a ∧ es_abc f b ∧ (f a = f b))"

(* marpoldia1 paupeddeg *)
fun es_abc2 :: "(arbol => 'a) => arbol => bool" where
"es_abc2 f H = True" |
"es_abc2 f (N i d) = ((f i = f d) ∧ (es_abc2 f i) ∧ (es_abc2 f d))"

(* anaprarod *)
(* Equivalencia de las definiciones *)
lemma "es_abc f a = es_abc2 f a"
by (induct a) auto

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.
---------------------------------------------------------------------
*}

(* fraortmoy marpoldia1 migtermor wilmorort *)

lemma abc_prof_num_hojas:
assumes "es_abc profundidad a"
shows "hojas a = 2^(profundidad a)"
using assms
by (induct a) auto

(* Otra forma de escribir lo mismo *)
(* anaprarod *)
lemma AUX7: "es_abc profundidad a ⟶ (hojas a = 2^(profundidad a))"
by (induct a) simp_all

(* fraortmoy marpoldia1 migtermor anaprarod wilmorort *)
(* También funciona con AUX7 *)

lemma lej7: "es_abc profundidad a = es_abc hojas a"
by (induct a) (auto simp add: abc_prof_num_hojas)

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

(* fraortmoy marpoldia1 migtermor wilmorort*)

lemma abc_hojas_num_nodos:
assumes "es_abc hojas a"
shows "Suc(size a) = hojas a"
using assms
by (induct a) auto

(* Otra forma de escribir lo mismo *)
(* anaprarod *)
lemma AUX8: "es_abc hojas a ⟶ (hojas a = (Suc (size a)))"
by (induct a) auto

(* fraortmoy marpoldia1 migtermor anaprarod wilmorort*)

lemma lej8: "es_abc hojas a = es_abc size a"
by (induct a) (auto simp add:abc_hojas_num_nodos [symmetric])

(* anaprarod *)
(* Usando AUX8 *)
lemma L8: "es_abc hojas a= es_abc size a"
by (induct a) (auto simp add: AUX8)

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

(* fraortmoy marpoldia1 migtermor anaprarod *)

lemma lej9: "es_abc profundidad a = es_abc size a"
by (simp add: lej7 lej8)

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

(* fraortmoy marpoldia1 migtermor *)

lemma lej10: "es_abc profundidad (abc n)"
by (induct n) auto

(* anaprarod *)
(* con un demostrador más débil *)
lemma L10: "es_abc f (abc a)"
by (induct a) simp_all

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

(* fraortmoy marpoldia1 migtermor *)

lemma lej11:
assumes " es_abc profundidad a"
shows "a = (abc (profundidad a))"
using assms
by (induct a) auto

(* Otra forma de escribir lo mismo *)
(* anaprarod *)
lemma "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).
---------------------------------------------------------------------
*}

(* migtermor *)
fun medida_nula :: "arbol => nat" where
"medida_nula H = 0"
| "medida_nula (N i d) = 0"

lemma "es_abc medida_nula a = es_abc size a"
quickcheck
oops
(* Quickcheck encuentra el siguiente contraejemplo:
a= N H (N H H)
Tras evaluar:
es_abc medida_nula a = True
es_abc size a = False*)

(* anaprarod wilmorort*)
lemma "es_abc f a = es_abc size a"
quickcheck
(* Quickcheck found a counterexample:
f = λx. a⇩1
a = N H (N H H)
Evaluated terms:
es_abc f a = True
es_abc size a = False *)
oops

end
</source>

Relación 6

2016-12-04T03:47:37Z

Pablucoto:

<source lang="isar">
chapter {* R6: Recorridos de árboles *}

theory R6_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))".
---------------------------------------------------------------------
*}

(* ivamenjim marpoldia1 manmorjim1 bowma *)

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]
---------------------------------------------------------------------
*}

(* ivamenjim marpoldia1 pablucoto bowma*)

fun preOrden :: "'a arbol ⇒ 'a list" where
"preOrden (H t) = [t]"
| "preOrden (N t i d) = [t] @ (preOrden i) @ (preOrden d)"

(* danrodcha crigomgom manmorjim1 bowma *)
fun preOrden1 :: "'a arbol ⇒ 'a list" where
"preOrden1 (H x) = [x]"
| "preOrden1 (N x i d) = x#preOrden1 i @ preOrden1 d"

value "preOrden (N e (N c (H a) (H d)) (N g (H f) (H h)))
= [e,c,a,d,g,f,h]"
value "preOrden1 (N e (N c (H a) (H d)) (N g (H f) (H h)))
= [e,c,a,d,g,f,h]"

(* danrodcha *)
lemma "preOrden a = preOrden1 a"
by (induct a) simp_all

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]
---------------------------------------------------------------------
*}

(* ivamenjim danrodcha crigomgom marpoldia1 manmorjim1 pablucoto bowma*)

fun postOrden :: "'a arbol ⇒ 'a list" where
"postOrden (H t) = [t]"
| "postOrden (N t i d) = (postOrden i) @ (postOrden d) @ [t]"

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]
---------------------------------------------------------------------
*}

(* ivamenjim crigomgom marpoldia1 pablucoto bowma *)

fun inOrden :: "'a arbol ⇒ 'a list" where
"inOrden (H t) = [t]"
| "inOrden (N t i d) = (inOrden i) @ [t] @ (inOrden d)"

(* danrodcha manmorjim1 *)
fun inOrden1 :: "'a arbol ⇒ 'a list" where
"inOrden1 (H t) = [t]"
| "inOrden1 (N t i d) = inOrden1 i @ t#inOrden1 d"

value "inOrden (N e (N c (H a) (H d)) (N g (H f) (H h)))
= [a,c,d,e,f,g,h]"
value "inOrden1 (N e (N c (H a) (H d)) (N g (H f) (H h)))
= [a,c,d,e,f,g,h]"

(* manmorjim1 *)
lemma "inOrden t = inOrden1 t"
apply (induct t)
apply auto
done

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))
---------------------------------------------------------------------
*}

(* ivamenjim danrodcha crigomgom marpoldia1 manmorjim1 pablucoto bowma *)

fun espejo :: "'a arbol ⇒ 'a arbol" where
"espejo (H t) = H t"
| "espejo (N t i d) = N t (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)
---------------------------------------------------------------------
*}

(* ivamenjim *)

lemma "preOrden (espejo a) = rev (postOrden a)" (is "?P a")
proof (induct a)
fix x
show "?P (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
finally show "preOrden (espejo (N x i d)) = rev (postOrden (N x i d))" by simp
qed

(* danrodcha *)
lemma "preOrden (espejo a) = rev (postOrden a)"
by (induct a) simp_all

(* danrodcha crigomgom*)
lemma "preOrden (espejo a) = rev (postOrden a)" (is "?P a")
proof (induct a)
fix x
show "?P (H x)" by simp
next
fix x i d
assume HIi: "?P i"
assume HId: "?P d"
have "preOrden (espejo (N x i d)) = preOrden (N x (espejo d) (espejo i))"
by (simp only: espejo.simps(2))
also have "… = x#preOrden (espejo d) @ preOrden (espejo i)"
by (simp only: preOrden.simps(2))
also have"… = x#rev (postOrden d) @ rev (postOrden i)"
using HIi HId by simp
also have "… = rev (postOrden (N x i d))" by simp
finally show "?P (N x i d)" by simp
qed

(* danrodcha *)
lemma "preOrden (espejo a) = rev (postOrden a)" (is "?P a")
proof (induct a)
fix x
show "?P (H x)" by simp
next
fix x i d
assume HIi: "?P i"
assume HId: "?P d"
show "?P (N x i d)" using HIi HId by simp
qed

(* bowma *)
lemma "preOrden (espejo a) = rev (postOrden a)"
apply (induct a)
apply simp_all
done

(* pablucoto *)

lemma "preOrden (espejo a) = rev (postOrden a)" (is "?P a")
proof (induct a)
fix x
show "?P (H x)" by simp
next
fix x i d
assume h1: "?P i"
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 "... = [x] @ rev (postOrden i @ postOrden d)" by simp
also have "... = rev ( postOrden i @ postOrden d @ [x] ) " by simp
also have "... = rev (postOrden (N x i d)) " by simp
finally show "?P (N x i d)" by simp
qed

(* bowma *)
lemma "preOrden (espejo a) = rev (postOrden a)" (is "?p a")
proof (induct a)
fix t
show "?p (H t)" by simp
(* Aquí si le diga "preOrden (espejo (H t)) = rev (postOrden (H t))",isabelle dice:
proof (prove)
goal (1 subgoal):
1. preOrden (espejo (H t)) = rev (postOrden (H t))
Introduced fixed type variable(s): 'b in "t__"
No entiendo porqué *)
next
fix t i d
assume H1: "?p i"
assume H2: "?p d"
have "preOrden (espejo (N t i d)) = preOrden (N t (espejo d) (espejo i))" by simp
also have "... = [t] @ (preOrden (espejo d)) @ (preOrden (espejo i))" by simp
also have "... = [t] @ rev (postOrden d) @ rev (postOrden i)" using H1 H2 by simp
finally show "?p (N t i d)" by simp
qed

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

(* ivamenjim crigomgom bowma *)

lemma "postOrden (espejo a) = rev (preOrden a)" (is "?P a")
proof (induct a)
fix x
show "?P (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
finally show "postOrden (espejo (N x i d)) = rev (preOrden (N x i d))" by simp
(* "?p (N x i d)" más corto *)
qed

(* danrodcha *)
lemma "postOrden (espejo a) = rev (preOrden a)" (is "?P a")
proof (induct a)
fix x
show "?P (H x)" by simp
next
fix x i d
assume HIi: "?P i"
assume HId: "?P d"
show "?P (N x i d)" using HIi HId by simp
qed

(* pablucoto *)

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

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

(* ivamenjim crigomgom bowma *)

theorem "inOrden (espejo a) = rev (inOrden a)" (is "?P a")
proof (induct a)
fix x
show "?P (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
finally show "inOrden (espejo (N x i d)) = rev (inOrden (N x i d))" by simp
qed

(* danrodcha *)
theorem "inOrden (espejo a) = rev (inOrden a)"
by (induct a) simp_all

(* danrodcha *)
theorem "inOrden (espejo a) = rev (inOrden a)" (is "?P a")
proof (induct a)
fix x
show "?P (H x)" by simp
next
fix x i d
assume HIi: "?P i"
assume HId: "?P d"
show "?P (N x i d)" using HIi HId by simp
qed

(* pablucoto *)
theorem "inOrden (espejo a) = rev (inOrden a)" (is "?P a")
proof (induct a)
fix x
show "?P (H x) " by simp
next
fix x i d
assume HI1: "?P i"
assume HI2: "?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 HI1 HI2 by simp
also have "... = rev (x # inOrden d ) @ rev (inOrden i)" by simp
also have "... = rev ( inOrden i @ x # inOrden d) " by simp
also have "... = rev (inOrden (N x i d))" by simp
finally show "?P (N x i d)" 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
---------------------------------------------------------------------
*}

(* danrodcha crigomgom manmorjim1 ivamenjim bowma pablucoto*)
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
---------------------------------------------------------------------
*}

(* danrodcha crigomgom manmorjim1 ivamenjim bowma pablucoto *)
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
---------------------------------------------------------------------
*}

(* danrodcha crigomgom manmorjim1 ivamenjim bowma pablucoto*)
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"

(* bowma *)
fun extremo_derecha_1 :: "'a arbol ⇒ 'a" where
"extremo_derecha_1 (H t) = t"
| "extremo_derecha_1 (N t i d) = last (inOrden (N t i d))"

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

(* danrodcha *)
lemma aux_ej12: "inOrden a ≠ []"
apply (induct a)
apply simp
apply simp
done

(* danrodcha pablucoto *)
theorem "last (inOrden a) = extremo_derecha a" (is "?P a")
proof (induct a)
fix x
show "?P (H x)" by simp
next
fix x i d
assume HIi: "?P i"
assume HId: "?P d"
have "last (inOrden (N x i d)) = last (inOrden i @ [x] @ inOrden d)"
by (simp only: inOrden.simps(2))
also have "… = last (inOrden d)" by (simp add: aux_ej12)
also have "… = extremo_derecha d" using HId by simp
also have "… = extremo_derecha (N x i d)" by simp
finally show "?P (N x i d)" by simp
qed

(* ivamenjim *)

lemma aux_ej12_1: "inOrden a ≠ []"
by (induct a) simp_all

(* ivamenjim *)
(* Igual que la anterior, pero poniendo solo by simp en el primer have *)

theorem "last (inOrden a) = extremo_derecha a" (is "?P a")
proof (induct a)
fix x
show "?P (H x)" by simp
next
fix x
fix i assume h1: "?P i"
fix d assume h2: "?P d"
have "last (inOrden (N x i d)) = last ((inOrden i) @ [x] @ (inOrden d))" by simp
also have "... = last (inOrden d)" by (simp add: aux_ej12_1)
also have "... = extremo_derecha d" using h2 by simp
finally show "last (inOrden (N x i d)) = extremo_derecha (N x i d)" by simp
qed

(* bowma *)
(* Casi lo mismo que el anterior,pero no hace falta suponer "?p i" *)
theorem "last (inOrden a) = extremo_derecha a" (is "?p a")
proof (induct a)
fix t
show "?p (H t)" by simp
next
fix t i d
assume HI: "?p d"
have "last (inOrden (N t i d)) = last (inOrden i @ [t] @ inOrden d)" by simp
also have "... = last (inOrden d)" by (simp add:aux_ej12)
also have "... = extremo_derecha d" using HI by simp
finally show "?p (N t i d)" by simp
qed

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

(* danrodcha pablucoto *)
theorem "hd (inOrden a) = extremo_izquierda a" (is "?P a")
proof (induct a)
fix x
show "?P (H x)" by simp
next
fix x i d
assume HIi: "?P i"
assume HId: "?P d"
have "hd (inOrden (N x i d)) = hd (inOrden i @ [x] @ inOrden d)"
by (simp only: inOrden.simps(2))
also have "… = hd (inOrden i)" by (simp add: aux_ej12)
also have "… = extremo_izquierda i" using HIi by simp
also have "… = extremo_izquierda (N x i d)" by simp
finally show "?P (N x i d)" by simp
qed

(* bowma *)
theorem "hd (inOrden a) = extremo_izquierda a" (is "?p a")
proof (induct a)
fix t
show "?p (H t)" by simp
next
fix t i d
assume HI: "?p i"
have "hd (inOrden (N t i d)) = hd (inOrden i @ [t] @ inOrden d)" by simp
also have "… = hd (inOrden i)" by (simp add: aux_ej12)
also have "… = extremo_izquierda i" using HI by simp
finally show "?p (N t i d)" by simp
qed

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

(* danrodcha *)
theorem "hd (preOrden a) = last (postOrden a)" (is "?P a")
proof (induct a)
fix x
show "?P (H x)" by simp
fix x i d
show "?P (N x i d)" by simp
qed

(* danrodcha *)
theorem "hd (preOrden a) = last (postOrden a)" (is "?P a")
proof (induct a)
fix x
show "?P (H x)" by simp
fix x i d
assume HIi: "?P i"
assume HId: "?P d"
have "hd (preOrden (N x i d)) = hd (x#preOrden i @ preOrden d)"
by (simp only: preOrden.simps(2))
also have "… = x" by simp
also have "… = last (postOrden i @ postOrden d @ [x])" by simp
also have "… = last (postOrden (N x i d))"
by (simp only: postOrden.simps(2))
finally show "?P (N x i d)" by simp
qed

(* pablucoto *) (*Similar al anterior*)
theorem "hd (preOrden a) = last (postOrden a)" (is "?P a")
proof (induct a)
fix x
show "?P (H x)" by simp
next
fix x i d
assume HI1: "?P i"
assume HI2: "?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 i @ postOrden d @ [x]) " by simp
also have "... = last ( postOrden (N x i d) )" by simp
finally show "?P (N x i d)" by simp
qed

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

(* danrodcha *)
theorem "hd (preOrden a) = raiz a" (is "?P a")
proof (induct a)
fix x
show "?P (H x)" by simp
fix x i d
assume HIi: "?P i"
assume HId: "?P d"
have "hd (preOrden (N x i d)) = hd (x#preOrden i @ preOrden d)"
by (simp only: preOrden.simps(2))
also have "… = x" by simp
also have "… = raiz (N x i d)" by simp
finally show "?P (N x i d)" by simp
qed

(* danrodcha *)
theorem "hd (preOrden a) = raiz a" (is "?P a")
proof (induct a)
fix x
show "?P (H x)" by simp
fix x i d
show "?P (N x i d)" by simp
qed

(* pablucoto *)
theorem "hd (preOrden a) = raiz a" (is "?P a")
proof (induct a)
fix x
show "?P (H x)" by simp
next
fix x i d
assume HI1: " ?P i"
assume HI2: " ?P d"
have " hd (preOrden (N x i d)) = hd ([x] @ preOrden i @ preOrden d) " by simp
also have "... = x" by simp
also have "... = raiz (N x i d)" by simp
finally show " ?P (N x i d)" by simp
qed

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

(*crigomgom pablucoto *)
theorem "hd (inOrden a) = raiz a"
quickcheck
oops

(* danrodcha:
Auto 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
---------------------------------------------------------------------
*}

(* danrodcha *)
theorem "last (postOrden a) = raiz a" (is "?P a")
proof (induct a)
fix x
show "?P (H x)" by simp
fix x i d
show "?P (N x i d)" by simp
qed

(* danrodcha *)
theorem "last (postOrden a) = raiz a" (is "?P a")
proof (induct a)
fix x
show "?P (H x)" by simp
fix x i d
assume HIi: "?P i"
assume HId: "?P d"
have "last (postOrden (N x i d)) = last (postOrden i @ postOrden d @ [x])"
by (simp only: postOrden.simps(2))
also have "… = x" by simp
also have "… = raiz (N x i d)" by (simp only: raiz.simps(2))
finally show "?P (N x i d)" by simp
qed

(* pablucoto *) (*Similar al anterior*)

theorem "last (postOrden a) = raiz a" (is "?P a")
proof (induct a )
fix x
show "?P (H x)" by simp
next
fix x i d
assume HI1: "?P i"
assume HI2: "?P d"
have "last (postOrden (N x i d)) = last ( postOrden i @ postOrden d @ [x])" by simp
also have "... = x" by simp
also have "... = raiz (N x i d) " by simp
finally show " ?P (N x i d)" by simp
qed

end
</source>

Relación 6

2016-12-04T03:33:16Z

Pablucoto:

<source lang="isar">
chapter {* R6: Recorridos de árboles *}

theory R6_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))".
---------------------------------------------------------------------
*}

(* ivamenjim marpoldia1 manmorjim1 bowma *)

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]
---------------------------------------------------------------------
*}

(* ivamenjim marpoldia1 pablucoto bowma*)

fun preOrden :: "'a arbol ⇒ 'a list" where
"preOrden (H t) = [t]"
| "preOrden (N t i d) = [t] @ (preOrden i) @ (preOrden d)"

(* danrodcha crigomgom manmorjim1 bowma *)
fun preOrden1 :: "'a arbol ⇒ 'a list" where
"preOrden1 (H x) = [x]"
| "preOrden1 (N x i d) = x#preOrden1 i @ preOrden1 d"

value "preOrden (N e (N c (H a) (H d)) (N g (H f) (H h)))
= [e,c,a,d,g,f,h]"
value "preOrden1 (N e (N c (H a) (H d)) (N g (H f) (H h)))
= [e,c,a,d,g,f,h]"

(* danrodcha *)
lemma "preOrden a = preOrden1 a"
by (induct a) simp_all

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]
---------------------------------------------------------------------
*}

(* ivamenjim danrodcha crigomgom marpoldia1 manmorjim1 pablucoto bowma*)

fun postOrden :: "'a arbol ⇒ 'a list" where
"postOrden (H t) = [t]"
| "postOrden (N t i d) = (postOrden i) @ (postOrden d) @ [t]"

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]
---------------------------------------------------------------------
*}

(* ivamenjim crigomgom marpoldia1 pablucoto bowma *)

fun inOrden :: "'a arbol ⇒ 'a list" where
"inOrden (H t) = [t]"
| "inOrden (N t i d) = (inOrden i) @ [t] @ (inOrden d)"

(* danrodcha manmorjim1 *)
fun inOrden1 :: "'a arbol ⇒ 'a list" where
"inOrden1 (H t) = [t]"
| "inOrden1 (N t i d) = inOrden1 i @ t#inOrden1 d"

value "inOrden (N e (N c (H a) (H d)) (N g (H f) (H h)))
= [a,c,d,e,f,g,h]"
value "inOrden1 (N e (N c (H a) (H d)) (N g (H f) (H h)))
= [a,c,d,e,f,g,h]"

(* manmorjim1 *)
lemma "inOrden t = inOrden1 t"
apply (induct t)
apply auto
done

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))
---------------------------------------------------------------------
*}

(* ivamenjim danrodcha crigomgom marpoldia1 manmorjim1 pablucoto bowma *)

fun espejo :: "'a arbol ⇒ 'a arbol" where
"espejo (H t) = H t"
| "espejo (N t i d) = N t (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)
---------------------------------------------------------------------
*}

(* ivamenjim *)

lemma "preOrden (espejo a) = rev (postOrden a)" (is "?P a")
proof (induct a)
fix x
show "?P (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
finally show "preOrden (espejo (N x i d)) = rev (postOrden (N x i d))" by simp
qed

(* danrodcha *)
lemma "preOrden (espejo a) = rev (postOrden a)"
by (induct a) simp_all

(* danrodcha crigomgom*)
lemma "preOrden (espejo a) = rev (postOrden a)" (is "?P a")
proof (induct a)
fix x
show "?P (H x)" by simp
next
fix x i d
assume HIi: "?P i"
assume HId: "?P d"
have "preOrden (espejo (N x i d)) = preOrden (N x (espejo d) (espejo i))"
by (simp only: espejo.simps(2))
also have "… = x#preOrden (espejo d) @ preOrden (espejo i)"
by (simp only: preOrden.simps(2))
also have"… = x#rev (postOrden d) @ rev (postOrden i)"
using HIi HId by simp
also have "… = rev (postOrden (N x i d))" by simp
finally show "?P (N x i d)" by simp
qed

(* danrodcha *)
lemma "preOrden (espejo a) = rev (postOrden a)" (is "?P a")
proof (induct a)
fix x
show "?P (H x)" by simp
next
fix x i d
assume HIi: "?P i"
assume HId: "?P d"
show "?P (N x i d)" using HIi HId by simp
qed

(* bowma *)
lemma "preOrden (espejo a) = rev (postOrden a)"
apply (induct a)
apply simp_all
done

(* pablucoto *)

lemma "preOrden (espejo a) = rev (postOrden a)" (is "?P a")
proof (induct a)
fix x
show "?P (H x)" by simp
next
fix x i d
assume h1: "?P i"
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 "... = [x] @ rev (postOrden i @ postOrden d)" by simp
also have "... = rev ( postOrden i @ postOrden d @ [x] ) " by simp
also have "... = rev (postOrden (N x i d)) " by simp
finally show "?P (N x i d)" by simp
qed

(* bowma *)
lemma "preOrden (espejo a) = rev (postOrden a)" (is "?p a")
proof (induct a)
fix t
show "?p (H t)" by simp
(* Aquí si le diga "preOrden (espejo (H t)) = rev (postOrden (H t))",isabelle dice:
proof (prove)
goal (1 subgoal):
1. preOrden (espejo (H t)) = rev (postOrden (H t))
Introduced fixed type variable(s): 'b in "t__"
No entiendo porqué *)
next
fix t i d
assume H1: "?p i"
assume H2: "?p d"
have "preOrden (espejo (N t i d)) = preOrden (N t (espejo d) (espejo i))" by simp
also have "... = [t] @ (preOrden (espejo d)) @ (preOrden (espejo i))" by simp
also have "... = [t] @ rev (postOrden d) @ rev (postOrden i)" using H1 H2 by simp
finally show "?p (N t i d)" by simp
qed

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

(* ivamenjim crigomgom bowma *)

lemma "postOrden (espejo a) = rev (preOrden a)" (is "?P a")
proof (induct a)
fix x
show "?P (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
finally show "postOrden (espejo (N x i d)) = rev (preOrden (N x i d))" by simp
(* "?p (N x i d)" más corto *)
qed

(* danrodcha *)
lemma "postOrden (espejo a) = rev (preOrden a)" (is "?P a")
proof (induct a)
fix x
show "?P (H x)" by simp
next
fix x i d
assume HIi: "?P i"
assume HId: "?P d"
show "?P (N x i d)" using HIi HId by simp
qed

(* pablucoto *)

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

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

(* ivamenjim crigomgom bowma *)

theorem "inOrden (espejo a) = rev (inOrden a)" (is "?P a")
proof (induct a)
fix x
show "?P (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
finally show "inOrden (espejo (N x i d)) = rev (inOrden (N x i d))" by simp
qed

(* danrodcha *)
theorem "inOrden (espejo a) = rev (inOrden a)"
by (induct a) simp_all

(* danrodcha *)
theorem "inOrden (espejo a) = rev (inOrden a)" (is "?P a")
proof (induct a)
fix x
show "?P (H x)" by simp
next
fix x i d
assume HIi: "?P i"
assume HId: "?P d"
show "?P (N x i d)" using HIi HId by simp
qed

(* pablucoto *)
theorem "inOrden (espejo a) = rev (inOrden a)" (is "?P a")
proof (induct a)
fix x
show "?P (H x) " by simp
next
fix x i d
assume HI1: "?P i"
assume HI2: "?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 HI1 HI2 by simp
also have "... = rev (x # inOrden d ) @ rev (inOrden i)" by simp
also have "... = rev ( inOrden i @ x # inOrden d) " by simp
also have "... = rev (inOrden (N x i d))" by simp
finally show "?P (N x i d)" 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
---------------------------------------------------------------------
*}

(* danrodcha crigomgom manmorjim1 ivamenjim bowma pablucoto*)
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
---------------------------------------------------------------------
*}

(* danrodcha crigomgom manmorjim1 ivamenjim bowma pablucoto *)
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
---------------------------------------------------------------------
*}

(* danrodcha crigomgom manmorjim1 ivamenjim bowma pablucoto*)
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"

(* bowma *)
fun extremo_derecha_1 :: "'a arbol ⇒ 'a" where
"extremo_derecha_1 (H t) = t"
| "extremo_derecha_1 (N t i d) = last (inOrden (N t i d))"

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

(* danrodcha *)
lemma aux_ej12: "inOrden a ≠ []"
apply (induct a)
apply simp
apply simp
done

(* danrodcha pablucoto *)
theorem "last (inOrden a) = extremo_derecha a" (is "?P a")
proof (induct a)
fix x
show "?P (H x)" by simp
next
fix x i d
assume HIi: "?P i"
assume HId: "?P d"
have "last (inOrden (N x i d)) = last (inOrden i @ [x] @ inOrden d)"
by (simp only: inOrden.simps(2))
also have "… = last (inOrden d)" by (simp add: aux_ej12)
also have "… = extremo_derecha d" using HId by simp
also have "… = extremo_derecha (N x i d)" by simp
finally show "?P (N x i d)" by simp
qed

(* ivamenjim *)

lemma aux_ej12_1: "inOrden a ≠ []"
by (induct a) simp_all

(* ivamenjim *)
(* Igual que la anterior, pero poniendo solo by simp en el primer have *)

theorem "last (inOrden a) = extremo_derecha a" (is "?P a")
proof (induct a)
fix x
show "?P (H x)" by simp
next
fix x
fix i assume h1: "?P i"
fix d assume h2: "?P d"
have "last (inOrden (N x i d)) = last ((inOrden i) @ [x] @ (inOrden d))" by simp
also have "... = last (inOrden d)" by (simp add: aux_ej12_1)
also have "... = extremo_derecha d" using h2 by simp
finally show "last (inOrden (N x i d)) = extremo_derecha (N x i d)" by simp
qed

(* bowma *)
(* Casi lo mismo que el anterior,pero no hace falta suponer "?p i" *)
theorem "last (inOrden a) = extremo_derecha a" (is "?p a")
proof (induct a)
fix t
show "?p (H t)" by simp
next
fix t i d
assume HI: "?p d"
have "last (inOrden (N t i d)) = last (inOrden i @ [t] @ inOrden d)" by simp
also have "... = last (inOrden d)" by (simp add:aux_ej12)
also have "... = extremo_derecha d" using HI by simp
finally show "?p (N t i d)" by simp
qed

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

(* danrodcha *)
theorem "hd (inOrden a) = extremo_izquierda a" (is "?P a")
proof (induct a)
fix x
show "?P (H x)" by simp
next
fix x i d
assume HIi: "?P i"
assume HId: "?P d"
have "hd (inOrden (N x i d)) = hd (inOrden i @ [x] @ inOrden d)"
by (simp only: inOrden.simps(2))
also have "… = hd (inOrden i)" by (simp add: aux_ej12)
also have "… = extremo_izquierda i" using HIi by simp
also have "… = extremo_izquierda (N x i d)" by simp
finally show "?P (N x i d)" by simp
qed

(* bowma *)
theorem "hd (inOrden a) = extremo_izquierda a" (is "?p a")
proof (induct a)
fix t
show "?p (H t)" by simp
next
fix t i d
assume HI: "?p i"
have "hd (inOrden (N t i d)) = hd (inOrden i @ [t] @ inOrden d)" by simp
also have "… = hd (inOrden i)" by (simp add: aux_ej12)
also have "… = extremo_izquierda i" using HI by simp
finally show "?p (N t i d)" by simp
qed

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

(* danrodcha *)
theorem "hd (preOrden a) = last (postOrden a)" (is "?P a")
proof (induct a)
fix x
show "?P (H x)" by simp
fix x i d
show "?P (N x i d)" by simp
qed

(* danrodcha *)
theorem "hd (preOrden a) = last (postOrden a)" (is "?P a")
proof (induct a)
fix x
show "?P (H x)" by simp
fix x i d
assume HIi: "?P i"
assume HId: "?P d"
have "hd (preOrden (N x i d)) = hd (x#preOrden i @ preOrden d)"
by (simp only: preOrden.simps(2))
also have "… = x" by simp
also have "… = last (postOrden i @ postOrden d @ [x])" by simp
also have "… = last (postOrden (N x i d))"
by (simp only: postOrden.simps(2))
finally show "?P (N x i d)" by simp
qed

(* pablucoto *) (*Similar al anterior*)
theorem "hd (preOrden a) = last (postOrden a)" (is "?P a")
proof (induct a)
fix x
show "?P (H x)" by simp
next
fix x i d
assume HI1: "?P i"
assume HI2: "?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 i @ postOrden d @ [x]) " by simp
also have "... = last ( postOrden (N x i d) )" by simp
finally show "?P (N x i d)" by simp
qed

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

(* danrodcha *)
theorem "hd (preOrden a) = raiz a" (is "?P a")
proof (induct a)
fix x
show "?P (H x)" by simp
fix x i d
assume HIi: "?P i"
assume HId: "?P d"
have "hd (preOrden (N x i d)) = hd (x#preOrden i @ preOrden d)"
by (simp only: preOrden.simps(2))
also have "… = x" by simp
also have "… = raiz (N x i d)" by simp
finally show "?P (N x i d)" by simp
qed

(* danrodcha *)
theorem "hd (preOrden a) = raiz a" (is "?P a")
proof (induct a)
fix x
show "?P (H x)" by simp
fix x i d
show "?P (N x i d)" by simp
qed

(* pablucoto *)
theorem "hd (preOrden a) = raiz a" (is "?P a")
proof (induct a)
fix x
show "?P (H x)" by simp
next
fix x i d
assume HI1: " ?P i"
assume HI2: " ?P d"
have " hd (preOrden (N x i d)) = hd ([x] @ preOrden i @ preOrden d) " by simp
also have "... = x" by simp
also have "... = raiz (N x i d)" by simp
finally show " ?P (N x i d)" by simp
qed

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

(*crigomgom pablucoto *)
theorem "hd (inOrden a) = raiz a"
quickcheck
oops

(* danrodcha:
Auto 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
---------------------------------------------------------------------
*}

(* danrodcha *)
theorem "last (postOrden a) = raiz a" (is "?P a")
proof (induct a)
fix x
show "?P (H x)" by simp
fix x i d
show "?P (N x i d)" by simp
qed

(* danrodcha *)
theorem "last (postOrden a) = raiz a" (is "?P a")
proof (induct a)
fix x
show "?P (H x)" by simp
fix x i d
assume HIi: "?P i"
assume HId: "?P d"
have "last (postOrden (N x i d)) = last (postOrden i @ postOrden d @ [x])"
by (simp only: postOrden.simps(2))
also have "… = x" by simp
also have "… = raiz (N x i d)" by (simp only: raiz.simps(2))
finally show "?P (N x i d)" by simp
qed

(* pablucoto *) (*Similar al anterior*)

theorem "last (postOrden a) = raiz a" (is "?P a")
proof (induct a )
fix x
show "?P (H x)" by simp
next
fix x i d
assume HI1: "?P i"
assume HI2: "?P d"
have "last (postOrden (N x i d)) = last ( postOrden i @ postOrden d @ [x])" by simp
also have "... = x" by simp
also have "... = raiz (N x i d) " by simp
finally show " ?P (N x i d)" by simp
qed

end
</source>

Relación 6

2016-12-04T03:31:51Z

Pablucoto:

<source lang="isar">
chapter {* R6: Recorridos de árboles *}

theory R6_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))".
---------------------------------------------------------------------
*}

(* ivamenjim marpoldia1 manmorjim1 bowma *)

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]
---------------------------------------------------------------------
*}

(* ivamenjim marpoldia1 pablucoto bowma*)

fun preOrden :: "'a arbol ⇒ 'a list" where
"preOrden (H t) = [t]"
| "preOrden (N t i d) = [t] @ (preOrden i) @ (preOrden d)"

(* danrodcha crigomgom manmorjim1 bowma *)
fun preOrden1 :: "'a arbol ⇒ 'a list" where
"preOrden1 (H x) = [x]"
| "preOrden1 (N x i d) = x#preOrden1 i @ preOrden1 d"

value "preOrden (N e (N c (H a) (H d)) (N g (H f) (H h)))
= [e,c,a,d,g,f,h]"
value "preOrden1 (N e (N c (H a) (H d)) (N g (H f) (H h)))
= [e,c,a,d,g,f,h]"

(* danrodcha *)
lemma "preOrden a = preOrden1 a"
by (induct a) simp_all

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]
---------------------------------------------------------------------
*}

(* ivamenjim danrodcha crigomgom marpoldia1 manmorjim1 pablucoto bowma*)

fun postOrden :: "'a arbol ⇒ 'a list" where
"postOrden (H t) = [t]"
| "postOrden (N t i d) = (postOrden i) @ (postOrden d) @ [t]"

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]
---------------------------------------------------------------------
*}

(* ivamenjim crigomgom marpoldia1 pablucoto bowma *)

fun inOrden :: "'a arbol ⇒ 'a list" where
"inOrden (H t) = [t]"
| "inOrden (N t i d) = (inOrden i) @ [t] @ (inOrden d)"

(* danrodcha manmorjim1 *)
fun inOrden1 :: "'a arbol ⇒ 'a list" where
"inOrden1 (H t) = [t]"
| "inOrden1 (N t i d) = inOrden1 i @ t#inOrden1 d"

value "inOrden (N e (N c (H a) (H d)) (N g (H f) (H h)))
= [a,c,d,e,f,g,h]"
value "inOrden1 (N e (N c (H a) (H d)) (N g (H f) (H h)))
= [a,c,d,e,f,g,h]"

(* manmorjim1 *)
lemma "inOrden t = inOrden1 t"
apply (induct t)
apply auto
done

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))
---------------------------------------------------------------------
*}

(* ivamenjim danrodcha crigomgom marpoldia1 manmorjim1 pablucoto bowma *)

fun espejo :: "'a arbol ⇒ 'a arbol" where
"espejo (H t) = H t"
| "espejo (N t i d) = N t (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)
---------------------------------------------------------------------
*}

(* ivamenjim *)

lemma "preOrden (espejo a) = rev (postOrden a)" (is "?P a")
proof (induct a)
fix x
show "?P (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
finally show "preOrden (espejo (N x i d)) = rev (postOrden (N x i d))" by simp
qed

(* danrodcha *)
lemma "preOrden (espejo a) = rev (postOrden a)"
by (induct a) simp_all

(* danrodcha crigomgom*)
lemma "preOrden (espejo a) = rev (postOrden a)" (is "?P a")
proof (induct a)
fix x
show "?P (H x)" by simp
next
fix x i d
assume HIi: "?P i"
assume HId: "?P d"
have "preOrden (espejo (N x i d)) = preOrden (N x (espejo d) (espejo i))"
by (simp only: espejo.simps(2))
also have "… = x#preOrden (espejo d) @ preOrden (espejo i)"
by (simp only: preOrden.simps(2))
also have"… = x#rev (postOrden d) @ rev (postOrden i)"
using HIi HId by simp
also have "… = rev (postOrden (N x i d))" by simp
finally show "?P (N x i d)" by simp
qed

(* danrodcha *)
lemma "preOrden (espejo a) = rev (postOrden a)" (is "?P a")
proof (induct a)
fix x
show "?P (H x)" by simp
next
fix x i d
assume HIi: "?P i"
assume HId: "?P d"
show "?P (N x i d)" using HIi HId by simp
qed

(* bowma *)
lemma "preOrden (espejo a) = rev (postOrden a)"
apply (induct a)
apply simp_all
done

(* pablucoto *)

lemma "preOrden (espejo a) = rev (postOrden a)" (is "?P a")
proof (induct a)
fix x
show "?P (H x)" by simp
next
fix x i d
assume h1: "?P i"
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 "... = [x] @ rev (postOrden i @ postOrden d)" by simp
also have "... = rev ( postOrden i @ postOrden d @ [x] ) " by simp
also have "... = rev (postOrden (N x i d)) " by simp
finally show "?P (N x i d)" by simp
qed

(* bowma *)
lemma "preOrden (espejo a) = rev (postOrden a)" (is "?p a")
proof (induct a)
fix t
show "?p (H t)" by simp
(* Aquí si le diga "preOrden (espejo (H t)) = rev (postOrden (H t))",isabelle dice:
proof (prove)
goal (1 subgoal):
1. preOrden (espejo (H t)) = rev (postOrden (H t))
Introduced fixed type variable(s): 'b in "t__"
No entiendo porqué *)
next
fix t i d
assume H1: "?p i"
assume H2: "?p d"
have "preOrden (espejo (N t i d)) = preOrden (N t (espejo d) (espejo i))" by simp
also have "... = [t] @ (preOrden (espejo d)) @ (preOrden (espejo i))" by simp
also have "... = [t] @ rev (postOrden d) @ rev (postOrden i)" using H1 H2 by simp
finally show "?p (N t i d)" by simp
qed

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

(* ivamenjim crigomgom bowma *)

lemma "postOrden (espejo a) = rev (preOrden a)" (is "?P a")
proof (induct a)
fix x
show "?P (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
finally show "postOrden (espejo (N x i d)) = rev (preOrden (N x i d))" by simp
(* "?p (N x i d)" más corto *)
qed

(* danrodcha *)
lemma "postOrden (espejo a) = rev (preOrden a)" (is "?P a")
proof (induct a)
fix x
show "?P (H x)" by simp
next
fix x i d
assume HIi: "?P i"
assume HId: "?P d"
show "?P (N x i d)" using HIi HId by simp
qed

(* pablucoto *)

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

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

(* ivamenjim crigomgom bowma *)

theorem "inOrden (espejo a) = rev (inOrden a)" (is "?P a")
proof (induct a)
fix x
show "?P (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
finally show "inOrden (espejo (N x i d)) = rev (inOrden (N x i d))" by simp
qed

(* danrodcha *)
theorem "inOrden (espejo a) = rev (inOrden a)"
by (induct a) simp_all

(* danrodcha *)
theorem "inOrden (espejo a) = rev (inOrden a)" (is "?P a")
proof (induct a)
fix x
show "?P (H x)" by simp
next
fix x i d
assume HIi: "?P i"
assume HId: "?P d"
show "?P (N x i d)" using HIi HId by simp
qed

(* pablucoto *)
theorem "inOrden (espejo a) = rev (inOrden a)" (is "?P a")
proof (induct a)
fix x
show "?P (H x) " by simp
next
fix x i d
assume HI1: "?P i"
assume HI2: "?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 HI1 HI2 by simp
also have "... = rev (x # inOrden d ) @ rev (inOrden i)" by simp
also have "... = rev ( inOrden i @ x # inOrden d) " by simp
also have "... = rev (inOrden (N x i d))" by simp
finally show "?P (N x i d)" 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
---------------------------------------------------------------------
*}

(* danrodcha crigomgom manmorjim1 ivamenjim bowma pablucoto*)
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
---------------------------------------------------------------------
*}

(* danrodcha crigomgom manmorjim1 ivamenjim bowma pablucoto *)
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
---------------------------------------------------------------------
*}

(* danrodcha crigomgom manmorjim1 ivamenjim bowma pablucoto*)
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"

(* bowma *)
fun extremo_derecha_1 :: "'a arbol ⇒ 'a" where
"extremo_derecha_1 (H t) = t"
| "extremo_derecha_1 (N t i d) = last (inOrden (N t i d))"

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

(* danrodcha *)
lemma aux_ej12: "inOrden a ≠ []"
apply (induct a)
apply simp
apply simp
done

(* danrodcha pablucoto *)
theorem "last (inOrden a) = extremo_derecha a" (is "?P a")
proof (induct a)
fix x
show "?P (H x)" by simp
next
fix x i d
assume HIi: "?P i"
assume HId: "?P d"
have "last (inOrden (N x i d)) = last (inOrden i @ [x] @ inOrden d)"
by (simp only: inOrden.simps(2))
also have "… = last (inOrden d)" by (simp add: aux_ej12)
also have "… = extremo_derecha d" using HId by simp
also have "… = extremo_derecha (N x i d)" by simp
finally show "?P (N x i d)" by simp
qed

(* ivamenjim *)

lemma aux_ej12_1: "inOrden a ≠ []"
by (induct a) simp_all

(* ivamenjim *)
(* Igual que la anterior, pero poniendo solo by simp en el primer have *)

theorem "last (inOrden a) = extremo_derecha a" (is "?P a")
proof (induct a)
fix x
show "?P (H x)" by simp
next
fix x
fix i assume h1: "?P i"
fix d assume h2: "?P d"
have "last (inOrden (N x i d)) = last ((inOrden i) @ [x] @ (inOrden d))" by simp
also have "... = last (inOrden d)" by (simp add: aux_ej12_1)
also have "... = extremo_derecha d" using h2 by simp
finally show "last (inOrden (N x i d)) = extremo_derecha (N x i d)" by simp
qed

(* bowma *)
(* Casi lo mismo que el anterior,pero no hace falta suponer "?p i" *)
theorem "last (inOrden a) = extremo_derecha a" (is "?p a")
proof (induct a)
fix t
show "?p (H t)" by simp
next
fix t i d
assume HI: "?p d"
have "last (inOrden (N t i d)) = last (inOrden i @ [t] @ inOrden d)" by simp
also have "... = last (inOrden d)" by (simp add:aux_ej12)
also have "... = extremo_derecha d" using HI by simp
finally show "?p (N t i d)" by simp
qed

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

(* danrodcha *)
theorem "hd (inOrden a) = extremo_izquierda a" (is "?P a")
proof (induct a)
fix x
show "?P (H x)" by simp
next
fix x i d
assume HIi: "?P i"
assume HId: "?P d"
have "hd (inOrden (N x i d)) = hd (inOrden i @ [x] @ inOrden d)"
by (simp only: inOrden.simps(2))
also have "… = hd (inOrden i)" by (simp add: aux_ej12)
also have "… = extremo_izquierda i" using HIi by simp
also have "… = extremo_izquierda (N x i d)" by simp
finally show "?P (N x i d)" by simp
qed

(* bowma *)
theorem "hd (inOrden a) = extremo_izquierda a" (is "?p a")
proof (induct a)
fix t
show "?p (H t)" by simp
next
fix t i d
assume HI: "?p i"
have "hd (inOrden (N t i d)) = hd (inOrden i @ [t] @ inOrden d)" by simp
also have "… = hd (inOrden i)" by (simp add: aux_ej12)
also have "… = extremo_izquierda i" using HI by simp
finally show "?p (N t i d)" by simp
qed

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

(* danrodcha *)
theorem "hd (preOrden a) = last (postOrden a)" (is "?P a")
proof (induct a)
fix x
show "?P (H x)" by simp
fix x i d
show "?P (N x i d)" by simp
qed

(* danrodcha *)
theorem "hd (preOrden a) = last (postOrden a)" (is "?P a")
proof (induct a)
fix x
show "?P (H x)" by simp
fix x i d
assume HIi: "?P i"
assume HId: "?P d"
have "hd (preOrden (N x i d)) = hd (x#preOrden i @ preOrden d)"
by (simp only: preOrden.simps(2))
also have "… = x" by simp
also have "… = last (postOrden i @ postOrden d @ [x])" by simp
also have "… = last (postOrden (N x i d))"
by (simp only: postOrden.simps(2))
finally show "?P (N x i d)" by simp
qed

(* pablucoto *) (*Similar a la anterior*)
theorem "hd (preOrden a) = last (postOrden a)" (is "?P a")
proof (induct a)
fix x
show "?P (H x)" by simp
next
fix x i d
assume HI1: "?P i"
assume HI2: "?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 i @ postOrden d @ [x]) " by simp
also have "... = last ( postOrden (N x i d) )" by simp
finally show "?P (N x i d)" by simp
qed

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

(* danrodcha *)
theorem "hd (preOrden a) = raiz a" (is "?P a")
proof (induct a)
fix x
show "?P (H x)" by simp
fix x i d
assume HIi: "?P i"
assume HId: "?P d"
have "hd (preOrden (N x i d)) = hd (x#preOrden i @ preOrden d)"
by (simp only: preOrden.simps(2))
also have "… = x" by simp
also have "… = raiz (N x i d)" by simp
finally show "?P (N x i d)" by simp
qed

(* danrodcha *)
theorem "hd (preOrden a) = raiz a" (is "?P a")
proof (induct a)
fix x
show "?P (H x)" by simp
fix x i d
show "?P (N x i d)" by simp
qed

(* pablucoto *)
theorem "hd (preOrden a) = raiz a" (is "?P a")
proof (induct a)
fix x
show "?P (H x)" by simp
next
fix x i d
assume HI1: " ?P i"
assume HI2: " ?P d"
have " hd (preOrden (N x i d)) = hd ([x] @ preOrden i @ preOrden d) " by simp
also have "... = x" by simp
also have "... = raiz (N x i d)" by simp
finally show " ?P (N x i d)" by simp
qed

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

(*crigomgom pablucoto *)
theorem "hd (inOrden a) = raiz a"
quickcheck
oops

(* danrodcha:
Auto 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
---------------------------------------------------------------------
*}

(* danrodcha *)
theorem "last (postOrden a) = raiz a" (is "?P a")
proof (induct a)
fix x
show "?P (H x)" by simp
fix x i d
show "?P (N x i d)" by simp
qed

(* danrodcha *)
theorem "last (postOrden a) = raiz a" (is "?P a")
proof (induct a)
fix x
show "?P (H x)" by simp
fix x i d
assume HIi: "?P i"
assume HId: "?P d"
have "last (postOrden (N x i d)) = last (postOrden i @ postOrden d @ [x])"
by (simp only: postOrden.simps(2))
also have "… = x" by simp
also have "… = raiz (N x i d)" by (simp only: raiz.simps(2))
finally show "?P (N x i d)" by simp
qed

(* pablucoto *) (*Similar al anterior*)

theorem "last (postOrden a) = raiz a" (is "?P a")
proof (induct a )
fix x
show "?P (H x)" by simp
next
fix x i d
assume HI1: "?P i"
assume HI2: "?P d"
have "last (postOrden (N x i d)) = last ( postOrden i @ postOrden d @ [x])" by simp
also have "... = x" by simp
also have "... = raiz (N x i d) " by simp
finally show " ?P (N x i d)" by simp
qed

end
</source>

Relación 6

2016-12-04T03:17:18Z

Pablucoto:

<source lang="isar">
chapter {* R6: Recorridos de árboles *}

theory R6_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))".
---------------------------------------------------------------------
*}

(* ivamenjim marpoldia1 manmorjim1 bowma *)

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]
---------------------------------------------------------------------
*}

(* ivamenjim marpoldia1 pablucoto bowma*)

fun preOrden :: "'a arbol ⇒ 'a list" where
"preOrden (H t) = [t]"
| "preOrden (N t i d) = [t] @ (preOrden i) @ (preOrden d)"

(* danrodcha crigomgom manmorjim1 bowma *)
fun preOrden1 :: "'a arbol ⇒ 'a list" where
"preOrden1 (H x) = [x]"
| "preOrden1 (N x i d) = x#preOrden1 i @ preOrden1 d"

value "preOrden (N e (N c (H a) (H d)) (N g (H f) (H h)))
= [e,c,a,d,g,f,h]"
value "preOrden1 (N e (N c (H a) (H d)) (N g (H f) (H h)))
= [e,c,a,d,g,f,h]"

(* danrodcha *)
lemma "preOrden a = preOrden1 a"
by (induct a) simp_all

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]
---------------------------------------------------------------------
*}

(* ivamenjim danrodcha crigomgom marpoldia1 manmorjim1 pablucoto bowma*)

fun postOrden :: "'a arbol ⇒ 'a list" where
"postOrden (H t) = [t]"
| "postOrden (N t i d) = (postOrden i) @ (postOrden d) @ [t]"

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]
---------------------------------------------------------------------
*}

(* ivamenjim crigomgom marpoldia1 pablucoto bowma *)

fun inOrden :: "'a arbol ⇒ 'a list" where
"inOrden (H t) = [t]"
| "inOrden (N t i d) = (inOrden i) @ [t] @ (inOrden d)"

(* danrodcha manmorjim1 *)
fun inOrden1 :: "'a arbol ⇒ 'a list" where
"inOrden1 (H t) = [t]"
| "inOrden1 (N t i d) = inOrden1 i @ t#inOrden1 d"

value "inOrden (N e (N c (H a) (H d)) (N g (H f) (H h)))
= [a,c,d,e,f,g,h]"
value "inOrden1 (N e (N c (H a) (H d)) (N g (H f) (H h)))
= [a,c,d,e,f,g,h]"

(* manmorjim1 *)
lemma "inOrden t = inOrden1 t"
apply (induct t)
apply auto
done

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))
---------------------------------------------------------------------
*}

(* ivamenjim danrodcha crigomgom marpoldia1 manmorjim1 pablucoto bowma *)

fun espejo :: "'a arbol ⇒ 'a arbol" where
"espejo (H t) = H t"
| "espejo (N t i d) = N t (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)
---------------------------------------------------------------------
*}

(* ivamenjim *)

lemma "preOrden (espejo a) = rev (postOrden a)" (is "?P a")
proof (induct a)
fix x
show "?P (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
finally show "preOrden (espejo (N x i d)) = rev (postOrden (N x i d))" by simp
qed

(* danrodcha *)
lemma "preOrden (espejo a) = rev (postOrden a)"
by (induct a) simp_all

(* danrodcha crigomgom*)
lemma "preOrden (espejo a) = rev (postOrden a)" (is "?P a")
proof (induct a)
fix x
show "?P (H x)" by simp
next
fix x i d
assume HIi: "?P i"
assume HId: "?P d"
have "preOrden (espejo (N x i d)) = preOrden (N x (espejo d) (espejo i))"
by (simp only: espejo.simps(2))
also have "… = x#preOrden (espejo d) @ preOrden (espejo i)"
by (simp only: preOrden.simps(2))
also have"… = x#rev (postOrden d) @ rev (postOrden i)"
using HIi HId by simp
also have "… = rev (postOrden (N x i d))" by simp
finally show "?P (N x i d)" by simp
qed

(* danrodcha *)
lemma "preOrden (espejo a) = rev (postOrden a)" (is "?P a")
proof (induct a)
fix x
show "?P (H x)" by simp
next
fix x i d
assume HIi: "?P i"
assume HId: "?P d"
show "?P (N x i d)" using HIi HId by simp
qed

(* bowma *)
lemma "preOrden (espejo a) = rev (postOrden a)"
apply (induct a)
apply simp_all
done

(* pablucoto *)

lemma "preOrden (espejo a) = rev (postOrden a)" (is "?P a")
proof (induct a)
fix x
show "?P (H x)" by simp
next
fix x i d
assume h1: "?P i"
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 "... = [x] @ rev (postOrden i @ postOrden d)" by simp
also have "... = rev ( postOrden i @ postOrden d @ [x] ) " by simp
also have "... = rev (postOrden (N x i d)) " by simp
finally show "?P (N x i d)" by simp
qed

(* bowma *)
lemma "preOrden (espejo a) = rev (postOrden a)" (is "?p a")
proof (induct a)
fix t
show "?p (H t)" by simp
(* Aquí si le diga "preOrden (espejo (H t)) = rev (postOrden (H t))",isabelle dice:
proof (prove)
goal (1 subgoal):
1. preOrden (espejo (H t)) = rev (postOrden (H t))
Introduced fixed type variable(s): 'b in "t__"
No entiendo porqué *)
next
fix t i d
assume H1: "?p i"
assume H2: "?p d"
have "preOrden (espejo (N t i d)) = preOrden (N t (espejo d) (espejo i))" by simp
also have "... = [t] @ (preOrden (espejo d)) @ (preOrden (espejo i))" by simp
also have "... = [t] @ rev (postOrden d) @ rev (postOrden i)" using H1 H2 by simp
finally show "?p (N t i d)" by simp
qed

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

(* ivamenjim crigomgom bowma *)

lemma "postOrden (espejo a) = rev (preOrden a)" (is "?P a")
proof (induct a)
fix x
show "?P (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
finally show "postOrden (espejo (N x i d)) = rev (preOrden (N x i d))" by simp
(* "?p (N x i d)" más corto *)
qed

(* danrodcha *)
lemma "postOrden (espejo a) = rev (preOrden a)" (is "?P a")
proof (induct a)
fix x
show "?P (H x)" by simp
next
fix x i d
assume HIi: "?P i"
assume HId: "?P d"
show "?P (N x i d)" using HIi HId by simp
qed

(* pablucoto *)

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

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

(* ivamenjim crigomgom bowma *)

theorem "inOrden (espejo a) = rev (inOrden a)" (is "?P a")
proof (induct a)
fix x
show "?P (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
finally show "inOrden (espejo (N x i d)) = rev (inOrden (N x i d))" by simp
qed

(* danrodcha *)
theorem "inOrden (espejo a) = rev (inOrden a)"
by (induct a) simp_all

(* danrodcha *)
theorem "inOrden (espejo a) = rev (inOrden a)" (is "?P a")
proof (induct a)
fix x
show "?P (H x)" by simp
next
fix x i d
assume HIi: "?P i"
assume HId: "?P d"
show "?P (N x i d)" using HIi HId by simp
qed

(* pablucoto *)
theorem "inOrden (espejo a) = rev (inOrden a)" (is "?P a")
proof (induct a)
fix x
show "?P (H x) " by simp
next
fix x i d
assume HI1: "?P i"
assume HI2: "?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 HI1 HI2 by simp
also have "... = rev (x # inOrden d ) @ rev (inOrden i)" by simp
also have "... = rev ( inOrden i @ x # inOrden d) " by simp
also have "... = rev (inOrden (N x i d))" by simp
finally show "?P (N x i d)" 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
---------------------------------------------------------------------
*}

(* danrodcha crigomgom manmorjim1 ivamenjim bowma pablucoto*)
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
---------------------------------------------------------------------
*}

(* danrodcha crigomgom manmorjim1 ivamenjim bowma pablucoto *)
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
---------------------------------------------------------------------
*}

(* danrodcha crigomgom manmorjim1 ivamenjim bowma pablucoto*)
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"

(* bowma *)
fun extremo_derecha_1 :: "'a arbol ⇒ 'a" where
"extremo_derecha_1 (H t) = t"
| "extremo_derecha_1 (N t i d) = last (inOrden (N t i d))"

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

(* danrodcha *)
lemma aux_ej12: "inOrden a ≠ []"
apply (induct a)
apply simp
apply simp
done

(* danrodcha pablucoto *)
theorem "last (inOrden a) = extremo_derecha a" (is "?P a")
proof (induct a)
fix x
show "?P (H x)" by simp
next
fix x i d
assume HIi: "?P i"
assume HId: "?P d"
have "last (inOrden (N x i d)) = last (inOrden i @ [x] @ inOrden d)"
by (simp only: inOrden.simps(2))
also have "… = last (inOrden d)" by (simp add: aux_ej12)
also have "… = extremo_derecha d" using HId by simp
also have "… = extremo_derecha (N x i d)" by simp
finally show "?P (N x i d)" by simp
qed

(* ivamenjim *)

lemma aux_ej12_1: "inOrden a ≠ []"
by (induct a) simp_all

(* ivamenjim *)
(* Igual que la anterior, pero poniendo solo by simp en el primer have *)

theorem "last (inOrden a) = extremo_derecha a" (is "?P a")
proof (induct a)
fix x
show "?P (H x)" by simp
next
fix x
fix i assume h1: "?P i"
fix d assume h2: "?P d"
have "last (inOrden (N x i d)) = last ((inOrden i) @ [x] @ (inOrden d))" by simp
also have "... = last (inOrden d)" by (simp add: aux_ej12_1)
also have "... = extremo_derecha d" using h2 by simp
finally show "last (inOrden (N x i d)) = extremo_derecha (N x i d)" by simp
qed

(* bowma *)
(* Casi lo mismo que el anterior,pero no hace falta suponer "?p i" *)
theorem "last (inOrden a) = extremo_derecha a" (is "?p a")
proof (induct a)
fix t
show "?p (H t)" by simp
next
fix t i d
assume HI: "?p d"
have "last (inOrden (N t i d)) = last (inOrden i @ [t] @ inOrden d)" by simp
also have "... = last (inOrden d)" by (simp add:aux_ej12)
also have "... = extremo_derecha d" using HI by simp
finally show "?p (N t i d)" by simp
qed

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

(* danrodcha *)
theorem "hd (inOrden a) = extremo_izquierda a" (is "?P a")
proof (induct a)
fix x
show "?P (H x)" by simp
next
fix x i d
assume HIi: "?P i"
assume HId: "?P d"
have "hd (inOrden (N x i d)) = hd (inOrden i @ [x] @ inOrden d)"
by (simp only: inOrden.simps(2))
also have "… = hd (inOrden i)" by (simp add: aux_ej12)
also have "… = extremo_izquierda i" using HIi by simp
also have "… = extremo_izquierda (N x i d)" by simp
finally show "?P (N x i d)" by simp
qed

(* bowma *)
theorem "hd (inOrden a) = extremo_izquierda a" (is "?p a")
proof (induct a)
fix t
show "?p (H t)" by simp
next
fix t i d
assume HI: "?p i"
have "hd (inOrden (N t i d)) = hd (inOrden i @ [t] @ inOrden d)" by simp
also have "… = hd (inOrden i)" by (simp add: aux_ej12)
also have "… = extremo_izquierda i" using HI by simp
finally show "?p (N t i d)" by simp
qed

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

(* danrodcha *)
theorem "hd (preOrden a) = last (postOrden a)" (is "?P a")
proof (induct a)
fix x
show "?P (H x)" by simp
fix x i d
show "?P (N x i d)" by simp
qed

(* danrodcha *)
theorem "hd (preOrden a) = last (postOrden a)" (is "?P a")
proof (induct a)
fix x
show "?P (H x)" by simp
fix x i d
assume HIi: "?P i"
assume HId: "?P d"
have "hd (preOrden (N x i d)) = hd (x#preOrden i @ preOrden d)"
by (simp only: preOrden.simps(2))
also have "… = x" by simp
also have "… = last (postOrden i @ postOrden d @ [x])" by simp
also have "… = last (postOrden (N x i d))"
by (simp only: postOrden.simps(2))
finally show "?P (N x i d)" by simp
qed

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

(* danrodcha *)
theorem "hd (preOrden a) = raiz a" (is "?P a")
proof (induct a)
fix x
show "?P (H x)" by simp
fix x i d
assume HIi: "?P i"
assume HId: "?P d"
have "hd (preOrden (N x i d)) = hd (x#preOrden i @ preOrden d)"
by (simp only: preOrden.simps(2))
also have "… = x" by simp
also have "… = raiz (N x i d)" by simp
finally show "?P (N x i d)" by simp
qed

(* danrodcha *)
theorem "hd (preOrden a) = raiz a" (is "?P a")
proof (induct a)
fix x
show "?P (H x)" by simp
fix x i d
show "?P (N x i d)" by simp
qed

(* pablucoto *)
theorem "hd (preOrden a) = raiz a" (is "?P a")
proof (induct a)
fix x
show "?P (H x)" by simp
next
fix x i d
assume HI1: " ?P i"
assume HI2: " ?P d"
have " hd (preOrden (N x i d)) = hd ([x] @ preOrden i @ preOrden d) " by simp
also have "... = x" by simp
also have "... = raiz (N x i d)" by simp
finally show " ?P (N x i d)" by simp
qed

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

(*crigomgom pablucoto *)
theorem "hd (inOrden a) = raiz a"
quickcheck
oops

(* danrodcha:
Auto 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
---------------------------------------------------------------------
*}

(* danrodcha *)
theorem "last (postOrden a) = raiz a" (is "?P a")
proof (induct a)
fix x
show "?P (H x)" by simp
fix x i d
show "?P (N x i d)" by simp
qed

(* danrodcha *)
theorem "last (postOrden a) = raiz a" (is "?P a")
proof (induct a)
fix x
show "?P (H x)" by simp
fix x i d
assume HIi: "?P i"
assume HId: "?P d"
have "last (postOrden (N x i d)) = last (postOrden i @ postOrden d @ [x])"
by (simp only: postOrden.simps(2))
also have "… = x" by simp
also have "… = raiz (N x i d)" by (simp only: raiz.simps(2))
finally show "?P (N x i d)" by simp
qed

(* pablucoto *) (*Similar al anterior*)

theorem "last (postOrden a) = raiz a" (is "?P a")
proof (induct a )
fix x
show "?P (H x)" by simp
next
fix x i d
assume HI1: "?P i"
assume HI2: "?P d"
have "last (postOrden (N x i d)) = last ( postOrden i @ postOrden d @ [x])" by simp
also have "... = x" by simp
also have "... = raiz (N x i d) " by simp
finally show " ?P (N x i d)" by simp
qed

end
</source>

Relación 6

2016-12-04T03:14:49Z

Pablucoto:

<source lang="isar">
chapter {* R6: Recorridos de árboles *}

theory R6_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))".
---------------------------------------------------------------------
*}

(* ivamenjim marpoldia1 manmorjim1 bowma *)

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]
---------------------------------------------------------------------
*}

(* ivamenjim marpoldia1 pablucoto bowma*)

fun preOrden :: "'a arbol ⇒ 'a list" where
"preOrden (H t) = [t]"
| "preOrden (N t i d) = [t] @ (preOrden i) @ (preOrden d)"

(* danrodcha crigomgom manmorjim1 bowma *)
fun preOrden1 :: "'a arbol ⇒ 'a list" where
"preOrden1 (H x) = [x]"
| "preOrden1 (N x i d) = x#preOrden1 i @ preOrden1 d"

value "preOrden (N e (N c (H a) (H d)) (N g (H f) (H h)))
= [e,c,a,d,g,f,h]"
value "preOrden1 (N e (N c (H a) (H d)) (N g (H f) (H h)))
= [e,c,a,d,g,f,h]"

(* danrodcha *)
lemma "preOrden a = preOrden1 a"
by (induct a) simp_all

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]
---------------------------------------------------------------------
*}

(* ivamenjim danrodcha crigomgom marpoldia1 manmorjim1 pablucoto bowma*)

fun postOrden :: "'a arbol ⇒ 'a list" where
"postOrden (H t) = [t]"
| "postOrden (N t i d) = (postOrden i) @ (postOrden d) @ [t]"

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]
---------------------------------------------------------------------
*}

(* ivamenjim crigomgom marpoldia1 pablucoto bowma *)

fun inOrden :: "'a arbol ⇒ 'a list" where
"inOrden (H t) = [t]"
| "inOrden (N t i d) = (inOrden i) @ [t] @ (inOrden d)"

(* danrodcha manmorjim1 *)
fun inOrden1 :: "'a arbol ⇒ 'a list" where
"inOrden1 (H t) = [t]"
| "inOrden1 (N t i d) = inOrden1 i @ t#inOrden1 d"

value "inOrden (N e (N c (H a) (H d)) (N g (H f) (H h)))
= [a,c,d,e,f,g,h]"
value "inOrden1 (N e (N c (H a) (H d)) (N g (H f) (H h)))
= [a,c,d,e,f,g,h]"

(* manmorjim1 *)
lemma "inOrden t = inOrden1 t"
apply (induct t)
apply auto
done

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))
---------------------------------------------------------------------
*}

(* ivamenjim danrodcha crigomgom marpoldia1 manmorjim1 pablucoto bowma *)

fun espejo :: "'a arbol ⇒ 'a arbol" where
"espejo (H t) = H t"
| "espejo (N t i d) = N t (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)
---------------------------------------------------------------------
*}

(* ivamenjim *)

lemma "preOrden (espejo a) = rev (postOrden a)" (is "?P a")
proof (induct a)
fix x
show "?P (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
finally show "preOrden (espejo (N x i d)) = rev (postOrden (N x i d))" by simp
qed

(* danrodcha *)
lemma "preOrden (espejo a) = rev (postOrden a)"
by (induct a) simp_all

(* danrodcha crigomgom*)
lemma "preOrden (espejo a) = rev (postOrden a)" (is "?P a")
proof (induct a)
fix x
show "?P (H x)" by simp
next
fix x i d
assume HIi: "?P i"
assume HId: "?P d"
have "preOrden (espejo (N x i d)) = preOrden (N x (espejo d) (espejo i))"
by (simp only: espejo.simps(2))
also have "… = x#preOrden (espejo d) @ preOrden (espejo i)"
by (simp only: preOrden.simps(2))
also have"… = x#rev (postOrden d) @ rev (postOrden i)"
using HIi HId by simp
also have "… = rev (postOrden (N x i d))" by simp
finally show "?P (N x i d)" by simp
qed

(* danrodcha *)
lemma "preOrden (espejo a) = rev (postOrden a)" (is "?P a")
proof (induct a)
fix x
show "?P (H x)" by simp
next
fix x i d
assume HIi: "?P i"
assume HId: "?P d"
show "?P (N x i d)" using HIi HId by simp
qed

(* bowma *)
lemma "preOrden (espejo a) = rev (postOrden a)"
apply (induct a)
apply simp_all
done

(* pablucoto *)

lemma "preOrden (espejo a) = rev (postOrden a)" (is "?P a")
proof (induct a)
fix x
show "?P (H x)" by simp
next
fix x i d
assume h1: "?P i"
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 "... = [x] @ rev (postOrden i @ postOrden d)" by simp
also have "... = rev ( postOrden i @ postOrden d @ [x] ) " by simp
also have "... = rev (postOrden (N x i d)) " by simp
finally show "?P (N x i d)" by simp
qed

(* bowma *)
lemma "preOrden (espejo a) = rev (postOrden a)" (is "?p a")
proof (induct a)
fix t
show "?p (H t)" by simp
(* Aquí si le diga "preOrden (espejo (H t)) = rev (postOrden (H t))",isabelle dice:
proof (prove)
goal (1 subgoal):
1. preOrden (espejo (H t)) = rev (postOrden (H t))
Introduced fixed type variable(s): 'b in "t__"
No entiendo porqué *)
next
fix t i d
assume H1: "?p i"
assume H2: "?p d"
have "preOrden (espejo (N t i d)) = preOrden (N t (espejo d) (espejo i))" by simp
also have "... = [t] @ (preOrden (espejo d)) @ (preOrden (espejo i))" by simp
also have "... = [t] @ rev (postOrden d) @ rev (postOrden i)" using H1 H2 by simp
finally show "?p (N t i d)" by simp
qed

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

(* ivamenjim crigomgom bowma *)

lemma "postOrden (espejo a) = rev (preOrden a)" (is "?P a")
proof (induct a)
fix x
show "?P (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
finally show "postOrden (espejo (N x i d)) = rev (preOrden (N x i d))" by simp
(* "?p (N x i d)" más corto *)
qed

(* danrodcha *)
lemma "postOrden (espejo a) = rev (preOrden a)" (is "?P a")
proof (induct a)
fix x
show "?P (H x)" by simp
next
fix x i d
assume HIi: "?P i"
assume HId: "?P d"
show "?P (N x i d)" using HIi HId by simp
qed

(* pablucoto *)

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

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

(* ivamenjim crigomgom bowma *)

theorem "inOrden (espejo a) = rev (inOrden a)" (is "?P a")
proof (induct a)
fix x
show "?P (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
finally show "inOrden (espejo (N x i d)) = rev (inOrden (N x i d))" by simp
qed

(* danrodcha *)
theorem "inOrden (espejo a) = rev (inOrden a)"
by (induct a) simp_all

(* danrodcha *)
theorem "inOrden (espejo a) = rev (inOrden a)" (is "?P a")
proof (induct a)
fix x
show "?P (H x)" by simp
next
fix x i d
assume HIi: "?P i"
assume HId: "?P d"
show "?P (N x i d)" using HIi HId by simp
qed

(* pablucoto *)
theorem "inOrden (espejo a) = rev (inOrden a)" (is "?P a")
proof (induct a)
fix x
show "?P (H x) " by simp
next
fix x i d
assume HI1: "?P i"
assume HI2: "?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 HI1 HI2 by simp
also have "... = rev (x # inOrden d ) @ rev (inOrden i)" by simp
also have "... = rev ( inOrden i @ x # inOrden d) " by simp
also have "... = rev (inOrden (N x i d))" by simp
finally show "?P (N x i d)" 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
---------------------------------------------------------------------
*}

(* danrodcha crigomgom manmorjim1 ivamenjim bowma pablucoto*)
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
---------------------------------------------------------------------
*}

(* danrodcha crigomgom manmorjim1 ivamenjim bowma pablucoto *)
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
---------------------------------------------------------------------
*}

(* danrodcha crigomgom manmorjim1 ivamenjim bowma pablucoto*)
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"

(* bowma *)
fun extremo_derecha_1 :: "'a arbol ⇒ 'a" where
"extremo_derecha_1 (H t) = t"
| "extremo_derecha_1 (N t i d) = last (inOrden (N t i d))"

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

(* danrodcha *)
lemma aux_ej12: "inOrden a ≠ []"
apply (induct a)
apply simp
apply simp
done

(* danrodcha pablucoto *)
theorem "last (inOrden a) = extremo_derecha a" (is "?P a")
proof (induct a)
fix x
show "?P (H x)" by simp
next
fix x i d
assume HIi: "?P i"
assume HId: "?P d"
have "last (inOrden (N x i d)) = last (inOrden i @ [x] @ inOrden d)"
by (simp only: inOrden.simps(2))
also have "… = last (inOrden d)" by (simp add: aux_ej12)
also have "… = extremo_derecha d" using HId by simp
also have "… = extremo_derecha (N x i d)" by simp
finally show "?P (N x i d)" by simp
qed

(* ivamenjim *)

lemma aux_ej12_1: "inOrden a ≠ []"
by (induct a) simp_all

(* ivamenjim *)
(* Igual que la anterior, pero poniendo solo by simp en el primer have *)

theorem "last (inOrden a) = extremo_derecha a" (is "?P a")
proof (induct a)
fix x
show "?P (H x)" by simp
next
fix x
fix i assume h1: "?P i"
fix d assume h2: "?P d"
have "last (inOrden (N x i d)) = last ((inOrden i) @ [x] @ (inOrden d))" by simp
also have "... = last (inOrden d)" by (simp add: aux_ej12_1)
also have "... = extremo_derecha d" using h2 by simp
finally show "last (inOrden (N x i d)) = extremo_derecha (N x i d)" by simp
qed

(* bowma *)
(* Casi lo mismo que el anterior,pero no hace falta suponer "?p i" *)
theorem "last (inOrden a) = extremo_derecha a" (is "?p a")
proof (induct a)
fix t
show "?p (H t)" by simp
next
fix t i d
assume HI: "?p d"
have "last (inOrden (N t i d)) = last (inOrden i @ [t] @ inOrden d)" by simp
also have "... = last (inOrden d)" by (simp add:aux_ej12)
also have "... = extremo_derecha d" using HI by simp
finally show "?p (N t i d)" by simp
qed

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

(* danrodcha *)
theorem "hd (inOrden a) = extremo_izquierda a" (is "?P a")
proof (induct a)
fix x
show "?P (H x)" by simp
next
fix x i d
assume HIi: "?P i"
assume HId: "?P d"
have "hd (inOrden (N x i d)) = hd (inOrden i @ [x] @ inOrden d)"
by (simp only: inOrden.simps(2))
also have "… = hd (inOrden i)" by (simp add: aux_ej12)
also have "… = extremo_izquierda i" using HIi by simp
also have "… = extremo_izquierda (N x i d)" by simp
finally show "?P (N x i d)" by simp
qed

(* bowma *)
theorem "hd (inOrden a) = extremo_izquierda a" (is "?p a")
proof (induct a)
fix t
show "?p (H t)" by simp
next
fix t i d
assume HI: "?p i"
have "hd (inOrden (N t i d)) = hd (inOrden i @ [t] @ inOrden d)" by simp
also have "… = hd (inOrden i)" by (simp add: aux_ej12)
also have "… = extremo_izquierda i" using HI by simp
finally show "?p (N t i d)" by simp
qed

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

(* danrodcha *)
theorem "hd (preOrden a) = last (postOrden a)" (is "?P a")
proof (induct a)
fix x
show "?P (H x)" by simp
fix x i d
show "?P (N x i d)" by simp
qed

(* danrodcha *)
theorem "hd (preOrden a) = last (postOrden a)" (is "?P a")
proof (induct a)
fix x
show "?P (H x)" by simp
fix x i d
assume HIi: "?P i"
assume HId: "?P d"
have "hd (preOrden (N x i d)) = hd (x#preOrden i @ preOrden d)"
by (simp only: preOrden.simps(2))
also have "… = x" by simp
also have "… = last (postOrden i @ postOrden d @ [x])" by simp
also have "… = last (postOrden (N x i d))"
by (simp only: postOrden.simps(2))
finally show "?P (N x i d)" by simp
qed

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

(* danrodcha *)
theorem "hd (preOrden a) = raiz a" (is "?P a")
proof (induct a)
fix x
show "?P (H x)" by simp
fix x i d
assume HIi: "?P i"
assume HId: "?P d"
have "hd (preOrden (N x i d)) = hd (x#preOrden i @ preOrden d)"
by (simp only: preOrden.simps(2))
also have "… = x" by simp
also have "… = raiz (N x i d)" by simp
finally show "?P (N x i d)" by simp
qed

(* danrodcha *)
theorem "hd (preOrden a) = raiz a" (is "?P a")
proof (induct a)
fix x
show "?P (H x)" by simp
fix x i d
show "?P (N x i d)" by simp
qed

(* pablucoto *)
theorem "hd (preOrden a) = raiz a" (is "?P a")
proof (induct a)
fix x
show "?P (H x)" by simp
next
fix x i d
assume HI1: " ?P i"
assume HI2: " ?P d"
have " hd (preOrden (N x i d)) = hd ([x] @ preOrden i @ preOrden d) " by simp
also have "... = x" by simp
also have "... = raiz (N x i d)" by simp
finally show " ?P (N x i d)" by simp
qed

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

(*crigomgom pablucoto *)
theorem "hd (inOrden a) = raiz a"
quickcheck
oops

(* danrodcha:
Auto 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
---------------------------------------------------------------------
*}

(* danrodcha *)
theorem "last (postOrden a) = raiz a" (is "?P a")
proof (induct a)
fix x
show "?P (H x)" by simp
fix x i d
show "?P (N x i d)" by simp
qed

(* danrodcha *)
theorem "last (postOrden a) = raiz a" (is "?P a")
proof (induct a)
fix x
show "?P (H x)" by simp
fix x i d
assume HIi: "?P i"
assume HId: "?P d"
have "last (postOrden (N x i d)) = last (postOrden i @ postOrden d @ [x])"
by (simp only: postOrden.simps(2))
also have "… = x" by simp
also have "… = raiz (N x i d)" by (simp only: raiz.simps(2))
finally show "?P (N x i d)" by simp
qed

(* pablucoto *) (*Similar al anterior*)

theorem "last (postOrden a) = raiz a" (is "?P a")
proof (induct a )
fix x
show "?P (H x)" by simp
next
fix x i d
assume HI1: "?P i"
assume HI1: "?P d"
have "last (postOrden (N x i d)) = last ( postOrden i @ postOrden d @ [x])" by simp
also have "... = x" by simp
also have "... = raiz (N x i d) " by simp
finally show " ?P (N x i d)" by simp
qed

end
</source>

Relación 6

2016-12-04T03:08:06Z

Pablucoto:

<source lang="isar">
chapter {* R6: Recorridos de árboles *}

theory R6_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))".
---------------------------------------------------------------------
*}

(* ivamenjim marpoldia1 manmorjim1 bowma *)

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]
---------------------------------------------------------------------
*}

(* ivamenjim marpoldia1 pablucoto bowma*)

fun preOrden :: "'a arbol ⇒ 'a list" where
"preOrden (H t) = [t]"
| "preOrden (N t i d) = [t] @ (preOrden i) @ (preOrden d)"

(* danrodcha crigomgom manmorjim1 bowma *)
fun preOrden1 :: "'a arbol ⇒ 'a list" where
"preOrden1 (H x) = [x]"
| "preOrden1 (N x i d) = x#preOrden1 i @ preOrden1 d"

value "preOrden (N e (N c (H a) (H d)) (N g (H f) (H h)))
= [e,c,a,d,g,f,h]"
value "preOrden1 (N e (N c (H a) (H d)) (N g (H f) (H h)))
= [e,c,a,d,g,f,h]"

(* danrodcha *)
lemma "preOrden a = preOrden1 a"
by (induct a) simp_all

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]
---------------------------------------------------------------------
*}

(* ivamenjim danrodcha crigomgom marpoldia1 manmorjim1 pablucoto bowma*)

fun postOrden :: "'a arbol ⇒ 'a list" where
"postOrden (H t) = [t]"
| "postOrden (N t i d) = (postOrden i) @ (postOrden d) @ [t]"

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]
---------------------------------------------------------------------
*}

(* ivamenjim crigomgom marpoldia1 pablucoto bowma *)

fun inOrden :: "'a arbol ⇒ 'a list" where
"inOrden (H t) = [t]"
| "inOrden (N t i d) = (inOrden i) @ [t] @ (inOrden d)"

(* danrodcha manmorjim1 *)
fun inOrden1 :: "'a arbol ⇒ 'a list" where
"inOrden1 (H t) = [t]"
| "inOrden1 (N t i d) = inOrden1 i @ t#inOrden1 d"

value "inOrden (N e (N c (H a) (H d)) (N g (H f) (H h)))
= [a,c,d,e,f,g,h]"
value "inOrden1 (N e (N c (H a) (H d)) (N g (H f) (H h)))
= [a,c,d,e,f,g,h]"

(* manmorjim1 *)
lemma "inOrden t = inOrden1 t"
apply (induct t)
apply auto
done

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))
---------------------------------------------------------------------
*}

(* ivamenjim danrodcha crigomgom marpoldia1 manmorjim1 pablucoto bowma *)

fun espejo :: "'a arbol ⇒ 'a arbol" where
"espejo (H t) = H t"
| "espejo (N t i d) = N t (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)
---------------------------------------------------------------------
*}

(* ivamenjim *)

lemma "preOrden (espejo a) = rev (postOrden a)" (is "?P a")
proof (induct a)
fix x
show "?P (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
finally show "preOrden (espejo (N x i d)) = rev (postOrden (N x i d))" by simp
qed

(* danrodcha *)
lemma "preOrden (espejo a) = rev (postOrden a)"
by (induct a) simp_all

(* danrodcha crigomgom*)
lemma "preOrden (espejo a) = rev (postOrden a)" (is "?P a")
proof (induct a)
fix x
show "?P (H x)" by simp
next
fix x i d
assume HIi: "?P i"
assume HId: "?P d"
have "preOrden (espejo (N x i d)) = preOrden (N x (espejo d) (espejo i))"
by (simp only: espejo.simps(2))
also have "… = x#preOrden (espejo d) @ preOrden (espejo i)"
by (simp only: preOrden.simps(2))
also have"… = x#rev (postOrden d) @ rev (postOrden i)"
using HIi HId by simp
also have "… = rev (postOrden (N x i d))" by simp
finally show "?P (N x i d)" by simp
qed

(* danrodcha *)
lemma "preOrden (espejo a) = rev (postOrden a)" (is "?P a")
proof (induct a)
fix x
show "?P (H x)" by simp
next
fix x i d
assume HIi: "?P i"
assume HId: "?P d"
show "?P (N x i d)" using HIi HId by simp
qed

(* bowma *)
lemma "preOrden (espejo a) = rev (postOrden a)"
apply (induct a)
apply simp_all
done

(* pablucoto *)

lemma "preOrden (espejo a) = rev (postOrden a)" (is "?P a")
proof (induct a)
fix x
show "?P (H x)" by simp
next
fix x i d
assume h1: "?P i"
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 "... = [x] @ rev (postOrden i @ postOrden d)" by simp
also have "... = rev ( postOrden i @ postOrden d @ [x] ) " by simp
also have "... = rev (postOrden (N x i d)) " by simp
finally show "?P (N x i d)" by simp
qed

(* bowma *)
lemma "preOrden (espejo a) = rev (postOrden a)" (is "?p a")
proof (induct a)
fix t
show "?p (H t)" by simp
(* Aquí si le diga "preOrden (espejo (H t)) = rev (postOrden (H t))",isabelle dice:
proof (prove)
goal (1 subgoal):
1. preOrden (espejo (H t)) = rev (postOrden (H t))
Introduced fixed type variable(s): 'b in "t__"
No entiendo porqué *)
next
fix t i d
assume H1: "?p i"
assume H2: "?p d"
have "preOrden (espejo (N t i d)) = preOrden (N t (espejo d) (espejo i))" by simp
also have "... = [t] @ (preOrden (espejo d)) @ (preOrden (espejo i))" by simp
also have "... = [t] @ rev (postOrden d) @ rev (postOrden i)" using H1 H2 by simp
finally show "?p (N t i d)" by simp
qed

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

(* ivamenjim crigomgom bowma *)

lemma "postOrden (espejo a) = rev (preOrden a)" (is "?P a")
proof (induct a)
fix x
show "?P (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
finally show "postOrden (espejo (N x i d)) = rev (preOrden (N x i d))" by simp
(* "?p (N x i d)" más corto *)
qed

(* danrodcha *)
lemma "postOrden (espejo a) = rev (preOrden a)" (is "?P a")
proof (induct a)
fix x
show "?P (H x)" by simp
next
fix x i d
assume HIi: "?P i"
assume HId: "?P d"
show "?P (N x i d)" using HIi HId by simp
qed

(* pablucoto *)

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

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

(* ivamenjim crigomgom bowma *)

theorem "inOrden (espejo a) = rev (inOrden a)" (is "?P a")
proof (induct a)
fix x
show "?P (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
finally show "inOrden (espejo (N x i d)) = rev (inOrden (N x i d))" by simp
qed

(* danrodcha *)
theorem "inOrden (espejo a) = rev (inOrden a)"
by (induct a) simp_all

(* danrodcha *)
theorem "inOrden (espejo a) = rev (inOrden a)" (is "?P a")
proof (induct a)
fix x
show "?P (H x)" by simp
next
fix x i d
assume HIi: "?P i"
assume HId: "?P d"
show "?P (N x i d)" using HIi HId by simp
qed

(* pablucoto *)
theorem "inOrden (espejo a) = rev (inOrden a)" (is "?P a")
proof (induct a)
fix x
show "?P (H x) " by simp
next
fix x i d
assume HI1: "?P i"
assume HI2: "?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 HI1 HI2 by simp
also have "... = rev (x # inOrden d ) @ rev (inOrden i)" by simp
also have "... = rev ( inOrden i @ x # inOrden d) " by simp
also have "... = rev (inOrden (N x i d))" by simp
finally show "?P (N x i d)" 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
---------------------------------------------------------------------
*}

(* danrodcha crigomgom manmorjim1 ivamenjim bowma pablucoto*)
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
---------------------------------------------------------------------
*}

(* danrodcha crigomgom manmorjim1 ivamenjim bowma pablucoto *)
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
---------------------------------------------------------------------
*}

(* danrodcha crigomgom manmorjim1 ivamenjim bowma pablucoto*)
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"

(* bowma *)
fun extremo_derecha_1 :: "'a arbol ⇒ 'a" where
"extremo_derecha_1 (H t) = t"
| "extremo_derecha_1 (N t i d) = last (inOrden (N t i d))"

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

(* danrodcha *)
lemma aux_ej12: "inOrden a ≠ []"
apply (induct a)
apply simp
apply simp
done

(* danrodcha pablucoto *)
theorem "last (inOrden a) = extremo_derecha a" (is "?P a")
proof (induct a)
fix x
show "?P (H x)" by simp
next
fix x i d
assume HIi: "?P i"
assume HId: "?P d"
have "last (inOrden (N x i d)) = last (inOrden i @ [x] @ inOrden d)"
by (simp only: inOrden.simps(2))
also have "… = last (inOrden d)" by (simp add: aux_ej12)
also have "… = extremo_derecha d" using HId by simp
also have "… = extremo_derecha (N x i d)" by simp
finally show "?P (N x i d)" by simp
qed

(* ivamenjim *)

lemma aux_ej12_1: "inOrden a ≠ []"
by (induct a) simp_all

(* ivamenjim *)
(* Igual que la anterior, pero poniendo solo by simp en el primer have *)

theorem "last (inOrden a) = extremo_derecha a" (is "?P a")
proof (induct a)
fix x
show "?P (H x)" by simp
next
fix x
fix i assume h1: "?P i"
fix d assume h2: "?P d"
have "last (inOrden (N x i d)) = last ((inOrden i) @ [x] @ (inOrden d))" by simp
also have "... = last (inOrden d)" by (simp add: aux_ej12_1)
also have "... = extremo_derecha d" using h2 by simp
finally show "last (inOrden (N x i d)) = extremo_derecha (N x i d)" by simp
qed

(* bowma *)
(* Casi lo mismo que el anterior,pero no hace falta suponer "?p i" *)
theorem "last (inOrden a) = extremo_derecha a" (is "?p a")
proof (induct a)
fix t
show "?p (H t)" by simp
next
fix t i d
assume HI: "?p d"
have "last (inOrden (N t i d)) = last (inOrden i @ [t] @ inOrden d)" by simp
also have "... = last (inOrden d)" by (simp add:aux_ej12)
also have "... = extremo_derecha d" using HI by simp
finally show "?p (N t i d)" by simp
qed

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

(* danrodcha *)
theorem "hd (inOrden a) = extremo_izquierda a" (is "?P a")
proof (induct a)
fix x
show "?P (H x)" by simp
next
fix x i d
assume HIi: "?P i"
assume HId: "?P d"
have "hd (inOrden (N x i d)) = hd (inOrden i @ [x] @ inOrden d)"
by (simp only: inOrden.simps(2))
also have "… = hd (inOrden i)" by (simp add: aux_ej12)
also have "… = extremo_izquierda i" using HIi by simp
also have "… = extremo_izquierda (N x i d)" by simp
finally show "?P (N x i d)" by simp
qed

(* bowma *)
theorem "hd (inOrden a) = extremo_izquierda a" (is "?p a")
proof (induct a)
fix t
show "?p (H t)" by simp
next
fix t i d
assume HI: "?p i"
have "hd (inOrden (N t i d)) = hd (inOrden i @ [t] @ inOrden d)" by simp
also have "… = hd (inOrden i)" by (simp add: aux_ej12)
also have "… = extremo_izquierda i" using HI by simp
finally show "?p (N t i d)" by simp
qed

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

(* danrodcha *)
theorem "hd (preOrden a) = last (postOrden a)" (is "?P a")
proof (induct a)
fix x
show "?P (H x)" by simp
fix x i d
show "?P (N x i d)" by simp
qed

(* danrodcha *)
theorem "hd (preOrden a) = last (postOrden a)" (is "?P a")
proof (induct a)
fix x
show "?P (H x)" by simp
fix x i d
assume HIi: "?P i"
assume HId: "?P d"
have "hd (preOrden (N x i d)) = hd (x#preOrden i @ preOrden d)"
by (simp only: preOrden.simps(2))
also have "… = x" by simp
also have "… = last (postOrden i @ postOrden d @ [x])" by simp
also have "… = last (postOrden (N x i d))"
by (simp only: postOrden.simps(2))
finally show "?P (N x i d)" by simp
qed

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

(* danrodcha *)
theorem "hd (preOrden a) = raiz a" (is "?P a")
proof (induct a)
fix x
show "?P (H x)" by simp
fix x i d
assume HIi: "?P i"
assume HId: "?P d"
have "hd (preOrden (N x i d)) = hd (x#preOrden i @ preOrden d)"
by (simp only: preOrden.simps(2))
also have "… = x" by simp
also have "… = raiz (N x i d)" by simp
finally show "?P (N x i d)" by simp
qed

(* danrodcha *)
theorem "hd (preOrden a) = raiz a" (is "?P a")
proof (induct a)
fix x
show "?P (H x)" by simp
fix x i d
show "?P (N x i d)" by simp
qed

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

(*crigomgom pablucoto *)
theorem "hd (inOrden a) = raiz a"
quickcheck
oops

(* danrodcha:
Auto 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
---------------------------------------------------------------------
*}

(* danrodcha *)
theorem "last (postOrden a) = raiz a" (is "?P a")
proof (induct a)
fix x
show "?P (H x)" by simp
fix x i d
show "?P (N x i d)" by simp
qed

(* danrodcha *)
theorem "last (postOrden a) = raiz a" (is "?P a")
proof (induct a)
fix x
show "?P (H x)" by simp
fix x i d
assume HIi: "?P i"
assume HId: "?P d"
have "last (postOrden (N x i d)) = last (postOrden i @ postOrden d @ [x])"
by (simp only: postOrden.simps(2))
also have "… = x" by simp
also have "… = raiz (N x i d)" by (simp only: raiz.simps(2))
finally show "?P (N x i d)" by simp
qed

(* pablucoto *) (*Similar al anterior*)

theorem "last (postOrden a) = raiz a" (is "?P a")
proof (induct a )
fix x
show "?P (H x)" by simp
next
fix x i d
assume HI1: "?P i"
assume HI1: "?P d"
have "last (postOrden (N x i d)) = last ( postOrden i @ postOrden d @ [x])" by simp
also have "... = x" by simp
also have "... = raiz (N x i d) " by simp
finally show " ?P (N x i d)" by simp
qed

end
</source>

Relación 6

2016-12-04T03:03:20Z

Pablucoto:

<source lang="isar">
chapter {* R6: Recorridos de árboles *}

theory R6_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))".
---------------------------------------------------------------------
*}

(* ivamenjim marpoldia1 manmorjim1 bowma *)

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]
---------------------------------------------------------------------
*}

(* ivamenjim marpoldia1 pablucoto bowma*)

fun preOrden :: "'a arbol ⇒ 'a list" where
"preOrden (H t) = [t]"
| "preOrden (N t i d) = [t] @ (preOrden i) @ (preOrden d)"

(* danrodcha crigomgom manmorjim1 bowma *)
fun preOrden1 :: "'a arbol ⇒ 'a list" where
"preOrden1 (H x) = [x]"
| "preOrden1 (N x i d) = x#preOrden1 i @ preOrden1 d"

value "preOrden (N e (N c (H a) (H d)) (N g (H f) (H h)))
= [e,c,a,d,g,f,h]"
value "preOrden1 (N e (N c (H a) (H d)) (N g (H f) (H h)))
= [e,c,a,d,g,f,h]"

(* danrodcha *)
lemma "preOrden a = preOrden1 a"
by (induct a) simp_all

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]
---------------------------------------------------------------------
*}

(* ivamenjim danrodcha crigomgom marpoldia1 manmorjim1 pablucoto bowma*)

fun postOrden :: "'a arbol ⇒ 'a list" where
"postOrden (H t) = [t]"
| "postOrden (N t i d) = (postOrden i) @ (postOrden d) @ [t]"

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]
---------------------------------------------------------------------
*}

(* ivamenjim crigomgom marpoldia1 pablucoto bowma *)

fun inOrden :: "'a arbol ⇒ 'a list" where
"inOrden (H t) = [t]"
| "inOrden (N t i d) = (inOrden i) @ [t] @ (inOrden d)"

(* danrodcha manmorjim1 *)
fun inOrden1 :: "'a arbol ⇒ 'a list" where
"inOrden1 (H t) = [t]"
| "inOrden1 (N t i d) = inOrden1 i @ t#inOrden1 d"

value "inOrden (N e (N c (H a) (H d)) (N g (H f) (H h)))
= [a,c,d,e,f,g,h]"
value "inOrden1 (N e (N c (H a) (H d)) (N g (H f) (H h)))
= [a,c,d,e,f,g,h]"

(* manmorjim1 *)
lemma "inOrden t = inOrden1 t"
apply (induct t)
apply auto
done

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))
---------------------------------------------------------------------
*}

(* ivamenjim danrodcha crigomgom marpoldia1 manmorjim1 pablucoto bowma *)

fun espejo :: "'a arbol ⇒ 'a arbol" where
"espejo (H t) = H t"
| "espejo (N t i d) = N t (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)
---------------------------------------------------------------------
*}

(* ivamenjim *)

lemma "preOrden (espejo a) = rev (postOrden a)" (is "?P a")
proof (induct a)
fix x
show "?P (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
finally show "preOrden (espejo (N x i d)) = rev (postOrden (N x i d))" by simp
qed

(* danrodcha *)
lemma "preOrden (espejo a) = rev (postOrden a)"
by (induct a) simp_all

(* danrodcha crigomgom*)
lemma "preOrden (espejo a) = rev (postOrden a)" (is "?P a")
proof (induct a)
fix x
show "?P (H x)" by simp
next
fix x i d
assume HIi: "?P i"
assume HId: "?P d"
have "preOrden (espejo (N x i d)) = preOrden (N x (espejo d) (espejo i))"
by (simp only: espejo.simps(2))
also have "… = x#preOrden (espejo d) @ preOrden (espejo i)"
by (simp only: preOrden.simps(2))
also have"… = x#rev (postOrden d) @ rev (postOrden i)"
using HIi HId by simp
also have "… = rev (postOrden (N x i d))" by simp
finally show "?P (N x i d)" by simp
qed

(* danrodcha *)
lemma "preOrden (espejo a) = rev (postOrden a)" (is "?P a")
proof (induct a)
fix x
show "?P (H x)" by simp
next
fix x i d
assume HIi: "?P i"
assume HId: "?P d"
show "?P (N x i d)" using HIi HId by simp
qed

(* bowma *)
lemma "preOrden (espejo a) = rev (postOrden a)"
apply (induct a)
apply simp_all
done

(* pablucoto *)

lemma "preOrden (espejo a) = rev (postOrden a)" (is "?P a")
proof (induct a)
fix x
show "?P (H x)" by simp
next
fix x i d
assume h1: "?P i"
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 "... = [x] @ rev (postOrden i @ postOrden d)" by simp
also have "... = rev ( postOrden i @ postOrden d @ [x] ) " by simp
also have "... = rev (postOrden (N x i d)) " by simp
finally show "?P (N x i d)" by simp
qed

(* bowma *)
lemma "preOrden (espejo a) = rev (postOrden a)" (is "?p a")
proof (induct a)
fix t
show "?p (H t)" by simp
(* Aquí si le diga "preOrden (espejo (H t)) = rev (postOrden (H t))",isabelle dice:
proof (prove)
goal (1 subgoal):
1. preOrden (espejo (H t)) = rev (postOrden (H t))
Introduced fixed type variable(s): 'b in "t__"
No entiendo porqué *)
next
fix t i d
assume H1: "?p i"
assume H2: "?p d"
have "preOrden (espejo (N t i d)) = preOrden (N t (espejo d) (espejo i))" by simp
also have "... = [t] @ (preOrden (espejo d)) @ (preOrden (espejo i))" by simp
also have "... = [t] @ rev (postOrden d) @ rev (postOrden i)" using H1 H2 by simp
finally show "?p (N t i d)" by simp
qed

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

(* ivamenjim crigomgom bowma *)

lemma "postOrden (espejo a) = rev (preOrden a)" (is "?P a")
proof (induct a)
fix x
show "?P (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
finally show "postOrden (espejo (N x i d)) = rev (preOrden (N x i d))" by simp
(* "?p (N x i d)" más corto *)
qed

(* danrodcha *)
lemma "postOrden (espejo a) = rev (preOrden a)" (is "?P a")
proof (induct a)
fix x
show "?P (H x)" by simp
next
fix x i d
assume HIi: "?P i"
assume HId: "?P d"
show "?P (N x i d)" using HIi HId by simp
qed

(* pablucoto *)

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

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

(* ivamenjim crigomgom bowma *)

theorem "inOrden (espejo a) = rev (inOrden a)" (is "?P a")
proof (induct a)
fix x
show "?P (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
finally show "inOrden (espejo (N x i d)) = rev (inOrden (N x i d))" by simp
qed

(* danrodcha *)
theorem "inOrden (espejo a) = rev (inOrden a)"
by (induct a) simp_all

(* danrodcha *)
theorem "inOrden (espejo a) = rev (inOrden a)" (is "?P a")
proof (induct a)
fix x
show "?P (H x)" by simp
next
fix x i d
assume HIi: "?P i"
assume HId: "?P d"
show "?P (N x i d)" using HIi HId by simp
qed

(* pablucoto *)
theorem "inOrden (espejo a) = rev (inOrden a)" (is "?P a")
proof (induct a)
fix x
show "?P (H x) " by simp
next
fix x i d
assume HI1: "?P i"
assume HI2: "?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 HI1 HI2 by simp
also have "... = rev (x # inOrden d ) @ rev (inOrden i)" by simp
also have "... = rev ( inOrden i @ x # inOrden d) " by simp
also have "... = rev (inOrden (N x i d))" by simp
finally show "?P (N x i d)" 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
---------------------------------------------------------------------
*}

(* danrodcha crigomgom manmorjim1 ivamenjim bowma pablucoto*)
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
---------------------------------------------------------------------
*}

(* danrodcha crigomgom manmorjim1 ivamenjim bowma pablucoto *)
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
---------------------------------------------------------------------
*}

(* danrodcha crigomgom manmorjim1 ivamenjim bowma pablucoto*)
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"

(* bowma *)
fun extremo_derecha_1 :: "'a arbol ⇒ 'a" where
"extremo_derecha_1 (H t) = t"
| "extremo_derecha_1 (N t i d) = last (inOrden (N t i d))"

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

(* danrodcha *)
lemma aux_ej12: "inOrden a ≠ []"
apply (induct a)
apply simp
apply simp
done

(* danrodcha pablucoto *)
theorem "last (inOrden a) = extremo_derecha a" (is "?P a")
proof (induct a)
fix x
show "?P (H x)" by simp
next
fix x i d
assume HIi: "?P i"
assume HId: "?P d"
have "last (inOrden (N x i d)) = last (inOrden i @ [x] @ inOrden d)"
by (simp only: inOrden.simps(2))
also have "… = last (inOrden d)" by (simp add: aux_ej12)
also have "… = extremo_derecha d" using HId by simp
also have "… = extremo_derecha (N x i d)" by simp
finally show "?P (N x i d)" by simp
qed

(* ivamenjim *)

lemma aux_ej12_1: "inOrden a ≠ []"
by (induct a) simp_all

(* ivamenjim *)
(* Igual que la anterior, pero poniendo solo by simp en el primer have *)

theorem "last (inOrden a) = extremo_derecha a" (is "?P a")
proof (induct a)
fix x
show "?P (H x)" by simp
next
fix x
fix i assume h1: "?P i"
fix d assume h2: "?P d"
have "last (inOrden (N x i d)) = last ((inOrden i) @ [x] @ (inOrden d))" by simp
also have "... = last (inOrden d)" by (simp add: aux_ej12_1)
also have "... = extremo_derecha d" using h2 by simp
finally show "last (inOrden (N x i d)) = extremo_derecha (N x i d)" by simp
qed

(* bowma *)
(* Casi lo mismo que el anterior,pero no hace falta suponer "?p i" *)
theorem "last (inOrden a) = extremo_derecha a" (is "?p a")
proof (induct a)
fix t
show "?p (H t)" by simp
next
fix t i d
assume HI: "?p d"
have "last (inOrden (N t i d)) = last (inOrden i @ [t] @ inOrden d)" by simp
also have "... = last (inOrden d)" by (simp add:aux_ej12)
also have "... = extremo_derecha d" using HI by simp
finally show "?p (N t i d)" by simp
qed

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

(* danrodcha *)
theorem "hd (inOrden a) = extremo_izquierda a" (is "?P a")
proof (induct a)
fix x
show "?P (H x)" by simp
next
fix x i d
assume HIi: "?P i"
assume HId: "?P d"
have "hd (inOrden (N x i d)) = hd (inOrden i @ [x] @ inOrden d)"
by (simp only: inOrden.simps(2))
also have "… = hd (inOrden i)" by (simp add: aux_ej12)
also have "… = extremo_izquierda i" using HIi by simp
also have "… = extremo_izquierda (N x i d)" by simp
finally show "?P (N x i d)" by simp
qed

(* bowma *)
theorem "hd (inOrden a) = extremo_izquierda a" (is "?p a")
proof (induct a)
fix t
show "?p (H t)" by simp
next
fix t i d
assume HI: "?p i"
have "hd (inOrden (N t i d)) = hd (inOrden i @ [t] @ inOrden d)" by simp
also have "… = hd (inOrden i)" by (simp add: aux_ej12)
also have "… = extremo_izquierda i" using HI by simp
finally show "?p (N t i d)" by simp
qed

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

(* danrodcha *)
theorem "hd (preOrden a) = last (postOrden a)" (is "?P a")
proof (induct a)
fix x
show "?P (H x)" by simp
fix x i d
show "?P (N x i d)" by simp
qed

(* danrodcha *)
theorem "hd (preOrden a) = last (postOrden a)" (is "?P a")
proof (induct a)
fix x
show "?P (H x)" by simp
fix x i d
assume HIi: "?P i"
assume HId: "?P d"
have "hd (preOrden (N x i d)) = hd (x#preOrden i @ preOrden d)"
by (simp only: preOrden.simps(2))
also have "… = x" by simp
also have "… = last (postOrden i @ postOrden d @ [x])" by simp
also have "… = last (postOrden (N x i d))"
by (simp only: postOrden.simps(2))
finally show "?P (N x i d)" by simp
qed

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

(* danrodcha *)
theorem "hd (preOrden a) = raiz a" (is "?P a")
proof (induct a)
fix x
show "?P (H x)" by simp
fix x i d
assume HIi: "?P i"
assume HId: "?P d"
have "hd (preOrden (N x i d)) = hd (x#preOrden i @ preOrden d)"
by (simp only: preOrden.simps(2))
also have "… = x" by simp
also have "… = raiz (N x i d)" by simp
finally show "?P (N x i d)" by simp
qed

(* danrodcha *)
theorem "hd (preOrden a) = raiz a" (is "?P a")
proof (induct a)
fix x
show "?P (H x)" by simp
fix x i d
show "?P (N x i d)" by simp
qed

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

(*crigomgom pablucoto *)
theorem "hd (inOrden a) = raiz a"
quickcheck
oops

(* danrodcha:
Auto 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
---------------------------------------------------------------------
*}

(* danrodcha *)
theorem "last (postOrden a) = raiz a" (is "?P a")
proof (induct a)
fix x
show "?P (H x)" by simp
fix x i d
show "?P (N x i d)" by simp
qed

(* danrodcha *)
theorem "last (postOrden a) = raiz a" (is "?P a")
proof (induct a)
fix x
show "?P (H x)" by simp
fix x i d
assume HIi: "?P i"
assume HId: "?P d"
have "last (postOrden (N x i d)) = last (postOrden i @ postOrden d @ [x])"
by (simp only: postOrden.simps(2))
also have "… = x" by simp
also have "… = raiz (N x i d)" by (simp only: raiz.simps(2))
finally show "?P (N x i d)" by simp
qed

(* pablucoto *) (*Similar a la anterior*)

theorem "last (postOrden a) = raiz a" (is "?P a")
proof (induct a )
fix x
show "?P (H x)" by simp
next
fix x i d
assume HI1: "?P i"
assume HI1: "?P d"
have "last (postOrden (N x i d)) = last ( postOrden i @ postOrden d @ [x])" by simp
also have "... = x" by simp
also have "... = raiz (N x i d) " by simp
finally show " ?P (N x i d)" by simp
qed

end
</source>

Relación 6

2016-12-04T02:53:18Z

Pablucoto:

<source lang="isar">
chapter {* R6: Recorridos de árboles *}

theory R6_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))".
---------------------------------------------------------------------
*}

(* ivamenjim marpoldia1 manmorjim1 bowma *)

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]
---------------------------------------------------------------------
*}

(* ivamenjim marpoldia1 pablucoto bowma*)

fun preOrden :: "'a arbol ⇒ 'a list" where
"preOrden (H t) = [t]"
| "preOrden (N t i d) = [t] @ (preOrden i) @ (preOrden d)"

(* danrodcha crigomgom manmorjim1 bowma *)
fun preOrden1 :: "'a arbol ⇒ 'a list" where
"preOrden1 (H x) = [x]"
| "preOrden1 (N x i d) = x#preOrden1 i @ preOrden1 d"

value "preOrden (N e (N c (H a) (H d)) (N g (H f) (H h)))
= [e,c,a,d,g,f,h]"
value "preOrden1 (N e (N c (H a) (H d)) (N g (H f) (H h)))
= [e,c,a,d,g,f,h]"

(* danrodcha *)
lemma "preOrden a = preOrden1 a"
by (induct a) simp_all

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]
---------------------------------------------------------------------
*}

(* ivamenjim danrodcha crigomgom marpoldia1 manmorjim1 pablucoto bowma*)

fun postOrden :: "'a arbol ⇒ 'a list" where
"postOrden (H t) = [t]"
| "postOrden (N t i d) = (postOrden i) @ (postOrden d) @ [t]"

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]
---------------------------------------------------------------------
*}

(* ivamenjim crigomgom marpoldia1 pablucoto bowma *)

fun inOrden :: "'a arbol ⇒ 'a list" where
"inOrden (H t) = [t]"
| "inOrden (N t i d) = (inOrden i) @ [t] @ (inOrden d)"

(* danrodcha manmorjim1 *)
fun inOrden1 :: "'a arbol ⇒ 'a list" where
"inOrden1 (H t) = [t]"
| "inOrden1 (N t i d) = inOrden1 i @ t#inOrden1 d"

value "inOrden (N e (N c (H a) (H d)) (N g (H f) (H h)))
= [a,c,d,e,f,g,h]"
value "inOrden1 (N e (N c (H a) (H d)) (N g (H f) (H h)))
= [a,c,d,e,f,g,h]"

(* manmorjim1 *)
lemma "inOrden t = inOrden1 t"
apply (induct t)
apply auto
done

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))
---------------------------------------------------------------------
*}

(* ivamenjim danrodcha crigomgom marpoldia1 manmorjim1 pablucoto bowma *)

fun espejo :: "'a arbol ⇒ 'a arbol" where
"espejo (H t) = H t"
| "espejo (N t i d) = N t (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)
---------------------------------------------------------------------
*}

(* ivamenjim *)

lemma "preOrden (espejo a) = rev (postOrden a)" (is "?P a")
proof (induct a)
fix x
show "?P (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
finally show "preOrden (espejo (N x i d)) = rev (postOrden (N x i d))" by simp
qed

(* danrodcha *)
lemma "preOrden (espejo a) = rev (postOrden a)"
by (induct a) simp_all

(* danrodcha crigomgom*)
lemma "preOrden (espejo a) = rev (postOrden a)" (is "?P a")
proof (induct a)
fix x
show "?P (H x)" by simp
next
fix x i d
assume HIi: "?P i"
assume HId: "?P d"
have "preOrden (espejo (N x i d)) = preOrden (N x (espejo d) (espejo i))"
by (simp only: espejo.simps(2))
also have "… = x#preOrden (espejo d) @ preOrden (espejo i)"
by (simp only: preOrden.simps(2))
also have"… = x#rev (postOrden d) @ rev (postOrden i)"
using HIi HId by simp
also have "… = rev (postOrden (N x i d))" by simp
finally show "?P (N x i d)" by simp
qed

(* danrodcha *)
lemma "preOrden (espejo a) = rev (postOrden a)" (is "?P a")
proof (induct a)
fix x
show "?P (H x)" by simp
next
fix x i d
assume HIi: "?P i"
assume HId: "?P d"
show "?P (N x i d)" using HIi HId by simp
qed

(* bowma *)
lemma "preOrden (espejo a) = rev (postOrden a)"
apply (induct a)
apply simp_all
done

(* pablucoto *)

lemma "preOrden (espejo a) = rev (postOrden a)" (is "?P a")
proof (induct a)
fix x
show "?P (H x)" by simp
next
fix x i d
assume h1: "?P i"
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 "... = [x] @ rev (postOrden i @ postOrden d)" by simp
also have "... = rev ( postOrden i @ postOrden d @ [x] ) " by simp
also have "... = rev (postOrden (N x i d)) " by simp
finally show "?P (N x i d)" by simp
qed

(* bowma *)
lemma "preOrden (espejo a) = rev (postOrden a)" (is "?p a")
proof (induct a)
fix t
show "?p (H t)" by simp
(* Aquí si le diga "preOrden (espejo (H t)) = rev (postOrden (H t))",isabelle dice:
proof (prove)
goal (1 subgoal):
1. preOrden (espejo (H t)) = rev (postOrden (H t))
Introduced fixed type variable(s): 'b in "t__"
No entiendo porqué *)
next
fix t i d
assume H1: "?p i"
assume H2: "?p d"
have "preOrden (espejo (N t i d)) = preOrden (N t (espejo d) (espejo i))" by simp
also have "... = [t] @ (preOrden (espejo d)) @ (preOrden (espejo i))" by simp
also have "... = [t] @ rev (postOrden d) @ rev (postOrden i)" using H1 H2 by simp
finally show "?p (N t i d)" by simp
qed

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

(* ivamenjim crigomgom bowma *)

lemma "postOrden (espejo a) = rev (preOrden a)" (is "?P a")
proof (induct a)
fix x
show "?P (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
finally show "postOrden (espejo (N x i d)) = rev (preOrden (N x i d))" by simp
(* "?p (N x i d)" más corto *)
qed

(* danrodcha *)
lemma "postOrden (espejo a) = rev (preOrden a)" (is "?P a")
proof (induct a)
fix x
show "?P (H x)" by simp
next
fix x i d
assume HIi: "?P i"
assume HId: "?P d"
show "?P (N x i d)" using HIi HId by simp
qed

(* pablucoto *)

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

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

(* ivamenjim crigomgom bowma *)

theorem "inOrden (espejo a) = rev (inOrden a)" (is "?P a")
proof (induct a)
fix x
show "?P (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
finally show "inOrden (espejo (N x i d)) = rev (inOrden (N x i d))" by simp
qed

(* danrodcha *)
theorem "inOrden (espejo a) = rev (inOrden a)"
by (induct a) simp_all

(* danrodcha *)
theorem "inOrden (espejo a) = rev (inOrden a)" (is "?P a")
proof (induct a)
fix x
show "?P (H x)" by simp
next
fix x i d
assume HIi: "?P i"
assume HId: "?P d"
show "?P (N x i d)" using HIi HId by simp
qed

(* pablucoto *)
theorem "inOrden (espejo a) = rev (inOrden a)" (is "?P a")
proof (induct a)
fix x
show "?P (H x) " by simp
next
fix x i d
assume HI1: "?P i"
assume HI2: "?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 HI1 HI2 by simp
also have "... = rev (x # inOrden d ) @ rev (inOrden i)" by simp
also have "... = rev ( inOrden i @ x # inOrden d) " by simp
also have "... = rev (inOrden (N x i d))" by simp
finally show "?P (N x i d)" 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
---------------------------------------------------------------------
*}

(* danrodcha crigomgom manmorjim1 ivamenjim bowma pablucoto*)
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
---------------------------------------------------------------------
*}

(* danrodcha crigomgom manmorjim1 ivamenjim bowma pablucoto *)
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
---------------------------------------------------------------------
*}

(* danrodcha crigomgom manmorjim1 ivamenjim bowma pablucoto*)
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"

(* bowma *)
fun extremo_derecha_1 :: "'a arbol ⇒ 'a" where
"extremo_derecha_1 (H t) = t"
| "extremo_derecha_1 (N t i d) = last (inOrden (N t i d))"

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

(* danrodcha *)
lemma aux_ej12: "inOrden a ≠ []"
apply (induct a)
apply simp
apply simp
done

(* danrodcha pablucoto *)
theorem "last (inOrden a) = extremo_derecha a" (is "?P a")
proof (induct a)
fix x
show "?P (H x)" by simp
next
fix x i d
assume HIi: "?P i"
assume HId: "?P d"
have "last (inOrden (N x i d)) = last (inOrden i @ [x] @ inOrden d)"
by (simp only: inOrden.simps(2))
also have "… = last (inOrden d)" by (simp add: aux_ej12)
also have "… = extremo_derecha d" using HId by simp
also have "… = extremo_derecha (N x i d)" by simp
finally show "?P (N x i d)" by simp
qed

(* ivamenjim *)

lemma aux_ej12_1: "inOrden a ≠ []"
by (induct a) simp_all

(* ivamenjim *)
(* Igual que la anterior, pero poniendo solo by simp en el primer have *)

theorem "last (inOrden a) = extremo_derecha a" (is "?P a")
proof (induct a)
fix x
show "?P (H x)" by simp
next
fix x
fix i assume h1: "?P i"
fix d assume h2: "?P d"
have "last (inOrden (N x i d)) = last ((inOrden i) @ [x] @ (inOrden d))" by simp
also have "... = last (inOrden d)" by (simp add: aux_ej12_1)
also have "... = extremo_derecha d" using h2 by simp
finally show "last (inOrden (N x i d)) = extremo_derecha (N x i d)" by simp
qed

(* bowma *)
(* Casi lo mismo que el anterior,pero no hace falta suponer "?p i" *)
theorem "last (inOrden a) = extremo_derecha a" (is "?p a")
proof (induct a)
fix t
show "?p (H t)" by simp
next
fix t i d
assume HI: "?p d"
have "last (inOrden (N t i d)) = last (inOrden i @ [t] @ inOrden d)" by simp
also have "... = last (inOrden d)" by (simp add:aux_ej12)
also have "... = extremo_derecha d" using HI by simp
finally show "?p (N t i d)" by simp
qed

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

(* danrodcha *)
theorem "hd (inOrden a) = extremo_izquierda a" (is "?P a")
proof (induct a)
fix x
show "?P (H x)" by simp
next
fix x i d
assume HIi: "?P i"
assume HId: "?P d"
have "hd (inOrden (N x i d)) = hd (inOrden i @ [x] @ inOrden d)"
by (simp only: inOrden.simps(2))
also have "… = hd (inOrden i)" by (simp add: aux_ej12)
also have "… = extremo_izquierda i" using HIi by simp
also have "… = extremo_izquierda (N x i d)" by simp
finally show "?P (N x i d)" by simp
qed

(* bowma *)
theorem "hd (inOrden a) = extremo_izquierda a" (is "?p a")
proof (induct a)
fix t
show "?p (H t)" by simp
next
fix t i d
assume HI: "?p i"
have "hd (inOrden (N t i d)) = hd (inOrden i @ [t] @ inOrden d)" by simp
also have "… = hd (inOrden i)" by (simp add: aux_ej12)
also have "… = extremo_izquierda i" using HI by simp
finally show "?p (N t i d)" by simp
qed

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

(* danrodcha *)
theorem "hd (preOrden a) = last (postOrden a)" (is "?P a")
proof (induct a)
fix x
show "?P (H x)" by simp
fix x i d
show "?P (N x i d)" by simp
qed

(* danrodcha *)
theorem "hd (preOrden a) = last (postOrden a)" (is "?P a")
proof (induct a)
fix x
show "?P (H x)" by simp
fix x i d
assume HIi: "?P i"
assume HId: "?P d"
have "hd (preOrden (N x i d)) = hd (x#preOrden i @ preOrden d)"
by (simp only: preOrden.simps(2))
also have "… = x" by simp
also have "… = last (postOrden i @ postOrden d @ [x])" by simp
also have "… = last (postOrden (N x i d))"
by (simp only: postOrden.simps(2))
finally show "?P (N x i d)" by simp
qed

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

(* danrodcha *)
theorem "hd (preOrden a) = raiz a" (is "?P a")
proof (induct a)
fix x
show "?P (H x)" by simp
fix x i d
assume HIi: "?P i"
assume HId: "?P d"
have "hd (preOrden (N x i d)) = hd (x#preOrden i @ preOrden d)"
by (simp only: preOrden.simps(2))
also have "… = x" by simp
also have "… = raiz (N x i d)" by simp
finally show "?P (N x i d)" by simp
qed

(* danrodcha *)
theorem "hd (preOrden a) = raiz a" (is "?P a")
proof (induct a)
fix x
show "?P (H x)" by simp
fix x i d
show "?P (N x i d)" by simp
qed

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

(*crigomgom pablucoto *)
theorem "hd (inOrden a) = raiz a"
quickcheck
oops

(* danrodcha:
Auto 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
---------------------------------------------------------------------
*}

(* danrodcha *)
theorem "last (postOrden a) = raiz a" (is "?P a")
proof (induct a)
fix x
show "?P (H x)" by simp
fix x i d
show "?P (N x i d)" by simp
qed

(* danrodcha *)
theorem "last (postOrden a) = raiz a" (is "?P a")
proof (induct a)
fix x
show "?P (H x)" by simp
fix x i d
assume HIi: "?P i"
assume HId: "?P d"
have "last (postOrden (N x i d)) = last (postOrden i @ postOrden d @ [x])"
by (simp only: postOrden.simps(2))
also have "… = x" by simp
also have "… = raiz (N x i d)" by (simp only: raiz.simps(2))
finally show "?P (N x i d)" by simp
qed

end
</source>

Relación 6

2016-12-04T02:52:19Z

Pablucoto:

<source lang="isar">
chapter {* R6: Recorridos de árboles *}

theory R6_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))".
---------------------------------------------------------------------
*}

(* ivamenjim marpoldia1 manmorjim1 bowma *)

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]
---------------------------------------------------------------------
*}

(* ivamenjim marpoldia1 pablucoto bowma*)

fun preOrden :: "'a arbol ⇒ 'a list" where
"preOrden (H t) = [t]"
| "preOrden (N t i d) = [t] @ (preOrden i) @ (preOrden d)"

(* danrodcha crigomgom manmorjim1 bowma *)
fun preOrden1 :: "'a arbol ⇒ 'a list" where
"preOrden1 (H x) = [x]"
| "preOrden1 (N x i d) = x#preOrden1 i @ preOrden1 d"

value "preOrden (N e (N c (H a) (H d)) (N g (H f) (H h)))
= [e,c,a,d,g,f,h]"
value "preOrden1 (N e (N c (H a) (H d)) (N g (H f) (H h)))
= [e,c,a,d,g,f,h]"

(* danrodcha *)
lemma "preOrden a = preOrden1 a"
by (induct a) simp_all

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]
---------------------------------------------------------------------
*}

(* ivamenjim danrodcha crigomgom marpoldia1 manmorjim1 pablucoto bowma*)

fun postOrden :: "'a arbol ⇒ 'a list" where
"postOrden (H t) = [t]"
| "postOrden (N t i d) = (postOrden i) @ (postOrden d) @ [t]"

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]
---------------------------------------------------------------------
*}

(* ivamenjim crigomgom marpoldia1 pablucoto bowma *)

fun inOrden :: "'a arbol ⇒ 'a list" where
"inOrden (H t) = [t]"
| "inOrden (N t i d) = (inOrden i) @ [t] @ (inOrden d)"

(* danrodcha manmorjim1 *)
fun inOrden1 :: "'a arbol ⇒ 'a list" where
"inOrden1 (H t) = [t]"
| "inOrden1 (N t i d) = inOrden1 i @ t#inOrden1 d"

value "inOrden (N e (N c (H a) (H d)) (N g (H f) (H h)))
= [a,c,d,e,f,g,h]"
value "inOrden1 (N e (N c (H a) (H d)) (N g (H f) (H h)))
= [a,c,d,e,f,g,h]"

(* manmorjim1 *)
lemma "inOrden t = inOrden1 t"
apply (induct t)
apply auto
done

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))
---------------------------------------------------------------------
*}

(* ivamenjim danrodcha crigomgom marpoldia1 manmorjim1 pablucoto bowma *)

fun espejo :: "'a arbol ⇒ 'a arbol" where
"espejo (H t) = H t"
| "espejo (N t i d) = N t (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)
---------------------------------------------------------------------
*}

(* ivamenjim *)

lemma "preOrden (espejo a) = rev (postOrden a)" (is "?P a")
proof (induct a)
fix x
show "?P (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
finally show "preOrden (espejo (N x i d)) = rev (postOrden (N x i d))" by simp
qed

(* danrodcha *)
lemma "preOrden (espejo a) = rev (postOrden a)"
by (induct a) simp_all

(* danrodcha crigomgom*)
lemma "preOrden (espejo a) = rev (postOrden a)" (is "?P a")
proof (induct a)
fix x
show "?P (H x)" by simp
next
fix x i d
assume HIi: "?P i"
assume HId: "?P d"
have "preOrden (espejo (N x i d)) = preOrden (N x (espejo d) (espejo i))"
by (simp only: espejo.simps(2))
also have "… = x#preOrden (espejo d) @ preOrden (espejo i)"
by (simp only: preOrden.simps(2))
also have"… = x#rev (postOrden d) @ rev (postOrden i)"
using HIi HId by simp
also have "… = rev (postOrden (N x i d))" by simp
finally show "?P (N x i d)" by simp
qed

(* danrodcha *)
lemma "preOrden (espejo a) = rev (postOrden a)" (is "?P a")
proof (induct a)
fix x
show "?P (H x)" by simp
next
fix x i d
assume HIi: "?P i"
assume HId: "?P d"
show "?P (N x i d)" using HIi HId by simp
qed

(* bowma *)
lemma "preOrden (espejo a) = rev (postOrden a)"
apply (induct a)
apply simp_all
done

(* pablucoto *)

lemma "preOrden (espejo a) = rev (postOrden a)" (is "?P a")
proof (induct a)
fix x
show "?P (H x)" by simp
next
fix x i d
assume h1: "?P i"
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 "... = [x] @ rev (postOrden i @ postOrden d)" by simp
also have "... = rev ( postOrden i @ postOrden d @ [x] ) " by simp
also have "... = rev (postOrden (N x i d)) " by simp
finally show "?P (N x i d)" by simp
qed

(* bowma *)
lemma "preOrden (espejo a) = rev (postOrden a)" (is "?p a")
proof (induct a)
fix t
show "?p (H t)" by simp
(* Aquí si le diga "preOrden (espejo (H t)) = rev (postOrden (H t))",isabelle dice:
proof (prove)
goal (1 subgoal):
1. preOrden (espejo (H t)) = rev (postOrden (H t))
Introduced fixed type variable(s): 'b in "t__"
No entiendo porqué *)
next
fix t i d
assume H1: "?p i"
assume H2: "?p d"
have "preOrden (espejo (N t i d)) = preOrden (N t (espejo d) (espejo i))" by simp
also have "... = [t] @ (preOrden (espejo d)) @ (preOrden (espejo i))" by simp
also have "... = [t] @ rev (postOrden d) @ rev (postOrden i)" using H1 H2 by simp
finally show "?p (N t i d)" by simp
qed

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

(* ivamenjim crigomgom bowma *)

lemma "postOrden (espejo a) = rev (preOrden a)" (is "?P a")
proof (induct a)
fix x
show "?P (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
finally show "postOrden (espejo (N x i d)) = rev (preOrden (N x i d))" by simp
(* "?p (N x i d)" más corto *)
qed

(* danrodcha *)
lemma "postOrden (espejo a) = rev (preOrden a)" (is "?P a")
proof (induct a)
fix x
show "?P (H x)" by simp
next
fix x i d
assume HIi: "?P i"
assume HId: "?P d"
show "?P (N x i d)" using HIi HId by simp
qed

(* pablucoto *)

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

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

(* ivamenjim crigomgom bowma *)

theorem "inOrden (espejo a) = rev (inOrden a)" (is "?P a")
proof (induct a)
fix x
show "?P (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
finally show "inOrden (espejo (N x i d)) = rev (inOrden (N x i d))" by simp
qed

(* danrodcha *)
theorem "inOrden (espejo a) = rev (inOrden a)"
by (induct a) simp_all

(* danrodcha *)
theorem "inOrden (espejo a) = rev (inOrden a)" (is "?P a")
proof (induct a)
fix x
show "?P (H x)" by simp
next
fix x i d
assume HIi: "?P i"
assume HId: "?P d"
show "?P (N x i d)" using HIi HId by simp
qed

(* pablucoto *)
theorem "inOrden (espejo a) = rev (inOrden a)" (is "?P a")
proof (induct a)
fix x
show "?P (H x) " by simp
next
fix x i d
assume HI1: "?P i"
assume HI2: "?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 HI1 HI2 by simp
also have "... = rev (x # inOrden d ) @ rev (inOrden i)" by simp
also have "... = rev ( inOrden i @ x # inOrden d) " by simp
also have "... = rev (inOrden (N x i d))" by simp
finally show "?P (N x i d)" 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
---------------------------------------------------------------------
*}

(* danrodcha crigomgom manmorjim1 ivamenjim bowma pablucoto*)
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
---------------------------------------------------------------------
*}

(* danrodcha crigomgom manmorjim1 ivamenjim bowma pablucoto *)
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
---------------------------------------------------------------------
*}

(* danrodcha crigomgom manmorjim1 ivamenjim bowma pablucoto*)
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"

(* bowma *)
fun extremo_derecha_1 :: "'a arbol ⇒ 'a" where
"extremo_derecha_1 (H t) = t"
| "extremo_derecha_1 (N t i d) = last (inOrden (N t i d))"

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

(* danrodcha *)
lemma aux_ej12: "inOrden a ≠ []"
apply (induct a)
apply simp
apply simp
done

(* danrodcha pablucoto *)
theorem "last (inOrden a) = extremo_derecha a" (is "?P a")
proof (induct a)
fix x
show "?P (H x)" by simp
next
fix x i d
assume HIi: "?P i"
assume HId: "?P d"
have "last (inOrden (N x i d)) = last (inOrden i @ [x] @ inOrden d)"
by (simp only: inOrden.simps(2))
also have "… = last (inOrden d)" by (simp add: aux_ej12)
also have "… = extremo_derecha d" using HId by simp
also have "… = extremo_derecha (N x i d)" by simp
finally show "?P (N x i d)" by simp
qed

(* ivamenjim *)

lemma aux_ej12_1: "inOrden a ≠ []"
by (induct a) simp_all

(* ivamenjim *)
(* Igual que la anterior, pero poniendo solo by simp en el primer have *)

theorem "last (inOrden a) = extremo_derecha a" (is "?P a")
proof (induct a)
fix x
show "?P (H x)" by simp
next
fix x
fix i assume h1: "?P i"
fix d assume h2: "?P d"
have "last (inOrden (N x i d)) = last ((inOrden i) @ [x] @ (inOrden d))" by simp
also have "... = last (inOrden d)" by (simp add: aux_ej12_1)
also have "... = extremo_derecha d" using h2 by simp
finally show "last (inOrden (N x i d)) = extremo_derecha (N x i d)" by simp
qed

(* bowma *)
(* Casi lo mismo que el anterior,pero no hace falta suponer "?p i" *)
theorem "last (inOrden a) = extremo_derecha a" (is "?p a")
proof (induct a)
fix t
show "?p (H t)" by simp
next
fix t i d
assume HI: "?p d"
have "last (inOrden (N t i d)) = last (inOrden i @ [t] @ inOrden d)" by simp
also have "... = last (inOrden d)" by (simp add:aux_ej12)
also have "... = extremo_derecha d" using HI by simp
finally show "?p (N t i d)" by simp
qed

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

(* danrodcha *)
theorem "hd (inOrden a) = extremo_izquierda a" (is "?P a")
proof (induct a)
fix x
show "?P (H x)" by simp
next
fix x i d
assume HIi: "?P i"
assume HId: "?P d"
have "hd (inOrden (N x i d)) = hd (inOrden i @ [x] @ inOrden d)"
by (simp only: inOrden.simps(2))
also have "… = hd (inOrden i)" by (simp add: aux_ej12)
also have "… = extremo_izquierda i" using HIi by simp
also have "… = extremo_izquierda (N x i d)" by simp
finally show "?P (N x i d)" by simp
qed

(* bowma *)
theorem "hd (inOrden a) = extremo_izquierda a" (is "?p a")
proof (induct a)
fix t
show "?p (H t)" by simp
next
fix t i d
assume HI: "?p i"
have "hd (inOrden (N t i d)) = hd (inOrden i @ [t] @ inOrden d)" by simp
also have "… = hd (inOrden i)" by (simp add: aux_ej12)
also have "… = extremo_izquierda i" using HI by simp
finally show "?p (N t i d)" by simp
qed

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

(* danrodcha *)
theorem "hd (preOrden a) = last (postOrden a)" (is "?P a")
proof (induct a)
fix x
show "?P (H x)" by simp
fix x i d
show "?P (N x i d)" by simp
qed

(* danrodcha *)
theorem "hd (preOrden a) = last (postOrden a)" (is "?P a")
proof (induct a)
fix x
show "?P (H x)" by simp
fix x i d
assume HIi: "?P i"
assume HId: "?P d"
have "hd (preOrden (N x i d)) = hd (x#preOrden i @ preOrden d)"
by (simp only: preOrden.simps(2))
also have "… = x" by simp
also have "… = last (postOrden i @ postOrden d @ [x])" by simp
also have "… = last (postOrden (N x i d))"
by (simp only: postOrden.simps(2))
finally show "?P (N x i d)" by simp
qed

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

(* danrodcha *)
theorem "hd (preOrden a) = raiz a" (is "?P a")
proof (induct a)
fix x
show "?P (H x)" by simp
fix x i d
assume HIi: "?P i"
assume HId: "?P d"
have "hd (preOrden (N x i d)) = hd (x#preOrden i @ preOrden d)"
by (simp only: preOrden.simps(2))
also have "… = x" by simp
also have "… = raiz (N x i d)" by simp
finally show "?P (N x i d)" by simp
qed

(* danrodcha *)
theorem "hd (preOrden a) = raiz a" (is "?P a")
proof (induct a)
fix x
show "?P (H x)" by simp
fix x i d
show "?P (N x i d)" by simp
qed

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

(*crigomgom*)
theorem "hd (inOrden a) = raiz a"
quickcheck
oops

(* danrodcha:
Auto 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
---------------------------------------------------------------------
*}

(* danrodcha *)
theorem "last (postOrden a) = raiz a" (is "?P a")
proof (induct a)
fix x
show "?P (H x)" by simp
fix x i d
show "?P (N x i d)" by simp
qed

(* danrodcha *)
theorem "last (postOrden a) = raiz a" (is "?P a")
proof (induct a)
fix x
show "?P (H x)" by simp
fix x i d
assume HIi: "?P i"
assume HId: "?P d"
have "last (postOrden (N x i d)) = last (postOrden i @ postOrden d @ [x])"
by (simp only: postOrden.simps(2))
also have "… = x" by simp
also have "… = raiz (N x i d)" by (simp only: raiz.simps(2))
finally show "?P (N x i d)" by simp
qed

end
</source>

Relación 6

2016-12-04T02:10:37Z

Pablucoto:

<source lang="isar">
chapter {* R6: Recorridos de árboles *}

theory R6_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))".
---------------------------------------------------------------------
*}

(* ivamenjim marpoldia1 manmorjim1 bowma *)

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]
---------------------------------------------------------------------
*}

(* ivamenjim marpoldia1 pablucoto bowma*)

fun preOrden :: "'a arbol ⇒ 'a list" where
"preOrden (H t) = [t]"
| "preOrden (N t i d) = [t] @ (preOrden i) @ (preOrden d)"

(* danrodcha crigomgom manmorjim1 bowma *)
fun preOrden1 :: "'a arbol ⇒ 'a list" where
"preOrden1 (H x) = [x]"
| "preOrden1 (N x i d) = x#preOrden1 i @ preOrden1 d"

value "preOrden (N e (N c (H a) (H d)) (N g (H f) (H h)))
= [e,c,a,d,g,f,h]"
value "preOrden1 (N e (N c (H a) (H d)) (N g (H f) (H h)))
= [e,c,a,d,g,f,h]"

(* danrodcha *)
lemma "preOrden a = preOrden1 a"
by (induct a) simp_all

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]
---------------------------------------------------------------------
*}

(* ivamenjim danrodcha crigomgom marpoldia1 manmorjim1 pablucoto bowma*)

fun postOrden :: "'a arbol ⇒ 'a list" where
"postOrden (H t) = [t]"
| "postOrden (N t i d) = (postOrden i) @ (postOrden d) @ [t]"

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]
---------------------------------------------------------------------
*}

(* ivamenjim crigomgom marpoldia1 pablucoto bowma *)

fun inOrden :: "'a arbol ⇒ 'a list" where
"inOrden (H t) = [t]"
| "inOrden (N t i d) = (inOrden i) @ [t] @ (inOrden d)"

(* danrodcha manmorjim1 *)
fun inOrden1 :: "'a arbol ⇒ 'a list" where
"inOrden1 (H t) = [t]"
| "inOrden1 (N t i d) = inOrden1 i @ t#inOrden1 d"

value "inOrden (N e (N c (H a) (H d)) (N g (H f) (H h)))
= [a,c,d,e,f,g,h]"
value "inOrden1 (N e (N c (H a) (H d)) (N g (H f) (H h)))
= [a,c,d,e,f,g,h]"

(* manmorjim1 *)
lemma "inOrden t = inOrden1 t"
apply (induct t)
apply auto
done

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))
---------------------------------------------------------------------
*}

(* ivamenjim danrodcha crigomgom marpoldia1 manmorjim1 pablucoto bowma *)

fun espejo :: "'a arbol ⇒ 'a arbol" where
"espejo (H t) = H t"
| "espejo (N t i d) = N t (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)
---------------------------------------------------------------------
*}

(* ivamenjim *)

lemma "preOrden (espejo a) = rev (postOrden a)" (is "?P a")
proof (induct a)
fix x
show "?P (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
finally show "preOrden (espejo (N x i d)) = rev (postOrden (N x i d))" by simp
qed

(* danrodcha *)
lemma "preOrden (espejo a) = rev (postOrden a)"
by (induct a) simp_all

(* danrodcha crigomgom*)
lemma "preOrden (espejo a) = rev (postOrden a)" (is "?P a")
proof (induct a)
fix x
show "?P (H x)" by simp
next
fix x i d
assume HIi: "?P i"
assume HId: "?P d"
have "preOrden (espejo (N x i d)) = preOrden (N x (espejo d) (espejo i))"
by (simp only: espejo.simps(2))
also have "… = x#preOrden (espejo d) @ preOrden (espejo i)"
by (simp only: preOrden.simps(2))
also have"… = x#rev (postOrden d) @ rev (postOrden i)"
using HIi HId by simp
also have "… = rev (postOrden (N x i d))" by simp
finally show "?P (N x i d)" by simp
qed

(* danrodcha *)
lemma "preOrden (espejo a) = rev (postOrden a)" (is "?P a")
proof (induct a)
fix x
show "?P (H x)" by simp
next
fix x i d
assume HIi: "?P i"
assume HId: "?P d"
show "?P (N x i d)" using HIi HId by simp
qed

(* bowma *)
lemma "preOrden (espejo a) = rev (postOrden a)"
apply (induct a)
apply simp_all
done

(* pablucoto *)

lemma "preOrden (espejo a) = rev (postOrden a)" (is "?P a")
proof (induct a)
fix x
show "?P (H x)" by simp
next
fix x i d
assume h1: "?P i"
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 "... = [x] @ rev (postOrden i @ postOrden d)" by simp
also have "... = rev ( postOrden i @ postOrden d @ [x] ) " by simp
also have "... = rev (postOrden (N x i d)) " by simp
finally show "?P (N x i d)" by simp
qed

(* bowma *)
lemma "preOrden (espejo a) = rev (postOrden a)" (is "?p a")
proof (induct a)
fix t
show "?p (H t)" by simp
(* Aquí si le diga "preOrden (espejo (H t)) = rev (postOrden (H t))",isabelle dice:
proof (prove)
goal (1 subgoal):
1. preOrden (espejo (H t)) = rev (postOrden (H t))
Introduced fixed type variable(s): 'b in "t__"
No entiendo porqué *)
next
fix t i d
assume H1: "?p i"
assume H2: "?p d"
have "preOrden (espejo (N t i d)) = preOrden (N t (espejo d) (espejo i))" by simp
also have "... = [t] @ (preOrden (espejo d)) @ (preOrden (espejo i))" by simp
also have "... = [t] @ rev (postOrden d) @ rev (postOrden i)" using H1 H2 by simp
finally show "?p (N t i d)" by simp
qed

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

(* ivamenjim crigomgom bowma *)

lemma "postOrden (espejo a) = rev (preOrden a)" (is "?P a")
proof (induct a)
fix x
show "?P (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
finally show "postOrden (espejo (N x i d)) = rev (preOrden (N x i d))" by simp
(* "?p (N x i d)" más corto *)
qed

(* danrodcha *)
lemma "postOrden (espejo a) = rev (preOrden a)" (is "?P a")
proof (induct a)
fix x
show "?P (H x)" by simp
next
fix x i d
assume HIi: "?P i"
assume HId: "?P d"
show "?P (N x i d)" using HIi HId by simp
qed

(* pablucoto *)

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

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

(* ivamenjim crigomgom bowma *)

theorem "inOrden (espejo a) = rev (inOrden a)" (is "?P a")
proof (induct a)
fix x
show "?P (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
finally show "inOrden (espejo (N x i d)) = rev (inOrden (N x i d))" by simp
qed

(* danrodcha *)
theorem "inOrden (espejo a) = rev (inOrden a)"
by (induct a) simp_all

(* danrodcha *)
theorem "inOrden (espejo a) = rev (inOrden a)" (is "?P a")
proof (induct a)
fix x
show "?P (H x)" by simp
next
fix x i d
assume HIi: "?P i"
assume HId: "?P d"
show "?P (N x i d)" using HIi HId by simp
qed

(* pablucoto *)
theorem "inOrden (espejo a) = rev (inOrden a)" (is "?P a")
proof (induct a)
fix x
show "?P (H x) " by simp
next
fix x i d
assume HI1: "?P i"
assume HI2: "?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 HI1 HI2 by simp
also have "... = rev (x # inOrden d ) @ rev (inOrden i)" by simp
also have "... = rev ( inOrden i @ x # inOrden d) " by simp
also have "... = rev (inOrden (N x i d))" by simp
finally show "?P (N x i d)" 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
---------------------------------------------------------------------
*}

(* danrodcha crigomgom manmorjim1 ivamenjim bowma pablucoto*)
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
---------------------------------------------------------------------
*}

(* danrodcha crigomgom manmorjim1 ivamenjim bowma pablucoto *)
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
---------------------------------------------------------------------
*}

(* danrodcha crigomgom manmorjim1 ivamenjim bowma pablucoto*)
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"

(* bowma *)
fun extremo_derecha_1 :: "'a arbol ⇒ 'a" where
"extremo_derecha_1 (H t) = t"
| "extremo_derecha_1 (N t i d) = last (inOrden (N t i d))"

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

(* danrodcha *)
lemma aux_ej12: "inOrden a ≠ []"
apply (induct a)
apply simp
apply simp
done

(* danrodcha *)
theorem "last (inOrden a) = extremo_derecha a" (is "?P a")
proof (induct a)
fix x
show "?P (H x)" by simp
next
fix x i d
assume HIi: "?P i"
assume HId: "?P d"
have "last (inOrden (N x i d)) = last (inOrden i @ [x] @ inOrden d)"
by (simp only: inOrden.simps(2))
also have "… = last (inOrden d)" by (simp add: aux_ej12)
also have "… = extremo_derecha d" using HId by simp
also have "… = extremo_derecha (N x i d)" by simp
finally show "?P (N x i d)" by simp
qed

(* ivamenjim *)

lemma aux_ej12_1: "inOrden a ≠ []"
by (induct a) simp_all

(* ivamenjim *)
(* Igual que la anterior, pero poniendo solo by simp en el primer have *)

theorem "last (inOrden a) = extremo_derecha a" (is "?P a")
proof (induct a)
fix x
show "?P (H x)" by simp
next
fix x
fix i assume h1: "?P i"
fix d assume h2: "?P d"
have "last (inOrden (N x i d)) = last ((inOrden i) @ [x] @ (inOrden d))" by simp
also have "... = last (inOrden d)" by (simp add: aux_ej12_1)
also have "... = extremo_derecha d" using h2 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
---------------------------------------------------------------------
*}

(* danrodcha *)
theorem "hd (inOrden a) = extremo_izquierda a" (is "?P a")
proof (induct a)
fix x
show "?P (H x)" by simp
next
fix x i d
assume HIi: "?P i"
assume HId: "?P d"
have "hd (inOrden (N x i d)) = hd (inOrden i @ [x] @ inOrden d)"
by (simp only: inOrden.simps(2))
also have "… = hd (inOrden i)" by (simp add: aux_ej12)
also have "… = extremo_izquierda i" using HIi by simp
also have "… = extremo_izquierda (N x i d)" by simp
finally show "?P (N x i d)" by simp
qed

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

(* danrodcha *)
theorem "hd (preOrden a) = last (postOrden a)" (is "?P a")
proof (induct a)
fix x
show "?P (H x)" by simp
fix x i d
show "?P (N x i d)" by simp
qed

(* danrodcha *)
theorem "hd (preOrden a) = last (postOrden a)" (is "?P a")
proof (induct a)
fix x
show "?P (H x)" by simp
fix x i d
assume HIi: "?P i"
assume HId: "?P d"
have "hd (preOrden (N x i d)) = hd (x#preOrden i @ preOrden d)"
by (simp only: preOrden.simps(2))
also have "… = x" by simp
also have "… = last (postOrden i @ postOrden d @ [x])" by simp
also have "… = last (postOrden (N x i d))"
by (simp only: postOrden.simps(2))
finally show "?P (N x i d)" by simp
qed

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

(* danrodcha *)
theorem "hd (preOrden a) = raiz a" (is "?P a")
proof (induct a)
fix x
show "?P (H x)" by simp
fix x i d
assume HIi: "?P i"
assume HId: "?P d"
have "hd (preOrden (N x i d)) = hd (x#preOrden i @ preOrden d)"
by (simp only: preOrden.simps(2))
also have "… = x" by simp
also have "… = raiz (N x i d)" by simp
finally show "?P (N x i d)" by simp
qed

(* danrodcha *)
theorem "hd (preOrden a) = raiz a" (is "?P a")
proof (induct a)
fix x
show "?P (H x)" by simp
fix x i d
show "?P (N x i d)" by simp
qed

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

(*crigomgom*)
theorem "hd (inOrden a) = raiz a"
quickcheck
oops

(* danrodcha:
Auto 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
---------------------------------------------------------------------
*}

(* danrodcha *)
theorem "last (postOrden a) = raiz a" (is "?P a")
proof (induct a)
fix x
show "?P (H x)" by simp
fix x i d
show "?P (N x i d)" by simp
qed

(* danrodcha *)
theorem "last (postOrden a) = raiz a" (is "?P a")
proof (induct a)
fix x
show "?P (H x)" by simp
fix x i d
assume HIi: "?P i"
assume HId: "?P d"
have "last (postOrden (N x i d)) = last (postOrden i @ postOrden d @ [x])"
by (simp only: postOrden.simps(2))
also have "… = x" by simp
also have "… = raiz (N x i d)" by (simp only: raiz.simps(2))
finally show "?P (N x i d)" by simp
qed

end
</source>

Relación 6

2016-12-04T02:02:31Z

Pablucoto:

<source lang="isar">
chapter {* R6: Recorridos de árboles *}

theory R6_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))".
---------------------------------------------------------------------
*}

(* ivamenjim marpoldia1 manmorjim1 bowma *)

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]
---------------------------------------------------------------------
*}

(* ivamenjim marpoldia1 pablucoto bowma*)

fun preOrden :: "'a arbol ⇒ 'a list" where
"preOrden (H t) = [t]"
| "preOrden (N t i d) = [t] @ (preOrden i) @ (preOrden d)"

(* danrodcha crigomgom manmorjim1 bowma *)
fun preOrden1 :: "'a arbol ⇒ 'a list" where
"preOrden1 (H x) = [x]"
| "preOrden1 (N x i d) = x#preOrden1 i @ preOrden1 d"

value "preOrden (N e (N c (H a) (H d)) (N g (H f) (H h)))
= [e,c,a,d,g,f,h]"
value "preOrden1 (N e (N c (H a) (H d)) (N g (H f) (H h)))
= [e,c,a,d,g,f,h]"

(* danrodcha *)
lemma "preOrden a = preOrden1 a"
by (induct a) simp_all

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]
---------------------------------------------------------------------
*}

(* ivamenjim danrodcha crigomgom marpoldia1 manmorjim1 pablucoto bowma*)

fun postOrden :: "'a arbol ⇒ 'a list" where
"postOrden (H t) = [t]"
| "postOrden (N t i d) = (postOrden i) @ (postOrden d) @ [t]"

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]
---------------------------------------------------------------------
*}

(* ivamenjim crigomgom marpoldia1 pablucoto bowma *)

fun inOrden :: "'a arbol ⇒ 'a list" where
"inOrden (H t) = [t]"
| "inOrden (N t i d) = (inOrden i) @ [t] @ (inOrden d)"

(* danrodcha manmorjim1 *)
fun inOrden1 :: "'a arbol ⇒ 'a list" where
"inOrden1 (H t) = [t]"
| "inOrden1 (N t i d) = inOrden1 i @ t#inOrden1 d"

value "inOrden (N e (N c (H a) (H d)) (N g (H f) (H h)))
= [a,c,d,e,f,g,h]"
value "inOrden1 (N e (N c (H a) (H d)) (N g (H f) (H h)))
= [a,c,d,e,f,g,h]"

(* manmorjim1 *)
lemma "inOrden t = inOrden1 t"
apply (induct t)
apply auto
done

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))
---------------------------------------------------------------------
*}

(* ivamenjim danrodcha crigomgom marpoldia1 manmorjim1 pablucoto bowma *)

fun espejo :: "'a arbol ⇒ 'a arbol" where
"espejo (H t) = H t"
| "espejo (N t i d) = N t (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)
---------------------------------------------------------------------
*}

(* ivamenjim *)

lemma "preOrden (espejo a) = rev (postOrden a)" (is "?P a")
proof (induct a)
fix x
show "?P (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
finally show "preOrden (espejo (N x i d)) = rev (postOrden (N x i d))" by simp
qed

(* danrodcha *)
lemma "preOrden (espejo a) = rev (postOrden a)"
by (induct a) simp_all

(* danrodcha crigomgom*)
lemma "preOrden (espejo a) = rev (postOrden a)" (is "?P a")
proof (induct a)
fix x
show "?P (H x)" by simp
next
fix x i d
assume HIi: "?P i"
assume HId: "?P d"
have "preOrden (espejo (N x i d)) = preOrden (N x (espejo d) (espejo i))"
by (simp only: espejo.simps(2))
also have "… = x#preOrden (espejo d) @ preOrden (espejo i)"
by (simp only: preOrden.simps(2))
also have"… = x#rev (postOrden d) @ rev (postOrden i)"
using HIi HId by simp
also have "… = rev (postOrden (N x i d))" by simp
finally show "?P (N x i d)" by simp
qed

(* danrodcha *)
lemma "preOrden (espejo a) = rev (postOrden a)" (is "?P a")
proof (induct a)
fix x
show "?P (H x)" by simp
next
fix x i d
assume HIi: "?P i"
assume HId: "?P d"
show "?P (N x i d)" using HIi HId by simp
qed

(* bowma *)
lemma "preOrden (espejo a) = rev (postOrden a)"
apply (induct a)
apply simp_all
done

(* pablucoto *)

lemma "preOrden (espejo a) = rev (postOrden a)" (is "?P a")
proof (induct a)
fix x
show "?P (H x)" by simp
next
fix x i d
assume h1: "?P i"
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 "... = [x] @ rev (postOrden i @ postOrden d)" by simp
also have "... = rev ( postOrden i @ postOrden d @ [x] ) " by simp
also have "... = rev (postOrden (N x i d)) " by simp
finally show "?P (N x i d)" by simp
qed

(* bowma *)
lemma "preOrden (espejo a) = rev (postOrden a)" (is "?p a")
proof (induct a)
fix t
show "?p (H t)" by simp
(* Aquí si le diga "preOrden (espejo (H t)) = rev (postOrden (H t))",isabelle dice:
proof (prove)
goal (1 subgoal):
1. preOrden (espejo (H t)) = rev (postOrden (H t))
Introduced fixed type variable(s): 'b in "t__"
No entiendo porqué *)
next
fix t i d
assume H1: "?p i"
assume H2: "?p d"
have "preOrden (espejo (N t i d)) = preOrden (N t (espejo d) (espejo i))" by simp
also have "... = [t] @ (preOrden (espejo d)) @ (preOrden (espejo i))" by simp
also have "... = [t] @ rev (postOrden d) @ rev (postOrden i)" using H1 H2 by simp
finally show "?p (N t i d)" by simp
qed

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

(* ivamenjim crigomgom bowma *)

lemma "postOrden (espejo a) = rev (preOrden a)" (is "?P a")
proof (induct a)
fix x
show "?P (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
finally show "postOrden (espejo (N x i d)) = rev (preOrden (N x i d))" by simp
(* "?p (N x i d)" más corto *)
qed

(* danrodcha *)
lemma "postOrden (espejo a) = rev (preOrden a)" (is "?P a")
proof (induct a)
fix x
show "?P (H x)" by simp
next
fix x i d
assume HIi: "?P i"
assume HId: "?P d"
show "?P (N x i d)" using HIi HId by simp
qed

(* pablucoto *)

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

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

(* ivamenjim crigomgom bowma *)

theorem "inOrden (espejo a) = rev (inOrden a)" (is "?P a")
proof (induct a)
fix x
show "?P (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
finally show "inOrden (espejo (N x i d)) = rev (inOrden (N x i d))" by simp
qed

(* danrodcha *)
theorem "inOrden (espejo a) = rev (inOrden a)"
by (induct a) simp_all

(* danrodcha *)
theorem "inOrden (espejo a) = rev (inOrden a)" (is "?P a")
proof (induct a)
fix x
show "?P (H x)" by simp
next
fix x i d
assume HIi: "?P i"
assume HId: "?P d"
show "?P (N x i d)" using HIi HId by simp
qed

(* pablucoto *)
theorem "inOrden (espejo a) = rev (inOrden a)" (is "?P a")
proof (induct a)
fix x
show "?P (H x) " by simp
next
fix x i d
assume HI1: "?P i"
assume HI2: "?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 HI1 HI2 by simp
also have "... = rev (x # inOrden d ) @ rev (inOrden i)" by simp
also have "... = rev ( inOrden i @ x # inOrden d) " by simp
also have "... = rev (inOrden (N x i d))" by simp
finally show "?P (N x i d)" 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
---------------------------------------------------------------------
*}

(* danrodcha crigomgom manmorjim1 ivamenjim bowma pablucoto*)
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
---------------------------------------------------------------------
*}

(* danrodcha crigomgom manmorjim1 ivamenjim bowma *)
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
---------------------------------------------------------------------
*}

(* danrodcha crigomgom manmorjim1 ivamenjim bowma *)
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"

(* bowma *)
fun extremo_derecha_1 :: "'a arbol ⇒ 'a" where
"extremo_derecha_1 (H t) = t"
| "extremo_derecha_1 (N t i d) = last (inOrden (N t i d))"

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

(* danrodcha *)
lemma aux_ej12: "inOrden a ≠ []"
apply (induct a)
apply simp
apply simp
done

(* danrodcha *)
theorem "last (inOrden a) = extremo_derecha a" (is "?P a")
proof (induct a)
fix x
show "?P (H x)" by simp
next
fix x i d
assume HIi: "?P i"
assume HId: "?P d"
have "last (inOrden (N x i d)) = last (inOrden i @ [x] @ inOrden d)"
by (simp only: inOrden.simps(2))
also have "… = last (inOrden d)" by (simp add: aux_ej12)
also have "… = extremo_derecha d" using HId by simp
also have "… = extremo_derecha (N x i d)" by simp
finally show "?P (N x i d)" by simp
qed

(* ivamenjim *)

lemma aux_ej12_1: "inOrden a ≠ []"
by (induct a) simp_all

(* ivamenjim *)
(* Igual que la anterior, pero poniendo solo by simp en el primer have *)

theorem "last (inOrden a) = extremo_derecha a" (is "?P a")
proof (induct a)
fix x
show "?P (H x)" by simp
next
fix x
fix i assume h1: "?P i"
fix d assume h2: "?P d"
have "last (inOrden (N x i d)) = last ((inOrden i) @ [x] @ (inOrden d))" by simp
also have "... = last (inOrden d)" by (simp add: aux_ej12_1)
also have "... = extremo_derecha d" using h2 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
---------------------------------------------------------------------
*}

(* danrodcha *)
theorem "hd (inOrden a) = extremo_izquierda a" (is "?P a")
proof (induct a)
fix x
show "?P (H x)" by simp
next
fix x i d
assume HIi: "?P i"
assume HId: "?P d"
have "hd (inOrden (N x i d)) = hd (inOrden i @ [x] @ inOrden d)"
by (simp only: inOrden.simps(2))
also have "… = hd (inOrden i)" by (simp add: aux_ej12)
also have "… = extremo_izquierda i" using HIi by simp
also have "… = extremo_izquierda (N x i d)" by simp
finally show "?P (N x i d)" by simp
qed

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

(* danrodcha *)
theorem "hd (preOrden a) = last (postOrden a)" (is "?P a")
proof (induct a)
fix x
show "?P (H x)" by simp
fix x i d
show "?P (N x i d)" by simp
qed

(* danrodcha *)
theorem "hd (preOrden a) = last (postOrden a)" (is "?P a")
proof (induct a)
fix x
show "?P (H x)" by simp
fix x i d
assume HIi: "?P i"
assume HId: "?P d"
have "hd (preOrden (N x i d)) = hd (x#preOrden i @ preOrden d)"
by (simp only: preOrden.simps(2))
also have "… = x" by simp
also have "… = last (postOrden i @ postOrden d @ [x])" by simp
also have "… = last (postOrden (N x i d))"
by (simp only: postOrden.simps(2))
finally show "?P (N x i d)" by simp
qed

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

(* danrodcha *)
theorem "hd (preOrden a) = raiz a" (is "?P a")
proof (induct a)
fix x
show "?P (H x)" by simp
fix x i d
assume HIi: "?P i"
assume HId: "?P d"
have "hd (preOrden (N x i d)) = hd (x#preOrden i @ preOrden d)"
by (simp only: preOrden.simps(2))
also have "… = x" by simp
also have "… = raiz (N x i d)" by simp
finally show "?P (N x i d)" by simp
qed

(* danrodcha *)
theorem "hd (preOrden a) = raiz a" (is "?P a")
proof (induct a)
fix x
show "?P (H x)" by simp
fix x i d
show "?P (N x i d)" by simp
qed

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

(*crigomgom*)
theorem "hd (inOrden a) = raiz a"
quickcheck
oops

(* danrodcha:
Auto 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
---------------------------------------------------------------------
*}

(* danrodcha *)
theorem "last (postOrden a) = raiz a" (is "?P a")
proof (induct a)
fix x
show "?P (H x)" by simp
fix x i d
show "?P (N x i d)" by simp
qed

(* danrodcha *)
theorem "last (postOrden a) = raiz a" (is "?P a")
proof (induct a)
fix x
show "?P (H x)" by simp
fix x i d
assume HIi: "?P i"
assume HId: "?P d"
have "last (postOrden (N x i d)) = last (postOrden i @ postOrden d @ [x])"
by (simp only: postOrden.simps(2))
also have "… = x" by simp
also have "… = raiz (N x i d)" by (simp only: raiz.simps(2))
finally show "?P (N x i d)" by simp
qed

end
</source>

Relación 6

2016-12-04T01:33:22Z

Pablucoto:

<source lang="isar">
chapter {* R6: Recorridos de árboles *}

theory R6_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))".
---------------------------------------------------------------------
*}

(* ivamenjim marpoldia1 manmorjim1 bowma *)

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]
---------------------------------------------------------------------
*}

(* ivamenjim marpoldia1 pablucoto bowma*)

fun preOrden :: "'a arbol ⇒ 'a list" where
"preOrden (H t) = [t]"
| "preOrden (N t i d) = [t] @ (preOrden i) @ (preOrden d)"

(* danrodcha crigomgom manmorjim1 bowma *)
fun preOrden1 :: "'a arbol ⇒ 'a list" where
"preOrden1 (H x) = [x]"
| "preOrden1 (N x i d) = x#preOrden1 i @ preOrden1 d"

value "preOrden (N e (N c (H a) (H d)) (N g (H f) (H h)))
= [e,c,a,d,g,f,h]"
value "preOrden1 (N e (N c (H a) (H d)) (N g (H f) (H h)))
= [e,c,a,d,g,f,h]"

(* danrodcha *)
lemma "preOrden a = preOrden1 a"
by (induct a) simp_all

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]
---------------------------------------------------------------------
*}

(* ivamenjim danrodcha crigomgom marpoldia1 manmorjim1 pablucoto bowma*)

fun postOrden :: "'a arbol ⇒ 'a list" where
"postOrden (H t) = [t]"
| "postOrden (N t i d) = (postOrden i) @ (postOrden d) @ [t]"

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]
---------------------------------------------------------------------
*}

(* ivamenjim crigomgom marpoldia1 pablucoto bowma *)

fun inOrden :: "'a arbol ⇒ 'a list" where
"inOrden (H t) = [t]"
| "inOrden (N t i d) = (inOrden i) @ [t] @ (inOrden d)"

(* danrodcha manmorjim1 *)
fun inOrden1 :: "'a arbol ⇒ 'a list" where
"inOrden1 (H t) = [t]"
| "inOrden1 (N t i d) = inOrden1 i @ t#inOrden1 d"

value "inOrden (N e (N c (H a) (H d)) (N g (H f) (H h)))
= [a,c,d,e,f,g,h]"
value "inOrden1 (N e (N c (H a) (H d)) (N g (H f) (H h)))
= [a,c,d,e,f,g,h]"

(* manmorjim1 *)
lemma "inOrden t = inOrden1 t"
apply (induct t)
apply auto
done

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))
---------------------------------------------------------------------
*}

(* ivamenjim danrodcha crigomgom marpoldia1 manmorjim1 pablucoto bowma *)

fun espejo :: "'a arbol ⇒ 'a arbol" where
"espejo (H t) = H t"
| "espejo (N t i d) = N t (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)
---------------------------------------------------------------------
*}

(* ivamenjim *)

lemma "preOrden (espejo a) = rev (postOrden a)" (is "?P a")
proof (induct a)
fix x
show "?P (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
finally show "preOrden (espejo (N x i d)) = rev (postOrden (N x i d))" by simp
qed

(* danrodcha *)
lemma "preOrden (espejo a) = rev (postOrden a)"
by (induct a) simp_all

(* danrodcha crigomgom*)
lemma "preOrden (espejo a) = rev (postOrden a)" (is "?P a")
proof (induct a)
fix x
show "?P (H x)" by simp
next
fix x i d
assume HIi: "?P i"
assume HId: "?P d"
have "preOrden (espejo (N x i d)) = preOrden (N x (espejo d) (espejo i))"
by (simp only: espejo.simps(2))
also have "… = x#preOrden (espejo d) @ preOrden (espejo i)"
by (simp only: preOrden.simps(2))
also have"… = x#rev (postOrden d) @ rev (postOrden i)"
using HIi HId by simp
also have "… = rev (postOrden (N x i d))" by simp
finally show "?P (N x i d)" by simp
qed

(* danrodcha *)
lemma "preOrden (espejo a) = rev (postOrden a)" (is "?P a")
proof (induct a)
fix x
show "?P (H x)" by simp
next
fix x i d
assume HIi: "?P i"
assume HId: "?P d"
show "?P (N x i d)" using HIi HId by simp
qed

(* bowma *)
lemma "preOrden (espejo a) = rev (postOrden a)"
apply (induct a)
apply simp_all
done

(* pablucoto *)

lemma "preOrden (espejo a) = rev (postOrden a)" (is "?P a")
proof (induct a)
fix x
show "?P (H x)" by simp
next
fix x i d
assume h1: "?P i"
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 "... = [x] @ rev (postOrden i @ postOrden d)" by simp
also have "... = rev ( postOrden i @ postOrden d @ [x] ) " by simp
also have "... = rev (postOrden (N x i d)) " by simp
finally show "?P (N x i d)" by simp
qed

(* bowma *)
lemma "preOrden (espejo a) = rev (postOrden a)" (is "?p a")
proof (induct a)
fix t
show "?p (H t)" by simp
(* Aquí si le diga "preOrden (espejo (H t)) = rev (postOrden (H t))",isabelle dice:
proof (prove)
goal (1 subgoal):
1. preOrden (espejo (H t)) = rev (postOrden (H t))
Introduced fixed type variable(s): 'b in "t__"
No entiendo porqué *)
next
fix t i d
assume H1: "?p i"
assume H2: "?p d"
have "preOrden (espejo (N t i d)) = preOrden (N t (espejo d) (espejo i))" by simp
also have "... = [t] @ (preOrden (espejo d)) @ (preOrden (espejo i))" by simp
also have "... = [t] @ rev (postOrden d) @ rev (postOrden i)" using H1 H2 by simp
finally show "?p (N t i d)" by simp
qed

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

(* ivamenjim crigomgom bowma *)

lemma "postOrden (espejo a) = rev (preOrden a)" (is "?P a")
proof (induct a)
fix x
show "?P (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
finally show "postOrden (espejo (N x i d)) = rev (preOrden (N x i d))" by simp
(* "?p (N t i d)" mas corto *)
qed

(* danrodcha *)
lemma "postOrden (espejo a) = rev (preOrden a)" (is "?P a")
proof (induct a)
fix x
show "?P (H x)" by simp
next
fix x i d
assume HIi: "?P i"
assume HId: "?P d"
show "?P (N x i d)" using HIi HId by simp
qed

(* pablucoto *)

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

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

(* ivamenjim crigomgom bowma *)

theorem "inOrden (espejo a) = rev (inOrden a)" (is "?P a")
proof (induct a)
fix x
show "?P (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
finally show "inOrden (espejo (N x i d)) = rev (inOrden (N x i d))" by simp
qed

(* danrodcha *)
theorem "inOrden (espejo a) = rev (inOrden a)"
by (induct a) simp_all

(* danrodcha *)
theorem "inOrden (espejo a) = rev (inOrden a)" (is "?P a")
proof (induct a)
fix x
show "?P (H x)" by simp
next
fix x i d
assume HIi: "?P i"
assume HId: "?P d"
show "?P (N x i d)" using HIi HId by simp
qed

(* pablucoto *)
theorem "inOrden (espejo a) = rev (inOrden a)" (is "?P a")
proof (induct a)
fix x
show "?P (H x) " by simp
next
fix x i d
assume HI1: "?P i"
assume HI2: "?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 HI1 HI2 by simp
also have "... = rev (x # inOrden d ) @ rev (inOrden i)" by simp
also have "... = rev ( inOrden i @ x # inOrden d) " by simp
also have "... = rev (inOrden (N x i d))" by simp
finally show "?P (N x i d)" 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
---------------------------------------------------------------------
*}

(* danrodcha crigomgom manmorjim1 ivamenjim bowma *)
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
---------------------------------------------------------------------
*}

(* danrodcha crigomgom manmorjim1 ivamenjim *)
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
---------------------------------------------------------------------
*}

(* danrodcha crigomgom manmorjim1 ivamenjim *)
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
---------------------------------------------------------------------
*}

(* danrodcha *)
lemma aux_ej12: "inOrden a ≠ []"
apply (induct a)
apply simp
apply simp
done

(* danrodcha *)
theorem "last (inOrden a) = extremo_derecha a" (is "?P a")
proof (induct a)
fix x
show "?P (H x)" by simp
next
fix x i d
assume HIi: "?P i"
assume HId: "?P d"
have "last (inOrden (N x i d)) = last (inOrden i @ [x] @ inOrden d)"
by (simp only: inOrden.simps(2))
also have "… = last (inOrden d)" by (simp add: aux_ej12)
also have "… = extremo_derecha d" using HId by simp
also have "… = extremo_derecha (N x i d)" by simp
finally show "?P (N x i d)" by simp
qed

(* ivamenjim *)

lemma aux_ej12_1: "inOrden a ≠ []"
by (induct a) simp_all

(* ivamenjim *)
(* Igual que la anterior, pero poniendo solo by simp en el primer have *)

theorem "last (inOrden a) = extremo_derecha a" (is "?P a")
proof (induct a)
fix x
show "?P (H x)" by simp
next
fix x
fix i assume h1: "?P i"
fix d assume h2: "?P d"
have "last (inOrden (N x i d)) = last ((inOrden i) @ [x] @ (inOrden d))" by simp
also have "... = last (inOrden d)" by (simp add: aux_ej12_1)
also have "... = extremo_derecha d" using h2 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
---------------------------------------------------------------------
*}

(* danrodcha *)
theorem "hd (inOrden a) = extremo_izquierda a" (is "?P a")
proof (induct a)
fix x
show "?P (H x)" by simp
next
fix x i d
assume HIi: "?P i"
assume HId: "?P d"
have "hd (inOrden (N x i d)) = hd (inOrden i @ [x] @ inOrden d)"
by (simp only: inOrden.simps(2))
also have "… = hd (inOrden i)" by (simp add: aux_ej12)
also have "… = extremo_izquierda i" using HIi by simp
also have "… = extremo_izquierda (N x i d)" by simp
finally show "?P (N x i d)" by simp
qed

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

(* danrodcha *)
theorem "hd (preOrden a) = last (postOrden a)" (is "?P a")
proof (induct a)
fix x
show "?P (H x)" by simp
fix x i d
show "?P (N x i d)" by simp
qed

(* danrodcha *)
theorem "hd (preOrden a) = last (postOrden a)" (is "?P a")
proof (induct a)
fix x
show "?P (H x)" by simp
fix x i d
assume HIi: "?P i"
assume HId: "?P d"
have "hd (preOrden (N x i d)) = hd (x#preOrden i @ preOrden d)"
by (simp only: preOrden.simps(2))
also have "… = x" by simp
also have "… = last (postOrden i @ postOrden d @ [x])" by simp
also have "… = last (postOrden (N x i d))"
by (simp only: postOrden.simps(2))
finally show "?P (N x i d)" by simp
qed

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

(* danrodcha *)
theorem "hd (preOrden a) = raiz a" (is "?P a")
proof (induct a)
fix x
show "?P (H x)" by simp
fix x i d
assume HIi: "?P i"
assume HId: "?P d"
have "hd (preOrden (N x i d)) = hd (x#preOrden i @ preOrden d)"
by (simp only: preOrden.simps(2))
also have "… = x" by simp
also have "… = raiz (N x i d)" by simp
finally show "?P (N x i d)" by simp
qed

(* danrodcha *)
theorem "hd (preOrden a) = raiz a" (is "?P a")
proof (induct a)
fix x
show "?P (H x)" by simp
fix x i d
show "?P (N x i d)" by simp
qed

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

(*crigomgom*)
theorem "hd (inOrden a) = raiz a"
quickcheck
oops

(* danrodcha:
Auto 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
---------------------------------------------------------------------
*}

(* danrodcha *)
theorem "last (postOrden a) = raiz a" (is "?P a")
proof (induct a)
fix x
show "?P (H x)" by simp
fix x i d
show "?P (N x i d)" by simp
qed

(* danrodcha *)
theorem "last (postOrden a) = raiz a" (is "?P a")
proof (induct a)
fix x
show "?P (H x)" by simp
fix x i d
assume HIi: "?P i"
assume HId: "?P d"
have "last (postOrden (N x i d)) = last (postOrden i @ postOrden d @ [x])"
by (simp only: postOrden.simps(2))
also have "… = x" by simp
also have "… = raiz (N x i d)" by (simp only: raiz.simps(2))
finally show "?P (N x i d)" by simp
qed

end
</source>

Relación 6

2016-12-04T01:16:58Z

Pablucoto:

<source lang="isar">
chapter {* R6: Recorridos de árboles *}

theory R6_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))".
---------------------------------------------------------------------
*}

(* ivamenjim marpoldia1 manmorjim1 bowma *)

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]
---------------------------------------------------------------------
*}

(* ivamenjim marpoldia1 pablucoto*)

fun preOrden :: "'a arbol ⇒ 'a list" where
"preOrden (H t) = [t]"
| "preOrden (N t i d) = [t] @ (preOrden i) @ (preOrden d)"

(* danrodcha crigomgom manmorjim1 bowma *)
fun preOrden1 :: "'a arbol ⇒ 'a list" where
"preOrden1 (H x) = [x]"
| "preOrden1 (N x i d) = x#preOrden1 i @ preOrden1 d"

value "preOrden (N e (N c (H a) (H d)) (N g (H f) (H h)))
= [e,c,a,d,g,f,h]"
value "preOrden1 (N e (N c (H a) (H d)) (N g (H f) (H h)))
= [e,c,a,d,g,f,h]"

(* danrodcha *)
lemma "preOrden a = preOrden1 a"
by (induct a) simp_all

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]
---------------------------------------------------------------------
*}

(* ivamenjim danrodcha crigomgom marpoldia1 manmorjim1 pablucoto bowma*)

fun postOrden :: "'a arbol ⇒ 'a list" where
"postOrden (H t) = [t]"
| "postOrden (N t i d) = (postOrden i) @ (postOrden d) @ [t]"

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]
---------------------------------------------------------------------
*}

(* ivamenjim crigomgom marpoldia1 pablucoto bowma *)

fun inOrden :: "'a arbol ⇒ 'a list" where
"inOrden (H t) = [t]"
| "inOrden (N t i d) = (inOrden i) @ [t] @ (inOrden d)"

(* danrodcha manmorjim1 *)
fun inOrden1 :: "'a arbol ⇒ 'a list" where
"inOrden1 (H t) = [t]"
| "inOrden1 (N t i d) = inOrden1 i @ t#inOrden1 d"

value "inOrden (N e (N c (H a) (H d)) (N g (H f) (H h)))
= [a,c,d,e,f,g,h]"
value "inOrden1 (N e (N c (H a) (H d)) (N g (H f) (H h)))
= [a,c,d,e,f,g,h]"

(* manmorjim1 *)
lemma "inOrden t = inOrden1 t"
apply (induct t)
apply auto
done

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))
---------------------------------------------------------------------
*}

(* ivamenjim danrodcha crigomgom marpoldia1 manmorjim1 pablucoto bowma *)

fun espejo :: "'a arbol ⇒ 'a arbol" where
"espejo (H t) = H t"
| "espejo (N t i d) = N t (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)
---------------------------------------------------------------------
*}

(* ivamenjim *)

lemma "preOrden (espejo a) = rev (postOrden a)" (is "?P a")
proof (induct a)
fix x
show "?P (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
finally show "preOrden (espejo (N x i d)) = rev (postOrden (N x i d))" by simp
qed

(* danrodcha *)
lemma "preOrden (espejo a) = rev (postOrden a)"
by (induct a) simp_all

(* danrodcha crigomgom*)
lemma "preOrden (espejo a) = rev (postOrden a)" (is "?P a")
proof (induct a)
fix x
show "?P (H x)" by simp
next
fix x i d
assume HIi: "?P i"
assume HId: "?P d"
have "preOrden (espejo (N x i d)) = preOrden (N x (espejo d) (espejo i))"
by (simp only: espejo.simps(2))
also have "… = x#preOrden (espejo d) @ preOrden (espejo i)"
by (simp only: preOrden.simps(2))
also have"… = x#rev (postOrden d) @ rev (postOrden i)"
using HIi HId by simp
also have "… = rev (postOrden (N x i d))" by simp
finally show "?P (N x i d)" by simp
qed

(* danrodcha *)
lemma "preOrden (espejo a) = rev (postOrden a)" (is "?P a")
proof (induct a)
fix x
show "?P (H x)" by simp
next
fix x i d
assume HIi: "?P i"
assume HId: "?P d"
show "?P (N x i d)" using HIi HId by simp
qed

(* bowma *)
lemma "preOrden (espejo a) = rev (postOrden a)"
apply (induct a)
apply simp_all
done

(* pablucoto *)

lemma "preOrden (espejo a) = rev (postOrden a)" (is "?P a")
proof (induct a)
fix x
show "?P (H x)" by simp
next
fix x i d
assume h1: "?P i"
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 "... = [x] @ rev (postOrden i @ postOrden d)" by simp
also have "... = rev ( postOrden i @ postOrden d @ [x] ) " by simp
also have "... = rev (postOrden (N x i d)) " by simp
finally show "?P (N x i d)" by simp
qed

(* bowma *)
lemma "preOrden (espejo a) = rev (postOrden a)" (is "?p a")
proof (induct a)
fix t
show "?p (H t)" by simp
(* Aquí si le diga "preOrden (espejo (H t)) = rev (postOrden (H t))",isabelle dice
"proof (prove)
goal (1 subgoal):
1. preOrden (espejo (H t)) = rev (postOrden (H t))
Introduced fixed type variable(s): 'b in "t__"
No entiendo porqué *)
next
fix t i d
assume H1: "?p i"
assume H2: "?p d"
have "preOrden (espejo (N t i d)) = preOrden (N t (espejo d) (espejo i))" by simp
also have "... = [t] @ (preOrden (espejo d)) @ (preOrden (espejo i))" by simp
also have "... = [t] @ rev (postOrden d) @ rev (postOrden i)" using H1 H2 by simp
finally show "?p (N t i d)" by simp
qed

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

(* ivamenjim crigomgom*)

lemma "postOrden (espejo a) = rev (preOrden a)" (is "?P a")
proof (induct a)
fix x
show "?P (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
finally show "postOrden (espejo (N x i d)) = rev (preOrden (N x i d))" by simp
qed

(* danrodcha *)
lemma "postOrden (espejo a) = rev (preOrden a)" (is "?P a")
proof (induct a)
fix x
show "?P (H x)" by simp
next
fix x i d
assume HIi: "?P i"
assume HId: "?P d"
show "?P (N x i d)" using HIi HId by simp
qed

(* pablucoto *)

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

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

(* ivamenjim crigomgom*)

theorem "inOrden (espejo a) = rev (inOrden a)" (is "?P a")
proof (induct a)
fix x
show "?P (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
finally show "inOrden (espejo (N x i d)) = rev (inOrden (N x i d))" by simp
qed

(* danrodcha *)
theorem "inOrden (espejo a) = rev (inOrden a)"
by (induct a) simp_all

(* danrodcha *)
theorem "inOrden (espejo a) = rev (inOrden a)" (is "?P a")
proof (induct a)
fix x
show "?P (H x)" by simp
next
fix x i d
assume HIi: "?P i"
assume HId: "?P d"
show "?P (N x i d)" using HIi HId 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
---------------------------------------------------------------------
*}

(* danrodcha crigomgom manmorjim1 ivamenjim bowma *)
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
---------------------------------------------------------------------
*}

(* danrodcha crigomgom manmorjim1 ivamenjim *)
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
---------------------------------------------------------------------
*}

(* danrodcha crigomgom manmorjim1 ivamenjim *)
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
---------------------------------------------------------------------
*}

(* danrodcha *)
lemma aux_ej12: "inOrden a ≠ []"
apply (induct a)
apply simp
apply simp
done

(* danrodcha *)
theorem "last (inOrden a) = extremo_derecha a" (is "?P a")
proof (induct a)
fix x
show "?P (H x)" by simp
next
fix x i d
assume HIi: "?P i"
assume HId: "?P d"
have "last (inOrden (N x i d)) = last (inOrden i @ [x] @ inOrden d)"
by (simp only: inOrden.simps(2))
also have "… = last (inOrden d)" by (simp add: aux_ej12)
also have "… = extremo_derecha d" using HId by simp
also have "… = extremo_derecha (N x i d)" by simp
finally show "?P (N x i d)" by simp
qed

(* ivamenjim *)

lemma aux_ej12_1: "inOrden a ≠ []"
by (induct a) simp_all

(* ivamenjim *)
(* Igual que la anterior, pero poniendo solo by simp en el primer have *)

theorem "last (inOrden a) = extremo_derecha a" (is "?P a")
proof (induct a)
fix x
show "?P (H x)" by simp
next
fix x
fix i assume h1: "?P i"
fix d assume h2: "?P d"
have "last (inOrden (N x i d)) = last ((inOrden i) @ [x] @ (inOrden d))" by simp
also have "... = last (inOrden d)" by (simp add: aux_ej12_1)
also have "... = extremo_derecha d" using h2 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
---------------------------------------------------------------------
*}

(* danrodcha *)
theorem "hd (inOrden a) = extremo_izquierda a" (is "?P a")
proof (induct a)
fix x
show "?P (H x)" by simp
next
fix x i d
assume HIi: "?P i"
assume HId: "?P d"
have "hd (inOrden (N x i d)) = hd (inOrden i @ [x] @ inOrden d)"
by (simp only: inOrden.simps(2))
also have "… = hd (inOrden i)" by (simp add: aux_ej12)
also have "… = extremo_izquierda i" using HIi by simp
also have "… = extremo_izquierda (N x i d)" by simp
finally show "?P (N x i d)" by simp
qed

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

(* danrodcha *)
theorem "hd (preOrden a) = last (postOrden a)" (is "?P a")
proof (induct a)
fix x
show "?P (H x)" by simp
fix x i d
show "?P (N x i d)" by simp
qed

(* danrodcha *)
theorem "hd (preOrden a) = last (postOrden a)" (is "?P a")
proof (induct a)
fix x
show "?P (H x)" by simp
fix x i d
assume HIi: "?P i"
assume HId: "?P d"
have "hd (preOrden (N x i d)) = hd (x#preOrden i @ preOrden d)"
by (simp only: preOrden.simps(2))
also have "… = x" by simp
also have "… = last (postOrden i @ postOrden d @ [x])" by simp
also have "… = last (postOrden (N x i d))"
by (simp only: postOrden.simps(2))
finally show "?P (N x i d)" by simp
qed

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

(* danrodcha *)
theorem "hd (preOrden a) = raiz a" (is "?P a")
proof (induct a)
fix x
show "?P (H x)" by simp
fix x i d
assume HIi: "?P i"
assume HId: "?P d"
have "hd (preOrden (N x i d)) = hd (x#preOrden i @ preOrden d)"
by (simp only: preOrden.simps(2))
also have "… = x" by simp
also have "… = raiz (N x i d)" by simp
finally show "?P (N x i d)" by simp
qed

(* danrodcha *)
theorem "hd (preOrden a) = raiz a" (is "?P a")
proof (induct a)
fix x
show "?P (H x)" by simp
fix x i d
show "?P (N x i d)" by simp
qed

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

(*crigomgom*)
theorem "hd (inOrden a) = raiz a"
quickcheck
oops

(* danrodcha:
Auto 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
---------------------------------------------------------------------
*}

(* danrodcha *)
theorem "last (postOrden a) = raiz a" (is "?P a")
proof (induct a)
fix x
show "?P (H x)" by simp
fix x i d
show "?P (N x i d)" by simp
qed

(* danrodcha *)
theorem "last (postOrden a) = raiz a" (is "?P a")
proof (induct a)
fix x
show "?P (H x)" by simp
fix x i d
assume HIi: "?P i"
assume HId: "?P d"
have "last (postOrden (N x i d)) = last (postOrden i @ postOrden d @ [x])"
by (simp only: postOrden.simps(2))
also have "… = x" by simp
also have "… = raiz (N x i d)" by (simp only: raiz.simps(2))
finally show "?P (N x i d)" by simp
qed

end
</source>

Relación 6

2016-12-04T00:42:28Z

Pablucoto:

<source lang="isar">
chapter {* R6: Recorridos de árboles *}

theory R6_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))".
---------------------------------------------------------------------
*}

(* ivamenjim marpoldia1 manmorjim1 bowma *)

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]
---------------------------------------------------------------------
*}

(* ivamenjim marpoldia1 pablucoto*)

fun preOrden :: "'a arbol ⇒ 'a list" where
"preOrden (H t) = [t]"
| "preOrden (N t i d) = [t] @ (preOrden i) @ (preOrden d)"

(* danrodcha crigomgom manmorjim1 bowma *)
fun preOrden1 :: "'a arbol ⇒ 'a list" where
"preOrden1 (H x) = [x]"
| "preOrden1 (N x i d) = x#preOrden1 i @ preOrden1 d"

value "preOrden (N e (N c (H a) (H d)) (N g (H f) (H h)))
= [e,c,a,d,g,f,h]"
value "preOrden1 (N e (N c (H a) (H d)) (N g (H f) (H h)))
= [e,c,a,d,g,f,h]"

(* danrodcha *)
lemma "preOrden a = preOrden1 a"
by (induct a) simp_all

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]
---------------------------------------------------------------------
*}

(* ivamenjim danrodcha crigomgom marpoldia1 manmorjim1 pablucoto bowma*)

fun postOrden :: "'a arbol ⇒ 'a list" where
"postOrden (H t) = [t]"
| "postOrden (N t i d) = (postOrden i) @ (postOrden d) @ [t]"

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]
---------------------------------------------------------------------
*}

(* ivamenjim crigomgom marpoldia1 pablucoto bowma *)

fun inOrden :: "'a arbol ⇒ 'a list" where
"inOrden (H t) = [t]"
| "inOrden (N t i d) = (inOrden i) @ [t] @ (inOrden d)"

(* danrodcha manmorjim1 *)
fun inOrden1 :: "'a arbol ⇒ 'a list" where
"inOrden1 (H t) = [t]"
| "inOrden1 (N t i d) = inOrden1 i @ t#inOrden1 d"

value "inOrden (N e (N c (H a) (H d)) (N g (H f) (H h)))
= [a,c,d,e,f,g,h]"
value "inOrden1 (N e (N c (H a) (H d)) (N g (H f) (H h)))
= [a,c,d,e,f,g,h]"

(* manmorjim1 *)
lemma "inOrden t = inOrden1 t"
apply (induct t)
apply auto
done

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))
---------------------------------------------------------------------
*}

(* ivamenjim danrodcha crigomgom marpoldia1 manmorjim1 pablucoto bowma *)

fun espejo :: "'a arbol ⇒ 'a arbol" where
"espejo (H t) = H t"
| "espejo (N t i d) = N t (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)
---------------------------------------------------------------------
*}

(* ivamenjim *)

lemma "preOrden (espejo a) = rev (postOrden a)" (is "?P a")
proof (induct a)
fix x
show "?P (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
finally show "preOrden (espejo (N x i d)) = rev (postOrden (N x i d))" by simp
qed

(* danrodcha *)
lemma "preOrden (espejo a) = rev (postOrden a)"
by (induct a) simp_all

(* danrodcha crigomgom*)
lemma "preOrden (espejo a) = rev (postOrden a)" (is "?P a")
proof (induct a)
fix x
show "?P (H x)" by simp
next
fix x i d
assume HIi: "?P i"
assume HId: "?P d"
have "preOrden (espejo (N x i d)) = preOrden (N x (espejo d) (espejo i))"
by (simp only: espejo.simps(2))
also have "… = x#preOrden (espejo d) @ preOrden (espejo i)"
by (simp only: preOrden.simps(2))
also have"… = x#rev (postOrden d) @ rev (postOrden i)"
using HIi HId by simp
also have "… = rev (postOrden (N x i d))" by simp
finally show "?P (N x i d)" by simp
qed

(* danrodcha *)
lemma "preOrden (espejo a) = rev (postOrden a)" (is "?P a")
proof (induct a)
fix x
show "?P (H x)" by simp
next
fix x i d
assume HIi: "?P i"
assume HId: "?P d"
show "?P (N x i d)" using HIi HId by simp
qed

(* bowma *)
lemma "preOrden (espejo a) = rev (postOrden a)"
apply (induct a)
apply simp_all
done

(* pablucoto*)

lemma "preOrden (espejo a) = rev (postOrden a)" (is "?P a")
proof (induct a)
fix x
show "?P (H x)" by simp
next
fix x i d
assume h1: "?P i"
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 "... = [x] @ rev (postOrden i @ postOrden d)" by simp
also have "... = rev ( postOrden i @ postOrden d @ [x] ) " by simp
also have "... = rev (postOrden (N x i d)) " by simp
finally show "?P (N x i d)" by simp
qed

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

(* ivamenjim crigomgom*)

lemma "postOrden (espejo a) = rev (preOrden a)" (is "?P a")
proof (induct a)
fix x
show "?P (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
finally show "postOrden (espejo (N x i d)) = rev (preOrden (N x i d))" by simp
qed

(* danrodcha *)
lemma "postOrden (espejo a) = rev (preOrden a)" (is "?P a")
proof (induct a)
fix x
show "?P (H x)" by simp
next
fix x i d
assume HIi: "?P i"
assume HId: "?P d"
show "?P (N x i d)" using HIi HId by simp
qed

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

(* ivamenjim crigomgom*)

theorem "inOrden (espejo a) = rev (inOrden a)" (is "?P a")
proof (induct a)
fix x
show "?P (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
finally show "inOrden (espejo (N x i d)) = rev (inOrden (N x i d))" by simp
qed

(* danrodcha *)
theorem "inOrden (espejo a) = rev (inOrden a)"
by (induct a) simp_all

(* danrodcha *)
theorem "inOrden (espejo a) = rev (inOrden a)" (is "?P a")
proof (induct a)
fix x
show "?P (H x)" by simp
next
fix x i d
assume HIi: "?P i"
assume HId: "?P d"
show "?P (N x i d)" using HIi HId 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
---------------------------------------------------------------------
*}

(* danrodcha crigomgom manmorjim1 ivamenjim bowma *)
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
---------------------------------------------------------------------
*}

(* danrodcha crigomgom manmorjim1 ivamenjim *)
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
---------------------------------------------------------------------
*}

(* danrodcha crigomgom manmorjim1 ivamenjim *)
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
---------------------------------------------------------------------
*}

(* danrodcha *)
lemma aux_ej12: "inOrden a ≠ []"
apply (induct a)
apply simp
apply simp
done

(* danrodcha *)
theorem "last (inOrden a) = extremo_derecha a" (is "?P a")
proof (induct a)
fix x
show "?P (H x)" by simp
next
fix x i d
assume HIi: "?P i"
assume HId: "?P d"
have "last (inOrden (N x i d)) = last (inOrden i @ [x] @ inOrden d)"
by (simp only: inOrden.simps(2))
also have "… = last (inOrden d)" by (simp add: aux_ej12)
also have "… = extremo_derecha d" using HId by simp
also have "… = extremo_derecha (N x i d)" by simp
finally show "?P (N x i d)" by simp
qed

(* ivamenjim *)

lemma aux_ej12_1: "inOrden a ≠ []"
by (induct a) simp_all

(* ivamenjim *)
(* Igual que la anterior, pero poniendo solo by simp en el primer have *)

theorem "last (inOrden a) = extremo_derecha a" (is "?P a")
proof (induct a)
fix x
show "?P (H x)" by simp
next
fix x
fix i assume h1: "?P i"
fix d assume h2: "?P d"
have "last (inOrden (N x i d)) = last ((inOrden i) @ [x] @ (inOrden d))" by simp
also have "... = last (inOrden d)" by (simp add: aux_ej12_1)
also have "... = extremo_derecha d" using h2 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
---------------------------------------------------------------------
*}

(* danrodcha *)
theorem "hd (inOrden a) = extremo_izquierda a" (is "?P a")
proof (induct a)
fix x
show "?P (H x)" by simp
next
fix x i d
assume HIi: "?P i"
assume HId: "?P d"
have "hd (inOrden (N x i d)) = hd (inOrden i @ [x] @ inOrden d)"
by (simp only: inOrden.simps(2))
also have "… = hd (inOrden i)" by (simp add: aux_ej12)
also have "… = extremo_izquierda i" using HIi by simp
also have "… = extremo_izquierda (N x i d)" by simp
finally show "?P (N x i d)" by simp
qed

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

(* danrodcha *)
theorem "hd (preOrden a) = last (postOrden a)" (is "?P a")
proof (induct a)
fix x
show "?P (H x)" by simp
fix x i d
show "?P (N x i d)" by simp
qed

(* danrodcha *)
theorem "hd (preOrden a) = last (postOrden a)" (is "?P a")
proof (induct a)
fix x
show "?P (H x)" by simp
fix x i d
assume HIi: "?P i"
assume HId: "?P d"
have "hd (preOrden (N x i d)) = hd (x#preOrden i @ preOrden d)"
by (simp only: preOrden.simps(2))
also have "… = x" by simp
also have "… = last (postOrden i @ postOrden d @ [x])" by simp
also have "… = last (postOrden (N x i d))"
by (simp only: postOrden.simps(2))
finally show "?P (N x i d)" by simp
qed

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

(* danrodcha *)
theorem "hd (preOrden a) = raiz a" (is "?P a")
proof (induct a)
fix x
show "?P (H x)" by simp
fix x i d
assume HIi: "?P i"
assume HId: "?P d"
have "hd (preOrden (N x i d)) = hd (x#preOrden i @ preOrden d)"
by (simp only: preOrden.simps(2))
also have "… = x" by simp
also have "… = raiz (N x i d)" by simp
finally show "?P (N x i d)" by simp
qed

(* danrodcha *)
theorem "hd (preOrden a) = raiz a" (is "?P a")
proof (induct a)
fix x
show "?P (H x)" by simp
fix x i d
show "?P (N x i d)" by simp
qed

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

(*crigomgom*)
theorem "hd (inOrden a) = raiz a"
quickcheck
oops

(* danrodcha:
Auto 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
---------------------------------------------------------------------
*}

(* danrodcha *)
theorem "last (postOrden a) = raiz a" (is "?P a")
proof (induct a)
fix x
show "?P (H x)" by simp
fix x i d
show "?P (N x i d)" by simp
qed

(* danrodcha *)
theorem "last (postOrden a) = raiz a" (is "?P a")
proof (induct a)
fix x
show "?P (H x)" by simp
fix x i d
assume HIi: "?P i"
assume HId: "?P d"
have "last (postOrden (N x i d)) = last (postOrden i @ postOrden d @ [x])"
by (simp only: postOrden.simps(2))
also have "… = x" by simp
also have "… = raiz (N x i d)" by (simp only: raiz.simps(2))
finally show "?P (N x i d)" by simp
qed

end
</source>

Relación 6

2016-12-03T22:23:56Z

Pablucoto:

<source lang="isar">
chapter {* R6: Recorridos de árboles *}

theory R6_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))".
---------------------------------------------------------------------
*}

(* ivamenjim marpoldia1 manmorjim1 *)

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]
---------------------------------------------------------------------
*}

(* ivamenjim marpoldia1 pablucoto*)

fun preOrden :: "'a arbol ⇒ 'a list" where
"preOrden (H t) = [t]"
| "preOrden (N t i d) = [t] @ (preOrden i) @ (preOrden d)"

(* danrodcha crigomgom manmorjim1 *)
fun preOrden1 :: "'a arbol ⇒ 'a list" where
"preOrden1 (H x) = [x]"
| "preOrden1 (N x i d) = x#preOrden1 i @ preOrden1 d"

value "preOrden (N e (N c (H a) (H d)) (N g (H f) (H h)))
= [e,c,a,d,g,f,h]"
value "preOrden1 (N e (N c (H a) (H d)) (N g (H f) (H h)))
= [e,c,a,d,g,f,h]"

(* danrodcha *)
lemma "preOrden a = preOrden1 a"
by (induct a) simp_all

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]
---------------------------------------------------------------------
*}

(* ivamenjim danrodcha crigomgom marpoldia1 manmorjim1 pablucoto*)

fun postOrden :: "'a arbol ⇒ 'a list" where
"postOrden (H t) = [t]"
| "postOrden (N t i d) = (postOrden i) @ (postOrden d) @ [t]"

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]
---------------------------------------------------------------------
*}

(* ivamenjim crigomgom marpoldia1 pablucoto*)

fun inOrden :: "'a arbol ⇒ 'a list" where
"inOrden (H t) = [t]"
| "inOrden (N t i d) = (inOrden i) @ [t] @ (inOrden d)"

(* danrodcha manmorjim1 *)
fun inOrden1 :: "'a arbol ⇒ 'a list" where
"inOrden1 (H t) = [t]"
| "inOrden1 (N t i d) = inOrden1 i @ t#inOrden1 d"

value "inOrden (N e (N c (H a) (H d)) (N g (H f) (H h)))
= [a,c,d,e,f,g,h]"
value "inOrden1 (N e (N c (H a) (H d)) (N g (H f) (H h)))
= [a,c,d,e,f,g,h]"

(* manmorjim1 *)
lemma "inOrden t = inOrden1 t"
apply (induct t)
apply auto
done

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))
---------------------------------------------------------------------
*}

(* ivamenjim danrodcha crigomgom marpoldia1 manmorjim1 pablucoto*)

fun espejo :: "'a arbol ⇒ 'a arbol" where
"espejo (H t) = H t"
| "espejo (N t i d) = N t (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)
---------------------------------------------------------------------
*}

(* ivamenjim *)

lemma "preOrden (espejo a) = rev (postOrden a)" (is "?P a")
proof (induct a)
fix x
show "?P (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
finally show "preOrden (espejo (N x i d)) = rev (postOrden (N x i d))" by simp
qed

(* danrodcha *)
lemma "preOrden (espejo a) = rev (postOrden a)"
by (induct a) simp_all

(* danrodcha crigomgom*)
lemma "preOrden (espejo a) = rev (postOrden a)" (is "?P a")
proof (induct a)
fix x
show "?P (H x)" by simp
next
fix x i d
assume HIi: "?P i"
assume HId: "?P d"
have "preOrden (espejo (N x i d)) = preOrden (N x (espejo d) (espejo i))"
by (simp only: espejo.simps(2))
also have "… = x#preOrden (espejo d) @ preOrden (espejo i)"
by (simp only: preOrden.simps(2))
also have"… = x#rev (postOrden d) @ rev (postOrden i)"
using HIi HId by simp
also have "… = rev (postOrden (N x i d))" by simp
finally show "?P (N x i d)" by simp
qed

(* danrodcha *)
lemma "preOrden (espejo a) = rev (postOrden a)" (is "?P a")
proof (induct a)
fix x
show "?P (H x)" by simp
next
fix x i d
assume HIi: "?P i"
assume HId: "?P d"
show "?P (N x i d)" using HIi HId by simp
qed

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

(* ivamenjim crigomgom*)

lemma "postOrden (espejo a) = rev (preOrden a)" (is "?P a")
proof (induct a)
fix x
show "?P (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
finally show "postOrden (espejo (N x i d)) = rev (preOrden (N x i d))" by simp
qed

(* danrodcha *)
lemma "postOrden (espejo a) = rev (preOrden a)" (is "?P a")
proof (induct a)
fix x
show "?P (H x)" by simp
next
fix x i d
assume HIi: "?P i"
assume HId: "?P d"
show "?P (N x i d)" using HIi HId by simp
qed

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

(* ivamenjim crigomgom*)

theorem "inOrden (espejo a) = rev (inOrden a)" (is "?P a")
proof (induct a)
fix x
show "?P (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
finally show "inOrden (espejo (N x i d)) = rev (inOrden (N x i d))" by simp
qed

(* danrodcha *)
theorem "inOrden (espejo a) = rev (inOrden a)"
by (induct a) simp_all

(* danrodcha *)
theorem "inOrden (espejo a) = rev (inOrden a)" (is "?P a")
proof (induct a)
fix x
show "?P (H x)" by simp
next
fix x i d
assume HIi: "?P i"
assume HId: "?P d"
show "?P (N x i d)" using HIi HId 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
---------------------------------------------------------------------
*}

(* danrodcha crigomgom manmorjim1 ivamenjim *)
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
---------------------------------------------------------------------
*}

(* danrodcha crigomgom manmorjim1 ivamenjim *)
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
---------------------------------------------------------------------
*}

(* danrodcha crigomgom manmorjim1 ivamenjim *)
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
---------------------------------------------------------------------
*}

(* danrodcha *)
lemma aux_ej12: "inOrden a ≠ []"
apply (induct a)
apply simp
apply simp
done

(* danrodcha *)
theorem "last (inOrden a) = extremo_derecha a" (is "?P a")
proof (induct a)
fix x
show "?P (H x)" by simp
next
fix x i d
assume HIi: "?P i"
assume HId: "?P d"
have "last (inOrden (N x i d)) = last (inOrden i @ [x] @ inOrden d)"
by (simp only: inOrden.simps(2))
also have "… = last (inOrden d)" by (simp add: aux_ej12)
also have "… = extremo_derecha d" using HId by simp
also have "… = extremo_derecha (N x i d)" by simp
finally show "?P (N x i d)" by simp
qed

(* ivamenjim *)

lemma aux_ej12_1: "inOrden a ≠ []"
by (induct a) simp_all

(* ivamenjim *)
(* Igual que la anterior, pero poniendo solo by simp en el primer have *)

theorem "last (inOrden a) = extremo_derecha a" (is "?P a")
proof (induct a)
fix x
show "?P (H x)" by simp
next
fix x
fix i assume h1: "?P i"
fix d assume h2: "?P d"
have "last (inOrden (N x i d)) = last ((inOrden i) @ [x] @ (inOrden d))" by simp
also have "... = last (inOrden d)" by (simp add: aux_ej12_1)
also have "... = extremo_derecha d" using h2 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
---------------------------------------------------------------------
*}

(* danrodcha *)
theorem "hd (inOrden a) = extremo_izquierda a" (is "?P a")
proof (induct a)
fix x
show "?P (H x)" by simp
next
fix x i d
assume HIi: "?P i"
assume HId: "?P d"
have "hd (inOrden (N x i d)) = hd (inOrden i @ [x] @ inOrden d)"
by (simp only: inOrden.simps(2))
also have "… = hd (inOrden i)" by (simp add: aux_ej12)
also have "… = extremo_izquierda i" using HIi by simp
also have "… = extremo_izquierda (N x i d)" by simp
finally show "?P (N x i d)" by simp
qed

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

(* danrodcha *)
theorem "hd (preOrden a) = last (postOrden a)" (is "?P a")
proof (induct a)
fix x
show "?P (H x)" by simp
fix x i d
show "?P (N x i d)" by simp
qed

(* danrodcha *)
theorem "hd (preOrden a) = last (postOrden a)" (is "?P a")
proof (induct a)
fix x
show "?P (H x)" by simp
fix x i d
assume HIi: "?P i"
assume HId: "?P d"
have "hd (preOrden (N x i d)) = hd (x#preOrden i @ preOrden d)"
by (simp only: preOrden.simps(2))
also have "… = x" by simp
also have "… = last (postOrden i @ postOrden d @ [x])" by simp
also have "… = last (postOrden (N x i d))"
by (simp only: postOrden.simps(2))
finally show "?P (N x i d)" by simp
qed

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

(* danrodcha *)
theorem "hd (preOrden a) = raiz a" (is "?P a")
proof (induct a)
fix x
show "?P (H x)" by simp
fix x i d
assume HIi: "?P i"
assume HId: "?P d"
have "hd (preOrden (N x i d)) = hd (x#preOrden i @ preOrden d)"
by (simp only: preOrden.simps(2))
also have "… = x" by simp
also have "… = raiz (N x i d)" by simp
finally show "?P (N x i d)" by simp
qed

(* danrodcha *)
theorem "hd (preOrden a) = raiz a" (is "?P a")
proof (induct a)
fix x
show "?P (H x)" by simp
fix x i d
show "?P (N x i d)" by simp
qed

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

(*crigomgom*)
theorem "hd (inOrden a) = raiz a"
quickcheck
oops

(* danrodcha:
Auto 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
---------------------------------------------------------------------
*}

(* danrodcha *)
theorem "last (postOrden a) = raiz a" (is "?P a")
proof (induct a)
fix x
show "?P (H x)" by simp
fix x i d
show "?P (N x i d)" by simp
qed

(* danrodcha *)
theorem "last (postOrden a) = raiz a" (is "?P a")
proof (induct a)
fix x
show "?P (H x)" by simp
fix x i d
assume HIi: "?P i"
assume HId: "?P d"
have "last (postOrden (N x i d)) = last (postOrden i @ postOrden d @ [x])"
by (simp only: postOrden.simps(2))
also have "… = x" by simp
also have "… = raiz (N x i d)" by (simp only: raiz.simps(2))
finally show "?P (N x i d)" by simp
qed

end
</source>

Relación 6

2016-12-03T21:49:00Z

Pablucoto:

<source lang="isar">
chapter {* R6: Recorridos de árboles *}

theory R6_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))".
---------------------------------------------------------------------
*}

(* ivamenjim marpoldia1 manmorjim1 *)

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]
---------------------------------------------------------------------
*}

(* ivamenjim marpoldia1 pablucoto*)

fun preOrden :: "'a arbol ⇒ 'a list" where
"preOrden (H t) = [t]"
| "preOrden (N t i d) = [t] @ (preOrden i) @ (preOrden d)"

(* danrodcha crigomgom manmorjim1 *)
fun preOrden1 :: "'a arbol ⇒ 'a list" where
"preOrden1 (H x) = [x]"
| "preOrden1 (N x i d) = x#preOrden1 i @ preOrden1 d"

value "preOrden (N e (N c (H a) (H d)) (N g (H f) (H h)))
= [e,c,a,d,g,f,h]"
value "preOrden1 (N e (N c (H a) (H d)) (N g (H f) (H h)))
= [e,c,a,d,g,f,h]"

(* danrodcha *)
lemma "preOrden a = preOrden1 a"
by (induct a) simp_all

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]
---------------------------------------------------------------------
*}

(* ivamenjim danrodcha crigomgom marpoldia1 manmorjim1 pablucoto*)

fun postOrden :: "'a arbol ⇒ 'a list" where
"postOrden (H t) = [t]"
| "postOrden (N t i d) = (postOrden i) @ (postOrden d) @ [t]"

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]
---------------------------------------------------------------------
*}

(* ivamenjim crigomgom marpoldia1 pablucoto*)

fun inOrden :: "'a arbol ⇒ 'a list" where
"inOrden (H t) = [t]"
| "inOrden (N t i d) = (inOrden i) @ [t] @ (inOrden d)"

(* danrodcha manmorjim1 *)
fun inOrden1 :: "'a arbol ⇒ 'a list" where
"inOrden1 (H t) = [t]"
| "inOrden1 (N t i d) = inOrden1 i @ t#inOrden1 d"

value "inOrden (N e (N c (H a) (H d)) (N g (H f) (H h)))
= [a,c,d,e,f,g,h]"
value "inOrden1 (N e (N c (H a) (H d)) (N g (H f) (H h)))
= [a,c,d,e,f,g,h]"

(* manmorjim1 *)
lemma "inOrden t = inOrden1 t"
apply (induct t)
apply auto
done

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))
---------------------------------------------------------------------
*}

(* ivamenjim danrodcha crigomgom marpoldia1 manmorjim1 *)

fun espejo :: "'a arbol ⇒ 'a arbol" where
"espejo (H t) = H t"
| "espejo (N t i d) = N t (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)
---------------------------------------------------------------------
*}

(* ivamenjim *)

lemma "preOrden (espejo a) = rev (postOrden a)" (is "?P a")
proof (induct a)
fix x
show "?P (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
finally show "preOrden (espejo (N x i d)) = rev (postOrden (N x i d))" by simp
qed

(* danrodcha *)
lemma "preOrden (espejo a) = rev (postOrden a)"
by (induct a) simp_all

(* danrodcha crigomgom*)
lemma "preOrden (espejo a) = rev (postOrden a)" (is "?P a")
proof (induct a)
fix x
show "?P (H x)" by simp
next
fix x i d
assume HIi: "?P i"
assume HId: "?P d"
have "preOrden (espejo (N x i d)) = preOrden (N x (espejo d) (espejo i))"
by (simp only: espejo.simps(2))
also have "… = x#preOrden (espejo d) @ preOrden (espejo i)"
by (simp only: preOrden.simps(2))
also have"… = x#rev (postOrden d) @ rev (postOrden i)"
using HIi HId by simp
also have "… = rev (postOrden (N x i d))" by simp
finally show "?P (N x i d)" by simp
qed

(* danrodcha *)
lemma "preOrden (espejo a) = rev (postOrden a)" (is "?P a")
proof (induct a)
fix x
show "?P (H x)" by simp
next
fix x i d
assume HIi: "?P i"
assume HId: "?P d"
show "?P (N x i d)" using HIi HId by simp
qed

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

(* ivamenjim crigomgom*)

lemma "postOrden (espejo a) = rev (preOrden a)" (is "?P a")
proof (induct a)
fix x
show "?P (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
finally show "postOrden (espejo (N x i d)) = rev (preOrden (N x i d))" by simp
qed

(* danrodcha *)
lemma "postOrden (espejo a) = rev (preOrden a)" (is "?P a")
proof (induct a)
fix x
show "?P (H x)" by simp
next
fix x i d
assume HIi: "?P i"
assume HId: "?P d"
show "?P (N x i d)" using HIi HId by simp
qed

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

(* ivamenjim crigomgom*)

theorem "inOrden (espejo a) = rev (inOrden a)" (is "?P a")
proof (induct a)
fix x
show "?P (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
finally show "inOrden (espejo (N x i d)) = rev (inOrden (N x i d))" by simp
qed

(* danrodcha *)
theorem "inOrden (espejo a) = rev (inOrden a)"
by (induct a) simp_all

(* danrodcha *)
theorem "inOrden (espejo a) = rev (inOrden a)" (is "?P a")
proof (induct a)
fix x
show "?P (H x)" by simp
next
fix x i d
assume HIi: "?P i"
assume HId: "?P d"
show "?P (N x i d)" using HIi HId 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
---------------------------------------------------------------------
*}

(* danrodcha crigomgom manmorjim1 ivamenjim *)
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
---------------------------------------------------------------------
*}

(* danrodcha crigomgom manmorjim1 ivamenjim *)
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
---------------------------------------------------------------------
*}

(* danrodcha crigomgom manmorjim1 ivamenjim *)
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
---------------------------------------------------------------------
*}

(* danrodcha *)
lemma aux_ej12: "inOrden a ≠ []"
apply (induct a)
apply simp
apply simp
done

(* danrodcha *)
theorem "last (inOrden a) = extremo_derecha a" (is "?P a")
proof (induct a)
fix x
show "?P (H x)" by simp
next
fix x i d
assume HIi: "?P i"
assume HId: "?P d"
have "last (inOrden (N x i d)) = last (inOrden i @ [x] @ inOrden d)"
by (simp only: inOrden.simps(2))
also have "… = last (inOrden d)" by (simp add: aux_ej12)
also have "… = extremo_derecha d" using HId by simp
also have "… = extremo_derecha (N x i d)" by simp
finally show "?P (N x i d)" by simp
qed

(* ivamenjim *)

lemma aux_ej12_1: "inOrden a ≠ []"
by (induct a) simp_all

(* ivamenjim *)
(* Igual que la anterior, pero poniendo solo by simp en el primer have *)

theorem "last (inOrden a) = extremo_derecha a" (is "?P a")
proof (induct a)
fix x
show "?P (H x)" by simp
next
fix x
fix i assume h1: "?P i"
fix d assume h2: "?P d"
have "last (inOrden (N x i d)) = last ((inOrden i) @ [x] @ (inOrden d))" by simp
also have "... = last (inOrden d)" by (simp add: aux_ej12_1)
also have "... = extremo_derecha d" using h2 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
---------------------------------------------------------------------
*}

(* danrodcha *)
theorem "hd (inOrden a) = extremo_izquierda a" (is "?P a")
proof (induct a)
fix x
show "?P (H x)" by simp
next
fix x i d
assume HIi: "?P i"
assume HId: "?P d"
have "hd (inOrden (N x i d)) = hd (inOrden i @ [x] @ inOrden d)"
by (simp only: inOrden.simps(2))
also have "… = hd (inOrden i)" by (simp add: aux_ej12)
also have "… = extremo_izquierda i" using HIi by simp
also have "… = extremo_izquierda (N x i d)" by simp
finally show "?P (N x i d)" by simp
qed

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

(* danrodcha *)
theorem "hd (preOrden a) = last (postOrden a)" (is "?P a")
proof (induct a)
fix x
show "?P (H x)" by simp
fix x i d
show "?P (N x i d)" by simp
qed

(* danrodcha *)
theorem "hd (preOrden a) = last (postOrden a)" (is "?P a")
proof (induct a)
fix x
show "?P (H x)" by simp
fix x i d
assume HIi: "?P i"
assume HId: "?P d"
have "hd (preOrden (N x i d)) = hd (x#preOrden i @ preOrden d)"
by (simp only: preOrden.simps(2))
also have "… = x" by simp
also have "… = last (postOrden i @ postOrden d @ [x])" by simp
also have "… = last (postOrden (N x i d))"
by (simp only: postOrden.simps(2))
finally show "?P (N x i d)" by simp
qed

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

(* danrodcha *)
theorem "hd (preOrden a) = raiz a" (is "?P a")
proof (induct a)
fix x
show "?P (H x)" by simp
fix x i d
assume HIi: "?P i"
assume HId: "?P d"
have "hd (preOrden (N x i d)) = hd (x#preOrden i @ preOrden d)"
by (simp only: preOrden.simps(2))
also have "… = x" by simp
also have "… = raiz (N x i d)" by simp
finally show "?P (N x i d)" by simp
qed

(* danrodcha *)
theorem "hd (preOrden a) = raiz a" (is "?P a")
proof (induct a)
fix x
show "?P (H x)" by simp
fix x i d
show "?P (N x i d)" by simp
qed

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

(*crigomgom*)
theorem "hd (inOrden a) = raiz a"
quickcheck
oops

(* danrodcha:
Auto 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
---------------------------------------------------------------------
*}

(* danrodcha *)
theorem "last (postOrden a) = raiz a" (is "?P a")
proof (induct a)
fix x
show "?P (H x)" by simp
fix x i d
show "?P (N x i d)" by simp
qed

(* danrodcha *)
theorem "last (postOrden a) = raiz a" (is "?P a")
proof (induct a)
fix x
show "?P (H x)" by simp
fix x i d
assume HIi: "?P i"
assume HId: "?P d"
have "last (postOrden (N x i d)) = last (postOrden i @ postOrden d @ [x])"
by (simp only: postOrden.simps(2))
also have "… = x" by simp
also have "… = raiz (N x i d)" by (simp only: raiz.simps(2))
finally show "?P (N x i d)" by simp
qed

end
</source>

Relación 6

2016-12-03T21:04:36Z

Pablucoto:

<source lang="isar">
chapter {* R6: Recorridos de árboles *}

theory R6_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))".
---------------------------------------------------------------------
*}

(* ivamenjim marpoldia1 manmorjim1 *)

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]
---------------------------------------------------------------------
*}

(* ivamenjim marpoldia1 pablucoto*)

fun preOrden :: "'a arbol ⇒ 'a list" where
"preOrden (H t) = [t]"
| "preOrden (N t i d) = [t] @ (preOrden i) @ (preOrden d)"

(* danrodcha crigomgom manmorjim1 *)
fun preOrden1 :: "'a arbol ⇒ 'a list" where
"preOrden1 (H x) = [x]"
| "preOrden1 (N x i d) = x#preOrden1 i @ preOrden1 d"

value "preOrden (N e (N c (H a) (H d)) (N g (H f) (H h)))
= [e,c,a,d,g,f,h]"
value "preOrden1 (N e (N c (H a) (H d)) (N g (H f) (H h)))
= [e,c,a,d,g,f,h]"

(* danrodcha *)
lemma "preOrden a = preOrden1 a"
by (induct a) simp_all

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]
---------------------------------------------------------------------
*}

(* ivamenjim danrodcha crigomgom marpoldia1 manmorjim1 pablucoto*)

fun postOrden :: "'a arbol ⇒ 'a list" where
"postOrden (H t) = [t]"
| "postOrden (N t i d) = (postOrden i) @ (postOrden d) @ [t]"

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]
---------------------------------------------------------------------
*}

(* ivamenjim crigomgom marpoldia1*)

fun inOrden :: "'a arbol ⇒ 'a list" where
"inOrden (H t) = [t]"
| "inOrden (N t i d) = (inOrden i) @ [t] @ (inOrden d)"

(* danrodcha manmorjim1 *)
fun inOrden1 :: "'a arbol ⇒ 'a list" where
"inOrden1 (H t) = [t]"
| "inOrden1 (N t i d) = inOrden1 i @ t#inOrden1 d"

value "inOrden (N e (N c (H a) (H d)) (N g (H f) (H h)))
= [a,c,d,e,f,g,h]"
value "inOrden1 (N e (N c (H a) (H d)) (N g (H f) (H h)))
= [a,c,d,e,f,g,h]"

(* manmorjim1 *)
lemma "inOrden t = inOrden1 t"
apply (induct t)
apply auto
done

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))
---------------------------------------------------------------------
*}

(* ivamenjim danrodcha crigomgom marpoldia1 manmorjim1 *)

fun espejo :: "'a arbol ⇒ 'a arbol" where
"espejo (H t) = H t"
| "espejo (N t i d) = N t (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)
---------------------------------------------------------------------
*}

(* ivamenjim *)

lemma "preOrden (espejo a) = rev (postOrden a)" (is "?P a")
proof (induct a)
fix x
show "?P (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
finally show "preOrden (espejo (N x i d)) = rev (postOrden (N x i d))" by simp
qed

(* danrodcha *)
lemma "preOrden (espejo a) = rev (postOrden a)"
by (induct a) simp_all

(* danrodcha crigomgom*)
lemma "preOrden (espejo a) = rev (postOrden a)" (is "?P a")
proof (induct a)
fix x
show "?P (H x)" by simp
next
fix x i d
assume HIi: "?P i"
assume HId: "?P d"
have "preOrden (espejo (N x i d)) = preOrden (N x (espejo d) (espejo i))"
by (simp only: espejo.simps(2))
also have "… = x#preOrden (espejo d) @ preOrden (espejo i)"
by (simp only: preOrden.simps(2))
also have"… = x#rev (postOrden d) @ rev (postOrden i)"
using HIi HId by simp
also have "… = rev (postOrden (N x i d))" by simp
finally show "?P (N x i d)" by simp
qed

(* danrodcha *)
lemma "preOrden (espejo a) = rev (postOrden a)" (is "?P a")
proof (induct a)
fix x
show "?P (H x)" by simp
next
fix x i d
assume HIi: "?P i"
assume HId: "?P d"
show "?P (N x i d)" using HIi HId by simp
qed

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

(* ivamenjim crigomgom*)

lemma "postOrden (espejo a) = rev (preOrden a)" (is "?P a")
proof (induct a)
fix x
show "?P (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
finally show "postOrden (espejo (N x i d)) = rev (preOrden (N x i d))" by simp
qed

(* danrodcha *)
lemma "postOrden (espejo a) = rev (preOrden a)" (is "?P a")
proof (induct a)
fix x
show "?P (H x)" by simp
next
fix x i d
assume HIi: "?P i"
assume HId: "?P d"
show "?P (N x i d)" using HIi HId by simp
qed

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

(* ivamenjim crigomgom*)

theorem "inOrden (espejo a) = rev (inOrden a)" (is "?P a")
proof (induct a)
fix x
show "?P (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
finally show "inOrden (espejo (N x i d)) = rev (inOrden (N x i d))" by simp
qed

(* danrodcha *)
theorem "inOrden (espejo a) = rev (inOrden a)"
by (induct a) simp_all

(* danrodcha *)
theorem "inOrden (espejo a) = rev (inOrden a)" (is "?P a")
proof (induct a)
fix x
show "?P (H x)" by simp
next
fix x i d
assume HIi: "?P i"
assume HId: "?P d"
show "?P (N x i d)" using HIi HId 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
---------------------------------------------------------------------
*}

(* danrodcha crigomgom manmorjim1 ivamenjim *)
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
---------------------------------------------------------------------
*}

(* danrodcha crigomgom manmorjim1 ivamenjim *)
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
---------------------------------------------------------------------
*}

(* danrodcha crigomgom manmorjim1 ivamenjim *)
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
---------------------------------------------------------------------
*}

(* danrodcha *)
lemma aux_ej12: "inOrden a ≠ []"
apply (induct a)
apply simp
apply simp
done

(* danrodcha *)
theorem "last (inOrden a) = extremo_derecha a" (is "?P a")
proof (induct a)
fix x
show "?P (H x)" by simp
next
fix x i d
assume HIi: "?P i"
assume HId: "?P d"
have "last (inOrden (N x i d)) = last (inOrden i @ [x] @ inOrden d)"
by (simp only: inOrden.simps(2))
also have "… = last (inOrden d)" by (simp add: aux_ej12)
also have "… = extremo_derecha d" using HId by simp
also have "… = extremo_derecha (N x i d)" by simp
finally show "?P (N x i d)" by simp
qed

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

(* danrodcha *)
theorem "hd (inOrden a) = extremo_izquierda a" (is "?P a")
proof (induct a)
fix x
show "?P (H x)" by simp
next
fix x i d
assume HIi: "?P i"
assume HId: "?P d"
have "hd (inOrden (N x i d)) = hd (inOrden i @ [x] @ inOrden d)"
by (simp only: inOrden.simps(2))
also have "… = hd (inOrden i)" by (simp add: aux_ej12)
also have "… = extremo_izquierda i" using HIi by simp
also have "… = extremo_izquierda (N x i d)" by simp
finally show "?P (N x i d)" by simp
qed

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

(* danrodcha *)
theorem "hd (preOrden a) = last (postOrden a)" (is "?P a")
proof (induct a)
fix x
show "?P (H x)" by simp
fix x i d
show "?P (N x i d)" by simp
qed

(* danrodcha *)
theorem "hd (preOrden a) = last (postOrden a)" (is "?P a")
proof (induct a)
fix x
show "?P (H x)" by simp
fix x i d
assume HIi: "?P i"
assume HId: "?P d"
have "hd (preOrden (N x i d)) = hd (x#preOrden i @ preOrden d)"
by (simp only: preOrden.simps(2))
also have "… = x" by simp
also have "… = last (postOrden i @ postOrden d @ [x])" by simp
also have "… = last (postOrden (N x i d))"
by (simp only: postOrden.simps(2))
finally show "?P (N x i d)" by simp
qed

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

(* danrodcha *)
theorem "hd (preOrden a) = raiz a" (is "?P a")
proof (induct a)
fix x
show "?P (H x)" by simp
fix x i d
assume HIi: "?P i"
assume HId: "?P d"
have "hd (preOrden (N x i d)) = hd (x#preOrden i @ preOrden d)"
by (simp only: preOrden.simps(2))
also have "… = x" by simp
also have "… = raiz (N x i d)" by simp
finally show "?P (N x i d)" by simp
qed

(* danrodcha *)
theorem "hd (preOrden a) = raiz a" (is "?P a")
proof (induct a)
fix x
show "?P (H x)" by simp
fix x i d
show "?P (N x i d)" by simp
qed

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

(*crigomgom*)
theorem "hd (inOrden a) = raiz a"
quickcheck
oops

(* danrodcha:
Auto 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
---------------------------------------------------------------------
*}

(* danrodcha *)
theorem "last (postOrden a) = raiz a" (is "?P a")
proof (induct a)
fix x
show "?P (H x)" by simp
fix x i d
show "?P (N x i d)" by simp
qed

(* danrodcha *)
theorem "last (postOrden a) = raiz a" (is "?P a")
proof (induct a)
fix x
show "?P (H x)" by simp
fix x i d
assume HIi: "?P i"
assume HId: "?P d"
have "last (postOrden (N x i d)) = last (postOrden i @ postOrden d @ [x])"
by (simp only: postOrden.simps(2))
also have "… = x" by simp
also have "… = raiz (N x i d)" by (simp only: raiz.simps(2))
finally show "?P (N x i d)" by simp
qed

end
</source>

Relación 5

2016-12-01T02:04:21Z

Pablucoto:

<source lang="isar">
chapter {* R5: Eliminación de duplicados *}

theory R5_Eliminacion_de_duplicados
imports Main
begin

text {*
---------------------------------------------------------------------
Ejercicio 1. Definir la funcion primitiva recursiva
estaEn :: 'a ⇒ 'a list ⇒ bool
tal que (estaEn x xs) se verifica si el elemento x está en la lista
xs. Por ejemplo,
estaEn (2::nat) [3,2,4] = True
estaEn (1::nat) [3,2,4] = False
---------------------------------------------------------------------
*}

(* crigomgom rubgonmar bowma wilmorort pablucoto serrodcal
anaprarod migtermor paupeddeg fraortmoy marpoldia1
danrodcha manmorjim1 jeamacpov *)
fun estaEn :: "'a ⇒ 'a list ⇒ bool" where
"estaEn _ [] = False"
| "estaEn x (a#xs) = ((a = x) ∨ (estaEn x xs))"

value "estaEn (2::nat) [3,2,4] = True"
value "estaEn (1::nat) [3,2,4] = False"

(* ivamenjim ferrenseg josgarsan juacabsou dancorgar pabrodmac lucnovdos
fracorjim1 *)
(* Igual que la anterior pero con x en lugar de _ en el caso base *)
fun estaEn1 :: "'a ⇒ 'a list ⇒ bool" where
"estaEn1 x [] = False"
| "estaEn1 x (a#xs) = ((x=a) ∨ estaEn1 x xs)"

value "estaEn1 (2::nat) [3,2,4] = True"
value "estaEn1 (1::nat) [3,2,4] = False"

(* wilmorort *)
(* reutilizando la funcion "algunos" de R4.thy*)
fun algunos :: "('a ⇒ bool) ⇒ 'a list ⇒ bool" where
"algunos p [] = False"
| "algunos p (x#xs) = (p x ∨ algunos p xs)"

fun estaEn2 :: "'a ⇒ 'a list ⇒ bool" where
"estaEn2 a xs = algunos (λx. x = a) xs"

value "estaEn2 (2::nat) [3,2,4] = True"
value "estaEn2 (1::nat) [3,2,4] = False"

text {*
---------------------------------------------------------------------
Ejercicio 2. Definir la función primitiva recursiva
sinDuplicados :: 'a list ⇒ bool
tal que (sinDuplicados xs) se verifica si la lista xs no contiene
duplicados. Por ejemplo,
sinDuplicados [1::nat,4,2] = True
sinDuplicados [1::nat,4,2,4] = False
---------------------------------------------------------------------
*}

(* crigomgom rubgonmar ivamenjim wilmorort bowma pablucoto
serrodcal anaprarod migtermor paupeddeg fraortmoy marpoldia1
ferrenseg josgarsan danrodcha manmorjim1 juacabsou dancorgar
pabrodmac lucnovdos jeamacpov *)
fun sinDuplicados :: "'a list ⇒ bool" where
"sinDuplicados [] = True"
| "sinDuplicados (x#xs) = (¬ estaEn x xs ∧ sinDuplicados xs)"

value "sinDuplicados [1::nat,4,2] = True"
value "sinDuplicados [1::nat,4,2,4] = False"

(* fracorjim1 - La versión anterior no prueba el segundo enunciado. La
que propongo demuestra ambos. *)
fun sinDuplicados1 :: "'a list ⇒ bool" where
"sinDuplicados1 [] = True"
| "sinDuplicados1 (x#xs) = (¬(estaEn x xs ∧ sinDuplicados xs))"

value "sinDuplicados1 [1::nat,4,2] = True"
value "sinDuplicados1 [1::nat,4,2,4] = False"

(* Comentario: La definición sinDuplicados1 no cumple el segundo
ejemplo. *)

(* paupeddeg *)
(* Utilizando la función ∉ de Isabelle *)
fun sinDuplicados2 :: "'a list ⇒ bool" where
"sinDuplicados2 [] = True"
| "sinDuplicados2 (a#xs) = ((a ∉ set xs) ∧ sinDuplicados2 xs ) "

(* Comentario: Uso de ∉ y set *)

text {*
---------------------------------------------------------------------
Ejercicio 3. Definir la función primitiva recursiva
borraDuplicados :: 'a list ⇒ bool
tal que (borraDuplicados xs) es la lista obtenida eliminando los
elementos duplicados de la lista xs. Por ejemplo,
borraDuplicados [1::nat,2,4,2,3] = [1,4,2,3]

Nota: La función borraDuplicados es equivalente a la predefinida
remdups.
---------------------------------------------------------------------
*}

(* crigomgom rubgonmar wilmorort bowma pablucoto serrodcal
anaprarod migtermor paupeddeg fraortmoy marpoldia1 ferrenseg
josgarsan danrodcha juacabsou dancorgar manmorjim1 pabrodmac
lucnovdos jeamacpov *)
fun borraDuplicados0 :: "'a list ⇒ 'a list" where
"borraDuplicados0 [] = []"
| "borraDuplicados0 (x#xs) = (if estaEn x xs
then borraDuplicados0 xs
else x#borraDuplicados0 xs)"

value "borraDuplicados0 [1::nat,2,4,2,3] = [1,4,2,3]"

(* rubgonmar *)
(* Otra forma Sin usar if
Utilizando case aunque se le sacaría más partido con más de 2 casos *)
fun borraDuplicados1 :: "'a list ⇒ 'a list" where
"borraDuplicados1 [] = []"
| "borraDuplicados1 (x#xs) = (case estaEn x xs of
False => x#borraDuplicados1 xs
| True => borraDuplicados1 xs )"

value "borraDuplicados1 [1::nat,2,4,2,3] = [1,4,2,3]"

(* rubgonmar *)
(* Otra forma utilizando let *)
fun borraDuplicados2 :: "'a list ⇒ 'a list" where
"borraDuplicados2 [] = []"
| "borraDuplicados2 (x#xs) =
(let condicion = estaEn x xs::bool in
if condicion
then borraDuplicados2 xs
else x # borraDuplicados2 xs)"

value "borraDuplicados2 [1::nat,2,4,2,3] = [1,4,2,3]"

(* ivamenjim *)
(* Utilizando la negación primero *)
fun borraDuplicados3 :: "'a list ⇒ 'a list" where
"borraDuplicados3 [] = []"
| "borraDuplicados3 (x#xs) = (if ¬(estaEn x xs)
then (x#(borraDuplicados3 xs))
else borraDuplicados3 xs)"

value "borraDuplicados3 [1::nat,2,4,2,3] = [1,4,2,3]"

fun borraDuplicados :: "'a list ⇒ 'a list" where
"borraDuplicados xs = borraDuplicados0 xs"

(* fracorjim1. Utilizo un acumulador para optimizar la eficiencia *)

fun borraDuplicadosAc :: "'a list ⇒ 'a list ⇒ 'a list" where
"borraDuplicadosAc [] ys = ys"
| "borraDuplicadosAc (x#xs) ys = (if (estaEn x ys) then (borraDuplicadosAc xs ys)
else (borraDuplicadosAc xs (x#ys)))"

(* fracorjim1. Uso un caso base *)
fun borraDuplicados4 :: "'a list ⇒ 'a list" where
"borraDuplicados4 xs = (if (sinDuplicados xs) then xs else borraDuplicadosAc xs [])"

text {*
---------------------------------------------------------------------
Ejercicio 4.1. Demostrar o refutar automáticamente
length (borraDuplicados xs) ≤ length xs
---------------------------------------------------------------------
*}

(* crigomgom anaprarod ferrenseg *)
lemma length_borraDuplicados:
"length (borraDuplicados xs) ≤ length xs"
by (induct xs, simp_all)

(* Comentario: Falla para borraDuplicados1. *)

(* rubgonmar wilmorort pablucoto serrodcal migtermor paupeddeg
fraortmoy marpoldia1 danrodcha juacabsou dancorgar josgarsan
pabrodmac lucnovdos *)
lemma length_borraDuplicados2:
"length ( borraDuplicados xs ) ≤ length xs"
by (induct xs) auto

(* manmorjim1 *)
(* soy de ponerlo mejor por pasos *)
lemma length_borraDuplicados3:
"length ( borraDuplicados xs ) ≤ length xs"
apply (induct xs)
apply auto
done

(* ivamenjim *)
(* Demostrando objetivo a objetivo *)
lemma length_borraDuplicados4:
"length (borraDuplicados xs) ≤ length xs"
apply (induct xs)
apply simp
apply auto
done

(* bowma anaprarod *)
lemma length_borraDuplicados5:
"length (borraDuplicados xs) ≤ length xs"
apply (induct xs)
apply (simp, simp) (* creo que es mejor poner aquí simp_all *)
done

(* ferrenseg *)
lemma length_borraDuplicados6:
"length (borraDuplicados xs) ≤ length xs"
by (induct xs) simp_all (* Creo que se puede poner simp_all fuera de
paréntesis *)

text {*
---------------------------------------------------------------------
Ejercicio 4.2. Demostrar o refutar detalladamente
length (borraDuplicados xs) ≤ length xs
---------------------------------------------------------------------
*}

(* crigomgom *)
lemma length_borraDuplicados_2:
"length (borraDuplicados xs) ≤ length xs"
proof (induct xs)
show "length (borraDuplicados []) ≤ length []" by simp
next
fix x xs
assume HI: "length (borraDuplicados xs) ≤ length xs"
show "length (borraDuplicados (x#xs)) ≤ length (x#xs)"
proof (cases)
assume "estaEn x xs"
then have "length (borraDuplicados (x#xs)) =
length (borraDuplicados xs)" by simp
also have "... ≤ length xs" using HI by simp
also have "... ≤ length (x#xs)" by simp
finally show "length (borraDuplicados (x#xs)) ≤ length (x#xs)"
by simp
next
assume "(¬ estaEn x xs)"
then have "length (borraDuplicados (x#xs)) =
length (x#borraDuplicados xs)" by simp
also have "... = 1 + length (borraDuplicados xs)" by simp
also have "... ≤ 1 + length xs" using HI by simp
also have "... = length (x#xs)" by simp
finally show "length (borraDuplicados (x#xs)) ≤ length (x#xs)"
by simp
qed
qed

(* ivamenjim wilmorort ferrenseg rubgonmar juacabsou dancorgar
josgarsan lucnovdos *)
lemma length_borraDuplicados_2b:
"length (borraDuplicados xs) ≤ length xs"
proof (induct xs)
show "length (borraDuplicados []) ≤ length []" by simp
next
fix a xs
assume HI: "length (borraDuplicados xs) ≤ length xs"
have "length (borraDuplicados (a # xs)) ≤
1+length (borraDuplicados xs)" by simp
also have "... ≤ 1+length xs" using HI by simp
finally show "length (borraDuplicados (a # xs)) ≤ length (a # xs)"
by simp
qed

(* serrodcal anaprarod danrodcha *)
lemma length_borraDuplicados_2c:
"length (borraDuplicados xs) ≤ length xs" (is "?P xs")
proof (induct xs)
show "?P []" by simp
next
fix a xs
assume HI: "?P xs"
have "length (borraDuplicados (a # xs)) ≤
1+length (borraDuplicados xs)" by simp
also have "... ≤ 1+length xs" using HI by simp
finally show "?P (a # xs)" by simp
qed

(* pablucoto jeamacpov *)
lemma length_borraDuplicados_2d:
"length (borraDuplicados xs) ≤ length xs"
proof(induct xs)
show "length (borraDuplicados []) ≤ length [] " by simp
next
fix a xs
assume HI: " length (borraDuplicados xs) ≤ length xs "
have "length (borraDuplicados (a # xs)) ≤
1 + length(borraDuplicados xs)" by simp
also have "... ≤ 1 + length xs" using HI by simp
also have "... ≤ length (a#xs)" by simp
finally show "length (borraDuplicados (a # xs)) ≤ length (a # xs) "
by simp
qed

(* bowma *)
lemma length_borraDuplicados_2f:
"length (borraDuplicados xs) ≤ length xs" (is "?p xs")
proof (induct xs)
show "length (borraDuplicados []) ≤ length []" by simp
next
fix a xs
assume HI: "?p xs"
have c1: "1+length xs = length (a#xs)" by simp
also have "length(borraDuplicados (a#xs)) ≤
1 + length(borraDuplicados xs)" by simp
also have "... ≤ 1+length xs" using HI by simp
then show "length(borraDuplicados (a#xs)) ≤ length (a#xs)"
using c1 by simp
(* ¿Aquí porqué no puedo usar "finally show "?p (a # xs)" using c1 by simp?
Y porque no puedo añadir "finally show "?p (a # xs)" by simp al final? *)
qed

(* Comentario: Responder la pregunta. *)

(* migtermor *)
lemma length_borraDuplicados_2g:
"length (borraDuplicados xs) ≤ length xs" (is "?P xs")
proof (induct xs)
show "?P []" by simp
next
fix a xs
assume HI: "?P xs"
have "length (borraDuplicados (a#xs)) ≤ (length (a#xs))"
proof (cases)
assume "(estaEn a xs)"
then have Aux: "length (borraDuplicados (a#xs)) =
length (borraDuplicados xs)" by simp
also have "… ≤ length (a#xs)" using HI by simp
then show "length (borraDuplicados (a#xs)) ≤ (length (a#xs))"
using Aux by simp
next
assume "¬ (estaEn a xs)"
then have Aux: "length (borraDuplicados (a#xs)) =
1+ length (borraDuplicados xs)" by simp
also have "… ≤ length (a#xs)" using HI by simp
then show "length (borraDuplicados (a#xs)) ≤ (length (a#xs))"
using Aux by simp
qed
then show "?P (a#xs)" by simp
qed

(* paupeddeg marpoldia1 pabrodmac*)
lemma length_borraDuplicados_2h:
"length (borraDuplicados xs) ≤ length xs"
proof (induct xs)
show "length (borraDuplicados []) ≤ length []" by simp
next
fix a xs
assume HI: "length (borraDuplicados xs) ≤ length xs"
have "length (borraDuplicados (a # xs)) ≤
length [a] + length (borraDuplicados xs) " by simp
also have "... ≤ 1 + length (borraDuplicados xs)" by simp
also have "... ≤ 1 + length xs" using HI by simp
finally show "length (borraDuplicados (a # xs)) ≤
length (a # xs)" by simp
qed

(* fraortmoy *)
(* muy parecida a alguna anterior, pero yo dí más pasos *)
lemma length_borraDuplicados_2i:
"length (borraDuplicados xs) ≤ length xs"
proof (induct xs)
show "length (borraDuplicados []) ≤ length []" by simp
next
fix a xs
assume H1: "length (borraDuplicados xs) ≤ length xs"
have "length (borraDuplicados (a # xs)) ≤
length(borraDuplicados [a])+length (borraDuplicados xs)"
by simp
also have "… ≤ 1 + length (borraDuplicados xs)" by simp
also have "… ≤ 1 + length xs" using H1 by simp
finally show "length (borraDuplicados (a # xs)) ≤ length (a # xs)"
by simp
qed

text {*
---------------------------------------------------------------------
Ejercicio 5.1. Demostrar o refutar automáticamente
estaEn a (borraDuplicados xs) = estaEn a xs
---------------------------------------------------------------------
*}

(* crigomgom rubgonmar wilmorort pablucoto serrodcal bowma
migtermor fraortmoy marpoldia1 ferrenseg danrodcha juacabsou paupeddeg pabrodmac lucnovdos dancorgar jeamacpov *)
lemma estaEn_borraDuplicados:
"estaEn a (borraDuplicados xs) = estaEn a xs"
by (induct xs) auto

(* ivamenjim manmorjim1 *)
lemma estaEn_borraDuplicados2:
"estaEn a (borraDuplicados xs) = estaEn a xs"
apply (induct xs)
apply auto
done

(* anaprarod *)
lemma estaEn_borraDuplicados3:
"estaEn a (borraDuplicados xs) = estaEn a xs"
by (induct xs, simp_all, blast)

(* anaprarod *)
lemma estaEn_borraDuplicados4:
"estaEn a (borraDuplicados xs) = estaEn a xs"
apply (induct xs)
apply (cases "estaEn x xs")
apply (simp_all)
apply blast
done

(* bowma *)
lemma estaEn_borraDuplicados5:
"estaEn a (borraDuplicados xs) = estaEn a xs"
apply (induct xs)
apply simp
apply (simp, blast)
done

text {*
---------------------------------------------------------------------
Ejercicio 5.2. Demostrar o refutar detalladamente
estaEn a (borraDuplicados xs) = estaEn a xs
Nota: Para la demostración de la equivalencia se puede usar
proof (rule iffI)
La regla iffI es
⟦P ⟹ Q ; Q ⟹ P⟧ ⟹ P = Q
---------------------------------------------------------------------
*}

(* wilmorort *)
lemma estaEn_borraDuplicados_2:
"estaEn a (borraDuplicados xs) = estaEn a xs"
proof (induct xs)
show "estaEn a (borraDuplicados []) = estaEn a []" by simp
next
fix b xs
assume HI: "estaEn a (borraDuplicados xs) = estaEn a xs"
show "estaEn a (borraDuplicados (b#xs)) = estaEn a (b#xs)"
proof (rule iffI)
assume H1: "estaEn a (borraDuplicados (b#xs))"
show "estaEn a (b#xs)"
proof (cases)
assume "estaEn b xs"
then have "estaEn a (borraDuplicados xs)" using H1 by simp
then have "estaEn a xs" using HI by simp
then show "estaEn a (b#xs)" by simp
next
assume "¬ estaEn b xs"
then have "estaEn a (b#(borraDuplicados xs))" using H1 by simp
then have "a=b ∨ (estaEn a (borraDuplicados xs))" by simp
then have " a=b ∨ (estaEn a xs)" using HI by simp
then show "estaEn a (b#xs)" by simp
qed
next
assume H2: "estaEn a (b#xs)"
show "estaEn a (borraDuplicados (b#xs))"
proof (cases)
assume "a=b"
then have "estaEn b (borraDuplicados xs) = estaEn b xs"
using HI by simp
then have "(estaEn b xs ⟶ estaEn b (borraDuplicados xs)) ∧
(¬ estaEn b xs ⟶ estaEn b (b # borraDuplicados xs))"
by simp
then have "estaEn b (borraDuplicados (b#xs))" by simp
then show "estaEn a (borraDuplicados (b#xs))" using `a=b` by simp
next
assume "a≠b"
then have "estaEn a (b#xs)" using H2 by simp
then have "a = b ∨ estaEn a xs" by simp
then have "False ∨ estaEn a xs " using `a≠b` by simp
then have "estaEn a xs" by simp
then have "estaEn a (borraDuplicados xs)" using HI by simp
then show "estaEn a (borraDuplicados (b#xs))" using `a≠b` by simp
qed
qed
qed

(* anaprarod marpoldia1 ferrenseg juacabsou *)
lemma estaEn_borraDuplicados_2b:
"estaEn a (borraDuplicados xs) = estaEn a xs" (is "?P xs")
proof (induct xs)
show "?P []" by simp
next
fix x xs
assume HI: "?P xs"
show "?P (x#xs)"
proof (cases)
assume "estaEn x xs"
then have "estaEn a (borraDuplicados (x#xs)) =
estaEn a (borraDuplicados xs)" by simp
also have "...= estaEn a xs" using HI by simp
finally show "?P (x#xs)" by auto
next
assume "¬estaEn x xs"
then have "estaEn a (borraDuplicados (x#xs)) =
estaEn a (x#borraDuplicados xs)" by simp
also have "...= (x = a ∨ estaEn a (borraDuplicados xs))" by simp
finally show "?P (x#xs)" using HI by simp
qed
qed

(* migtermor *)
lemma estaEn_borraDuplicados_2c:
"estaEn a (borraDuplicados xs) = estaEn a xs" (is "?P xs")
proof (induct xs)
show "?P []" by simp
next
fix aa xs
assume HI: "?P xs"
have P1: "estaEn a (borraDuplicados (aa#xs)) = estaEn a (aa#xs)"
proof (cases)
assume C1: "(estaEn aa xs)"
have "estaEn a (borraDuplicados (aa#xs)) =
estaEn a (borraDuplicados xs)"
using C1 by simp
also have P3: "… = estaEn a xs" using HI by simp
also have "… = estaEn a (aa#xs)"
proof (cases)
assume "(a=aa)"
then show "estaEn a xs = estaEn a (aa#xs)" using C1 by simp
next
assume "¬(a=aa)"
then show "estaEn a xs = estaEn a (aa#xs)" by simp
qed
then show "estaEn a (borraDuplicados (aa#xs)) = estaEn a (aa#xs)"
using P3 by simp
next
assume C2: "¬(estaEn aa xs)"
then show "estaEn a (borraDuplicados (aa#xs)) =
estaEn a (aa#xs)" using HI by simp
qed
also have Conc: "estaEn a (borraDuplicados (aa#xs)) = estaEn a (aa#xs)"
using P1 by simp
finally show "?P (aa#xs)" using Conc by simp
qed

(* crigomgom paupeddeg pabrodmac*)
lemma estaEn_borraDuplicados_2d:
"estaEn a (borraDuplicados xs) = estaEn a xs"
proof (induct xs)
show "estaEn a (borraDuplicados []) = estaEn a []" by simp
next
fix x xs
assume HI: "estaEn a (borraDuplicados xs) = estaEn a xs"
show "estaEn a (borraDuplicados (x#xs)) = estaEn a (x#xs)"
proof (rule iffI)
assume a1: "estaEn a (borraDuplicados (x#xs))"
show "estaEn a (x#xs)"
proof (cases)
assume "estaEn x xs"
then have "estaEn a (borraDuplicados xs)" using a1 by simp
then have "estaEn a xs" using HI by simp
then show "estaEn a (x#xs)" by simp
next
assume "¬ estaEn x xs"
then have "estaEn a (x#(borraDuplicados xs))" using a1 by simp
then have " x=a ∨ (estaEn a (borraDuplicados xs))" by simp
then have " x=a ∨ (estaEn a xs)" using HI by simp
then show "estaEn a (x#xs)" by simp
qed
next
assume a2: "estaEn a (x#xs)"
show "estaEn a (borraDuplicados (x#xs))"
proof (cases)
assume "a=x"
then show "estaEn a (borraDuplicados (x#xs))" using HI by simp
next
assume b1: "a≠x"
then have "estaEn a (x#xs)" using a2 by simp
then have "x = a ∨ estaEn a xs" by simp
then have "estaEn a xs " using b1 by simp
then have "estaEn a (borraDuplicados xs)" using HI by simp
then show "estaEn a (borraDuplicados (x#xs))" using b1 by simp
qed
qed
qed

(* rubgonmar jeamacpov *)
lemma estaEn_borraDuplicados_2e:
"estaEn a ( borraDuplicados xs ) = estaEn a xs"
proof (induct xs)
show "estaEn a (borraDuplicados []) = estaEn a []" by simp
next
fix x
fix xs
assume HI: "estaEn a (borraDuplicados xs) = estaEn a xs"
show "estaEn a (borraDuplicados (x#xs)) = estaEn a (x#xs)"
proof (rule iffI) (* usamos proof de la regla dada iffI*)
assume cprim: "estaEn a (borraDuplicados (x#xs))"
show "estaEn a (x#xs)"
proof (cases)
assume "estaEn x xs"
then show "estaEn a (x#xs)" using cprim HI by simp
next
assume "¬ estaEn x xs"
then show "estaEn a (x#xs)" using cprim HI by simp
qed
next
assume cseg: "estaEn a (x#xs)"
show "estaEn a (borraDuplicados (x#xs))"
proof (cases)
assume "a=x"
then show "estaEn a (borraDuplicados (x#xs))" using HI by auto
next
assume "a≠x"
then show "estaEn a (borraDuplicados (x#xs))" using `a≠x`
cseg HI by simp
qed
qed
qed

(*
Aplico la regla iffI:
⟦P ⟹ Q ; Q ⟹ P⟧ ⟹ P = Q
Así:
[estaEn a (borraDuplicados (x # xs))
⟹ estaEn a (x # xs); estaEn a (x # xs)
⟹ estaEn a (borraDuplicados (x # xs))]
⟹ estaEn a (borraDuplicados (x # xs)) = estaEn a (x # xs)
*)

(* bowma ivamenjim *)
lemma estaEn_borraDuplicados_2f:
"estaEn a (borraDuplicados xs) = estaEn a xs" (is "?p xs")
proof (induct xs)
show "?p []" by simp
next
fix x xs
assume HI: "?p xs"
show "?p (x#xs)"
proof (cases)
assume H1:"estaEn x xs"
then have "estaEn a (borraDuplicados (x#xs)) =
estaEn a (borraDuplicados xs)" by simp
also have "... = estaEn a xs" using HI by simp
also have "... = estaEn a (x#xs)"
proof(cases)
assume "x=a"
then show "estaEn a xs = estaEn a (x#xs)" using H1 by simp
next
assume "x≠a"
then show "estaEn a xs = estaEn a (x#xs)" by simp
qed
finally show "?p (x#xs)" by simp
next
assume H2:"¬estaEn x xs"
then have "estaEn a (borraDuplicados (x#xs)) =
estaEn a (x#borraDuplicados xs)" by simp
also have "... = ((x=a) ∨ estaEn a (borraDuplicados xs))" by simp
also have "... = ((x=a) ∨ estaEn a xs)" using HI by simp
also have "... = estaEn a (x#xs)" by simp
finally show "?p (x#xs)" by simp
qed

(* danrodcha pablucoto*)
(* es como la de ruben pero con diferencias de estilo *)
lemma estaEn_borraDuplicados_2g:
"estaEn a (borraDuplicados xs) = estaEn a xs" (is "?P xs")
proof (induct xs)
show "?P []" by simp
next
fix x xs assume HI: "?P xs"
show "?P (x#xs)"
proof (rule iffI)
assume H1: "estaEn a (borraDuplicados (x # xs))"
show "estaEn a (x#xs)"
proof (cases "estaEn x xs")
case True
then show "estaEn a (x#xs)" using H1 HI by simp
next
case False
then show "estaEn a (x#xs)" using H1 HI by simp
qed
next
assume H2: "estaEn a (x#xs)"
show "estaEn a (borraDuplicados (x # xs))"
proof (cases "x=a")
case True
then show "estaEn a (borraDuplicados (x # xs))" using HI by simp
next
case False
then show "estaEn a (borraDuplicados (x # xs))" using H2 HI
by simp
qed
qed
qed

text {*
---------------------------------------------------------------------
Ejercicio 6.1. Demostrar o refutar automáticamente
sinDuplicados (borraDuplicados xs)
---------------------------------------------------------------------
*}

(* ivamenjim wilmorort serrodcal crigomgom anaprarod fraortmoy
marpoldia1 ferrenseg danrodcha juacabsou paupeddeg josgarsan pabrodmac dancorgar jeamacpov *)
lemma sinDuplicados_borraDuplicados:
"sinDuplicados (borraDuplicados xs)"
by (induct xs) (auto simp add: estaEn_borraDuplicados)

(* migtermor bowma *)
lemma sinDuplicados_borraDuplicados:
"sinDuplicados (borraDuplicados xs)"
by (induct xs, simp_all add: estaEn_borraDuplicados_2)

(* manmorjim1 no caí en usar la demostración anterior y he realizado
la demostración de que si un elemento no estaba en una lista seguirá sin estar
después de eliminar los duplicados en esa lista... *)
lemma noEsta_tras_borrarDuplicados:
"(¬estaEn x xs) ⟶ (¬estaEn x (borraDuplicados xs))"
apply (induct xs)
apply auto
done

lemma sinDuplicados_borraDuplicados:
"sinDuplicados (borraDuplicados xs)"
apply (induct xs)
apply simp
apply (induct xs)
apply auto
apply (simp add: noEsta_tras_borrarDuplicados)
done

text {*
---------------------------------------------------------------------
Ejercicio 6.2. Demostrar o refutar detalladamente
sinDuplicados (borraDuplicados xs)
---------------------------------------------------------------------
*}

(*wilmorort pablucoto*)

-- "La demostración estructurada es"
lemma sinDuplicados_borraDuplicados_2:
"sinDuplicados (borraDuplicados xs)"
proof (induct xs)
show " sinDuplicados (borraDuplicados [])" by simp
next
fix a xs
assume HI: "sinDuplicados (borraDuplicados xs)"
show "sinDuplicados (borraDuplicados (a # xs))"
proof (cases)
assume "estaEn a xs"
then show "sinDuplicados (borraDuplicados (a#xs))" using HI by simp
next
assume"¬ estaEn a xs"
then have "¬ (estaEn a xs) ∧ sinDuplicados (borraDuplicados xs)" using HI by simp
then have "¬ estaEn a (borraDuplicados xs) ∧ sinDuplicados (borraDuplicados xs)"
by (simp add: estaEn_borraDuplicados)
then have " sinDuplicados (a#borraDuplicados xs)" by simp
then show " sinDuplicados (borraDuplicados(a #xs))" by simp
qed
qed

(* ivamenjim migtermor crigomgom rubgonmar fraortmoy marpoldia1 ferrenseg bowma juacabsou serrodcal josgarsan pabrodmac dancorgar jeamacpov lucnovdos *)

lemma sinDuplicados_borraDuplicados_2:
"sinDuplicados (borraDuplicados xs)"
proof (induct xs)
show "sinDuplicados (borraDuplicados [])" by simp
next
fix a xs
assume HI: "sinDuplicados (borraDuplicados xs)"
show "sinDuplicados (borraDuplicados (a # xs))"
proof (cases)
assume "estaEn a xs"
then show "sinDuplicados (borraDuplicados (a # xs))" using HI by simp
next
assume "¬(estaEn a xs)"
then show "sinDuplicados (borraDuplicados (a # xs))" using HI by (simp add: estaEn_borraDuplicados)
qed
qed

(* anaprarod paupeddeg*)
lemma sinDuplicados_borraDuplicados_2:
"sinDuplicados (borraDuplicados xs)" (is "?P xs")
proof (induct xs)
show "?P []" by simp
next
fix x xs
assume HI: "?P xs"
show "?P (x#xs)"
proof (cases)
assume c1: "estaEn x xs"
then show "sinDuplicados (borraDuplicados (x#xs))" using HI by simp
next
assume c2: "¬ estaEn x xs"
then have "sinDuplicados (borraDuplicados (x#xs)) =sinDuplicados (x#borraDuplicados xs)" by simp
also have "…= (¬estaEn x (borraDuplicados xs) ∧ sinDuplicados (borraDuplicados xs))" by simp
also have "… = (¬estaEn x (borraDuplicados xs))" using HI by simp
also have "… = (¬(estaEn x xs))" by (simp add:estaEn_borraDuplicados)
also have "… = True" using c2 by simp
finally show "?P (x#xs)" by simp
qed
qed

(* danrodcha *)
lemma sinDuplicados_borraDuplicados_2:
"sinDuplicados (borraDuplicados xs)" (is "?P xs")
proof (induct xs)
show "?P []" by simp
next
fix x xs assume HI: "?P xs"
show "?P (x#xs)"
proof (cases "estaEn x xs")
case True
then have 1: "sinDuplicados (borraDuplicados (x#xs))
= sinDuplicados (borraDuplicados xs)" by (simp add: estaEn_borraDuplicados_2)
show "?P (x#xs)" using HI 1 by simp
next
case False
then have "sinDuplicados (borraDuplicados (x#xs))
= sinDuplicados (x#borraDuplicados xs)" by simp
also have "… = (¬ (estaEn x (borraDuplicados xs)) ∧
sinDuplicados (borraDuplicados xs))" by simp
also have "… = True" using `¬ estaEn x xs` HI by (simp add:estaEn_borraDuplicados)
finally show "?P (x#xs)" by simp
qed
qed

text {*
---------------------------------------------------------------------
Ejercicio 7. Demostrar o refutar:
borraDuplicados (rev xs) = rev (borraDuplicados xs)
---------------------------------------------------------------------
*}

(*crigomgom rubgonmar ivamenjim wilmorort pablucoto migtermor
anaprarod fraortmoy ferrenseg marpoldia1 bowma danrodcha juacabsou paupeddeg manmorjim1 serrodcal josgarsan pabrodmac lucnovdos dancorgar jeamacpov *)
lemma "borraDuplicados (rev xs) = rev (borraDuplicados xs)"
quickcheck
oops

(* ivamenjim: Quickcheck encuentra el siguiente contraejemplo:
xs = [a1, a2, a1]
Por lo que:
· "borraDuplicados (rev xs) = [a2, a1]"
· "rev (borraDuplicados xs) = [a1, a2]" *)

end
</source>

Relación 5

2016-11-27T23:01:22Z

Pablucoto:

Relación 5

2016-11-27T22:36:42Z

Pablucoto:

Relación 5

2016-11-27T22:20:57Z

Pablucoto:

Relación 5

2016-11-27T21:21:57Z

Pablucoto:

Relación 5

2016-11-27T19:58:56Z

Pablucoto:

Relación 5

2016-11-27T19:40:51Z

Pablucoto:

Relación 4

2016-11-23T04:41:48Z

Pablucoto:

<source lang="isar">
chapter {* R4: Cuantificadores sobre listas *}

theory R4_Cuantificadores_sobre_listas
imports Main
begin

text {*
---------------------------------------------------------------------
Ejercicio 1. Definir la función
todos :: ('a ⇒ bool) ⇒ 'a list ⇒ bool
tal que (todos p xs) se verifica si todos los elementos de la lista
xs cumplen la propiedad p. Por ejemplo, se verifica
todos (λx. 1<length x) [[2,1,4],[1,3]]
¬todos (λx. 1<length x) [[2,1,4],[3]]

Nota: La función todos es equivalente a la predefinida list_all.
---------------------------------------------------------------------
*}

{*danrodcha*}
fun todos :: "('a ⇒ bool) ⇒ 'a list ⇒ bool" where
"todos p xs = "todos p [] = True"
| "todos p (x#xs) = (p x ∧ todos p xs)""

(* ivamenjim migtermor dancorgar wilmorort marpoldia1 ferrenseg paupeddeg pablucoto crigomgom anaprarod serrodcal juacabsou rubgonmar josgarsan fraortmoy lucnovdos*)
fun todos :: "('a ⇒ bool) ⇒ 'a list ⇒ bool" where
"todos p [] = True"
| "todos p (x#xs) = (p x ∧ todos p xs)"

text {*
---------------------------------------------------------------------
Ejercicio 2. Definir la función
algunos :: ('a ⇒ bool) ⇒ 'a list ⇒ bool
tal que (algunos p xs) se verifica si algunos elementos de la lista
xs cumplen la propiedad p. Por ejemplo, se verifica
algunos (λx. 1<length x) [[2,1,4],[3]]
¬algunos (λx. 1<length x) [[],[3]]"

Nota: La función algunos es equivalente a la predefinida list_ex.
---------------------------------------------------------------------
*}

{*danrodcha ivamenjim migtermor dancorgar marpoldia1 ferrenseg wilmorort paupeddeg pablucoto crigomgom anaprarod serrodcal juacabsou rubgonmar josgarsan fraortmoy lucnovdos*}
fun algunos :: "('a ⇒ bool) ⇒ 'a list ⇒ bool" where
"algunos p [] = False"
| "algunos p (x#xs) = (p x ∨ algunos p xs)"

text {*
---------------------------------------------------------------------
Ejercicio 3.1. Demostrar o refutar automáticamente
todos (λx. P x ∧ Q x) xs = (todos P xs ∧ todos Q xs)
---------------------------------------------------------------------
*}

{*danrodcha ivamenjim migtermor dancorgar marpoldia1 ferrenseg wilmorort paupeddeg pablucoto anaprarod serrodcal juacabsou rubgonmar josgarsan fraortmoy lucnovdos*}
lemma "todos (λx. P x ∧ Q x) xs = (todos P xs ∧ todos Q xs)"
by (induct xs) auto

text {*
---------------------------------------------------------------------
Ejercicio 3.2. Demostrar o refutar detalladamente
todos (λx. P x ∧ Q x) xs = (todos P xs ∧ todos Q xs)
---------------------------------------------------------------------
*}

{*danrodcha*}
lemma "todos (λx. P x ∧ Q x) xs = (todos P xs ∧ todos Q xs)"
proof (induct xs)
show "?R []" by simp
next
fix a xs
assume HI: "?R xs"
have "todos (λx. P x ∧ Q x) (a#xs) = (P a ∧ Q a ∧ todos (λx. P x ∧ Q x) xs)" by simp
also have "… = (P a ∧ Q a ∧ (todos P xs ∧ todos Q xs))" using HI by simp
also have "… = ((P a ∧ todos P xs) ∧ (Q a ∧ todos Q xs))" by blast
also have "… = (todos P (a#xs) ∧ todos Q (a#xs))" by simp
finally show "?R (a#xs)" by simp
qed

(* ivamenjim wilmorort serrodcal josgarsan*)

lemma "todos (λx. P x ∧ Q x) xs = (todos P xs ∧ todos Q xs)"
proof (induct xs)
show "todos (λx. P x ∧ Q x) [] = (todos P [] ∧ todos Q [])" by simp
next
fix a xs
assume HI: "todos (λx. P x ∧ Q x) xs = (todos P xs ∧ todos Q xs)"
have "todos (λx. P x ∧ Q x) (a # xs) = (P a ∧ Q a ∧ todos (λx. P x ∧ Q x) xs)" by simp
also have "... = (P a ∧ Q a ∧ todos P xs ∧ todos Q xs)" using HI by simp
finally show "todos (λx. P x ∧ Q x) (a # xs) = (todos P (a # xs) ∧ todos Q (a # xs))" by auto
qed

(* dancorgar paupeddeg *)
lemma "todos (λx. P x ∧ Q x) xs = (todos P xs ∧ todos Q xs)"
proof (induct xs)
show "todos (λx. P x ∧ Q x) [] = (todos P [] ∧ todos Q [])" by simp
next
fix y xs
assume HI: "todos (λx. P x ∧ Q x) xs = (todos P xs ∧ todos Q xs)"
have "todos (λx. P x ∧ Q x) (y#xs) = ((P y ∧ Q y) ∧ (todos (λx. P x ∧ Q x) xs))" by simp
also have "… = ((P y ∧ Q y) ∧ (todos P xs ∧ todos Q xs))" using HI by simp
also have "… = ((P y ∧ todos P xs) ∧ (Q y ∧ todos Q xs))" by blast
also have "… = ((todos P (y#xs)) ∧ (todos Q (y#xs)))" by simp
finally show "todos (λx. P x ∧ Q x) (y#xs) = (todos P (y#xs) ∧ todos Q (y#xs))" by simp
qed

(* migtermor *)
lemma "todos (λx. P x ∧ Q x) xs = (todos P xs ∧ todos Q xs)"
proof (induct xs)
show "todos (λx. P x ∧ Q x) [] = (todos P [] ∧ todos Q [])" by simp
next
fix x xs
assume HI: "todos (λx. P x ∧ Q x) xs = (todos P xs ∧ todos Q xs)"
have "(todos (λx. P x ∧ Q x) (x#xs)) = (( P x ∧ Q x) ∧ (todos (λx. P x ∧ Q x) xs))"
by (simp only: todos.simps(2))
also have "… = ((P x ∧ Q x) ∧ (todos P xs ∧ todos Q xs))" using HI by simp
also have "… = (((P x)∧(todos P xs)) ∧ ((Q x) ∧ (todos Q xs)))" by arith
also have "((P x ∧ Q x) ∧ (todos P xs ∧ todos Q xs)) = (((P x)∧(todos P xs)) ∧ ((Q x) ∧ (todos Q xs)))"
by arith (* Este paso es exactamente el mismo que el anterior, pero sin cualquiera de los dos no funciona el "finally show" *)
have "… = (((P x)∧(todos P xs))∧((Q x)∧(todos Q xs)))" by simp
have "… = ((todos P (x#xs))∧(todos Q (x#xs)))" by simp
finally show "(todos (λx. P x ∧ Q x) (x#xs)) = ((todos P (x#xs)) ∧ (todos Q (x#xs)))" by simp
qed

(* marpoldia1 *)
lemma "todos (λx. P x ∧ Q x) xs = (todos P xs ∧ todos Q xs)"
proof (induct xs)
show "todos (λx. P x ∧ Q x) [] = (todos P [] ∧ todos Q [])" by simp
next
fix a xs
assume HI: "todos (λx. P x ∧ Q x) xs = (todos P xs ∧ todos Q xs)"
have "todos (λx. P x ∧ Q x) (a # xs) = ((P a ∧ Q a) ∧ (todos P xs ∧ todos Q xs))" using HI by simp
also have "... = ((P a ∧ todos P xs) ∧ (Q a ∧ todos Q xs))" by blast
also have "... = (todos P (a#xs) ∧ todos Q (a#xs)) " by simp
finally show "todos (λx. P x ∧ Q x) (a#xs) = (todos P (a#xs) ∧ todos Q (a#xs))" by simp
qed

(* ferrenseg *)

lemma "todos (λx. P x ∧ Q x) xs = (todos P xs ∧ todos Q xs)"
proof (induct xs)
show "todos (λx. P x ∧ Q x) [] = (todos P [] ∧ todos Q [])" by simp
next
fix n xs
assume HI: "todos (λx. P x ∧ Q x) xs = (todos P xs ∧ todos Q xs)"
have "todos (λx. P x ∧ Q x) (n # xs) = (P n ∧ Q n ∧ todos P xs ∧ todos Q xs)" using HI by simp
also have "… = P n ∧ Q n ∧ todos P xs ∧ todos Q xs" by blast
also have "⋯ = todos P (n # xs) ∧ todos Q (n # xs)" by simp
finally show "todos (λx. P x ∧ Q x) (n # xs) = todos P (n # xs) ∧ todos Q (n # xs)" by simp
qed

(* lucnovdos *)

lemma "todos (λx. P x ∧ Q x) xs = (todos P xs ∧ todos Q xs)"
proof (induct xs)
show "todos (λx. P x ∧ Q x) [] = (todos P [] ∧ todos Q [])" by simp
next
fix n xs
assume HI: "todos (λx. P x ∧ Q x) xs = (todos P xs ∧ todos Q xs)"
have "todos (λx. P x ∧ Q x) (n # xs) = ((P n ∧ Q n) ∧ (todos P xs ∧ todos Q xs))" using HI by simp
also have "... = ((P n ∧ todos P xs) ∧ (Q n ∧ todos Q xs))" by arith
also have "... = ((todos P(n#xs)) ∧ (todos Q(n#xs)))" by simp

finally show "todos (λx. P x ∧ Q x) (n#xs) = (todos P (n#xs) ∧ todos Q (n#xs))" by simp
qed

{* wilmorort pablucoto crigomgom anaprarod juacabsou rubgonmar fraortmoy *}

lemma "todos (λx. P x ∧ Q x) xs = (todos P xs ∧ todos Q xs)"
proof (induct xs)
show "todos (λx. P x ∧ Q x) [] = (todos P [] ∧ todos Q [])" by simp
next
fix a xs
assume HI: "todos (λx. P x ∧ Q x) xs = (todos P xs ∧ todos Q xs)"
have "todos (λx. P x ∧ Q x) (a # xs) = (P a ∧ Q a ∧ todos (λx. P x ∧ Q x) xs)" by simp
also have "... = (P a ∧ Q a ∧ todos P xs ∧ todos Q xs)" using HI by simp
also have "... = (P a ∧ todos P xs ∧ Q a ∧ todos Q xs)" by arith
also have "... = (todos P (a # xs) ∧ todos Q (a # xs))" by simp (* Este paso se puede obviar*)
finally show "todos (λx. P x ∧ Q x) (a # xs) = (todos P (a # xs) ∧ todos Q (a # xs))" by simp
qed

text {*
---------------------------------------------------------------------
Ejercicio 4.1. Demostrar o refutar automáticamente
todos P (x @ y) = (todos P x ∧ todos P y)
---------------------------------------------------------------------
*}
{*danrodcha ivamenjim marpoldia1 migtermor ferrenseg wilmorort paupeddeg crigomgom anaprarod serrodcal juacabsou rubgonmar pablucoto fraortmoy josgarsan lucnovdos*}
lemma "todos P (x @ y) = (todos P x ∧ todos P y)"
by (induct x) auto

{* anaprarod dancorgar *}
lemma "todos P (x @ y) = (todos P x ∧ todos P y)"
apply (induct x)
apply auto
done

text {*
---------------------------------------------------------------------
Ejercicio 4.2. Demostrar o refutar detalladamente
todos P (x @ y) = (todos P x ∧ todos P y)
---------------------------------------------------------------------
*}

(* ivamenjim ferrenseg wilmorort dancorgar fraortmoy josgarsan lucnovdos*)

lemma todos_append:
"todos P (x @ y) = (todos P x ∧ todos P y)"
proof (induct x)
show "todos P ([] @ y) = (todos P [] ∧ todos P y)" by simp
next
fix a x
assume HI: "todos P (x @ y) = (todos P x ∧ todos P y)"
have "todos P ((a#x) @ y) = (P a ∧ (todos P (x @ y)))" by simp
also have "... = (P a ∧ (todos P x ∧ todos P y))" using HI by simp
finally show "todos P ((a # x) @ y) = (todos P (a # x) ∧ todos P y)" by simp
qed

(* migtermor serrodcal *)

lemma todos_append1:
"todos P (x @ y) = (todos P x ∧ todos P y)" (is "?P x")
proof (induct x)
show "?P []" by simp
next
fix a x
assume HI: "?P x"
have "todos P ((a#x) @ y) = (P a ∧ (todos P (x @ y)))" by simp
also have "… = (P a ∧ (todos P x ∧ todos P y))" using HI by simp
finally show "?P (a#x)" by simp
qed

(* marpoldia1 *)
lemma todos_append:
"todos P (x @ y) = (todos P x ∧ todos P y)"
proof (induct x)
show "todos P ([] @ y) = (todos P [] ∧ todos P y)" by simp
next
fix a x
assume HI: "todos P (x @ y) = (todos P x ∧ todos P y)"
have "todos P ((a # x) @ y) = todos P (a#(x@y))" by simp
also have "... = (P a ∧ todos P (x @ y))" by simp
also have "... = (P a ∧ (todos P x ∧ todos P y))" using HI by simp
finally show "todos P ((a # x) @ y) = (todos P (a # x) ∧ todos P y)" by simp
qed

(* paupeddeg *)
lemma todos_append3:
"todos P (x @ y) = (todos P x ∧ todos P y)"
proof (induct x)
show "todos P ([] @ y) = (todos P [] ∧ todos P y)" by simp
next
fix a x
assume HI: "todos P (x @ y) = (todos P x ∧ todos P y)"
have "todos P ((a # x) @ y) = todos P (a # (x @ y))" by simp
also have "... = (todos P [a] ∧ todos P (x @ y)) " by simp
also have "... = (todos P [a] ∧ (todos P x ∧ todos P y))" using HI by simp
finally show "todos P ((a # x) @ y) = (todos P (a # x) ∧ todos P y)" by simp
qed

(* crigomgom juacabsou anaprarod*)

lemma todos_append4:
"todos P (x @ y) = (todos P x ∧ todos P y)"
proof (induct x)
show "todos P ([] @ y) = (todos P [] ∧ todos P y)" by simp
next
fix x xs
assume HI:"todos P (xs @ y) = (todos P xs ∧ todos P y)"
have "todos P ((x # xs) @ y) = (P x ∧ todos P (xs @ y ))" by simp
also have "... = (P x ∧ (todos P xs ∧ todos P y)) " using HI by simp
also have "... = ((P x ∧ todos P xs) ∧ todos P y)" by simp
finally show "todos P ((x # xs) @ y) = (todos P (x # xs) ∧ todos P y)" by simp
qed

(* pablucoto *)
lemma todos_append5:
"todos P (x @ y) = (todos P x ∧ todos P y)"
proof (induct x )
show "todos P ([] @ y) = (todos P [] ∧ todos P y)" by simp
next
fix a x
assume HI:"todos P (x @ y) = (todos P x ∧ todos P y) "
have "todos P ((a # x) @ y) = ( P a ∧ todos P (x @ y))" by simp
also have " ... =( P a ∧ (todos P x ∧ todos P y))" using HI by simp
also have "... = (todos P (a#x) ∧ todos P y)" by simp
finally show "todos P ((a # x) @ y) = (todos P (a # x) ∧ todos P y)" by auto
qed

{*danrodcha*}
lemma todos_append6:
"todos P (x @ y) = (todos P x ∧ todos P y)" (is "?Q x")
proof (induct x)
show "?Q []" by simp
next
fix x a assume HI: "?Q x"
have "todos P ((a # x) @ y) = todos P (a # x @ y)" by simp
also have "… = ((P a) ∧ todos P (x @ y))" by simp
also have "… = ((P a) ∧ (todos P x ∧ todos P y))" using HI by simp
also have "… = (todos P (a # x) ∧ todos P y)" by simp
finally show "?Q (a # x)" by simp
qed

text {*
---------------------------------------------------------------------
Ejercicio 5.1. Demostrar o refutar automáticamente
todos P (rev xs) = todos P xs
---------------------------------------------------------------------
*}

lemma "todos P (rev xs) = todos P xs"
oops

(* migtermor ivamenjim marpoldia1 serrodcal anaprarod paupeddeg dancorgar *)
lemma "todos P (rev xs) = todos P xs"
apply (induct xs)
apply simp
apply (simp add: todos_append)
apply auto
done

(* ferrenseg crigomgom rubgonmar fraortmoy josgarsan danrodcha lucnovdos*)

lemma "todos P (rev xs) = todos P xs"
by (induct xs) (auto simp add: todos_append)

(* wilmorort *)
lemma "todos P (rev xs) = todos P xs"
by (induct xs,simp,simp add: todos_append,auto)

(* juacabsou *)
lemma "todos P (rev xs) = todos P xs"
apply (induct xs,simp,simp add: todos_append,auto) done

(* pablucoto * )
lemma "todos P (rev xs) = todos P xs"
by (induct xs) (auto, simp_all add: todos_append)
lemma "todos P (rev xs) = todos P xs"
by (induct xs) ( simp_all add: todos_append, auto)

text {*
---------------------------------------------------------------------
Ejercicio 5.2. Demostrar o refutar detalladamente
todos P (rev xs) = todos P xs
---------------------------------------------------------------------
*}

lemma "todos P (rev xs) = todos P xs"
oops

(* migtermor *)

lemma auxiliar:
"rev (a#xs) = rev xs @ [a]"
by auto

lemma "todos P (rev xs) = todos P xs" (is "?Q xs")
proof (induct xs)
show "?Q []" by (simp only: rev.simps(1))
next
fix a xs
assume HI: "?Q xs"
have "todos P (rev (a#xs)) = (todos P (rev xs @ [a]))" by (simp add: auxiliar)
have "… = ((todos P (rev xs)) ∧ (todos P [a]))" by (simp add: todos_append)
have "… = (todos P (rev xs) ∧ P a)" by simp
also have Aux: "… = (todos P xs ∧ P a)" using HI by simp
have Aux1: "… = (P a ∧ todos P xs)" by arith
have "(todos P (rev xs) ∧ P a) = (P a ∧ todos P xs)" using Aux Aux1 by simp
finally show "todos P (rev (a#xs)) = todos P (a#xs)" by (simp add: todos_append)
qed

(* marpoldia1 ferrenseg crigomgom serrodcal juacabsou rubgonmar josgarsan pablucoto*)
lemma "todos P (rev xs) = todos P xs"
proof (induct xs)
show "todos P (rev []) = todos P []" by simp
next
fix a xs
assume HI: "todos P (rev xs) = todos P xs"
have "todos P (rev (a # xs)) = (todos P ((rev xs)@[a]))" by simp
also have "... = (todos P (rev xs) ∧ todos P [a])" by (simp add:todos_append)
also have "... = (todos P xs ∧ P a)" using HI by simp
also have "... = (P a ∧ todos P xs)" by arith
also have "... = (todos P (a#xs))" by simp
finally show "todos P (rev (a # xs)) = (todos P (a#xs))" by simp
qed

(* ivamenjim *)

lemma "todos P (rev xs) = todos P xs" (is "?P xs")
proof (induct xs)
show "?P []" by simp
next
fix a xs
assume HI: "?P xs"
have "todos P (rev (a # xs)) = (todos P ((rev xs)@[a]))" by simp
also have "... = ((todos P (rev xs)) ∧ todos P [a])" by (simp add: todos_append)
also have "... = (todos P xs ∧ todos P [a])" using HI by simp
also have "... = (todos P [a] ∧ todos P xs)" by arith
finally show "todos P (rev (a # xs)) = todos P (a # xs)" by simp
qed

(* fraortmoy *)

lemma "todos P (rev xs) = todos P xs"
proof (induct xs)
show "todos P (rev []) = todos P []" by simp
next
fix a xs
assume H1: "todos P (rev xs) = todos P xs"
have " todos P (rev (a # xs)) = todos P (rev xs @ [a])" by simp
also have "… = (todos P (rev xs) ∧ todos P [a])" by (simp add:todos_append)
also have "… = (todos P xs ∧ todos P [a])" using H1 by simp
also have "… = (todos P [a] ∧ todos P xs)" by arith
finally show "todos P (rev (a # xs)) = todos P (a#xs)" by simp
qed

(* paupeddeg dancorgar *)
lemma "todos P (rev xs) = todos P xs"
proof (induct xs)
show " todos P (rev []) = todos P []" by simp
next
fix a xs
assume HI: "todos P (rev xs) = todos P xs "
have "todos P (rev (a # xs)) = todos P (rev(xs) @ [a])" by simp
also have "... = (todos P (rev xs) ∧ todos P [a]) " by (simp add: todos_append)
also have "... = (todos P xs ∧ todos P [a])" using HI by simp
also have "... = (todos P [a] ∧ todos P xs)" by arith
also have "... = todos P ([a] @ xs)" by (simp add: todos_append)
finally show "todos P (rev (a # xs)) = todos P (a # xs)" by simp
qed

(* wilmorort *)
lemma "todos P (rev xs) = todos P xs" (is "?P xs")
proof (induct xs)
show "?P []" by simp
next
fix a xs
assume HI: "?P xs"
have "todos P (rev (a#xs)) = todos P ((rev xs) @ [a])" by simp
also have "... = (todos P(rev xs) ∧ todos P [a])" by (simp add: todos_append)
also have "... = (todos P xs ∧ todos P [a])" using HI by simp
also have "... = (todos P [a] ∧ todos P xs)" by (simp add: HOL.conj_commute)
also have "... = todos P([a]@(xs))" by (simp)
finally show "todos P (rev (a#xs))= todos P (a # xs)" by simp
qed

(* anaprarod *)
(* es igual que las anteriores pero con el final también con patrones *)

lemma "todos P (rev xs) = todos P xs" (is "?P xs")
proof (induct xs)
show "?P []" by auto
next
fix a xs
assume HI: "?P xs"
have "todos P (rev (a#xs)) = todos P (rev xs @ [a])" by simp
also have "… = (todos P (rev xs) ∧ todos P [a])" by (simp add: todos_append)
also have "… = (todos P xs ∧ todos P [a])" using HI by simp
also have "… = (todos P [a] ∧ todos P xs)" by arith
also have "… = todos P (a#xs)" by simp
finally show "?P (a # xs)" by simp
qed

{*danrodcha*}
lemma "todos P (rev xs) = todos P xs" (is "?Q xs")
proof (induct xs)
show "?Q []" by simp
next
fix a xs assume HI: "?Q xs"
have "todos P (rev (a#xs)) = todos P ((rev xs) @ [a])" by simp
also have "… = (todos P (rev xs) ∧ todos P [a])" by (simp add:todos_append)
also have "… = (todos P xs ∧ todos P [a])" using HI by simp
also have "… = (todos P [a] ∧ todos P xs)" by (simp add: HOL.conj_commute)
also have "… = todos P (a # xs)" by simp
finally show "?Q (a # xs)" by simp
qed

text {*
---------------------------------------------------------------------
Ejercicio 6. Demostrar o refutar:
algunos (λx. P x ∧ Q x) xs = (algunos P xs ∧ algunos Q xs)
---------------------------------------------------------------------
*}

lemma "algunos (λx. P x ∧ Q x) xs = (algunos P xs ∧ algunos Q xs)"
oops

(* migtermor ivamenjim ferrenseg paupeddeg crigomgom serrodcal wilmorort juacabsou rubgonmar anaprarod marpoldia1 fraortmoy josgarsan danrodcha pablucoto*)
lemma "algunos (λx. P x ∧ Q x) xs = (algunos P xs ∧ algunos Q xs)"
quickcheck
oops
(* Quickcheck encuentra el siguiente contraejemplo: P={a1}, Q={a2}, xs={a1,a2}. En este ejemplo:
· "algunos (λx. P x ∧ Q x) xs = False"
· "(algunos P xs ∧ algunos Q xs) = True" *)

(* dancorgar *)
lemma "algunos (λx. P x ∧ Q x) xs = (algunos P xs ∧ algunos Q xs)"
proof -
assume H1: "xs = [a, b]"
assume H2: "P a = True"
assume H3: "Q a = False"
assume H4: "P b = False"
assume H5: "Q b = True"
have F1: "(algunos P xs ∧ algunos Q xs) = True" using H1 H2 H3 H4 H5 by simp
have F2: "algunos (λx. P x ∧ Q x) xs = False" using H1 H2 H3 H4 H5 by simp
have "algunos (λx. P x ∧ Q x) xs ≠ (algunos P xs ∧ algunos Q xs)" using F1 F2 by simp
oops

text {*
---------------------------------------------------------------------
Ejercicio 7.1. Demostrar o refutar automáticamente
algunos P (map f xs) = algunos (P ∘ f) xs
---------------------------------------------------------------------
*}

(* ivamenjim migtermor marpoldia1 crigomgom rubgonmar wilmorort anaprarod
fraortmoy juacabsou paupeddeg josgarsan danrodcha pablucoto *)

lemma "algunos P (map f xs) = algunos (P o f) xs"
by (induct xs) auto

(* ferrenseg *)

lemma "algunos P (map f xs) = algunos (P o f) xs"
by (induct xs) simp_all

(* anaprarod dancorgar *)
lemma "algunos P (map f xs) = algunos (P o f) xs"
apply (induct xs)
apply auto
done

text {*
---------------------------------------------------------------------
Ejercicio 7.2. Demostrar o refutar detalladamente
algunos P (map f xs) = algunos (P ∘ f) xs
---------------------------------------------------------------------
*}

lemma "algunos P (map f xs) = algunos (P o f) xs"
oops

(* migtermor *)

lemma AUX: "algunos (λa. P (f a)) xs = algunos (P o f) xs"
by (induct xs) auto

lemma "algunos P (map f xs) = algunos (P o f) xs" (is "?Q xs")
proof (induct xs)
show "?Q []" by simp
next
fix x xs
assume HI: "?Q xs"
have "algunos P (map f (x#xs)) = (algunos P ((f x)#(map f xs)))" by simp
also have "… = ((P (f x)) ∨ (algunos P (map f xs)))" by (simp only: algunos.simps(2))
also have "… = ((P (f x)) ∨ (algunos (P o f) xs))"
proof (cases)
assume C1: "(P (f x))"
have Aux: "((P (f x)) ∨ (algunos P (map f xs))) = True" using C1 by simp
have Aux1: "… = ((P (f x)) ∨ (algunos (P o f) xs))" using C1 by simp
then show "((P (f x)) ∨ (algunos P (map f xs))) = ((P (f x)) ∨ (algunos (P o f) xs))"
using Aux Aux1 by simp
next
assume C2: "¬(P (f x))"
have Aux2: "((P (f x)) ∨ (algunos P (map f xs))) = (algunos P (map f xs))" using C2 by simp
have Aux3: "… = (algunos (P o f) xs)" using HI by (simp add: AUX)
also have "… = ((P (f x)) ∨ (algunos (P o f) xs))" using C2 by simp
then show "((P (f x)) ∨ (algunos P (map f xs))) = ((P (f x)) ∨ (algunos (P o f) xs))"
using Aux2 Aux3 by simp
qed
also have "… = (((P o f) x) ∨ (algunos (P o f) xs))" by simp
also have "… = (algunos (P o f) (x#xs))" by simp
finally show "algunos P (map f (x#xs)) = (algunos (P o f) (x#xs))" by simp
qed

(* ferrenseg *)

lemma "algunos P (map f xs) = algunos (P ∘ f) xs"
proof (induct xs)
show "algunos P (map f []) = algunos (P ∘ f) []" by simp
next
fix x xs
assume HI: "algunos P (map f xs) = algunos (P ∘ f) xs"
show "algunos P (map f (x # xs)) = algunos (P ∘ f) (x # xs)"
proof -
have "algunos P (map f (x # xs)) = algunos P ((f x) # (map f xs))" by simp
also have "… = ((P (f x)) ∨ (algunos P (map f xs)))" by simp
also have "… = (((P ∘ f) x) ∨ (algunos (P ∘ f) xs))" using HI by simp
also have "… = algunos (P ∘ f) (x # xs)" by simp
finally show ?thesis
qed
qed

(* ivamenjim *)

lemma "algunos P (map f xs) = algunos (P o f) xs"
proof (induct xs)
show "algunos P (map f []) = algunos (P ∘ f) []" by simp
next
fix a xs
assume HI: "algunos P (map f xs) = algunos (P ∘ f) xs"
have "algunos P (map f (a # xs)) = algunos P ((f a)#map f xs)" by simp
also have "... = (algunos P (map f [a]) ∨ algunos P (map f xs))" by simp
also have "... = (algunos P (map f [a]) ∨ algunos (P ∘ f) xs)" using HI by simp
finally show "algunos P (map f (a # xs)) = algunos (P ∘ f) (a # xs)" by simp
qed

(* crigomgom rubgonmar anaprarod marpoldia1 juacabsou danrodcha paupeddeg *)
lemma "algunos P (map f xs) = algunos (P o f) xs"
proof (induct xs)
show "algunos P (map f []) = algunos (P ∘ f) []" by simp
next
fix x xs
assume HI: "algunos P (map f xs) = algunos (P ∘ f) xs"
have "algunos P (map f (x # xs)) = algunos P ((f x) # (map f xs))" by simp
also have "... = (P (f x) ∨ algunos P (map f xs))" by simp
also have "... = (P (f x) ∨ algunos (P ∘ f) xs)" using HI by simp
also have "... = ((P ∘ f) x ∨ algunos (P ∘ f) xs)" by simp
finally show "algunos P (map f (x # xs)) = algunos (P ∘ f) (x # xs)" by simp
qed

(* wilmorort pablucoto*)
lemma "algunos P (map f xs) = algunos (P o f) xs" (is "?P xs" )
proof (induct xs)
show "?P []" by simp
next
fix a xs
assume HI: "algunos P (map f xs) = algunos (P ∘ f) xs"
have "algunos P (map f (a # xs)) = (P (f a) ∨ algunos P (map f xs))" by simp
also have "... = (P (f a) ∨ algunos (P ∘ f) xs )" using HI by simp
also have "... = ((P ∘ f) a ∨ algunos (P ∘ f)xs)" by simp
finally show " algunos P (map f (a # xs)) = algunos (P ∘ f) (a # xs)" by simp
qed

(* fraortmoy *)

lemma "algunos P (map f xs) = algunos (P o f) xs"
proof (induct xs)
show "algunos P (map f []) = algunos (P ∘ f) []" by simp
next
fix a xs
assume H1: "algunos P (map f xs) = algunos (P ∘ f) xs"
have "algunos P (map f (a # xs)) = (algunos P (map f [a]) ∨ algunos P (map f xs))" by simp
also have "… = (algunos (P ∘ f) [a] ∨ algunos (P ∘ f) xs )" using H1 by simp
also have "… = algunos (P ∘ f) (a#xs)" by simp
finally show "algunos P (map f (a # xs)) = algunos (P ∘ f) (a#xs)" by simp
qed

{*danrodcha*}
lemma "algunos P (map f xs) = algunos (P o f) xs" (is "?Q xs")
proof (induct xs)
show "?Q []" by simp
next
fix a xs assume HI: "?Q xs"
have " algunos P (map f (a # xs)) = algunos P ((f a)#(map f xs))" by simp
also have "… = (P (f a) ∨ algunos P (map f xs))" by simp
also have "… = (P (f a) ∨ algunos (P ∘ f) xs)" using HI by blast
also have "… = ((P ∘ f) a ∨ algunos (P ∘ f) xs)" by simp
also have "… = algunos (P ∘ f) (a # xs)" by simp
finally show "?Q (a # xs)" by blast
qed

(* dancorgar *)
lemma "algunos P (map f xs) = algunos (P o f) xs"
proof (induct xs)
show "algunos P (map f []) = algunos (P o f) []" by simp
next
fix x xs
assume HI: "algunos P (map f xs) = algunos (P o f) xs"
have "algunos P (map f (x#xs)) = ((P (f x)) ∨ (algunos P (map f xs)))" by simp
also have "… = ((P (f x)) ∨ (algunos (P o f) xs))" using HI by simp
finally show "algunos P (map f (x#xs)) = algunos (P o f) (x#xs)" by simp
qed

text {*
---------------------------------------------------------------------
Ejercicio 8.1. Demostrar o refutar automáticamente
algunos P (xs @ ys) = (algunos P xs ∨ algunos P ys)
---------------------------------------------------------------------
*}

(* ivamenjim marpoldia1 paupeddeg crigomgom rubgonmar wilmorort fraortmoy danrodcha pablucoto dancorgar *)

lemma "algunos P (xs @ ys) = (algunos P xs ∨ algunos P ys)"
by (induct xs) auto

(* ferrenseg *)
lemma "algunos P (xs @ ys) = (algunos P xs ∨ algunos P ys)"
by (induct xs) simp_all

text {*
---------------------------------------------------------------------
Ejercicio 8.2. Demostrar o refutar detalladamente
algunos P (xs @ ys) = (algunos P xs ∨ algunos P ys)
---------------------------------------------------------------------
*}

(* ivamenjim ferrenseg marpoldia1 wilmorort dancorgar *)

lemma algunos_append:
"algunos P (xs @ ys) = (algunos P xs ∨ algunos P ys)"
proof (induct xs)
show "algunos P ([] @ ys) = (algunos P [] ∨ algunos P ys)" by simp
next
fix a xs
assume HI: "algunos P (xs @ ys) = (algunos P xs ∨ algunos P ys)"
have "algunos P ((a # xs) @ ys) = (P a ∨ (algunos P (xs @ ys)))" by simp
also have "... = (P a ∨ (algunos P xs ∨ algunos P ys))" using HI by simp
finally show "algunos P ((a # xs) @ ys) = (algunos P (a # xs) ∨ algunos P ys)" by simp
qed

(* migtermor crigomgom *)

lemma algunos_append2:
"algunos P (xs @ ys) = (algunos P xs ∨ algunos P ys)" (is "?Q xs")
proof (induct xs)
show "?Q []" by simp
next
fix x xs
assume HI: "?Q xs"
have "algunos P ((x#xs) @ ys) = algunos P (x#(xs @ ys))" by simp
also have "… = ((P x) ∨ (algunos P (xs @ ys)))" by simp
also have "… = ((P x) ∨ (algunos P xs) ∨ (algunos P ys))" using HI by simp
also have "… = ((algunos P (x#xs)) ∨ (algunos P ys))" by simp
finally show "algunos P ((x#xs) @ ys) = (algunos P (x#xs) ∨ algunos P ys)" by simp
qed

(* paupeddeg*)
lemma algunos_append3:
"algunos P (xs @ ys) = (algunos P xs ∨ algunos P ys)"
proof (induct xs)
show "algunos P ([] @ ys) = (algunos P [] ∨ algunos P ys)" by simp
next
fix a xs
assume HI: "algunos P (xs @ ys) = (algunos P xs ∨ algunos P ys)"
have "algunos P ((a # xs) @ ys) = (algunos P [a] ∨ ( algunos P (xs @ ys)))" by simp
also have "... = (algunos P [a] ∨ (algunos P xs ∨ algunos P ys))" using HI by simp
finally show "algunos P ((a # xs) @ ys) = (algunos P (a # xs) ∨ algunos P ys)" by simp
qed

(* fraortmoy *)
lemma algunos_append4:
"algunos P (xs @ ys) = (algunos P xs ∨ algunos P ys)"
proof (induct xs)
show " algunos P ([] @ ys) = (algunos P [] ∨ algunos P ys)" by simp
next
fix a xs
assume H1: "algunos P (xs @ ys) = (algunos P xs ∨ algunos P ys)"
have "algunos P ((a # xs) @ ys) = (algunos P [a] ∨ algunos P (xs @ ys))" by simp
also have "… = (algunos P [a] ∨ algunos P xs ∨ algunos P ys)" using H1 by simp
also have "… = ((algunos P [a] ∨ algunos P xs) ∨ algunos P ys)" by simp
also have "… = (algunos P (a#xs) ∨ algunos P ys)" by simp
finally show "algunos P ((a # xs) @ ys) = (algunos P (a#xs) ∨ algunos P ys)" by simp
qed

{* danrodcha pablucoto*}
lemma algunos_append5:
"algunos P (xs @ ys) = (algunos P xs ∨ algunos P ys)" (is "?Q xs")
proof (induct xs)
show "?Q []" by simp
next
fix a xs assume HI: "?Q xs"
have "algunos P ((a#xs) @ ys) = algunos P (a#(xs @ ys))" by simp
also have "… = (algunos P [a] ∨ algunos P (xs @ ys))" by simp
also have "… = (algunos P [a] ∨ algunos P xs ∨ algunos P ys)" using HI by simp
also have "… = (algunos P (a # xs) ∨ algunos P ys)" by simp
finally show "?Q (a # xs)" by simp
qed

text {*
---------------------------------------------------------------------
Ejercicio 9.1. Demostrar o refutar automáticamente
algunos P (rev xs) = algunos P xs
---------------------------------------------------------------------
*}

(* ivamenjim migtermor marpoldia1 rubgonmar paupeddeg dancorgar *)

lemma "algunos P (rev xs) = algunos P xs"
apply (induct xs)
apply simp
apply (simp add: algunos_append)
apply auto
done

(* ferrenseg crigomgom danrodcha pablucoto*)
lemma "algunos P (rev xs) = algunos P xs"
by (induct xs) (auto simp add: algunos_append)

(* wilmorort *)
lemma "algunos P (rev xs) = algunos P xs"
by (induct xs,simp,simp add: algunos_append,auto)

text {*
---------------------------------------------------------------------
Ejercicio 9.2. Demostrar o refutar detalladamente
algunos P (rev xs) = algunos P xs
---------------------------------------------------------------------
*}

lemma "algunos P (rev xs) = algunos P xs"
oops

(* migtermor *)
lemma auxiliar1:
"rev (a#xs) = rev xs @ [a]"
by auto

lemma "algunos P (rev xs) = algunos P xs" (is "?Q xs")
proof (induct xs)
show "?Q []" by simp
next
fix x xs
assume HI: "?Q xs"
have "algunos P (rev (x#xs)) = (algunos P (rev xs @ [x]))" using auxiliar1 by simp
also have "… = ((algunos P (rev xs)) ∨ (algunos P [x]))" by (simp add: algunos_append)
also have "… = ((P x) ∨ (algunos P (rev xs)))" by simp arith
also have "… = ((P x) ∨ (algunos P xs))" using HI by simp
finally show "algunos P (rev (x#xs)) = (algunos P (x#xs))" by simp
qed

(* ferrenseg *)

lemma "algunos P (rev xs) = algunos P xs"
proof (induct xs)
show "algunos P (rev []) = algunos P []" by simp
next
fix x xs
assume HI: "algunos P (rev xs) = algunos P xs"
show "algunos P (rev (x # xs)) = algunos P (x # xs)"
proof -
have "algunos P (rev (x # xs)) = algunos P ((rev xs) @ [x])" by simp
also have "… = (algunos P (rev xs) ∨ algunos P [x])"
by (simp add: algunos_append)
also have "… = (algunos P xs ∨ algunos P [x])" using HI by simp
also have "… = (algunos P xs ∨ P x)" by simp
also have "… = (P x ∨ algunos P xs)" by auto
also have "… = algunos P (x # xs)" by simp
finally show ?thesis
qed
qed

(* ivamenjim marpoldia1*)

lemma "algunos P (rev xs) = algunos P xs" (is "?P xs")
proof (induct xs)
show "?P []" by simp
next
fix a xs
assume HI: "?P xs"
have "algunos P (rev (a # xs)) = (algunos P ((rev xs)@[a]))" by simp
also have "... = ((algunos P (rev xs)) ∨ algunos P [a])" by (simp add: algunos_append)
also have "... = (algunos P xs ∨ algunos P [a])" using HI by simp
also have "... = (algunos P [a] ∨ algunos P xs)" by arith
finally show "algunos P (rev (a # xs)) = algunos P (a # xs)" by simp
qed

(* paupeddeg dancorgar *)
lemma "algunos P (rev xs) = algunos P xs"
proof (induct xs)
show "algunos P (rev []) = algunos P []" by simp
next
fix a xs
assume HI:"algunos P (rev xs) = algunos P xs"
have "algunos P (rev (a # xs)) = algunos P ((rev xs) @ [a])" by simp
also have "... = ((algunos P (rev xs)) ∨ (algunos P [a]))" by (simp add: algunos_append)
also have "... = ((algunos P xs) ∨ (algunos P [a]))" using HI by simp
also have "... = ((algunos P [a]) ∨ (algunos P xs))" by arith
finally show "algunos P (rev (a # xs)) = algunos P (a # xs)" by simp
qed

(*crigomgom pablucoto*)
lemma "algunos P (rev xs) = algunos P xs"
proof (induct xs)
show "algunos P (rev []) = algunos P []" by simp
next
fix x xs
assume HI: " algunos P (rev xs) = algunos P xs"
have "algunos P (rev (x # xs)) = algunos P ((rev xs) @ [x])" by simp
also have "... = (algunos P (rev xs) ∨ algunos P [x])" by (simp add: algunos_append)
also have "... = (algunos P xs ∨ algunos P [x])" using HI by simp
also have "... = (algunos P xs ∨ P x)" by simp
also have "... = (P x ∨ algunos P xs)" by arith
also have "... = algunos P (x#xs)" by simp
finally show "algunos P (rev (x # xs)) = algunos P (x # xs)" by simp
qed

(*wilmorort*)
lemma "algunos P (rev xs) = algunos P xs" (is "?P xs")
proof (induct xs)
show "?P []" by simp
next
fix a xs
assume HI: "?P xs"
have "algunos P (rev (a#xs)) = algunos P ((rev xs) @ [a])" by simp
also have "... = (algunos P(rev xs) ∨ algunos P [a])" by (simp add: algunos_append)
also have "... = (algunos P xs ∨ algunos P [a])" using HI by simp
also have "... = (algunos P [a] ∨ algunos P xs)" by (simp add: HOL.disj_commute)
also have "... = algunos P([a]@(xs))" by (simp)
finally show "algunos P (rev (a#xs))= algunos P (a # xs)" by simp
qed

{*danrodcha *}
lemma "algunos P (rev xs) = algunos P xs" (is "?Q xs")
proof (induct xs)
show "?Q []" by simp
next
fix a xs assume HI: "?Q xs"
have "algunos P (rev (a#xs)) = algunos P ((rev xs) @ [a])" by simp
also have "… = (algunos P (rev xs) ∨ algunos P [a])" by (simp add:algunos_append)
also have "… = (algunos P xs ∨ algunos P [a])" using HI by simp
also have "… = (algunos P [a] ∨ algunos P xs)" by (simp add: HOL.disj_commute)
also have "… = algunos P (a # xs)" by simp
finally show "?Q (a # xs)" by simp
qed

text {*
---------------------------------------------------------------------
Ejercicio 10. Encontrar un término no trivial Z tal que sea cierta la
siguiente ecuación:
algunos (λx. P x ∨ Q x) xs = Z
y demostrar la equivalencia de forma automática y detallada.
---------------------------------------------------------------------
*}

(* migtermor *)

lemma "algunos (λx. P x ∨ Q x) xs = ((algunos (λx. P x) xs) ∨ (algunos (λx. Q x) xs))"
by (induct xs) auto

lemma "algunos (λx. P x ∨ Q x) xs = ((algunos (λx. P x) xs) ∨ (algunos (λx. Q x) xs))" (is "?R xs")
proof (induct xs)
show "?R []" by simp
next
fix x xs
assume HI: "?R xs"
have "algunos (λx. P x ∨ Q x) (x#xs) = (((λx. P x ∨ Q x) x) ∨ (algunos (λx. P x ∨ Q x) xs))"
by simp
also have "… = ((P x ∨ Q x) ∨ (algunos (λx. P x ∨ Q x) xs))" by simp
also have H1: "… = ((((λx. P x) x) ∨ (algunos (λx. P x) xs)) ∨ (((λx. Q x) x) ∨ (algunos (λx. Q x) xs)))"
using HI by simp arith
have H2: "… = ((algunos (λx. P x) (x#xs)) ∨ (algunos (λx. Q x) (x#xs)))"
by simp
have C: "(algunos (λx. P x ∨ Q x) (x#xs)) =
((algunos (λx. P x) (x#xs)) ∨ (algunos (λx. Q x) (x#xs)))"
using H1 H2 by simp
finally show "(algunos (λx. P x ∨ Q x) (x#xs)) =
((algunos (λx. P x) (x#xs)) ∨ (algunos (λx. Q x) (x#xs)))"
using C by simp
qed

(* ferrenseg *)

lemma "algunos (λx. P x ∨ Q x) xs = (algunos P xs ∨ algunos Q xs)"
proof (induct xs)
show "algunos (λx. P x ∨ Q x) [] = (algunos P [] ∨ algunos Q [])" by simp
next
fix x xs
assume HI: "algunos (λx. (P x ∨ Q x)) xs = (algunos P xs ∨ algunos Q xs)"
show "algunos (λx. P x ∨ Q x) (x # xs) = (algunos P (x # xs) ∨ algunos Q (x # xs))"
proof -
have "algunos (λx. P x ∨ Q x) (x # xs) =
((P x) ∨ (Q x) ∨ algunos (λx. P x ∨ Q x) xs)" by simp
also have "… = ((P x) ∨ (Q x) ∨ algunos P xs ∨ algunos Q xs)"
using HI by simp
also have "… = (((P x) ∨ algunos P xs) ∨ ((Q x) ∨ algunos Q xs))" by auto
also have "… = (algunos P (x # xs) ∨ algunos Q (x # xs))" by simp
finally show ?thesis
qed
qed

(* ivamenjim marpoldia1 paupeddeg *)

lemma "algunos (λx. P x ∨ Q x) xs = (algunos P xs ∨ algunos Q xs)"
by (induct xs) auto

lemma "algunos (λx. P x ∨ Q x) xs = (algunos P xs ∨ algunos Q xs)"
proof (induct xs)
show "algunos (λx. P x ∨ Q x) [] = (algunos P [] ∨ algunos Q [])" by simp
next
fix a xs
assume HI: "algunos (λx. P x ∨ Q x) xs = (algunos P xs ∨ algunos Q xs)"
have "algunos (λx. P x ∨ Q x) (a # xs) = (P a ∨ Q a ∨ algunos (λx. P x ∨ Q x) xs)" by simp
also have "... = (P a ∨ Q a ∨ algunos P xs ∨ algunos Q xs)" using HI by simp
finally show "algunos (λx. P x ∨ Q x) (a # xs) = (algunos P (a # xs) ∨ algunos Q (a # xs))" by auto
qed

(* wilmorort danrodcha*)
lemma "algunos (λx. P x ∨ Q x) xs = (¬todos(λx. ¬P x)xs ∨ ¬todos(λx. ¬Q x)xs)"
by (induct xs) auto

(* wilmorort *)
lemma "algunos (λx. P x ∨ Q x) xs = (¬todos(λx. ¬P x)xs ∨ ¬todos(λx. ¬Q x)xs)"
(is "?P xs")
proof (induct xs)
show "?P []" by simp
next
fix a xs
assume HI: "?P xs"
have "algunos (λx. P x ∨ Q x) (a # xs) = (P a ∨ Q a ∨ algunos (λx. P x ∨ Q x) xs)" by simp
also have "... = (P a ∨ Q a ∨ (¬ todos (λx. ¬ P x) xs ∨ ¬ todos (λx. ¬ Q x) xs))" using HI by simp
also have "... = (P a ∨ ¬ todos (λx. ¬ P x) xs ∨ Q a ∨ ¬ todos (λx. ¬ Q x) xs )" by arith
finally show "algunos (λx. P x ∨ Q x) (a # xs) = (~ todos (λx. ¬ P x) (a # xs) ∨ ~ todos (λx. ¬ Q x) (a # xs))" by simp
qed

{* danrodcha *}
lemma "algunos (λx. P x ∨ Q x) xs = (algunos P xs ∨ algunos Q xs)" (is "?R xs")
proof (induct xs)
show "?R []" by simp
next
fix a xs assume HI: "?R xs"
have 1:" (Q a ∨ algunos P xs) = (algunos P xs ∨ Q a)" by (simp add: HOL.disj_commute)
have "algunos (λx. P x ∨ Q x) (a # xs) =
(algunos P [a] ∨ algunos Q [a] ∨ algunos (λx. P x ∨ Q x) xs)" by simp
also have "… = (P a ∨ (Q a ∨ algunos P xs) ∨ algunos Q xs)" using HI by simp
also have "… = (P a ∨ algunos P xs ∨ Q a ∨ algunos Q xs)" using 1 by simp
also have "… = (algunos P (a # xs) ∨ algunos Q (a # xs))" by simp
finally show "?R (a # xs)" by simp
qed

(* dancorgar *)
lemma "algunos (λx. P x ∧ Q x) xs ⟹ (algunos P xs ∧ algunos Q xs)"
apply (induct xs)
apply auto
done

(* dancorgar *)
lemma "algunos (λx. P x ∧ Q x) xs ⟹ (algunos P xs ∧ algunos Q xs)"
proof (induct xs)
show "algunos (λx. P x ∧ Q x) [] ⟹ (algunos P [] ∧ algunos Q [])" by simp
next
fix xf xs
assume HI: "algunos (λx. P x ∧ Q x) xs ⟹ (algunos P xs ∧ algunos Q xs)"
have F1: "algunos (λx. P x ∧ Q x) (xf#xs) ⟹ ((P xf ∧ Q xf) ∨ algunos (λx. P x ∧ Q x) xs)" by simp
also have F2: "… ⟹ (P xf ∧ Q xf ∨ (algunos P xs ∧ algunos Q xs))" using HI by simp
have F3: "((P xf ∧ Q xf) ∨ (algunos P xs ∧ algunos Q xs)) ⟹
((P xf ∨ algunos P xs) ∧ (Q xf ∨ algunos Q xs))" by blast
have F4: "((P xf ∨ algunos P xs) ∧ (Q xf ∨ algunos Q xs)) ⟹
(algunos P (xf#xs) ∧ algunos Q (xf#xs))" by simp
show "algunos (λx. P x ∧ Q x) (xf#xs) ⟹ (algunos P (xf#xs) ∧ algunos Q (xf#xs))"
using F1 F2 F3 F4 by blast
qed

(*pablucoto*)
-- "Automatica"
lemma "algunos (λx. P x ∨ Q x) xs = (algunos P xs ∨ algunos Q xs)"
by (induct xs) auto
-- "Detallada"
lemma "algunos (λx. P x ∨ Q x) xs = (algunos P xs ∨ algunos Q xs) "
proof (induct xs)
show "algunos (λx. P x ∨ Q x) [] = (algunos P [] ∨ algunos Q []) " by simp
next
fix a xs
assume HI: " algunos (λx. P x ∨ Q x) xs = (algunos P xs ∨ algunos Q xs)"
have " algunos (λx. P x ∨ Q x) (a # xs) = ( P a ∨ Q a ∨ algunos (λx. P x ∨ Q x) xs)" by simp
also have "... = (P a ∨ Q a ∨ (algunos P xs ∨ algunos Q xs))" using HI by simp
also have "… = (((P a) ∨ algunos P xs) ∨ ((Q a) ∨ algunos Q xs))" by arith
also have "… = (algunos P (a#xs) ∨ algunos Q (a#xs))" by simp
finally show "algunos (λx. P x ∨ Q x) (a # xs) = (algunos P (a # xs) ∨ algunos Q (a # xs)) " by simp
qed

text {*
---------------------------------------------------------------------
Ejercicio 11.1. Demostrar o refutar automáticamente
algunos P xs = (¬ todos (λx. (¬ P x)) xs)
---------------------------------------------------------------------
*}

(* ivamenjim wilmorort migtermor marpoldia1 crigomgom rubgonmar paupeddeg danrodcha pablucoto dancorgar *)

lemma "algunos P xs = (¬ todos (λx. (¬ P x)) xs)"
by (induct xs) auto

(* ferrenseg *)

lemma "algunos P xs = (¬ todos (λx. (¬ P x)) xs)"
by (induct xs) simp_all

text {*
---------------------------------------------------------------------
Ejercicio 11.2. Demostrar o refutar datalladamente
algunos P xs = (¬ todos (λx. (¬ P x)) xs)
---------------------------------------------------------------------
*}

(* ivamenjim migtermor marpoldia1 crigomgom paupeddeg*)

lemma "algunos P xs = (¬ todos (λx. (¬ P x)) xs)"
proof (induct xs)
show "algunos P [] = (¬ todos (λx. ¬ P x) [])" by simp
next
fix a xs
assume HI: "algunos P xs = (¬ todos (λx. ¬ P x) xs)"
have "algunos P (a # xs) = (P a ∨ (algunos P xs))" by simp
also have "... = (P a ∨ (¬ todos (λx. ¬ P x) xs))" using HI by simp
finally show "algunos P (a # xs) = (¬ todos (λx. ¬ P x) (a # xs))" by simp
qed

(* ferrenseg *)

lemma "algunos P xs = (¬ todos (λx. (¬ P x)) xs)"
proof (induct xs)
show "algunos P [] = (¬ todos (λx. (¬ P x)) [])" by simp
next
fix x xs
assume HI: "algunos P xs = (¬ todos (λx. (¬ P x)) xs)"
show "algunos P (x # xs) = (¬ todos (λx. (¬ P x)) (x # xs))"
proof -
have "algunos P (x # xs) = ((P x) ∨ algunos P xs)" by simp
also have "… = ((P x) ∨ ¬ todos (λx. (¬ P x)) xs)" using HI by simp
also have "… = (¬ (¬ (P x) ∧ todos (λx. (¬ P x)) xs))" by simp
also have "… = (¬ todos (λx. (¬ P x)) (x # xs))" by simp
finally show ?thesis
qed
qed

{* danrodcha pablucoto*}
lemma "algunos P xs = (¬ todos (λx. (¬ P x)) xs)" (is "?Q xs")
proof (induct xs)
show "?Q []" by simp
next
fix a xs assume HI: "?Q xs"
have "algunos P (a # xs) = (P a ∨ algunos P xs)" by simp
also have "… = (P a ∨ (¬ todos (λx. (¬ P x)) xs))" using HI by simp
also have "… = (¬(¬(P a) ∧ todos(λx. (¬ P x)) xs))" by simp
also have "… = (¬ todos(λx. (¬ P x)) (a # xs))" by simp
finally show "?Q (a # xs)" by simp
qed

(* dancorgar *)
lemma "algunos P xs = (¬ todos (λx. (¬ P x)) xs)"
proof (induct xs)
show "algunos P [] = (¬ todos (λx. (¬ P x)) [])" by simp
next
fix xf xs
assume HI: "algunos P xs = (¬ todos (λx. (¬ P x)) xs)"
have "algunos P (xf#xs) = ((P xf) ∨ algunos P xs)" by simp
also have "… = ( (P xf) ∨ (¬ todos (λx. (¬ P x)) xs) )" using HI by simp
also have "… = (¬((¬P xf) ∧ (todos (λx. (¬ P x)) xs)))" by simp
finally show "algunos P (xf#xs) = (¬todos (λx. (¬ P x)) (xf#xs))" by simp
qed

text {*
---------------------------------------------------------------------
Ejercicio 12. Definir la funcion primitiva recursiva
estaEn :: 'a ⇒ 'a list ⇒ bool
tal que (estaEn x xs) se verifica si el elemento x está en la lista
xs. Por ejemplo,
estaEn (2::nat) [3,2,4] = True
estaEn (1::nat) [3,2,4] = False
---------------------------------------------------------------------
*}

(* ivamenjim ferrenseg migtermor serrodcal crigomgom rubgonmar marpoldia1 paupeddeg danrodcha dancorgar pablucoto*)

fun estaEn :: "'a ⇒ 'a list ⇒ bool" where
"estaEn x [] = False"
| "estaEn x (a#xs) = ((a = x) ∨ (estaEn x xs))"

value "estaEn (2::nat) [3,2,4] = True"
value "estaEn (1::nat) [3,2,4] = False"

text {*
---------------------------------------------------------------------
Ejercicio 13. Expresar la relación existente entre estaEn y algunos.
Demostrar dicha relación de forma automática y detallada.
---------------------------------------------------------------------
*}

(* migtermor crigomgom *)

lemma "estaEn x xs = (algunos (λa. a=x) xs)"
by (induct xs) auto

lemma auxiliar13:
"(x=a) = ((λa. a=x) a)"
by auto

lemma "estaEn x xs = (algunos (λa. a=x) xs)" (is "?P xs")
proof (induct xs)
show "?P []" by simp
next
fix a xs
assume HI: "?P xs"
have "estaEn x (a#xs) = ((a=x) ∨ (estaEn x xs))" by simp
also have H: "… = ((a=x) ∨ (algunos (λa. a=x) xs))" using HI by simp
also have "… = (((λa. a=x) a) ∨ (algunos (λa. a=x) xs))" using auxiliar13 by simp
also have "… = (algunos (λa. a=x) (a#xs))" by simp
have C: "estaEn x (a#xs) = (algunos (λa. a=x) (a#xs))" using H by simp
finally show "estaEn x (a#xs) = (algunos (λa. a=x) (a#xs))" using C by simp
qed

(* ferrenseg *)

lemma estaEn_algunos:
"estaEn y xs = algunos (λx. x=y) xs"
by (induct xs) auto

lemma estaEn_algunos_2:
"estaEn y xs = algunos (λx. x=y) xs"
proof (induct xs)
show "estaEn y [] = algunos (λx. x=y) []" by simp
next
fix x xs
assume HI: "estaEn y xs = algunos (λx. x=y) xs"
show "estaEn y (x # xs) = algunos (λx. x=y) (x # xs)"
proof -
have "estaEn y (x # xs) = (y=x ∨ estaEn y xs)" by simp
also have "… = (y=x ∨ algunos (λx. x=y) xs)" using HI by simp
also have "… = (x=y ∨ algunos (λx. x=y) xs)" by auto
also have "… = algunos (λx. x=y) (x # xs)" by simp
finally show ?thesis
qed
qed

(* crigomgom paupeddeg dancorgar *)

lemma "estaEn a xs = algunos (λx. x = a) xs"
by (induct xs) auto

lemma "estaEn x xs = (algunos (λa. a=x) xs)"
proof (induct xs)
show " estaEn x [] = algunos (λa. a = x) []" by simp
next
fix a xs
assume HI: "estaEn x xs = algunos (λa. a = x) xs"
have "estaEn x (a # xs) = (a = x ∨ estaEn x xs)" by simp
also have "... = (a = x ∨ algunos (λa. a = x) xs)" using HI by simp
finally show " estaEn x (a # xs) = algunos (λa. a = x) (a # xs)" by simp
qed

(* ivamenjim marpoldia1 *)

lemma "estaEn x xs = algunos (λy. x=y) xs"
by (induct xs) auto

lemma "estaEn x xs = algunos (λy. x=y) xs"
proof (induct xs)
show "estaEn x [] = algunos (op = x) []" by simp
next
fix a xs
assume HI: "estaEn x xs = algunos (op = x) xs"
have "estaEn x (a # xs) = ((a = x) ∨ (estaEn x xs))" by simp
also have "... = (a = x ∨ algunos (op = x) xs)" using HI by simp
finally show " estaEn x (a # xs) = algunos (op = x) (a # xs)" by auto
qed

{* danrodcha *}
lemma "estaEn a xs = algunos (op = a) xs"
by (induct xs) auto

{* danrodcha *}
lemma "estaEn a xs = algunos (op = a) xs"
proof (induct xs)
show "estaEn a [] = algunos (op = a) []" by simp
next
fix x xs
assume HI: "estaEn a xs = algunos (op = a) xs"
have "estaEn a (x # xs) = ((x = a) ∨ (estaEn a xs))" by simp
also have "… = (x = a ∨ algunos (op = a) xs)" using HI by simp
also have "… = ((op = a) x ∨ algunos (op = a) xs)" by (simp add:HOL.eq_commute)
also have "… = algunos (op = a) (x # xs)" by simp
finally show "estaEn a (x # xs) = algunos (op = a) (x # xs)" by simp
qed

(* pablucoto *)

-- "Automatica"
lemma " estaEn y xs = algunos (λx. x=y) xs"
by (induct xs) auto

-- "Detallada"
lemma " estaEn y xs = algunos (λx. x=y) xs"
proof (induct xs)
show "estaEn y [] = algunos (λx. x = y) []" by simp
next
fix a xs
assume HI: " estaEn y xs = algunos (λx. x = y) xs "
have " estaEn y (a # xs) = (y = a ∨ estaEn y xs)" by simp
also have "... = (y = a ∨ algunos (λx. x = y) xs) " using HI by simp
also have "… = (a=y ∨ algunos (λx. x=y) xs)" by auto
also have "... = (algunos (λx. x = y) (a#xs))" by simp
finally show "estaEn y (a # xs) = algunos (λx. x = y) (a # xs) " by simp
qed

end
</source>

Relación 4

2016-11-23T04:06:33Z

Pablucoto:

Relación 4

2016-11-23T03:58:03Z

Pablucoto:

<source lang="isar">
chapter {* R4: Cuantificadores sobre listas *}

theory R4_Cuantificadores_sobre_listas
imports Main
begin

text {*
---------------------------------------------------------------------
Ejercicio 1. Definir la función
todos :: ('a ⇒ bool) ⇒ 'a list ⇒ bool
tal que (todos p xs) se verifica si todos los elementos de la lista
xs cumplen la propiedad p. Por ejemplo, se verifica
todos (λx. 1<length x) [[2,1,4],[1,3]]
¬todos (λx. 1<length x) [[2,1,4],[3]]

Nota: La función todos es equivalente a la predefinida list_all.
---------------------------------------------------------------------
*}

{*danrodcha*}
fun todos :: "('a ⇒ bool) ⇒ 'a list ⇒ bool" where
"todos p xs = "todos p [] = True"
| "todos p (x#xs) = (p x ∧ todos p xs)""

(* ivamenjim migtermor dancorgar wilmorort marpoldia1 ferrenseg paupeddeg pablucoto crigomgom anaprarod serrodcal juacabsou rubgonmar josgarsan fraortmoy lucnovdos*)
fun todos :: "('a ⇒ bool) ⇒ 'a list ⇒ bool" where
"todos p [] = True"
| "todos p (x#xs) = (p x ∧ todos p xs)"

text {*
---------------------------------------------------------------------
Ejercicio 2. Definir la función
algunos :: ('a ⇒ bool) ⇒ 'a list ⇒ bool
tal que (algunos p xs) se verifica si algunos elementos de la lista
xs cumplen la propiedad p. Por ejemplo, se verifica
algunos (λx. 1<length x) [[2,1,4],[3]]
¬algunos (λx. 1<length x) [[],[3]]"

Nota: La función algunos es equivalente a la predefinida list_ex.
---------------------------------------------------------------------
*}

{*danrodcha ivamenjim migtermor dancorgar marpoldia1 ferrenseg wilmorort paupeddeg pablucoto crigomgom anaprarod serrodcal juacabsou rubgonmar josgarsan fraortmoy lucnovdos*}
fun algunos :: "('a ⇒ bool) ⇒ 'a list ⇒ bool" where
"algunos p [] = False"
| "algunos p (x#xs) = (p x ∨ algunos p xs)"

text {*
---------------------------------------------------------------------
Ejercicio 3.1. Demostrar o refutar automáticamente
todos (λx. P x ∧ Q x) xs = (todos P xs ∧ todos Q xs)
---------------------------------------------------------------------
*}

{*danrodcha ivamenjim migtermor dancorgar marpoldia1 ferrenseg wilmorort paupeddeg pablucoto anaprarod serrodcal juacabsou rubgonmar josgarsan fraortmoy lucnovdos*}
lemma "todos (λx. P x ∧ Q x) xs = (todos P xs ∧ todos Q xs)"
by (induct xs) auto

text {*
---------------------------------------------------------------------
Ejercicio 3.2. Demostrar o refutar detalladamente
todos (λx. P x ∧ Q x) xs = (todos P xs ∧ todos Q xs)
---------------------------------------------------------------------
*}

{*danrodcha*}
lemma "todos (λx. P x ∧ Q x) xs = (todos P xs ∧ todos Q xs)"
proof (induct xs)
show "?R []" by simp
next
fix a xs
assume HI: "?R xs"
have "todos (λx. P x ∧ Q x) (a#xs) = (P a ∧ Q a ∧ todos (λx. P x ∧ Q x) xs)" by simp
also have "… = (P a ∧ Q a ∧ (todos P xs ∧ todos Q xs))" using HI by simp
also have "… = ((P a ∧ todos P xs) ∧ (Q a ∧ todos Q xs))" by blast
also have "… = (todos P (a#xs) ∧ todos Q (a#xs))" by simp
finally show "?R (a#xs)" by simp
qed

(* ivamenjim wilmorort serrodcal josgarsan*)

lemma "todos (λx. P x ∧ Q x) xs = (todos P xs ∧ todos Q xs)"
proof (induct xs)
show "todos (λx. P x ∧ Q x) [] = (todos P [] ∧ todos Q [])" by simp
next
fix a xs
assume HI: "todos (λx. P x ∧ Q x) xs = (todos P xs ∧ todos Q xs)"
have "todos (λx. P x ∧ Q x) (a # xs) = (P a ∧ Q a ∧ todos (λx. P x ∧ Q x) xs)" by simp
also have "... = (P a ∧ Q a ∧ todos P xs ∧ todos Q xs)" using HI by simp
finally show "todos (λx. P x ∧ Q x) (a # xs) = (todos P (a # xs) ∧ todos Q (a # xs))" by auto
qed

(* dancorgar paupeddeg *)
lemma "todos (λx. P x ∧ Q x) xs = (todos P xs ∧ todos Q xs)"
proof (induct xs)
show "todos (λx. P x ∧ Q x) [] = (todos P [] ∧ todos Q [])" by simp
next
fix y xs
assume HI: "todos (λx. P x ∧ Q x) xs = (todos P xs ∧ todos Q xs)"
have "todos (λx. P x ∧ Q x) (y#xs) = ((P y ∧ Q y) ∧ (todos (λx. P x ∧ Q x) xs))" by simp
also have "… = ((P y ∧ Q y) ∧ (todos P xs ∧ todos Q xs))" using HI by simp
also have "… = ((P y ∧ todos P xs) ∧ (Q y ∧ todos Q xs))" by blast
also have "… = ((todos P (y#xs)) ∧ (todos Q (y#xs)))" by simp
finally show "todos (λx. P x ∧ Q x) (y#xs) = (todos P (y#xs) ∧ todos Q (y#xs))" by simp
qed

(* migtermor *)
lemma "todos (λx. P x ∧ Q x) xs = (todos P xs ∧ todos Q xs)"
proof (induct xs)
show "todos (λx. P x ∧ Q x) [] = (todos P [] ∧ todos Q [])" by simp
next
fix x xs
assume HI: "todos (λx. P x ∧ Q x) xs = (todos P xs ∧ todos Q xs)"
have "(todos (λx. P x ∧ Q x) (x#xs)) = (( P x ∧ Q x) ∧ (todos (λx. P x ∧ Q x) xs))"
by (simp only: todos.simps(2))
also have "… = ((P x ∧ Q x) ∧ (todos P xs ∧ todos Q xs))" using HI by simp
also have "… = (((P x)∧(todos P xs)) ∧ ((Q x) ∧ (todos Q xs)))" by arith
also have "((P x ∧ Q x) ∧ (todos P xs ∧ todos Q xs)) = (((P x)∧(todos P xs)) ∧ ((Q x) ∧ (todos Q xs)))"
by arith (* Este paso es exactamente el mismo que el anterior, pero sin cualquiera de los dos no funciona el "finally show" *)
have "… = (((P x)∧(todos P xs))∧((Q x)∧(todos Q xs)))" by simp
have "… = ((todos P (x#xs))∧(todos Q (x#xs)))" by simp
finally show "(todos (λx. P x ∧ Q x) (x#xs)) = ((todos P (x#xs)) ∧ (todos Q (x#xs)))" by simp
qed

(* marpoldia1 *)
lemma "todos (λx. P x ∧ Q x) xs = (todos P xs ∧ todos Q xs)"
proof (induct xs)
show "todos (λx. P x ∧ Q x) [] = (todos P [] ∧ todos Q [])" by simp
next
fix a xs
assume HI: "todos (λx. P x ∧ Q x) xs = (todos P xs ∧ todos Q xs)"
have "todos (λx. P x ∧ Q x) (a # xs) = ((P a ∧ Q a) ∧ (todos P xs ∧ todos Q xs))" using HI by simp
also have "... = ((P a ∧ todos P xs) ∧ (Q a ∧ todos Q xs))" by blast
also have "... = (todos P (a#xs) ∧ todos Q (a#xs)) " by simp
finally show "todos (λx. P x ∧ Q x) (a#xs) = (todos P (a#xs) ∧ todos Q (a#xs))" by simp
qed

(* ferrenseg *)

lemma "todos (λx. P x ∧ Q x) xs = (todos P xs ∧ todos Q xs)"
proof (induct xs)
show "todos (λx. P x ∧ Q x) [] = (todos P [] ∧ todos Q [])" by simp
next
fix n xs
assume HI: "todos (λx. P x ∧ Q x) xs = (todos P xs ∧ todos Q xs)"
have "todos (λx. P x ∧ Q x) (n # xs) = (P n ∧ Q n ∧ todos P xs ∧ todos Q xs)" using HI by simp
also have "… = P n ∧ Q n ∧ todos P xs ∧ todos Q xs" by blast
also have "⋯ = todos P (n # xs) ∧ todos Q (n # xs)" by simp
finally show "todos (λx. P x ∧ Q x) (n # xs) = todos P (n # xs) ∧ todos Q (n # xs)" by simp
qed

(* lucnovdos *)

lemma "todos (λx. P x ∧ Q x) xs = (todos P xs ∧ todos Q xs)"
proof (induct xs)
show "todos (λx. P x ∧ Q x) [] = (todos P [] ∧ todos Q [])" by simp
next
fix n xs
assume HI: "todos (λx. P x ∧ Q x) xs = (todos P xs ∧ todos Q xs)"
have "todos (λx. P x ∧ Q x) (n # xs) = ((P n ∧ Q n) ∧ (todos P xs ∧ todos Q xs))" using HI by simp
also have "... = ((P n ∧ todos P xs) ∧ (Q n ∧ todos Q xs))" by arith
also have "... = ((todos P(n#xs)) ∧ (todos Q(n#xs)))" by simp

finally show "todos (λx. P x ∧ Q x) (n#xs) = (todos P (n#xs) ∧ todos Q (n#xs))" by simp
qed

{* wilmorort pablucoto crigomgom anaprarod juacabsou rubgonmar fraortmoy *}

lemma "todos (λx. P x ∧ Q x) xs = (todos P xs ∧ todos Q xs)"
proof (induct xs)
show "todos (λx. P x ∧ Q x) [] = (todos P [] ∧ todos Q [])" by simp
next
fix a xs
assume HI: "todos (λx. P x ∧ Q x) xs = (todos P xs ∧ todos Q xs)"
have "todos (λx. P x ∧ Q x) (a # xs) = (P a ∧ Q a ∧ todos (λx. P x ∧ Q x) xs)" by simp
also have "... = (P a ∧ Q a ∧ todos P xs ∧ todos Q xs)" using HI by simp
also have "... = (P a ∧ todos P xs ∧ Q a ∧ todos Q xs)" by arith
also have "... = (todos P (a # xs) ∧ todos Q (a # xs))" by simp (* Este paso se puede obviar*)
finally show "todos (λx. P x ∧ Q x) (a # xs) = (todos P (a # xs) ∧ todos Q (a # xs))" by simp
qed

text {*
---------------------------------------------------------------------
Ejercicio 4.1. Demostrar o refutar automáticamente
todos P (x @ y) = (todos P x ∧ todos P y)
---------------------------------------------------------------------
*}
{*danrodcha ivamenjim marpoldia1 migtermor ferrenseg wilmorort paupeddeg crigomgom anaprarod serrodcal juacabsou rubgonmar pablucoto fraortmoy josgarsan lucnovdos*}
lemma "todos P (x @ y) = (todos P x ∧ todos P y)"
by (induct x) auto

{* anaprarod dancorgar *}
lemma "todos P (x @ y) = (todos P x ∧ todos P y)"
apply (induct x)
apply auto
done

text {*
---------------------------------------------------------------------
Ejercicio 4.2. Demostrar o refutar detalladamente
todos P (x @ y) = (todos P x ∧ todos P y)
---------------------------------------------------------------------
*}

(* ivamenjim ferrenseg wilmorort dancorgar fraortmoy josgarsan lucnovdos*)

lemma todos_append:
"todos P (x @ y) = (todos P x ∧ todos P y)"
proof (induct x)
show "todos P ([] @ y) = (todos P [] ∧ todos P y)" by simp
next
fix a x
assume HI: "todos P (x @ y) = (todos P x ∧ todos P y)"
have "todos P ((a#x) @ y) = (P a ∧ (todos P (x @ y)))" by simp
also have "... = (P a ∧ (todos P x ∧ todos P y))" using HI by simp
finally show "todos P ((a # x) @ y) = (todos P (a # x) ∧ todos P y)" by simp
qed

(* migtermor serrodcal *)

lemma todos_append1:
"todos P (x @ y) = (todos P x ∧ todos P y)" (is "?P x")
proof (induct x)
show "?P []" by simp
next
fix a x
assume HI: "?P x"
have "todos P ((a#x) @ y) = (P a ∧ (todos P (x @ y)))" by simp
also have "… = (P a ∧ (todos P x ∧ todos P y))" using HI by simp
finally show "?P (a#x)" by simp
qed

(* marpoldia1 *)
lemma todos_append:
"todos P (x @ y) = (todos P x ∧ todos P y)"
proof (induct x)
show "todos P ([] @ y) = (todos P [] ∧ todos P y)" by simp
next
fix a x
assume HI: "todos P (x @ y) = (todos P x ∧ todos P y)"
have "todos P ((a # x) @ y) = todos P (a#(x@y))" by simp
also have "... = (P a ∧ todos P (x @ y))" by simp
also have "... = (P a ∧ (todos P x ∧ todos P y))" using HI by simp
finally show "todos P ((a # x) @ y) = (todos P (a # x) ∧ todos P y)" by simp
qed

(* paupeddeg *)
lemma todos_append3:
"todos P (x @ y) = (todos P x ∧ todos P y)"
proof (induct x)
show "todos P ([] @ y) = (todos P [] ∧ todos P y)" by simp
next
fix a x
assume HI: "todos P (x @ y) = (todos P x ∧ todos P y)"
have "todos P ((a # x) @ y) = todos P (a # (x @ y))" by simp
also have "... = (todos P [a] ∧ todos P (x @ y)) " by simp
also have "... = (todos P [a] ∧ (todos P x ∧ todos P y))" using HI by simp
finally show "todos P ((a # x) @ y) = (todos P (a # x) ∧ todos P y)" by simp
qed

(* crigomgom juacabsou anaprarod*)

lemma todos_append4:
"todos P (x @ y) = (todos P x ∧ todos P y)"
proof (induct x)
show "todos P ([] @ y) = (todos P [] ∧ todos P y)" by simp
next
fix x xs
assume HI:"todos P (xs @ y) = (todos P xs ∧ todos P y)"
have "todos P ((x # xs) @ y) = (P x ∧ todos P (xs @ y ))" by simp
also have "... = (P x ∧ (todos P xs ∧ todos P y)) " using HI by simp
also have "... = ((P x ∧ todos P xs) ∧ todos P y)" by simp
finally show "todos P ((x # xs) @ y) = (todos P (x # xs) ∧ todos P y)" by simp
qed

(* pablucoto *)
lemma todos_append5:
"todos P (x @ y) = (todos P x ∧ todos P y)"
proof (induct x )
show "todos P ([] @ y) = (todos P [] ∧ todos P y)" by simp
next
fix a x
assume HI:"todos P (x @ y) = (todos P x ∧ todos P y) "
have "todos P ((a # x) @ y) = ( P a ∧ todos P (x @ y))" by simp
also have " ... =( P a ∧ (todos P x ∧ todos P y))" using HI by simp
also have "... = (todos P (a#x) ∧ todos P y)" by simp
finally show "todos P ((a # x) @ y) = (todos P (a # x) ∧ todos P y)" by auto
qed

{*danrodcha*}
lemma todos_append6:
"todos P (x @ y) = (todos P x ∧ todos P y)" (is "?Q x")
proof (induct x)
show "?Q []" by simp
next
fix x a assume HI: "?Q x"
have "todos P ((a # x) @ y) = todos P (a # x @ y)" by simp
also have "… = ((P a) ∧ todos P (x @ y))" by simp
also have "… = ((P a) ∧ (todos P x ∧ todos P y))" using HI by simp
also have "… = (todos P (a # x) ∧ todos P y)" by simp
finally show "?Q (a # x)" by simp
qed

text {*
---------------------------------------------------------------------
Ejercicio 5.1. Demostrar o refutar automáticamente
todos P (rev xs) = todos P xs
---------------------------------------------------------------------
*}

lemma "todos P (rev xs) = todos P xs"
oops

(* migtermor ivamenjim marpoldia1 serrodcal anaprarod paupeddeg dancorgar *)
lemma "todos P (rev xs) = todos P xs"
apply (induct xs)
apply simp
apply (simp add: todos_append)
apply auto
done

(* ferrenseg crigomgom rubgonmar fraortmoy josgarsan danrodcha lucnovdos*)

lemma "todos P (rev xs) = todos P xs"
by (induct xs) (auto simp add: todos_append)

(* wilmorort *)
lemma "todos P (rev xs) = todos P xs"
by (induct xs,simp,simp add: todos_append,auto)

(* juacabsou *)
lemma "todos P (rev xs) = todos P xs"
apply (induct xs,simp,simp add: todos_append,auto) done

(* pablucoto * )
lemma "todos P (rev xs) = todos P xs"
by (induct xs) (auto, simp_all add: todos_append)
lemma "todos P (rev xs) = todos P xs"
by (induct xs) ( simp_all add: todos_append, auto)

text {*
---------------------------------------------------------------------
Ejercicio 5.2. Demostrar o refutar detalladamente
todos P (rev xs) = todos P xs
---------------------------------------------------------------------
*}

lemma "todos P (rev xs) = todos P xs"
oops

(* migtermor *)

lemma auxiliar:
"rev (a#xs) = rev xs @ [a]"
by auto

lemma "todos P (rev xs) = todos P xs" (is "?Q xs")
proof (induct xs)
show "?Q []" by (simp only: rev.simps(1))
next
fix a xs
assume HI: "?Q xs"
have "todos P (rev (a#xs)) = (todos P (rev xs @ [a]))" by (simp add: auxiliar)
have "… = ((todos P (rev xs)) ∧ (todos P [a]))" by (simp add: todos_append)
have "… = (todos P (rev xs) ∧ P a)" by simp
also have Aux: "… = (todos P xs ∧ P a)" using HI by simp
have Aux1: "… = (P a ∧ todos P xs)" by arith
have "(todos P (rev xs) ∧ P a) = (P a ∧ todos P xs)" using Aux Aux1 by simp
finally show "todos P (rev (a#xs)) = todos P (a#xs)" by (simp add: todos_append)
qed

(* marpoldia1 ferrenseg crigomgom serrodcal juacabsou rubgonmar josgarsan pablucoto*)
lemma "todos P (rev xs) = todos P xs"
proof (induct xs)
show "todos P (rev []) = todos P []" by simp
next
fix a xs
assume HI: "todos P (rev xs) = todos P xs"
have "todos P (rev (a # xs)) = (todos P ((rev xs)@[a]))" by simp
also have "... = (todos P (rev xs) ∧ todos P [a])" by (simp add:todos_append)
also have "... = (todos P xs ∧ P a)" using HI by simp
also have "... = (P a ∧ todos P xs)" by arith
also have "... = (todos P (a#xs))" by simp
finally show "todos P (rev (a # xs)) = (todos P (a#xs))" by simp
qed

(* ivamenjim *)

lemma "todos P (rev xs) = todos P xs" (is "?P xs")
proof (induct xs)
show "?P []" by simp
next
fix a xs
assume HI: "?P xs"
have "todos P (rev (a # xs)) = (todos P ((rev xs)@[a]))" by simp
also have "... = ((todos P (rev xs)) ∧ todos P [a])" by (simp add: todos_append)
also have "... = (todos P xs ∧ todos P [a])" using HI by simp
also have "... = (todos P [a] ∧ todos P xs)" by arith
finally show "todos P (rev (a # xs)) = todos P (a # xs)" by simp
qed

(* fraortmoy *)

lemma "todos P (rev xs) = todos P xs"
proof (induct xs)
show "todos P (rev []) = todos P []" by simp
next
fix a xs
assume H1: "todos P (rev xs) = todos P xs"
have " todos P (rev (a # xs)) = todos P (rev xs @ [a])" by simp
also have "… = (todos P (rev xs) ∧ todos P [a])" by (simp add:todos_append)
also have "… = (todos P xs ∧ todos P [a])" using H1 by simp
also have "… = (todos P [a] ∧ todos P xs)" by arith
finally show "todos P (rev (a # xs)) = todos P (a#xs)" by simp
qed

(* paupeddeg dancorgar *)
lemma "todos P (rev xs) = todos P xs"
proof (induct xs)
show " todos P (rev []) = todos P []" by simp
next
fix a xs
assume HI: "todos P (rev xs) = todos P xs "
have "todos P (rev (a # xs)) = todos P (rev(xs) @ [a])" by simp
also have "... = (todos P (rev xs) ∧ todos P [a]) " by (simp add: todos_append)
also have "... = (todos P xs ∧ todos P [a])" using HI by simp
also have "... = (todos P [a] ∧ todos P xs)" by arith
also have "... = todos P ([a] @ xs)" by (simp add: todos_append)
finally show "todos P (rev (a # xs)) = todos P (a # xs)" by simp
qed

(* wilmorort *)
lemma "todos P (rev xs) = todos P xs" (is "?P xs")
proof (induct xs)
show "?P []" by simp
next
fix a xs
assume HI: "?P xs"
have "todos P (rev (a#xs)) = todos P ((rev xs) @ [a])" by simp
also have "... = (todos P(rev xs) ∧ todos P [a])" by (simp add: todos_append)
also have "... = (todos P xs ∧ todos P [a])" using HI by simp
also have "... = (todos P [a] ∧ todos P xs)" by (simp add: HOL.conj_commute)
also have "... = todos P([a]@(xs))" by (simp)
finally show "todos P (rev (a#xs))= todos P (a # xs)" by simp
qed

(* anaprarod *)
(* es igual que las anteriores pero con el final también con patrones *)

lemma "todos P (rev xs) = todos P xs" (is "?P xs")
proof (induct xs)
show "?P []" by auto
next
fix a xs
assume HI: "?P xs"
have "todos P (rev (a#xs)) = todos P (rev xs @ [a])" by simp
also have "… = (todos P (rev xs) ∧ todos P [a])" by (simp add: todos_append)
also have "… = (todos P xs ∧ todos P [a])" using HI by simp
also have "… = (todos P [a] ∧ todos P xs)" by arith
also have "… = todos P (a#xs)" by simp
finally show "?P (a # xs)" by simp
qed

{*danrodcha*}
lemma "todos P (rev xs) = todos P xs" (is "?Q xs")
proof (induct xs)
show "?Q []" by simp
next
fix a xs assume HI: "?Q xs"
have "todos P (rev (a#xs)) = todos P ((rev xs) @ [a])" by simp
also have "… = (todos P (rev xs) ∧ todos P [a])" by (simp add:todos_append)
also have "… = (todos P xs ∧ todos P [a])" using HI by simp
also have "… = (todos P [a] ∧ todos P xs)" by (simp add: HOL.conj_commute)
also have "… = todos P (a # xs)" by simp
finally show "?Q (a # xs)" by simp
qed

text {*
---------------------------------------------------------------------
Ejercicio 6. Demostrar o refutar:
algunos (λx. P x ∧ Q x) xs = (algunos P xs ∧ algunos Q xs)
---------------------------------------------------------------------
*}

lemma "algunos (λx. P x ∧ Q x) xs = (algunos P xs ∧ algunos Q xs)"
oops

(* migtermor ivamenjim ferrenseg paupeddeg crigomgom serrodcal wilmorort juacabsou rubgonmar anaprarod marpoldia1 fraortmoy josgarsan danrodcha pablucoto*)
lemma "algunos (λx. P x ∧ Q x) xs = (algunos P xs ∧ algunos Q xs)"
quickcheck
oops
(* Quickcheck encuentra el siguiente contraejemplo: P={a1}, Q={a2}, xs={a1,a2}. En este ejemplo:
· "algunos (λx. P x ∧ Q x) xs = False"
· "(algunos P xs ∧ algunos Q xs) = True" *)

(* dancorgar *)
lemma "algunos (λx. P x ∧ Q x) xs = (algunos P xs ∧ algunos Q xs)"
proof -
assume H1: "xs = [a, b]"
assume H2: "P a = True"
assume H3: "Q a = False"
assume H4: "P b = False"
assume H5: "Q b = True"
have F1: "(algunos P xs ∧ algunos Q xs) = True" using H1 H2 H3 H4 H5 by simp
have F2: "algunos (λx. P x ∧ Q x) xs = False" using H1 H2 H3 H4 H5 by simp
have "algunos (λx. P x ∧ Q x) xs ≠ (algunos P xs ∧ algunos Q xs)" using F1 F2 by simp
oops

text {*
---------------------------------------------------------------------
Ejercicio 7.1. Demostrar o refutar automáticamente
algunos P (map f xs) = algunos (P ∘ f) xs
---------------------------------------------------------------------
*}

(* ivamenjim migtermor marpoldia1 crigomgom rubgonmar wilmorort anaprarod
fraortmoy juacabsou paupeddeg josgarsan danrodcha pablucoto *)

lemma "algunos P (map f xs) = algunos (P o f) xs"
by (induct xs) auto

(* ferrenseg *)

lemma "algunos P (map f xs) = algunos (P o f) xs"
by (induct xs) simp_all

(* anaprarod dancorgar *)
lemma "algunos P (map f xs) = algunos (P o f) xs"
apply (induct xs)
apply auto
done

text {*
---------------------------------------------------------------------
Ejercicio 7.2. Demostrar o refutar detalladamente
algunos P (map f xs) = algunos (P ∘ f) xs
---------------------------------------------------------------------
*}

lemma "algunos P (map f xs) = algunos (P o f) xs"
oops

(* migtermor *)

lemma AUX: "algunos (λa. P (f a)) xs = algunos (P o f) xs"
by (induct xs) auto

lemma "algunos P (map f xs) = algunos (P o f) xs" (is "?Q xs")
proof (induct xs)
show "?Q []" by simp
next
fix x xs
assume HI: "?Q xs"
have "algunos P (map f (x#xs)) = (algunos P ((f x)#(map f xs)))" by simp
also have "… = ((P (f x)) ∨ (algunos P (map f xs)))" by (simp only: algunos.simps(2))
also have "… = ((P (f x)) ∨ (algunos (P o f) xs))"
proof (cases)
assume C1: "(P (f x))"
have Aux: "((P (f x)) ∨ (algunos P (map f xs))) = True" using C1 by simp
have Aux1: "… = ((P (f x)) ∨ (algunos (P o f) xs))" using C1 by simp
then show "((P (f x)) ∨ (algunos P (map f xs))) = ((P (f x)) ∨ (algunos (P o f) xs))"
using Aux Aux1 by simp
next
assume C2: "¬(P (f x))"
have Aux2: "((P (f x)) ∨ (algunos P (map f xs))) = (algunos P (map f xs))" using C2 by simp
have Aux3: "… = (algunos (P o f) xs)" using HI by (simp add: AUX)
also have "… = ((P (f x)) ∨ (algunos (P o f) xs))" using C2 by simp
then show "((P (f x)) ∨ (algunos P (map f xs))) = ((P (f x)) ∨ (algunos (P o f) xs))"
using Aux2 Aux3 by simp
qed
also have "… = (((P o f) x) ∨ (algunos (P o f) xs))" by simp
also have "… = (algunos (P o f) (x#xs))" by simp
finally show "algunos P (map f (x#xs)) = (algunos (P o f) (x#xs))" by simp
qed

(* ferrenseg *)

lemma "algunos P (map f xs) = algunos (P ∘ f) xs"
proof (induct xs)
show "algunos P (map f []) = algunos (P ∘ f) []" by simp
next
fix x xs
assume HI: "algunos P (map f xs) = algunos (P ∘ f) xs"
show "algunos P (map f (x # xs)) = algunos (P ∘ f) (x # xs)"
proof -
have "algunos P (map f (x # xs)) = algunos P ((f x) # (map f xs))" by simp
also have "… = ((P (f x)) ∨ (algunos P (map f xs)))" by simp
also have "… = (((P ∘ f) x) ∨ (algunos (P ∘ f) xs))" using HI by simp
also have "… = algunos (P ∘ f) (x # xs)" by simp
finally show ?thesis
qed
qed

(* ivamenjim *)

lemma "algunos P (map f xs) = algunos (P o f) xs"
proof (induct xs)
show "algunos P (map f []) = algunos (P ∘ f) []" by simp
next
fix a xs
assume HI: "algunos P (map f xs) = algunos (P ∘ f) xs"
have "algunos P (map f (a # xs)) = algunos P ((f a)#map f xs)" by simp
also have "... = (algunos P (map f [a]) ∨ algunos P (map f xs))" by simp
also have "... = (algunos P (map f [a]) ∨ algunos (P ∘ f) xs)" using HI by simp
finally show "algunos P (map f (a # xs)) = algunos (P ∘ f) (a # xs)" by simp
qed

(* crigomgom rubgonmar anaprarod marpoldia1 juacabsou danrodcha paupeddeg *)
lemma "algunos P (map f xs) = algunos (P o f) xs"
proof (induct xs)
show "algunos P (map f []) = algunos (P ∘ f) []" by simp
next
fix x xs
assume HI: "algunos P (map f xs) = algunos (P ∘ f) xs"
have "algunos P (map f (x # xs)) = algunos P ((f x) # (map f xs))" by simp
also have "... = (P (f x) ∨ algunos P (map f xs))" by simp
also have "... = (P (f x) ∨ algunos (P ∘ f) xs)" using HI by simp
also have "... = ((P ∘ f) x ∨ algunos (P ∘ f) xs)" by simp
finally show "algunos P (map f (x # xs)) = algunos (P ∘ f) (x # xs)" by simp
qed

(* wilmorort pablucoto*)
lemma "algunos P (map f xs) = algunos (P o f) xs" (is "?P xs" )
proof (induct xs)
show "?P []" by simp
next
fix a xs
assume HI: "algunos P (map f xs) = algunos (P ∘ f) xs"
have "algunos P (map f (a # xs)) = (P (f a) ∨ algunos P (map f xs))" by simp
also have "... = (P (f a) ∨ algunos (P ∘ f) xs )" using HI by simp
also have "... = ((P ∘ f) a ∨ algunos (P ∘ f)xs)" by simp
finally show " algunos P (map f (a # xs)) = algunos (P ∘ f) (a # xs)" by simp
qed

(* fraortmoy *)

lemma "algunos P (map f xs) = algunos (P o f) xs"
proof (induct xs)
show "algunos P (map f []) = algunos (P ∘ f) []" by simp
next
fix a xs
assume H1: "algunos P (map f xs) = algunos (P ∘ f) xs"
have "algunos P (map f (a # xs)) = (algunos P (map f [a]) ∨ algunos P (map f xs))" by simp
also have "… = (algunos (P ∘ f) [a] ∨ algunos (P ∘ f) xs )" using H1 by simp
also have "… = algunos (P ∘ f) (a#xs)" by simp
finally show "algunos P (map f (a # xs)) = algunos (P ∘ f) (a#xs)" by simp
qed

{*danrodcha*}
lemma "algunos P (map f xs) = algunos (P o f) xs" (is "?Q xs")
proof (induct xs)
show "?Q []" by simp
next
fix a xs assume HI: "?Q xs"
have " algunos P (map f (a # xs)) = algunos P ((f a)#(map f xs))" by simp
also have "… = (P (f a) ∨ algunos P (map f xs))" by simp
also have "… = (P (f a) ∨ algunos (P ∘ f) xs)" using HI by blast
also have "… = ((P ∘ f) a ∨ algunos (P ∘ f) xs)" by simp
also have "… = algunos (P ∘ f) (a # xs)" by simp
finally show "?Q (a # xs)" by blast
qed

(* dancorgar *)
lemma "algunos P (map f xs) = algunos (P o f) xs"
proof (induct xs)
show "algunos P (map f []) = algunos (P o f) []" by simp
next
fix x xs
assume HI: "algunos P (map f xs) = algunos (P o f) xs"
have "algunos P (map f (x#xs)) = ((P (f x)) ∨ (algunos P (map f xs)))" by simp
also have "… = ((P (f x)) ∨ (algunos (P o f) xs))" using HI by simp
finally show "algunos P (map f (x#xs)) = algunos (P o f) (x#xs)" by simp
qed

text {*
---------------------------------------------------------------------
Ejercicio 8.1. Demostrar o refutar automáticamente
algunos P (xs @ ys) = (algunos P xs ∨ algunos P ys)
---------------------------------------------------------------------
*}

(* ivamenjim marpoldia1 paupeddeg crigomgom rubgonmar wilmorort fraortmoy danrodcha pablucoto dancorgar *)

lemma "algunos P (xs @ ys) = (algunos P xs ∨ algunos P ys)"
by (induct xs) auto

(* ferrenseg *)
lemma "algunos P (xs @ ys) = (algunos P xs ∨ algunos P ys)"
by (induct xs) simp_all

text {*
---------------------------------------------------------------------
Ejercicio 8.2. Demostrar o refutar detalladamente
algunos P (xs @ ys) = (algunos P xs ∨ algunos P ys)
---------------------------------------------------------------------
*}

(* ivamenjim ferrenseg marpoldia1 wilmorort dancorgar *)

lemma algunos_append:
"algunos P (xs @ ys) = (algunos P xs ∨ algunos P ys)"
proof (induct xs)
show "algunos P ([] @ ys) = (algunos P [] ∨ algunos P ys)" by simp
next
fix a xs
assume HI: "algunos P (xs @ ys) = (algunos P xs ∨ algunos P ys)"
have "algunos P ((a # xs) @ ys) = (P a ∨ (algunos P (xs @ ys)))" by simp
also have "... = (P a ∨ (algunos P xs ∨ algunos P ys))" using HI by simp
finally show "algunos P ((a # xs) @ ys) = (algunos P (a # xs) ∨ algunos P ys)" by simp
qed

(* migtermor crigomgom *)

lemma algunos_append2:
"algunos P (xs @ ys) = (algunos P xs ∨ algunos P ys)" (is "?Q xs")
proof (induct xs)
show "?Q []" by simp
next
fix x xs
assume HI: "?Q xs"
have "algunos P ((x#xs) @ ys) = algunos P (x#(xs @ ys))" by simp
also have "… = ((P x) ∨ (algunos P (xs @ ys)))" by simp
also have "… = ((P x) ∨ (algunos P xs) ∨ (algunos P ys))" using HI by simp
also have "… = ((algunos P (x#xs)) ∨ (algunos P ys))" by simp
finally show "algunos P ((x#xs) @ ys) = (algunos P (x#xs) ∨ algunos P ys)" by simp
qed

(* paupeddeg*)
lemma algunos_append3:
"algunos P (xs @ ys) = (algunos P xs ∨ algunos P ys)"
proof (induct xs)
show "algunos P ([] @ ys) = (algunos P [] ∨ algunos P ys)" by simp
next
fix a xs
assume HI: "algunos P (xs @ ys) = (algunos P xs ∨ algunos P ys)"
have "algunos P ((a # xs) @ ys) = (algunos P [a] ∨ ( algunos P (xs @ ys)))" by simp
also have "... = (algunos P [a] ∨ (algunos P xs ∨ algunos P ys))" using HI by simp
finally show "algunos P ((a # xs) @ ys) = (algunos P (a # xs) ∨ algunos P ys)" by simp
qed

(* fraortmoy *)
lemma algunos_append4:
"algunos P (xs @ ys) = (algunos P xs ∨ algunos P ys)"
proof (induct xs)
show " algunos P ([] @ ys) = (algunos P [] ∨ algunos P ys)" by simp
next
fix a xs
assume H1: "algunos P (xs @ ys) = (algunos P xs ∨ algunos P ys)"
have "algunos P ((a # xs) @ ys) = (algunos P [a] ∨ algunos P (xs @ ys))" by simp
also have "… = (algunos P [a] ∨ algunos P xs ∨ algunos P ys)" using H1 by simp
also have "… = ((algunos P [a] ∨ algunos P xs) ∨ algunos P ys)" by simp
also have "… = (algunos P (a#xs) ∨ algunos P ys)" by simp
finally show "algunos P ((a # xs) @ ys) = (algunos P (a#xs) ∨ algunos P ys)" by simp
qed

{* danrodcha pablucoto*}
lemma algunos_append5:
"algunos P (xs @ ys) = (algunos P xs ∨ algunos P ys)" (is "?Q xs")
proof (induct xs)
show "?Q []" by simp
next
fix a xs assume HI: "?Q xs"
have "algunos P ((a#xs) @ ys) = algunos P (a#(xs @ ys))" by simp
also have "… = (algunos P [a] ∨ algunos P (xs @ ys))" by simp
also have "… = (algunos P [a] ∨ algunos P xs ∨ algunos P ys)" using HI by simp
also have "… = (algunos P (a # xs) ∨ algunos P ys)" by simp
finally show "?Q (a # xs)" by simp
qed

text {*
---------------------------------------------------------------------
Ejercicio 9.1. Demostrar o refutar automáticamente
algunos P (rev xs) = algunos P xs
---------------------------------------------------------------------
*}

(* ivamenjim migtermor marpoldia1 rubgonmar paupeddeg dancorgar *)

lemma "algunos P (rev xs) = algunos P xs"
apply (induct xs)
apply simp
apply (simp add: algunos_append)
apply auto
done

(* ferrenseg crigomgom danrodcha pablucoto*)
lemma "algunos P (rev xs) = algunos P xs"
by (induct xs) (auto simp add: algunos_append)

(* wilmorort *)
lemma "algunos P (rev xs) = algunos P xs"
by (induct xs,simp,simp add: algunos_append,auto)

text {*
---------------------------------------------------------------------
Ejercicio 9.2. Demostrar o refutar detalladamente
algunos P (rev xs) = algunos P xs
---------------------------------------------------------------------
*}

lemma "algunos P (rev xs) = algunos P xs"
oops

(* migtermor *)
lemma auxiliar1:
"rev (a#xs) = rev xs @ [a]"
by auto

lemma "algunos P (rev xs) = algunos P xs" (is "?Q xs")
proof (induct xs)
show "?Q []" by simp
next
fix x xs
assume HI: "?Q xs"
have "algunos P (rev (x#xs)) = (algunos P (rev xs @ [x]))" using auxiliar1 by simp
also have "… = ((algunos P (rev xs)) ∨ (algunos P [x]))" by (simp add: algunos_append)
also have "… = ((P x) ∨ (algunos P (rev xs)))" by simp arith
also have "… = ((P x) ∨ (algunos P xs))" using HI by simp
finally show "algunos P (rev (x#xs)) = (algunos P (x#xs))" by simp
qed

(* ferrenseg *)

lemma "algunos P (rev xs) = algunos P xs"
proof (induct xs)
show "algunos P (rev []) = algunos P []" by simp
next
fix x xs
assume HI: "algunos P (rev xs) = algunos P xs"
show "algunos P (rev (x # xs)) = algunos P (x # xs)"
proof -
have "algunos P (rev (x # xs)) = algunos P ((rev xs) @ [x])" by simp
also have "… = (algunos P (rev xs) ∨ algunos P [x])"
by (simp add: algunos_append)
also have "… = (algunos P xs ∨ algunos P [x])" using HI by simp
also have "… = (algunos P xs ∨ P x)" by simp
also have "… = (P x ∨ algunos P xs)" by auto
also have "… = algunos P (x # xs)" by simp
finally show ?thesis
qed
qed

(* ivamenjim marpoldia1*)

lemma "algunos P (rev xs) = algunos P xs" (is "?P xs")
proof (induct xs)
show "?P []" by simp
next
fix a xs
assume HI: "?P xs"
have "algunos P (rev (a # xs)) = (algunos P ((rev xs)@[a]))" by simp
also have "... = ((algunos P (rev xs)) ∨ algunos P [a])" by (simp add: algunos_append)
also have "... = (algunos P xs ∨ algunos P [a])" using HI by simp
also have "... = (algunos P [a] ∨ algunos P xs)" by arith
finally show "algunos P (rev (a # xs)) = algunos P (a # xs)" by simp
qed

(* paupeddeg dancorgar *)
lemma "algunos P (rev xs) = algunos P xs"
proof (induct xs)
show "algunos P (rev []) = algunos P []" by simp
next
fix a xs
assume HI:"algunos P (rev xs) = algunos P xs"
have "algunos P (rev (a # xs)) = algunos P ((rev xs) @ [a])" by simp
also have "... = ((algunos P (rev xs)) ∨ (algunos P [a]))" by (simp add: algunos_append)
also have "... = ((algunos P xs) ∨ (algunos P [a]))" using HI by simp
also have "... = ((algunos P [a]) ∨ (algunos P xs))" by arith
finally show "algunos P (rev (a # xs)) = algunos P (a # xs)" by simp
qed

(*crigomgom pablucoto*)
lemma "algunos P (rev xs) = algunos P xs"
proof (induct xs)
show "algunos P (rev []) = algunos P []" by simp
next
fix x xs
assume HI: " algunos P (rev xs) = algunos P xs"
have "algunos P (rev (x # xs)) = algunos P ((rev xs) @ [x])" by simp
also have "... = (algunos P (rev xs) ∨ algunos P [x])" by (simp add: algunos_append)
also have "... = (algunos P xs ∨ algunos P [x])" using HI by simp
also have "... = (algunos P xs ∨ P x)" by simp
also have "... = (P x ∨ algunos P xs)" by arith
also have "... = algunos P (x#xs)" by simp
finally show "algunos P (rev (x # xs)) = algunos P (x # xs)" by simp
qed

(*wilmorort*)
lemma "algunos P (rev xs) = algunos P xs" (is "?P xs")
proof (induct xs)
show "?P []" by simp
next
fix a xs
assume HI: "?P xs"
have "algunos P (rev (a#xs)) = algunos P ((rev xs) @ [a])" by simp
also have "... = (algunos P(rev xs) ∨ algunos P [a])" by (simp add: algunos_append)
also have "... = (algunos P xs ∨ algunos P [a])" using HI by simp
also have "... = (algunos P [a] ∨ algunos P xs)" by (simp add: HOL.disj_commute)
also have "... = algunos P([a]@(xs))" by (simp)
finally show "algunos P (rev (a#xs))= algunos P (a # xs)" by simp
qed

{*danrodcha *}
lemma "algunos P (rev xs) = algunos P xs" (is "?Q xs")
proof (induct xs)
show "?Q []" by simp
next
fix a xs assume HI: "?Q xs"
have "algunos P (rev (a#xs)) = algunos P ((rev xs) @ [a])" by simp
also have "… = (algunos P (rev xs) ∨ algunos P [a])" by (simp add:algunos_append)
also have "… = (algunos P xs ∨ algunos P [a])" using HI by simp
also have "… = (algunos P [a] ∨ algunos P xs)" by (simp add: HOL.disj_commute)
also have "… = algunos P (a # xs)" by simp
finally show "?Q (a # xs)" by simp
qed

text {*
---------------------------------------------------------------------
Ejercicio 10. Encontrar un término no trivial Z tal que sea cierta la
siguiente ecuación:
algunos (λx. P x ∨ Q x) xs = Z
y demostrar la equivalencia de forma automática y detallada.
---------------------------------------------------------------------
*}

(* migtermor *)

lemma "algunos (λx. P x ∨ Q x) xs = ((algunos (λx. P x) xs) ∨ (algunos (λx. Q x) xs))"
by (induct xs) auto

lemma "algunos (λx. P x ∨ Q x) xs = ((algunos (λx. P x) xs) ∨ (algunos (λx. Q x) xs))" (is "?R xs")
proof (induct xs)
show "?R []" by simp
next
fix x xs
assume HI: "?R xs"
have "algunos (λx. P x ∨ Q x) (x#xs) = (((λx. P x ∨ Q x) x) ∨ (algunos (λx. P x ∨ Q x) xs))"
by simp
also have "… = ((P x ∨ Q x) ∨ (algunos (λx. P x ∨ Q x) xs))" by simp
also have H1: "… = ((((λx. P x) x) ∨ (algunos (λx. P x) xs)) ∨ (((λx. Q x) x) ∨ (algunos (λx. Q x) xs)))"
using HI by simp arith
have H2: "… = ((algunos (λx. P x) (x#xs)) ∨ (algunos (λx. Q x) (x#xs)))"
by simp
have C: "(algunos (λx. P x ∨ Q x) (x#xs)) =
((algunos (λx. P x) (x#xs)) ∨ (algunos (λx. Q x) (x#xs)))"
using H1 H2 by simp
finally show "(algunos (λx. P x ∨ Q x) (x#xs)) =
((algunos (λx. P x) (x#xs)) ∨ (algunos (λx. Q x) (x#xs)))"
using C by simp
qed

(* ferrenseg *)

lemma "algunos (λx. P x ∨ Q x) xs = (algunos P xs ∨ algunos Q xs)"
proof (induct xs)
show "algunos (λx. P x ∨ Q x) [] = (algunos P [] ∨ algunos Q [])" by simp
next
fix x xs
assume HI: "algunos (λx. (P x ∨ Q x)) xs = (algunos P xs ∨ algunos Q xs)"
show "algunos (λx. P x ∨ Q x) (x # xs) = (algunos P (x # xs) ∨ algunos Q (x # xs))"
proof -
have "algunos (λx. P x ∨ Q x) (x # xs) =
((P x) ∨ (Q x) ∨ algunos (λx. P x ∨ Q x) xs)" by simp
also have "… = ((P x) ∨ (Q x) ∨ algunos P xs ∨ algunos Q xs)"
using HI by simp
also have "… = (((P x) ∨ algunos P xs) ∨ ((Q x) ∨ algunos Q xs))" by auto
also have "… = (algunos P (x # xs) ∨ algunos Q (x # xs))" by simp
finally show ?thesis
qed
qed

(* ivamenjim marpoldia1 paupeddeg *)

lemma "algunos (λx. P x ∨ Q x) xs = (algunos P xs ∨ algunos Q xs)"
by (induct xs) auto

lemma "algunos (λx. P x ∨ Q x) xs = (algunos P xs ∨ algunos Q xs)"
proof (induct xs)
show "algunos (λx. P x ∨ Q x) [] = (algunos P [] ∨ algunos Q [])" by simp
next
fix a xs
assume HI: "algunos (λx. P x ∨ Q x) xs = (algunos P xs ∨ algunos Q xs)"
have "algunos (λx. P x ∨ Q x) (a # xs) = (P a ∨ Q a ∨ algunos (λx. P x ∨ Q x) xs)" by simp
also have "... = (P a ∨ Q a ∨ algunos P xs ∨ algunos Q xs)" using HI by simp
finally show "algunos (λx. P x ∨ Q x) (a # xs) = (algunos P (a # xs) ∨ algunos Q (a # xs))" by auto
qed

(* wilmorort danrodcha*)
lemma "algunos (λx. P x ∨ Q x) xs = (¬todos(λx. ¬P x)xs ∨ ¬todos(λx. ¬Q x)xs)"
by (induct xs) auto

(* wilmorort *)
lemma "algunos (λx. P x ∨ Q x) xs = (¬todos(λx. ¬P x)xs ∨ ¬todos(λx. ¬Q x)xs)"
(is "?P xs")
proof (induct xs)
show "?P []" by simp
next
fix a xs
assume HI: "?P xs"
have "algunos (λx. P x ∨ Q x) (a # xs) = (P a ∨ Q a ∨ algunos (λx. P x ∨ Q x) xs)" by simp
also have "... = (P a ∨ Q a ∨ (¬ todos (λx. ¬ P x) xs ∨ ¬ todos (λx. ¬ Q x) xs))" using HI by simp
also have "... = (P a ∨ ¬ todos (λx. ¬ P x) xs ∨ Q a ∨ ¬ todos (λx. ¬ Q x) xs )" by arith
finally show "algunos (λx. P x ∨ Q x) (a # xs) = (~ todos (λx. ¬ P x) (a # xs) ∨ ~ todos (λx. ¬ Q x) (a # xs))" by simp
qed

{* danrodcha *}
lemma "algunos (λx. P x ∨ Q x) xs = (algunos P xs ∨ algunos Q xs)" (is "?R xs")
proof (induct xs)
show "?R []" by simp
next
fix a xs assume HI: "?R xs"
have 1:" (Q a ∨ algunos P xs) = (algunos P xs ∨ Q a)" by (simp add: HOL.disj_commute)
have "algunos (λx. P x ∨ Q x) (a # xs) =
(algunos P [a] ∨ algunos Q [a] ∨ algunos (λx. P x ∨ Q x) xs)" by simp
also have "… = (P a ∨ (Q a ∨ algunos P xs) ∨ algunos Q xs)" using HI by simp
also have "… = (P a ∨ algunos P xs ∨ Q a ∨ algunos Q xs)" using 1 by simp
also have "… = (algunos P (a # xs) ∨ algunos Q (a # xs))" by simp
finally show "?R (a # xs)" by simp
qed

(* dancorgar *)
lemma "algunos (λx. P x ∧ Q x) xs ⟹ (algunos P xs ∧ algunos Q xs)"
apply (induct xs)
apply auto
done

(* dancorgar *)
lemma "algunos (λx. P x ∧ Q x) xs ⟹ (algunos P xs ∧ algunos Q xs)"
proof (induct xs)
show "algunos (λx. P x ∧ Q x) [] ⟹ (algunos P [] ∧ algunos Q [])" by simp
next
fix xf xs
assume HI: "algunos (λx. P x ∧ Q x) xs ⟹ (algunos P xs ∧ algunos Q xs)"
have F1: "algunos (λx. P x ∧ Q x) (xf#xs) ⟹ ((P xf ∧ Q xf) ∨ algunos (λx. P x ∧ Q x) xs)" by simp
also have F2: "… ⟹ (P xf ∧ Q xf ∨ (algunos P xs ∧ algunos Q xs))" using HI by simp
have F3: "((P xf ∧ Q xf) ∨ (algunos P xs ∧ algunos Q xs)) ⟹
((P xf ∨ algunos P xs) ∧ (Q xf ∨ algunos Q xs))" by blast
have F4: "((P xf ∨ algunos P xs) ∧ (Q xf ∨ algunos Q xs)) ⟹
(algunos P (xf#xs) ∧ algunos Q (xf#xs))" by simp
show "algunos (λx. P x ∧ Q x) (xf#xs) ⟹ (algunos P (xf#xs) ∧ algunos Q (xf#xs))"
using F1 F2 F3 F4 by blast
qed

(*pablucoto*)
-- "Automatica"
lemma "algunos (λx. P x ∨ Q x) xs = (algunos P xs ∨ algunos Q xs)"
by (induct xs) auto
-- "Detallada"
lemma "algunos (λx. P x ∨ Q x) xs = (algunos P xs ∨ algunos Q xs) "
proof (induct xs)
show "algunos (λx. P x ∨ Q x) [] = (algunos P [] ∨ algunos Q []) " by simp
next
fix a xs
assume HI: " algunos (λx. P x ∨ Q x) xs = (algunos P xs ∨ algunos Q xs)"
have " algunos (λx. P x ∨ Q x) (a # xs) = ( P a ∨ Q a ∨ algunos (λx. P x ∨ Q x) xs)" by simp
also have "... = (P a ∨ Q a ∨ (algunos P xs ∨ algunos Q xs))" using HI by simp
also have "… = (((P a) ∨ algunos P xs) ∨ ((Q a) ∨ algunos Q xs))" by arith
also have "… = (algunos P (a#xs) ∨ algunos Q (a#xs))" by simp
finally show "algunos (λx. P x ∨ Q x) (a # xs) = (algunos P (a # xs) ∨ algunos Q (a # xs)) " by simp
qed

text {*
---------------------------------------------------------------------
Ejercicio 11.1. Demostrar o refutar automáticamente
algunos P xs = (¬ todos (λx. (¬ P x)) xs)
---------------------------------------------------------------------
*}

(* ivamenjim wilmorort migtermor marpoldia1 crigomgom rubgonmar paupeddeg danrodcha pablucoto dancorgar *)

lemma "algunos P xs = (¬ todos (λx. (¬ P x)) xs)"
by (induct xs) auto

(* ferrenseg *)

lemma "algunos P xs = (¬ todos (λx. (¬ P x)) xs)"
by (induct xs) simp_all

text {*
---------------------------------------------------------------------
Ejercicio 11.2. Demostrar o refutar datalladamente
algunos P xs = (¬ todos (λx. (¬ P x)) xs)
---------------------------------------------------------------------
*}

(* ivamenjim migtermor marpoldia1 crigomgom paupeddeg*)

lemma "algunos P xs = (¬ todos (λx. (¬ P x)) xs)"
proof (induct xs)
show "algunos P [] = (¬ todos (λx. ¬ P x) [])" by simp
next
fix a xs
assume HI: "algunos P xs = (¬ todos (λx. ¬ P x) xs)"
have "algunos P (a # xs) = (P a ∨ (algunos P xs))" by simp
also have "... = (P a ∨ (¬ todos (λx. ¬ P x) xs))" using HI by simp
finally show "algunos P (a # xs) = (¬ todos (λx. ¬ P x) (a # xs))" by simp
qed

(* ferrenseg *)

lemma "algunos P xs = (¬ todos (λx. (¬ P x)) xs)"
proof (induct xs)
show "algunos P [] = (¬ todos (λx. (¬ P x)) [])" by simp
next
fix x xs
assume HI: "algunos P xs = (¬ todos (λx. (¬ P x)) xs)"
show "algunos P (x # xs) = (¬ todos (λx. (¬ P x)) (x # xs))"
proof -
have "algunos P (x # xs) = ((P x) ∨ algunos P xs)" by simp
also have "… = ((P x) ∨ ¬ todos (λx. (¬ P x)) xs)" using HI by simp
also have "… = (¬ (¬ (P x) ∧ todos (λx. (¬ P x)) xs))" by simp
also have "… = (¬ todos (λx. (¬ P x)) (x # xs))" by simp
finally show ?thesis
qed
qed

{* danrodcha pablucoto*}
lemma "algunos P xs = (¬ todos (λx. (¬ P x)) xs)" (is "?Q xs")
proof (induct xs)
show "?Q []" by simp
next
fix a xs assume HI: "?Q xs"
have "algunos P (a # xs) = (P a ∨ algunos P xs)" by simp
also have "… = (P a ∨ (¬ todos (λx. (¬ P x)) xs))" using HI by simp
also have "… = (¬(¬(P a) ∧ todos(λx. (¬ P x)) xs))" by simp
also have "… = (¬ todos(λx. (¬ P x)) (a # xs))" by simp
finally show "?Q (a # xs)" by simp
qed

(* dancorgar *)
lemma "algunos P xs = (¬ todos (λx. (¬ P x)) xs)"
proof (induct xs)
show "algunos P [] = (¬ todos (λx. (¬ P x)) [])" by simp
next
fix xf xs
assume HI: "algunos P xs = (¬ todos (λx. (¬ P x)) xs)"
have "algunos P (xf#xs) = ((P xf) ∨ algunos P xs)" by simp
also have "… = ( (P xf) ∨ (¬ todos (λx. (¬ P x)) xs) )" using HI by simp
also have "… = (¬((¬P xf) ∧ (todos (λx. (¬ P x)) xs)))" by simp
finally show "algunos P (xf#xs) = (¬todos (λx. (¬ P x)) (xf#xs))" by simp
qed

text {*
---------------------------------------------------------------------
Ejercicio 12. Definir la funcion primitiva recursiva
estaEn :: 'a ⇒ 'a list ⇒ bool
tal que (estaEn x xs) se verifica si el elemento x está en la lista
xs. Por ejemplo,
estaEn (2::nat) [3,2,4] = True
estaEn (1::nat) [3,2,4] = False
---------------------------------------------------------------------
*}

(* ivamenjim ferrenseg migtermor serrodcal crigomgom rubgonmar marpoldia1 paupeddeg danrodcha dancorgar *)

fun estaEn :: "'a ⇒ 'a list ⇒ bool" where
"estaEn x [] = False"
| "estaEn x (a#xs) = ((a = x) ∨ (estaEn x xs))"

value "estaEn (2::nat) [3,2,4] = True"
value "estaEn (1::nat) [3,2,4] = False"

text {*
---------------------------------------------------------------------
Ejercicio 13. Expresar la relación existente entre estaEn y algunos.
Demostrar dicha relación de forma automática y detallada.
---------------------------------------------------------------------
*}

(* migtermor crigomgom *)

lemma "estaEn x xs = (algunos (λa. a=x) xs)"
by (induct xs) auto

lemma auxiliar13:
"(x=a) = ((λa. a=x) a)"
by auto

lemma "estaEn x xs = (algunos (λa. a=x) xs)" (is "?P xs")
proof (induct xs)
show "?P []" by simp
next
fix a xs
assume HI: "?P xs"
have "estaEn x (a#xs) = ((a=x) ∨ (estaEn x xs))" by simp
also have H: "… = ((a=x) ∨ (algunos (λa. a=x) xs))" using HI by simp
also have "… = (((λa. a=x) a) ∨ (algunos (λa. a=x) xs))" using auxiliar13 by simp
also have "… = (algunos (λa. a=x) (a#xs))" by simp
have C: "estaEn x (a#xs) = (algunos (λa. a=x) (a#xs))" using H by simp
finally show "estaEn x (a#xs) = (algunos (λa. a=x) (a#xs))" using C by simp
qed

(* ferrenseg *)

lemma estaEn_algunos:
"estaEn y xs = algunos (λx. x=y) xs"
by (induct xs) auto

lemma estaEn_algunos_2:
"estaEn y xs = algunos (λx. x=y) xs"
proof (induct xs)
show "estaEn y [] = algunos (λx. x=y) []" by simp
next
fix x xs
assume HI: "estaEn y xs = algunos (λx. x=y) xs"
show "estaEn y (x # xs) = algunos (λx. x=y) (x # xs)"
proof -
have "estaEn y (x # xs) = (y=x ∨ estaEn y xs)" by simp
also have "… = (y=x ∨ algunos (λx. x=y) xs)" using HI by simp
also have "… = (x=y ∨ algunos (λx. x=y) xs)" by auto
also have "… = algunos (λx. x=y) (x # xs)" by simp
finally show ?thesis
qed
qed

(* crigomgom paupeddeg dancorgar *)

lemma "estaEn a xs = algunos (λx. x = a) xs"
by (induct xs) auto

lemma "estaEn x xs = (algunos (λa. a=x) xs)"
proof (induct xs)
show " estaEn x [] = algunos (λa. a = x) []" by simp
next
fix a xs
assume HI: "estaEn x xs = algunos (λa. a = x) xs"
have "estaEn x (a # xs) = (a = x ∨ estaEn x xs)" by simp
also have "... = (a = x ∨ algunos (λa. a = x) xs)" using HI by simp
finally show " estaEn x (a # xs) = algunos (λa. a = x) (a # xs)" by simp
qed

(* ivamenjim marpoldia1 *)

lemma "estaEn x xs = algunos (λy. x=y) xs"
by (induct xs) auto

lemma "estaEn x xs = algunos (λy. x=y) xs"
proof (induct xs)
show "estaEn x [] = algunos (op = x) []" by simp
next
fix a xs
assume HI: "estaEn x xs = algunos (op = x) xs"
have "estaEn x (a # xs) = ((a = x) ∨ (estaEn x xs))" by simp
also have "... = (a = x ∨ algunos (op = x) xs)" using HI by simp
finally show " estaEn x (a # xs) = algunos (op = x) (a # xs)" by auto
qed

{* danrodcha *}
lemma "estaEn a xs = algunos (op = a) xs"
by (induct xs) auto

{* danrodcha *}
lemma "estaEn a xs = algunos (op = a) xs"
proof (induct xs)
show "estaEn a [] = algunos (op = a) []" by simp
next
fix x xs
assume HI: "estaEn a xs = algunos (op = a) xs"
have "estaEn a (x # xs) = ((x = a) ∨ (estaEn a xs))" by simp
also have "… = (x = a ∨ algunos (op = a) xs)" using HI by simp
also have "… = ((op = a) x ∨ algunos (op = a) xs)" by (simp add:HOL.eq_commute)
also have "… = algunos (op = a) (x # xs)" by simp
finally show "estaEn a (x # xs) = algunos (op = a) (x # xs)" by simp
qed

end
</source>

Relación 4

2016-11-23T03:12:37Z

Pablucoto:

<source lang="isar">
chapter {* R4: Cuantificadores sobre listas *}

theory R4_Cuantificadores_sobre_listas
imports Main
begin

text {*
---------------------------------------------------------------------
Ejercicio 1. Definir la función
todos :: ('a ⇒ bool) ⇒ 'a list ⇒ bool
tal que (todos p xs) se verifica si todos los elementos de la lista
xs cumplen la propiedad p. Por ejemplo, se verifica
todos (λx. 1<length x) [[2,1,4],[1,3]]
¬todos (λx. 1<length x) [[2,1,4],[3]]

Nota: La función todos es equivalente a la predefinida list_all.
---------------------------------------------------------------------
*}

{*danrodcha*}
fun todos :: "('a ⇒ bool) ⇒ 'a list ⇒ bool" where
"todos p xs = "todos p [] = True"
| "todos p (x#xs) = (p x ∧ todos p xs)""

(* ivamenjim migtermor dancorgar wilmorort marpoldia1 ferrenseg paupeddeg pablucoto crigomgom anaprarod serrodcal juacabsou rubgonmar josgarsan fraortmoy lucnovdos*)
fun todos :: "('a ⇒ bool) ⇒ 'a list ⇒ bool" where
"todos p [] = True"
| "todos p (x#xs) = (p x ∧ todos p xs)"

text {*
---------------------------------------------------------------------
Ejercicio 2. Definir la función
algunos :: ('a ⇒ bool) ⇒ 'a list ⇒ bool
tal que (algunos p xs) se verifica si algunos elementos de la lista
xs cumplen la propiedad p. Por ejemplo, se verifica
algunos (λx. 1<length x) [[2,1,4],[3]]
¬algunos (λx. 1<length x) [[],[3]]"

Nota: La función algunos es equivalente a la predefinida list_ex.
---------------------------------------------------------------------
*}

{*danrodcha ivamenjim migtermor dancorgar marpoldia1 ferrenseg wilmorort paupeddeg pablucoto crigomgom anaprarod serrodcal juacabsou rubgonmar josgarsan fraortmoy lucnovdos*}
fun algunos :: "('a ⇒ bool) ⇒ 'a list ⇒ bool" where
"algunos p [] = False"
| "algunos p (x#xs) = (p x ∨ algunos p xs)"

text {*
---------------------------------------------------------------------
Ejercicio 3.1. Demostrar o refutar automáticamente
todos (λx. P x ∧ Q x) xs = (todos P xs ∧ todos Q xs)
---------------------------------------------------------------------
*}

{*danrodcha ivamenjim migtermor dancorgar marpoldia1 ferrenseg wilmorort paupeddeg pablucoto anaprarod serrodcal juacabsou rubgonmar josgarsan fraortmoy lucnovdos*}
lemma "todos (λx. P x ∧ Q x) xs = (todos P xs ∧ todos Q xs)"
by (induct xs) auto

text {*
---------------------------------------------------------------------
Ejercicio 3.2. Demostrar o refutar detalladamente
todos (λx. P x ∧ Q x) xs = (todos P xs ∧ todos Q xs)
---------------------------------------------------------------------
*}

{*danrodcha*}
lemma "todos (λx. P x ∧ Q x) xs = (todos P xs ∧ todos Q xs)"
proof (induct xs)
show "?R []" by simp
next
fix a xs
assume HI: "?R xs"
have "todos (λx. P x ∧ Q x) (a#xs) = (P a ∧ Q a ∧ todos (λx. P x ∧ Q x) xs)" by simp
also have "… = (P a ∧ Q a ∧ (todos P xs ∧ todos Q xs))" using HI by simp
also have "… = ((P a ∧ todos P xs) ∧ (Q a ∧ todos Q xs))" by blast
also have "… = (todos P (a#xs) ∧ todos Q (a#xs))" by simp
finally show "?R (a#xs)" by simp
qed

(* ivamenjim wilmorort serrodcal josgarsan*)

lemma "todos (λx. P x ∧ Q x) xs = (todos P xs ∧ todos Q xs)"
proof (induct xs)
show "todos (λx. P x ∧ Q x) [] = (todos P [] ∧ todos Q [])" by simp
next
fix a xs
assume HI: "todos (λx. P x ∧ Q x) xs = (todos P xs ∧ todos Q xs)"
have "todos (λx. P x ∧ Q x) (a # xs) = (P a ∧ Q a ∧ todos (λx. P x ∧ Q x) xs)" by simp
also have "... = (P a ∧ Q a ∧ todos P xs ∧ todos Q xs)" using HI by simp
finally show "todos (λx. P x ∧ Q x) (a # xs) = (todos P (a # xs) ∧ todos Q (a # xs))" by auto
qed

(* dancorgar paupeddeg *)
lemma "todos (λx. P x ∧ Q x) xs = (todos P xs ∧ todos Q xs)"
proof (induct xs)
show "todos (λx. P x ∧ Q x) [] = (todos P [] ∧ todos Q [])" by simp
next
fix y xs
assume HI: "todos (λx. P x ∧ Q x) xs = (todos P xs ∧ todos Q xs)"
have "todos (λx. P x ∧ Q x) (y#xs) = ((P y ∧ Q y) ∧ (todos (λx. P x ∧ Q x) xs))" by simp
also have "… = ((P y ∧ Q y) ∧ (todos P xs ∧ todos Q xs))" using HI by simp
also have "… = ((P y ∧ todos P xs) ∧ (Q y ∧ todos Q xs))" by blast
also have "… = ((todos P (y#xs)) ∧ (todos Q (y#xs)))" by simp
finally show "todos (λx. P x ∧ Q x) (y#xs) = (todos P (y#xs) ∧ todos Q (y#xs))" by simp
qed

(* migtermor *)
lemma "todos (λx. P x ∧ Q x) xs = (todos P xs ∧ todos Q xs)"
proof (induct xs)
show "todos (λx. P x ∧ Q x) [] = (todos P [] ∧ todos Q [])" by simp
next
fix x xs
assume HI: "todos (λx. P x ∧ Q x) xs = (todos P xs ∧ todos Q xs)"
have "(todos (λx. P x ∧ Q x) (x#xs)) = (( P x ∧ Q x) ∧ (todos (λx. P x ∧ Q x) xs))"
by (simp only: todos.simps(2))
also have "… = ((P x ∧ Q x) ∧ (todos P xs ∧ todos Q xs))" using HI by simp
also have "… = (((P x)∧(todos P xs)) ∧ ((Q x) ∧ (todos Q xs)))" by arith
also have "((P x ∧ Q x) ∧ (todos P xs ∧ todos Q xs)) = (((P x)∧(todos P xs)) ∧ ((Q x) ∧ (todos Q xs)))"
by arith (* Este paso es exactamente el mismo que el anterior, pero sin cualquiera de los dos no funciona el "finally show" *)
have "… = (((P x)∧(todos P xs))∧((Q x)∧(todos Q xs)))" by simp
have "… = ((todos P (x#xs))∧(todos Q (x#xs)))" by simp
finally show "(todos (λx. P x ∧ Q x) (x#xs)) = ((todos P (x#xs)) ∧ (todos Q (x#xs)))" by simp
qed

(* marpoldia1 *)
lemma "todos (λx. P x ∧ Q x) xs = (todos P xs ∧ todos Q xs)"
proof (induct xs)
show "todos (λx. P x ∧ Q x) [] = (todos P [] ∧ todos Q [])" by simp
next
fix a xs
assume HI: "todos (λx. P x ∧ Q x) xs = (todos P xs ∧ todos Q xs)"
have "todos (λx. P x ∧ Q x) (a # xs) = ((P a ∧ Q a) ∧ (todos P xs ∧ todos Q xs))" using HI by simp
also have "... = ((P a ∧ todos P xs) ∧ (Q a ∧ todos Q xs))" by blast
also have "... = (todos P (a#xs) ∧ todos Q (a#xs)) " by simp
finally show "todos (λx. P x ∧ Q x) (a#xs) = (todos P (a#xs) ∧ todos Q (a#xs))" by simp
qed

(* ferrenseg *)

lemma "todos (λx. P x ∧ Q x) xs = (todos P xs ∧ todos Q xs)"
proof (induct xs)
show "todos (λx. P x ∧ Q x) [] = (todos P [] ∧ todos Q [])" by simp
next
fix n xs
assume HI: "todos (λx. P x ∧ Q x) xs = (todos P xs ∧ todos Q xs)"
have "todos (λx. P x ∧ Q x) (n # xs) = (P n ∧ Q n ∧ todos P xs ∧ todos Q xs)" using HI by simp
also have "… = P n ∧ Q n ∧ todos P xs ∧ todos Q xs" by blast
also have "⋯ = todos P (n # xs) ∧ todos Q (n # xs)" by simp
finally show "todos (λx. P x ∧ Q x) (n # xs) = todos P (n # xs) ∧ todos Q (n # xs)" by simp
qed

(* lucnovdos *)

lemma "todos (λx. P x ∧ Q x) xs = (todos P xs ∧ todos Q xs)"
proof (induct xs)
show "todos (λx. P x ∧ Q x) [] = (todos P [] ∧ todos Q [])" by simp
next
fix n xs
assume HI: "todos (λx. P x ∧ Q x) xs = (todos P xs ∧ todos Q xs)"
have "todos (λx. P x ∧ Q x) (n # xs) = ((P n ∧ Q n) ∧ (todos P xs ∧ todos Q xs))" using HI by simp
also have "... = ((P n ∧ todos P xs) ∧ (Q n ∧ todos Q xs))" by arith
also have "... = ((todos P(n#xs)) ∧ (todos Q(n#xs)))" by simp

finally show "todos (λx. P x ∧ Q x) (n#xs) = (todos P (n#xs) ∧ todos Q (n#xs))" by simp
qed

{* wilmorort pablucoto crigomgom anaprarod juacabsou rubgonmar fraortmoy *}

lemma "todos (λx. P x ∧ Q x) xs = (todos P xs ∧ todos Q xs)"
proof (induct xs)
show "todos (λx. P x ∧ Q x) [] = (todos P [] ∧ todos Q [])" by simp
next
fix a xs
assume HI: "todos (λx. P x ∧ Q x) xs = (todos P xs ∧ todos Q xs)"
have "todos (λx. P x ∧ Q x) (a # xs) = (P a ∧ Q a ∧ todos (λx. P x ∧ Q x) xs)" by simp
also have "... = (P a ∧ Q a ∧ todos P xs ∧ todos Q xs)" using HI by simp
also have "... = (P a ∧ todos P xs ∧ Q a ∧ todos Q xs)" by arith
also have "... = (todos P (a # xs) ∧ todos Q (a # xs))" by simp (* Este paso se puede obviar*)
finally show "todos (λx. P x ∧ Q x) (a # xs) = (todos P (a # xs) ∧ todos Q (a # xs))" by simp
qed

text {*
---------------------------------------------------------------------
Ejercicio 4.1. Demostrar o refutar automáticamente
todos P (x @ y) = (todos P x ∧ todos P y)
---------------------------------------------------------------------
*}
{*danrodcha ivamenjim marpoldia1 migtermor ferrenseg wilmorort paupeddeg crigomgom anaprarod serrodcal juacabsou rubgonmar pablucoto fraortmoy josgarsan lucnovdos*}
lemma "todos P (x @ y) = (todos P x ∧ todos P y)"
by (induct x) auto

{* anaprarod dancorgar *}
lemma "todos P (x @ y) = (todos P x ∧ todos P y)"
apply (induct x)
apply auto
done

text {*
---------------------------------------------------------------------
Ejercicio 4.2. Demostrar o refutar detalladamente
todos P (x @ y) = (todos P x ∧ todos P y)
---------------------------------------------------------------------
*}

(* ivamenjim ferrenseg wilmorort dancorgar fraortmoy josgarsan lucnovdos*)

lemma todos_append:
"todos P (x @ y) = (todos P x ∧ todos P y)"
proof (induct x)
show "todos P ([] @ y) = (todos P [] ∧ todos P y)" by simp
next
fix a x
assume HI: "todos P (x @ y) = (todos P x ∧ todos P y)"
have "todos P ((a#x) @ y) = (P a ∧ (todos P (x @ y)))" by simp
also have "... = (P a ∧ (todos P x ∧ todos P y))" using HI by simp
finally show "todos P ((a # x) @ y) = (todos P (a # x) ∧ todos P y)" by simp
qed

(* migtermor serrodcal *)

lemma todos_append1:
"todos P (x @ y) = (todos P x ∧ todos P y)" (is "?P x")
proof (induct x)
show "?P []" by simp
next
fix a x
assume HI: "?P x"
have "todos P ((a#x) @ y) = (P a ∧ (todos P (x @ y)))" by simp
also have "… = (P a ∧ (todos P x ∧ todos P y))" using HI by simp
finally show "?P (a#x)" by simp
qed

(* marpoldia1 *)
lemma todos_append:
"todos P (x @ y) = (todos P x ∧ todos P y)"
proof (induct x)
show "todos P ([] @ y) = (todos P [] ∧ todos P y)" by simp
next
fix a x
assume HI: "todos P (x @ y) = (todos P x ∧ todos P y)"
have "todos P ((a # x) @ y) = todos P (a#(x@y))" by simp
also have "... = (P a ∧ todos P (x @ y))" by simp
also have "... = (P a ∧ (todos P x ∧ todos P y))" using HI by simp
finally show "todos P ((a # x) @ y) = (todos P (a # x) ∧ todos P y)" by simp
qed

(* paupeddeg *)
lemma todos_append3:
"todos P (x @ y) = (todos P x ∧ todos P y)"
proof (induct x)
show "todos P ([] @ y) = (todos P [] ∧ todos P y)" by simp
next
fix a x
assume HI: "todos P (x @ y) = (todos P x ∧ todos P y)"
have "todos P ((a # x) @ y) = todos P (a # (x @ y))" by simp
also have "... = (todos P [a] ∧ todos P (x @ y)) " by simp
also have "... = (todos P [a] ∧ (todos P x ∧ todos P y))" using HI by simp
finally show "todos P ((a # x) @ y) = (todos P (a # x) ∧ todos P y)" by simp
qed

(* crigomgom juacabsou anaprarod*)

lemma todos_append4:
"todos P (x @ y) = (todos P x ∧ todos P y)"
proof (induct x)
show "todos P ([] @ y) = (todos P [] ∧ todos P y)" by simp
next
fix x xs
assume HI:"todos P (xs @ y) = (todos P xs ∧ todos P y)"
have "todos P ((x # xs) @ y) = (P x ∧ todos P (xs @ y ))" by simp
also have "... = (P x ∧ (todos P xs ∧ todos P y)) " using HI by simp
also have "... = ((P x ∧ todos P xs) ∧ todos P y)" by simp
finally show "todos P ((x # xs) @ y) = (todos P (x # xs) ∧ todos P y)" by simp
qed

(* pablucoto *)
lemma todos_append5:
"todos P (x @ y) = (todos P x ∧ todos P y)"
proof (induct x )
show "todos P ([] @ y) = (todos P [] ∧ todos P y)" by simp
next
fix a x
assume HI:"todos P (x @ y) = (todos P x ∧ todos P y) "
have "todos P ((a # x) @ y) = ( P a ∧ todos P (x @ y))" by simp
also have " ... =( P a ∧ (todos P x ∧ todos P y))" using HI by simp
also have "... = (todos P (a#x) ∧ todos P y)" by simp
finally show "todos P ((a # x) @ y) = (todos P (a # x) ∧ todos P y)" by auto
qed

{*danrodcha*}
lemma todos_append6:
"todos P (x @ y) = (todos P x ∧ todos P y)" (is "?Q x")
proof (induct x)
show "?Q []" by simp
next
fix x a assume HI: "?Q x"
have "todos P ((a # x) @ y) = todos P (a # x @ y)" by simp
also have "… = ((P a) ∧ todos P (x @ y))" by simp
also have "… = ((P a) ∧ (todos P x ∧ todos P y))" using HI by simp
also have "… = (todos P (a # x) ∧ todos P y)" by simp
finally show "?Q (a # x)" by simp
qed

text {*
---------------------------------------------------------------------
Ejercicio 5.1. Demostrar o refutar automáticamente
todos P (rev xs) = todos P xs
---------------------------------------------------------------------
*}

lemma "todos P (rev xs) = todos P xs"
oops

(* migtermor ivamenjim marpoldia1 serrodcal anaprarod paupeddeg dancorgar *)
lemma "todos P (rev xs) = todos P xs"
apply (induct xs)
apply simp
apply (simp add: todos_append)
apply auto
done

(* ferrenseg crigomgom rubgonmar fraortmoy josgarsan danrodcha lucnovdos*)

lemma "todos P (rev xs) = todos P xs"
by (induct xs) (auto simp add: todos_append)

(* wilmorort *)
lemma "todos P (rev xs) = todos P xs"
by (induct xs,simp,simp add: todos_append,auto)

(* juacabsou *)
lemma "todos P (rev xs) = todos P xs"
apply (induct xs,simp,simp add: todos_append,auto) done

(* pablucoto * )
lemma "todos P (rev xs) = todos P xs"
by (induct xs) (auto, simp_all add: todos_append)
lemma "todos P (rev xs) = todos P xs"
by (induct xs) ( simp_all add: todos_append, auto)

text {*
---------------------------------------------------------------------
Ejercicio 5.2. Demostrar o refutar detalladamente
todos P (rev xs) = todos P xs
---------------------------------------------------------------------
*}

lemma "todos P (rev xs) = todos P xs"
oops

(* migtermor *)

lemma auxiliar:
"rev (a#xs) = rev xs @ [a]"
by auto

lemma "todos P (rev xs) = todos P xs" (is "?Q xs")
proof (induct xs)
show "?Q []" by (simp only: rev.simps(1))
next
fix a xs
assume HI: "?Q xs"
have "todos P (rev (a#xs)) = (todos P (rev xs @ [a]))" by (simp add: auxiliar)
have "… = ((todos P (rev xs)) ∧ (todos P [a]))" by (simp add: todos_append)
have "… = (todos P (rev xs) ∧ P a)" by simp
also have Aux: "… = (todos P xs ∧ P a)" using HI by simp
have Aux1: "… = (P a ∧ todos P xs)" by arith
have "(todos P (rev xs) ∧ P a) = (P a ∧ todos P xs)" using Aux Aux1 by simp
finally show "todos P (rev (a#xs)) = todos P (a#xs)" by (simp add: todos_append)
qed

(* marpoldia1 ferrenseg crigomgom serrodcal juacabsou rubgonmar josgarsan pablucoto*)
lemma "todos P (rev xs) = todos P xs"
proof (induct xs)
show "todos P (rev []) = todos P []" by simp
next
fix a xs
assume HI: "todos P (rev xs) = todos P xs"
have "todos P (rev (a # xs)) = (todos P ((rev xs)@[a]))" by simp
also have "... = (todos P (rev xs) ∧ todos P [a])" by (simp add:todos_append)
also have "... = (todos P xs ∧ P a)" using HI by simp
also have "... = (P a ∧ todos P xs)" by arith
also have "... = (todos P (a#xs))" by simp
finally show "todos P (rev (a # xs)) = (todos P (a#xs))" by simp
qed

(* ivamenjim *)

lemma "todos P (rev xs) = todos P xs" (is "?P xs")
proof (induct xs)
show "?P []" by simp
next
fix a xs
assume HI: "?P xs"
have "todos P (rev (a # xs)) = (todos P ((rev xs)@[a]))" by simp
also have "... = ((todos P (rev xs)) ∧ todos P [a])" by (simp add: todos_append)
also have "... = (todos P xs ∧ todos P [a])" using HI by simp
also have "... = (todos P [a] ∧ todos P xs)" by arith
finally show "todos P (rev (a # xs)) = todos P (a # xs)" by simp
qed

(* fraortmoy *)

lemma "todos P (rev xs) = todos P xs"
proof (induct xs)
show "todos P (rev []) = todos P []" by simp
next
fix a xs
assume H1: "todos P (rev xs) = todos P xs"
have " todos P (rev (a # xs)) = todos P (rev xs @ [a])" by simp
also have "… = (todos P (rev xs) ∧ todos P [a])" by (simp add:todos_append)
also have "… = (todos P xs ∧ todos P [a])" using H1 by simp
also have "… = (todos P [a] ∧ todos P xs)" by arith
finally show "todos P (rev (a # xs)) = todos P (a#xs)" by simp
qed

(* paupeddeg dancorgar *)
lemma "todos P (rev xs) = todos P xs"
proof (induct xs)
show " todos P (rev []) = todos P []" by simp
next
fix a xs
assume HI: "todos P (rev xs) = todos P xs "
have "todos P (rev (a # xs)) = todos P (rev(xs) @ [a])" by simp
also have "... = (todos P (rev xs) ∧ todos P [a]) " by (simp add: todos_append)
also have "... = (todos P xs ∧ todos P [a])" using HI by simp
also have "... = (todos P [a] ∧ todos P xs)" by arith
also have "... = todos P ([a] @ xs)" by (simp add: todos_append)
finally show "todos P (rev (a # xs)) = todos P (a # xs)" by simp
qed

(* wilmorort *)
lemma "todos P (rev xs) = todos P xs" (is "?P xs")
proof (induct xs)
show "?P []" by simp
next
fix a xs
assume HI: "?P xs"
have "todos P (rev (a#xs)) = todos P ((rev xs) @ [a])" by simp
also have "... = (todos P(rev xs) ∧ todos P [a])" by (simp add: todos_append)
also have "... = (todos P xs ∧ todos P [a])" using HI by simp
also have "... = (todos P [a] ∧ todos P xs)" by (simp add: HOL.conj_commute)
also have "... = todos P([a]@(xs))" by (simp)
finally show "todos P (rev (a#xs))= todos P (a # xs)" by simp
qed

(* anaprarod *)
(* es igual que las anteriores pero con el final también con patrones *)

lemma "todos P (rev xs) = todos P xs" (is "?P xs")
proof (induct xs)
show "?P []" by auto
next
fix a xs
assume HI: "?P xs"
have "todos P (rev (a#xs)) = todos P (rev xs @ [a])" by simp
also have "… = (todos P (rev xs) ∧ todos P [a])" by (simp add: todos_append)
also have "… = (todos P xs ∧ todos P [a])" using HI by simp
also have "… = (todos P [a] ∧ todos P xs)" by arith
also have "… = todos P (a#xs)" by simp
finally show "?P (a # xs)" by simp
qed

{*danrodcha*}
lemma "todos P (rev xs) = todos P xs" (is "?Q xs")
proof (induct xs)
show "?Q []" by simp
next
fix a xs assume HI: "?Q xs"
have "todos P (rev (a#xs)) = todos P ((rev xs) @ [a])" by simp
also have "… = (todos P (rev xs) ∧ todos P [a])" by (simp add:todos_append)
also have "… = (todos P xs ∧ todos P [a])" using HI by simp
also have "… = (todos P [a] ∧ todos P xs)" by (simp add: HOL.conj_commute)
also have "… = todos P (a # xs)" by simp
finally show "?Q (a # xs)" by simp
qed

text {*
---------------------------------------------------------------------
Ejercicio 6. Demostrar o refutar:
algunos (λx. P x ∧ Q x) xs = (algunos P xs ∧ algunos Q xs)
---------------------------------------------------------------------
*}

lemma "algunos (λx. P x ∧ Q x) xs = (algunos P xs ∧ algunos Q xs)"
oops

(* migtermor ivamenjim ferrenseg paupeddeg crigomgom serrodcal wilmorort juacabsou rubgonmar anaprarod marpoldia1 fraortmoy josgarsan danrodcha pablucoto*)
lemma "algunos (λx. P x ∧ Q x) xs = (algunos P xs ∧ algunos Q xs)"
quickcheck
oops
(* Quickcheck encuentra el siguiente contraejemplo: P={a1}, Q={a2}, xs={a1,a2}. En este ejemplo:
· "algunos (λx. P x ∧ Q x) xs = False"
· "(algunos P xs ∧ algunos Q xs) = True" *)

(* dancorgar *)
lemma "algunos (λx. P x ∧ Q x) xs = (algunos P xs ∧ algunos Q xs)"
proof -
assume H1: "xs = [a, b]"
assume H2: "P a = True"
assume H3: "Q a = False"
assume H4: "P b = False"
assume H5: "Q b = True"
have F1: "(algunos P xs ∧ algunos Q xs) = True" using H1 H2 H3 H4 H5 by simp
have F2: "algunos (λx. P x ∧ Q x) xs = False" using H1 H2 H3 H4 H5 by simp
have "algunos (λx. P x ∧ Q x) xs ≠ (algunos P xs ∧ algunos Q xs)" using F1 F2 by simp
oops

text {*
---------------------------------------------------------------------
Ejercicio 7.1. Demostrar o refutar automáticamente
algunos P (map f xs) = algunos (P ∘ f) xs
---------------------------------------------------------------------
*}

(* ivamenjim migtermor marpoldia1 crigomgom rubgonmar wilmorort anaprarod
fraortmoy juacabsou paupeddeg josgarsan danrodcha pablucoto *)

lemma "algunos P (map f xs) = algunos (P o f) xs"
by (induct xs) auto

(* ferrenseg *)

lemma "algunos P (map f xs) = algunos (P o f) xs"
by (induct xs) simp_all

(* anaprarod dancorgar *)
lemma "algunos P (map f xs) = algunos (P o f) xs"
apply (induct xs)
apply auto
done

text {*
---------------------------------------------------------------------
Ejercicio 7.2. Demostrar o refutar detalladamente
algunos P (map f xs) = algunos (P ∘ f) xs
---------------------------------------------------------------------
*}

lemma "algunos P (map f xs) = algunos (P o f) xs"
oops

(* migtermor *)

lemma AUX: "algunos (λa. P (f a)) xs = algunos (P o f) xs"
by (induct xs) auto

lemma "algunos P (map f xs) = algunos (P o f) xs" (is "?Q xs")
proof (induct xs)
show "?Q []" by simp
next
fix x xs
assume HI: "?Q xs"
have "algunos P (map f (x#xs)) = (algunos P ((f x)#(map f xs)))" by simp
also have "… = ((P (f x)) ∨ (algunos P (map f xs)))" by (simp only: algunos.simps(2))
also have "… = ((P (f x)) ∨ (algunos (P o f) xs))"
proof (cases)
assume C1: "(P (f x))"
have Aux: "((P (f x)) ∨ (algunos P (map f xs))) = True" using C1 by simp
have Aux1: "… = ((P (f x)) ∨ (algunos (P o f) xs))" using C1 by simp
then show "((P (f x)) ∨ (algunos P (map f xs))) = ((P (f x)) ∨ (algunos (P o f) xs))"
using Aux Aux1 by simp
next
assume C2: "¬(P (f x))"
have Aux2: "((P (f x)) ∨ (algunos P (map f xs))) = (algunos P (map f xs))" using C2 by simp
have Aux3: "… = (algunos (P o f) xs)" using HI by (simp add: AUX)
also have "… = ((P (f x)) ∨ (algunos (P o f) xs))" using C2 by simp
then show "((P (f x)) ∨ (algunos P (map f xs))) = ((P (f x)) ∨ (algunos (P o f) xs))"
using Aux2 Aux3 by simp
qed
also have "… = (((P o f) x) ∨ (algunos (P o f) xs))" by simp
also have "… = (algunos (P o f) (x#xs))" by simp
finally show "algunos P (map f (x#xs)) = (algunos (P o f) (x#xs))" by simp
qed

(* ferrenseg *)

lemma "algunos P (map f xs) = algunos (P ∘ f) xs"
proof (induct xs)
show "algunos P (map f []) = algunos (P ∘ f) []" by simp
next
fix x xs
assume HI: "algunos P (map f xs) = algunos (P ∘ f) xs"
show "algunos P (map f (x # xs)) = algunos (P ∘ f) (x # xs)"
proof -
have "algunos P (map f (x # xs)) = algunos P ((f x) # (map f xs))" by simp
also have "… = ((P (f x)) ∨ (algunos P (map f xs)))" by simp
also have "… = (((P ∘ f) x) ∨ (algunos (P ∘ f) xs))" using HI by simp
also have "… = algunos (P ∘ f) (x # xs)" by simp
finally show ?thesis
qed
qed

(* ivamenjim *)

lemma "algunos P (map f xs) = algunos (P o f) xs"
proof (induct xs)
show "algunos P (map f []) = algunos (P ∘ f) []" by simp
next
fix a xs
assume HI: "algunos P (map f xs) = algunos (P ∘ f) xs"
have "algunos P (map f (a # xs)) = algunos P ((f a)#map f xs)" by simp
also have "... = (algunos P (map f [a]) ∨ algunos P (map f xs))" by simp
also have "... = (algunos P (map f [a]) ∨ algunos (P ∘ f) xs)" using HI by simp
finally show "algunos P (map f (a # xs)) = algunos (P ∘ f) (a # xs)" by simp
qed

(* crigomgom rubgonmar anaprarod marpoldia1 juacabsou danrodcha paupeddeg *)
lemma "algunos P (map f xs) = algunos (P o f) xs"
proof (induct xs)
show "algunos P (map f []) = algunos (P ∘ f) []" by simp
next
fix x xs
assume HI: "algunos P (map f xs) = algunos (P ∘ f) xs"
have "algunos P (map f (x # xs)) = algunos P ((f x) # (map f xs))" by simp
also have "... = (P (f x) ∨ algunos P (map f xs))" by simp
also have "... = (P (f x) ∨ algunos (P ∘ f) xs)" using HI by simp
also have "... = ((P ∘ f) x ∨ algunos (P ∘ f) xs)" by simp
finally show "algunos P (map f (x # xs)) = algunos (P ∘ f) (x # xs)" by simp
qed

(* wilmorort pablucoto*)
lemma "algunos P (map f xs) = algunos (P o f) xs" (is "?P xs" )
proof (induct xs)
show "?P []" by simp
next
fix a xs
assume HI: "algunos P (map f xs) = algunos (P ∘ f) xs"
have "algunos P (map f (a # xs)) = (P (f a) ∨ algunos P (map f xs))" by simp
also have "... = (P (f a) ∨ algunos (P ∘ f) xs )" using HI by simp
also have "... = ((P ∘ f) a ∨ algunos (P ∘ f)xs)" by simp
finally show " algunos P (map f (a # xs)) = algunos (P ∘ f) (a # xs)" by simp
qed

(* fraortmoy *)

lemma "algunos P (map f xs) = algunos (P o f) xs"
proof (induct xs)
show "algunos P (map f []) = algunos (P ∘ f) []" by simp
next
fix a xs
assume H1: "algunos P (map f xs) = algunos (P ∘ f) xs"
have "algunos P (map f (a # xs)) = (algunos P (map f [a]) ∨ algunos P (map f xs))" by simp
also have "… = (algunos (P ∘ f) [a] ∨ algunos (P ∘ f) xs )" using H1 by simp
also have "… = algunos (P ∘ f) (a#xs)" by simp
finally show "algunos P (map f (a # xs)) = algunos (P ∘ f) (a#xs)" by simp
qed

{*danrodcha*}
lemma "algunos P (map f xs) = algunos (P o f) xs" (is "?Q xs")
proof (induct xs)
show "?Q []" by simp
next
fix a xs assume HI: "?Q xs"
have " algunos P (map f (a # xs)) = algunos P ((f a)#(map f xs))" by simp
also have "… = (P (f a) ∨ algunos P (map f xs))" by simp
also have "… = (P (f a) ∨ algunos (P ∘ f) xs)" using HI by blast
also have "… = ((P ∘ f) a ∨ algunos (P ∘ f) xs)" by simp
also have "… = algunos (P ∘ f) (a # xs)" by simp
finally show "?Q (a # xs)" by blast
qed

(* dancorgar *)
lemma "algunos P (map f xs) = algunos (P o f) xs"
proof (induct xs)
show "algunos P (map f []) = algunos (P o f) []" by simp
next
fix x xs
assume HI: "algunos P (map f xs) = algunos (P o f) xs"
have "algunos P (map f (x#xs)) = ((P (f x)) ∨ (algunos P (map f xs)))" by simp
also have "… = ((P (f x)) ∨ (algunos (P o f) xs))" using HI by simp
finally show "algunos P (map f (x#xs)) = algunos (P o f) (x#xs)" by simp
qed

text {*
---------------------------------------------------------------------
Ejercicio 8.1. Demostrar o refutar automáticamente
algunos P (xs @ ys) = (algunos P xs ∨ algunos P ys)
---------------------------------------------------------------------
*}

(* ivamenjim marpoldia1 paupeddeg crigomgom rubgonmar wilmorort fraortmoy danrodcha pablucoto dancorgar *)

lemma "algunos P (xs @ ys) = (algunos P xs ∨ algunos P ys)"
by (induct xs) auto

(* ferrenseg *)
lemma "algunos P (xs @ ys) = (algunos P xs ∨ algunos P ys)"
by (induct xs) simp_all

text {*
---------------------------------------------------------------------
Ejercicio 8.2. Demostrar o refutar detalladamente
algunos P (xs @ ys) = (algunos P xs ∨ algunos P ys)
---------------------------------------------------------------------
*}

(* ivamenjim ferrenseg marpoldia1 wilmorort dancorgar *)

lemma algunos_append:
"algunos P (xs @ ys) = (algunos P xs ∨ algunos P ys)"
proof (induct xs)
show "algunos P ([] @ ys) = (algunos P [] ∨ algunos P ys)" by simp
next
fix a xs
assume HI: "algunos P (xs @ ys) = (algunos P xs ∨ algunos P ys)"
have "algunos P ((a # xs) @ ys) = (P a ∨ (algunos P (xs @ ys)))" by simp
also have "... = (P a ∨ (algunos P xs ∨ algunos P ys))" using HI by simp
finally show "algunos P ((a # xs) @ ys) = (algunos P (a # xs) ∨ algunos P ys)" by simp
qed

(* migtermor crigomgom *)

lemma algunos_append2:
"algunos P (xs @ ys) = (algunos P xs ∨ algunos P ys)" (is "?Q xs")
proof (induct xs)
show "?Q []" by simp
next
fix x xs
assume HI: "?Q xs"
have "algunos P ((x#xs) @ ys) = algunos P (x#(xs @ ys))" by simp
also have "… = ((P x) ∨ (algunos P (xs @ ys)))" by simp
also have "… = ((P x) ∨ (algunos P xs) ∨ (algunos P ys))" using HI by simp
also have "… = ((algunos P (x#xs)) ∨ (algunos P ys))" by simp
finally show "algunos P ((x#xs) @ ys) = (algunos P (x#xs) ∨ algunos P ys)" by simp
qed

(* paupeddeg*)
lemma algunos_append3:
"algunos P (xs @ ys) = (algunos P xs ∨ algunos P ys)"
proof (induct xs)
show "algunos P ([] @ ys) = (algunos P [] ∨ algunos P ys)" by simp
next
fix a xs
assume HI: "algunos P (xs @ ys) = (algunos P xs ∨ algunos P ys)"
have "algunos P ((a # xs) @ ys) = (algunos P [a] ∨ ( algunos P (xs @ ys)))" by simp
also have "... = (algunos P [a] ∨ (algunos P xs ∨ algunos P ys))" using HI by simp
finally show "algunos P ((a # xs) @ ys) = (algunos P (a # xs) ∨ algunos P ys)" by simp
qed

(* fraortmoy *)
lemma algunos_append4:
"algunos P (xs @ ys) = (algunos P xs ∨ algunos P ys)"
proof (induct xs)
show " algunos P ([] @ ys) = (algunos P [] ∨ algunos P ys)" by simp
next
fix a xs
assume H1: "algunos P (xs @ ys) = (algunos P xs ∨ algunos P ys)"
have "algunos P ((a # xs) @ ys) = (algunos P [a] ∨ algunos P (xs @ ys))" by simp
also have "… = (algunos P [a] ∨ algunos P xs ∨ algunos P ys)" using H1 by simp
also have "… = ((algunos P [a] ∨ algunos P xs) ∨ algunos P ys)" by simp
also have "… = (algunos P (a#xs) ∨ algunos P ys)" by simp
finally show "algunos P ((a # xs) @ ys) = (algunos P (a#xs) ∨ algunos P ys)" by simp
qed

{* danrodcha pablucoto*}
lemma algunos_append5:
"algunos P (xs @ ys) = (algunos P xs ∨ algunos P ys)" (is "?Q xs")
proof (induct xs)
show "?Q []" by simp
next
fix a xs assume HI: "?Q xs"
have "algunos P ((a#xs) @ ys) = algunos P (a#(xs @ ys))" by simp
also have "… = (algunos P [a] ∨ algunos P (xs @ ys))" by simp
also have "… = (algunos P [a] ∨ algunos P xs ∨ algunos P ys)" using HI by simp
also have "… = (algunos P (a # xs) ∨ algunos P ys)" by simp
finally show "?Q (a # xs)" by simp
qed

text {*
---------------------------------------------------------------------
Ejercicio 9.1. Demostrar o refutar automáticamente
algunos P (rev xs) = algunos P xs
---------------------------------------------------------------------
*}

(* ivamenjim migtermor marpoldia1 rubgonmar paupeddeg dancorgar *)

lemma "algunos P (rev xs) = algunos P xs"
apply (induct xs)
apply simp
apply (simp add: algunos_append)
apply auto
done

(* ferrenseg crigomgom danrodcha pablucoto*)
lemma "algunos P (rev xs) = algunos P xs"
by (induct xs) (auto simp add: algunos_append)

(* wilmorort *)
lemma "algunos P (rev xs) = algunos P xs"
by (induct xs,simp,simp add: algunos_append,auto)

text {*
---------------------------------------------------------------------
Ejercicio 9.2. Demostrar o refutar detalladamente
algunos P (rev xs) = algunos P xs
---------------------------------------------------------------------
*}

lemma "algunos P (rev xs) = algunos P xs"
oops

(* migtermor *)
lemma auxiliar1:
"rev (a#xs) = rev xs @ [a]"
by auto

lemma "algunos P (rev xs) = algunos P xs" (is "?Q xs")
proof (induct xs)
show "?Q []" by simp
next
fix x xs
assume HI: "?Q xs"
have "algunos P (rev (x#xs)) = (algunos P (rev xs @ [x]))" using auxiliar1 by simp
also have "… = ((algunos P (rev xs)) ∨ (algunos P [x]))" by (simp add: algunos_append)
also have "… = ((P x) ∨ (algunos P (rev xs)))" by simp arith
also have "… = ((P x) ∨ (algunos P xs))" using HI by simp
finally show "algunos P (rev (x#xs)) = (algunos P (x#xs))" by simp
qed

(* ferrenseg *)

lemma "algunos P (rev xs) = algunos P xs"
proof (induct xs)
show "algunos P (rev []) = algunos P []" by simp
next
fix x xs
assume HI: "algunos P (rev xs) = algunos P xs"
show "algunos P (rev (x # xs)) = algunos P (x # xs)"
proof -
have "algunos P (rev (x # xs)) = algunos P ((rev xs) @ [x])" by simp
also have "… = (algunos P (rev xs) ∨ algunos P [x])"
by (simp add: algunos_append)
also have "… = (algunos P xs ∨ algunos P [x])" using HI by simp
also have "… = (algunos P xs ∨ P x)" by simp
also have "… = (P x ∨ algunos P xs)" by auto
also have "… = algunos P (x # xs)" by simp
finally show ?thesis
qed
qed

(* ivamenjim marpoldia1*)

lemma "algunos P (rev xs) = algunos P xs" (is "?P xs")
proof (induct xs)
show "?P []" by simp
next
fix a xs
assume HI: "?P xs"
have "algunos P (rev (a # xs)) = (algunos P ((rev xs)@[a]))" by simp
also have "... = ((algunos P (rev xs)) ∨ algunos P [a])" by (simp add: algunos_append)
also have "... = (algunos P xs ∨ algunos P [a])" using HI by simp
also have "... = (algunos P [a] ∨ algunos P xs)" by arith
finally show "algunos P (rev (a # xs)) = algunos P (a # xs)" by simp
qed

(* paupeddeg dancorgar *)
lemma "algunos P (rev xs) = algunos P xs"
proof (induct xs)
show "algunos P (rev []) = algunos P []" by simp
next
fix a xs
assume HI:"algunos P (rev xs) = algunos P xs"
have "algunos P (rev (a # xs)) = algunos P ((rev xs) @ [a])" by simp
also have "... = ((algunos P (rev xs)) ∨ (algunos P [a]))" by (simp add: algunos_append)
also have "... = ((algunos P xs) ∨ (algunos P [a]))" using HI by simp
also have "... = ((algunos P [a]) ∨ (algunos P xs))" by arith
finally show "algunos P (rev (a # xs)) = algunos P (a # xs)" by simp
qed

(*crigomgom pablucoto*)
lemma "algunos P (rev xs) = algunos P xs"
proof (induct xs)
show "algunos P (rev []) = algunos P []" by simp
next
fix x xs
assume HI: " algunos P (rev xs) = algunos P xs"
have "algunos P (rev (x # xs)) = algunos P ((rev xs) @ [x])" by simp
also have "... = (algunos P (rev xs) ∨ algunos P [x])" by (simp add: algunos_append)
also have "... = (algunos P xs ∨ algunos P [x])" using HI by simp
also have "... = (algunos P xs ∨ P x)" by simp
also have "... = (P x ∨ algunos P xs)" by arith
also have "... = algunos P (x#xs)" by simp
finally show "algunos P (rev (x # xs)) = algunos P (x # xs)" by simp
qed

(*wilmorort*)
lemma "algunos P (rev xs) = algunos P xs" (is "?P xs")
proof (induct xs)
show "?P []" by simp
next
fix a xs
assume HI: "?P xs"
have "algunos P (rev (a#xs)) = algunos P ((rev xs) @ [a])" by simp
also have "... = (algunos P(rev xs) ∨ algunos P [a])" by (simp add: algunos_append)
also have "... = (algunos P xs ∨ algunos P [a])" using HI by simp
also have "... = (algunos P [a] ∨ algunos P xs)" by (simp add: HOL.disj_commute)
also have "... = algunos P([a]@(xs))" by (simp)
finally show "algunos P (rev (a#xs))= algunos P (a # xs)" by simp
qed

{*danrodcha *}
lemma "algunos P (rev xs) = algunos P xs" (is "?Q xs")
proof (induct xs)
show "?Q []" by simp
next
fix a xs assume HI: "?Q xs"
have "algunos P (rev (a#xs)) = algunos P ((rev xs) @ [a])" by simp
also have "… = (algunos P (rev xs) ∨ algunos P [a])" by (simp add:algunos_append)
also have "… = (algunos P xs ∨ algunos P [a])" using HI by simp
also have "… = (algunos P [a] ∨ algunos P xs)" by (simp add: HOL.disj_commute)
also have "… = algunos P (a # xs)" by simp
finally show "?Q (a # xs)" by simp
qed

text {*
---------------------------------------------------------------------
Ejercicio 10. Encontrar un término no trivial Z tal que sea cierta la
siguiente ecuación:
algunos (λx. P x ∨ Q x) xs = Z
y demostrar la equivalencia de forma automática y detallada.
---------------------------------------------------------------------
*}

(* migtermor *)

lemma "algunos (λx. P x ∨ Q x) xs = ((algunos (λx. P x) xs) ∨ (algunos (λx. Q x) xs))"
by (induct xs) auto

lemma "algunos (λx. P x ∨ Q x) xs = ((algunos (λx. P x) xs) ∨ (algunos (λx. Q x) xs))" (is "?R xs")
proof (induct xs)
show "?R []" by simp
next
fix x xs
assume HI: "?R xs"
have "algunos (λx. P x ∨ Q x) (x#xs) = (((λx. P x ∨ Q x) x) ∨ (algunos (λx. P x ∨ Q x) xs))"
by simp
also have "… = ((P x ∨ Q x) ∨ (algunos (λx. P x ∨ Q x) xs))" by simp
also have H1: "… = ((((λx. P x) x) ∨ (algunos (λx. P x) xs)) ∨ (((λx. Q x) x) ∨ (algunos (λx. Q x) xs)))"
using HI by simp arith
have H2: "… = ((algunos (λx. P x) (x#xs)) ∨ (algunos (λx. Q x) (x#xs)))"
by simp
have C: "(algunos (λx. P x ∨ Q x) (x#xs)) =
((algunos (λx. P x) (x#xs)) ∨ (algunos (λx. Q x) (x#xs)))"
using H1 H2 by simp
finally show "(algunos (λx. P x ∨ Q x) (x#xs)) =
((algunos (λx. P x) (x#xs)) ∨ (algunos (λx. Q x) (x#xs)))"
using C by simp
qed

(* ferrenseg *)

lemma "algunos (λx. P x ∨ Q x) xs = (algunos P xs ∨ algunos Q xs)"
proof (induct xs)
show "algunos (λx. P x ∨ Q x) [] = (algunos P [] ∨ algunos Q [])" by simp
next
fix x xs
assume HI: "algunos (λx. (P x ∨ Q x)) xs = (algunos P xs ∨ algunos Q xs)"
show "algunos (λx. P x ∨ Q x) (x # xs) = (algunos P (x # xs) ∨ algunos Q (x # xs))"
proof -
have "algunos (λx. P x ∨ Q x) (x # xs) =
((P x) ∨ (Q x) ∨ algunos (λx. P x ∨ Q x) xs)" by simp
also have "… = ((P x) ∨ (Q x) ∨ algunos P xs ∨ algunos Q xs)"
using HI by simp
also have "… = (((P x) ∨ algunos P xs) ∨ ((Q x) ∨ algunos Q xs))" by auto
also have "… = (algunos P (x # xs) ∨ algunos Q (x # xs))" by simp
finally show ?thesis
qed
qed

(* ivamenjim marpoldia1 paupeddeg *)

lemma "algunos (λx. P x ∨ Q x) xs = (algunos P xs ∨ algunos Q xs)"
by (induct xs) auto

lemma "algunos (λx. P x ∨ Q x) xs = (algunos P xs ∨ algunos Q xs)"
proof (induct xs)
show "algunos (λx. P x ∨ Q x) [] = (algunos P [] ∨ algunos Q [])" by simp
next
fix a xs
assume HI: "algunos (λx. P x ∨ Q x) xs = (algunos P xs ∨ algunos Q xs)"
have "algunos (λx. P x ∨ Q x) (a # xs) = (P a ∨ Q a ∨ algunos (λx. P x ∨ Q x) xs)" by simp
also have "... = (P a ∨ Q a ∨ algunos P xs ∨ algunos Q xs)" using HI by simp
finally show "algunos (λx. P x ∨ Q x) (a # xs) = (algunos P (a # xs) ∨ algunos Q (a # xs))" by auto
qed

(* wilmorort danrodcha*)
lemma "algunos (λx. P x ∨ Q x) xs = (¬todos(λx. ¬P x)xs ∨ ¬todos(λx. ¬Q x)xs)"
by (induct xs) auto

(* wilmorort *)
lemma "algunos (λx. P x ∨ Q x) xs = (¬todos(λx. ¬P x)xs ∨ ¬todos(λx. ¬Q x)xs)"
(is "?P xs")
proof (induct xs)
show "?P []" by simp
next
fix a xs
assume HI: "?P xs"
have "algunos (λx. P x ∨ Q x) (a # xs) = (P a ∨ Q a ∨ algunos (λx. P x ∨ Q x) xs)" by simp
also have "... = (P a ∨ Q a ∨ (¬ todos (λx. ¬ P x) xs ∨ ¬ todos (λx. ¬ Q x) xs))" using HI by simp
also have "... = (P a ∨ ¬ todos (λx. ¬ P x) xs ∨ Q a ∨ ¬ todos (λx. ¬ Q x) xs )" by arith
finally show "algunos (λx. P x ∨ Q x) (a # xs) = (~ todos (λx. ¬ P x) (a # xs) ∨ ~ todos (λx. ¬ Q x) (a # xs))" by simp
qed

{* danrodcha *}
lemma "algunos (λx. P x ∨ Q x) xs = (algunos P xs ∨ algunos Q xs)" (is "?R xs")
proof (induct xs)
show "?R []" by simp
next
fix a xs assume HI: "?R xs"
have 1:" (Q a ∨ algunos P xs) = (algunos P xs ∨ Q a)" by (simp add: HOL.disj_commute)
have "algunos (λx. P x ∨ Q x) (a # xs) =
(algunos P [a] ∨ algunos Q [a] ∨ algunos (λx. P x ∨ Q x) xs)" by simp
also have "… = (P a ∨ (Q a ∨ algunos P xs) ∨ algunos Q xs)" using HI by simp
also have "… = (P a ∨ algunos P xs ∨ Q a ∨ algunos Q xs)" using 1 by simp
also have "… = (algunos P (a # xs) ∨ algunos Q (a # xs))" by simp
finally show "?R (a # xs)" by simp
qed

(* dancorgar *)
lemma "algunos (λx. P x ∧ Q x) xs ⟹ (algunos P xs ∧ algunos Q xs)"
apply (induct xs)
apply auto
done

(* dancorgar *)
lemma "algunos (λx. P x ∧ Q x) xs ⟹ (algunos P xs ∧ algunos Q xs)"
proof (induct xs)
show "algunos (λx. P x ∧ Q x) [] ⟹ (algunos P [] ∧ algunos Q [])" by simp
next
fix xf xs
assume HI: "algunos (λx. P x ∧ Q x) xs ⟹ (algunos P xs ∧ algunos Q xs)"
have F1: "algunos (λx. P x ∧ Q x) (xf#xs) ⟹ ((P xf ∧ Q xf) ∨ algunos (λx. P x ∧ Q x) xs)" by simp
also have F2: "… ⟹ (P xf ∧ Q xf ∨ (algunos P xs ∧ algunos Q xs))" using HI by simp
have F3: "((P xf ∧ Q xf) ∨ (algunos P xs ∧ algunos Q xs)) ⟹
((P xf ∨ algunos P xs) ∧ (Q xf ∨ algunos Q xs))" by blast
have F4: "((P xf ∨ algunos P xs) ∧ (Q xf ∨ algunos Q xs)) ⟹
(algunos P (xf#xs) ∧ algunos Q (xf#xs))" by simp
show "algunos (λx. P x ∧ Q x) (xf#xs) ⟹ (algunos P (xf#xs) ∧ algunos Q (xf#xs))"
using F1 F2 F3 F4 by blast
qed

(*pablucoto*)
-- "Automatica"
lemma "algunos (λx. P x ∨ Q x) xs = (algunos P xs ∨ algunos Q xs)"
by (induct xs) auto
-- "Detallada"
lemma "algunos (λx. P x ∨ Q x) xs = (algunos P xs ∨ algunos Q xs) "
proof (induct xs)
show "algunos (λx. P x ∨ Q x) [] = (algunos P [] ∨ algunos Q []) " by simp
next
fix a xs
assume HI: " algunos (λx. P x ∨ Q x) xs = (algunos P xs ∨ algunos Q xs)"
have " algunos (λx. P x ∨ Q x) (a # xs) = ( P a ∨ Q a ∨ algunos (λx. P x ∨ Q x) xs)" by simp
also have "... = (P a ∨ Q a ∨ (algunos P xs ∨ algunos Q xs))" using HI by simp
also have "… = (((P a) ∨ algunos P xs) ∨ ((Q a) ∨ algunos Q xs))" by arith
also have "… = (algunos P (a#xs) ∨ algunos Q (a#xs))" by simp
finally show "algunos (λx. P x ∨ Q x) (a # xs) = (algunos P (a # xs) ∨ algunos Q (a # xs)) " by simp
qed

text {*
---------------------------------------------------------------------
Ejercicio 11.1. Demostrar o refutar automáticamente
algunos P xs = (¬ todos (λx. (¬ P x)) xs)
---------------------------------------------------------------------
*}

(* ivamenjim wilmorort migtermor marpoldia1 crigomgom rubgonmar paupeddeg danrodcha pablucoto dancorgar *)

lemma "algunos P xs = (¬ todos (λx. (¬ P x)) xs)"
by (induct xs) auto

(* ferrenseg *)

lemma "algunos P xs = (¬ todos (λx. (¬ P x)) xs)"
by (induct xs) simp_all

text {*
---------------------------------------------------------------------
Ejercicio 11.2. Demostrar o refutar datalladamente
algunos P xs = (¬ todos (λx. (¬ P x)) xs)
---------------------------------------------------------------------
*}

(* ivamenjim migtermor marpoldia1 crigomgom paupeddeg*)

lemma "algunos P xs = (¬ todos (λx. (¬ P x)) xs)"
proof (induct xs)
show "algunos P [] = (¬ todos (λx. ¬ P x) [])" by simp
next
fix a xs
assume HI: "algunos P xs = (¬ todos (λx. ¬ P x) xs)"
have "algunos P (a # xs) = (P a ∨ (algunos P xs))" by simp
also have "... = (P a ∨ (¬ todos (λx. ¬ P x) xs))" using HI by simp
finally show "algunos P (a # xs) = (¬ todos (λx. ¬ P x) (a # xs))" by simp
qed

(* ferrenseg *)

lemma "algunos P xs = (¬ todos (λx. (¬ P x)) xs)"
proof (induct xs)
show "algunos P [] = (¬ todos (λx. (¬ P x)) [])" by simp
next
fix x xs
assume HI: "algunos P xs = (¬ todos (λx. (¬ P x)) xs)"
show "algunos P (x # xs) = (¬ todos (λx. (¬ P x)) (x # xs))"
proof -
have "algunos P (x # xs) = ((P x) ∨ algunos P xs)" by simp
also have "… = ((P x) ∨ ¬ todos (λx. (¬ P x)) xs)" using HI by simp
also have "… = (¬ (¬ (P x) ∧ todos (λx. (¬ P x)) xs))" by simp
also have "… = (¬ todos (λx. (¬ P x)) (x # xs))" by simp
finally show ?thesis
qed
qed

{* danrodcha *}
lemma "algunos P xs = (¬ todos (λx. (¬ P x)) xs)" (is "?Q xs")
proof (induct xs)
show "?Q []" by simp
next
fix a xs assume HI: "?Q xs"
have "algunos P (a # xs) = (P a ∨ algunos P xs)" by simp
also have "… = (P a ∨ (¬ todos (λx. (¬ P x)) xs))" using HI by simp
also have "… = (¬(¬(P a) ∧ todos(λx. (¬ P x)) xs))" by simp
also have "… = (¬ todos(λx. (¬ P x)) (a # xs))" by simp
finally show "?Q (a # xs)" by simp
qed

(* dancorgar *)
lemma "algunos P xs = (¬ todos (λx. (¬ P x)) xs)"
proof (induct xs)
show "algunos P [] = (¬ todos (λx. (¬ P x)) [])" by simp
next
fix xf xs
assume HI: "algunos P xs = (¬ todos (λx. (¬ P x)) xs)"
have "algunos P (xf#xs) = ((P xf) ∨ algunos P xs)" by simp
also have "… = ( (P xf) ∨ (¬ todos (λx. (¬ P x)) xs) )" using HI by simp
also have "… = (¬((¬P xf) ∧ (todos (λx. (¬ P x)) xs)))" by simp
finally show "algunos P (xf#xs) = (¬todos (λx. (¬ P x)) (xf#xs))" by simp
qed

text {*
---------------------------------------------------------------------
Ejercicio 12. Definir la funcion primitiva recursiva
estaEn :: 'a ⇒ 'a list ⇒ bool
tal que (estaEn x xs) se verifica si el elemento x está en la lista
xs. Por ejemplo,
estaEn (2::nat) [3,2,4] = True
estaEn (1::nat) [3,2,4] = False
---------------------------------------------------------------------
*}

(* ivamenjim ferrenseg migtermor serrodcal crigomgom rubgonmar marpoldia1 paupeddeg danrodcha dancorgar *)

fun estaEn :: "'a ⇒ 'a list ⇒ bool" where
"estaEn x [] = False"
| "estaEn x (a#xs) = ((a = x) ∨ (estaEn x xs))"

value "estaEn (2::nat) [3,2,4] = True"
value "estaEn (1::nat) [3,2,4] = False"

text {*
---------------------------------------------------------------------
Ejercicio 13. Expresar la relación existente entre estaEn y algunos.
Demostrar dicha relación de forma automática y detallada.
---------------------------------------------------------------------
*}

(* migtermor crigomgom *)

lemma "estaEn x xs = (algunos (λa. a=x) xs)"
by (induct xs) auto

lemma auxiliar13:
"(x=a) = ((λa. a=x) a)"
by auto

lemma "estaEn x xs = (algunos (λa. a=x) xs)" (is "?P xs")
proof (induct xs)
show "?P []" by simp
next
fix a xs
assume HI: "?P xs"
have "estaEn x (a#xs) = ((a=x) ∨ (estaEn x xs))" by simp
also have H: "… = ((a=x) ∨ (algunos (λa. a=x) xs))" using HI by simp
also have "… = (((λa. a=x) a) ∨ (algunos (λa. a=x) xs))" using auxiliar13 by simp
also have "… = (algunos (λa. a=x) (a#xs))" by simp
have C: "estaEn x (a#xs) = (algunos (λa. a=x) (a#xs))" using H by simp
finally show "estaEn x (a#xs) = (algunos (λa. a=x) (a#xs))" using C by simp
qed

(* ferrenseg *)

lemma estaEn_algunos:
"estaEn y xs = algunos (λx. x=y) xs"
by (induct xs) auto

lemma estaEn_algunos_2:
"estaEn y xs = algunos (λx. x=y) xs"
proof (induct xs)
show "estaEn y [] = algunos (λx. x=y) []" by simp
next
fix x xs
assume HI: "estaEn y xs = algunos (λx. x=y) xs"
show "estaEn y (x # xs) = algunos (λx. x=y) (x # xs)"
proof -
have "estaEn y (x # xs) = (y=x ∨ estaEn y xs)" by simp
also have "… = (y=x ∨ algunos (λx. x=y) xs)" using HI by simp
also have "… = (x=y ∨ algunos (λx. x=y) xs)" by auto
also have "… = algunos (λx. x=y) (x # xs)" by simp
finally show ?thesis
qed
qed

(* crigomgom paupeddeg dancorgar *)

lemma "estaEn a xs = algunos (λx. x = a) xs"
by (induct xs) auto

lemma "estaEn x xs = (algunos (λa. a=x) xs)"
proof (induct xs)
show " estaEn x [] = algunos (λa. a = x) []" by simp
next
fix a xs
assume HI: "estaEn x xs = algunos (λa. a = x) xs"
have "estaEn x (a # xs) = (a = x ∨ estaEn x xs)" by simp
also have "... = (a = x ∨ algunos (λa. a = x) xs)" using HI by simp
finally show " estaEn x (a # xs) = algunos (λa. a = x) (a # xs)" by simp
qed

(* ivamenjim marpoldia1 *)

lemma "estaEn x xs = algunos (λy. x=y) xs"
by (induct xs) auto

lemma "estaEn x xs = algunos (λy. x=y) xs"
proof (induct xs)
show "estaEn x [] = algunos (op = x) []" by simp
next
fix a xs
assume HI: "estaEn x xs = algunos (op = x) xs"
have "estaEn x (a # xs) = ((a = x) ∨ (estaEn x xs))" by simp
also have "... = (a = x ∨ algunos (op = x) xs)" using HI by simp
finally show " estaEn x (a # xs) = algunos (op = x) (a # xs)" by auto
qed

{* danrodcha *}
lemma "estaEn a xs = algunos (op = a) xs"
by (induct xs) auto

{* danrodcha *}
lemma "estaEn a xs = algunos (op = a) xs"
proof (induct xs)
show "estaEn a [] = algunos (op = a) []" by simp
next
fix x xs
assume HI: "estaEn a xs = algunos (op = a) xs"
have "estaEn a (x # xs) = ((x = a) ∨ (estaEn a xs))" by simp
also have "… = (x = a ∨ algunos (op = a) xs)" using HI by simp
also have "… = ((op = a) x ∨ algunos (op = a) xs)" by (simp add:HOL.eq_commute)
also have "… = algunos (op = a) (x # xs)" by simp
finally show "estaEn a (x # xs) = algunos (op = a) (x # xs)" by simp
qed

end
</source>

Relación 4

2016-11-23T00:42:50Z

Pablucoto:

<source lang="isar">
chapter {* R4: Cuantificadores sobre listas *}

theory R4_Cuantificadores_sobre_listas
imports Main
begin

text {*
---------------------------------------------------------------------
Ejercicio 1. Definir la función
todos :: ('a ⇒ bool) ⇒ 'a list ⇒ bool
tal que (todos p xs) se verifica si todos los elementos de la lista
xs cumplen la propiedad p. Por ejemplo, se verifica
todos (λx. 1<length x) [[2,1,4],[1,3]]
¬todos (λx. 1<length x) [[2,1,4],[3]]

Nota: La función todos es equivalente a la predefinida list_all.
---------------------------------------------------------------------
*}

{*danrodcha*}
fun todos :: "('a ⇒ bool) ⇒ 'a list ⇒ bool" where
"todos p xs = "todos p [] = True"
| "todos p (x#xs) = (p x ∧ todos p xs)""

(* ivamenjim migtermor dancorgar wilmorort marpoldia1 ferrenseg paupeddeg pablucoto crigomgom anaprarod serrodcal juacabsou rubgonmar josgarsan fraortmoy lucnovdos*)
fun todos :: "('a ⇒ bool) ⇒ 'a list ⇒ bool" where
"todos p [] = True"
| "todos p (x#xs) = (p x ∧ todos p xs)"

text {*
---------------------------------------------------------------------
Ejercicio 2. Definir la función
algunos :: ('a ⇒ bool) ⇒ 'a list ⇒ bool
tal que (algunos p xs) se verifica si algunos elementos de la lista
xs cumplen la propiedad p. Por ejemplo, se verifica
algunos (λx. 1<length x) [[2,1,4],[3]]
¬algunos (λx. 1<length x) [[],[3]]"

Nota: La función algunos es equivalente a la predefinida list_ex.
---------------------------------------------------------------------
*}

{*danrodcha ivamenjim migtermor dancorgar marpoldia1 ferrenseg wilmorort paupeddeg pablucoto crigomgom anaprarod serrodcal juacabsou rubgonmar josgarsan fraortmoy lucnovdos*}
fun algunos :: "('a ⇒ bool) ⇒ 'a list ⇒ bool" where
"algunos p [] = False"
| "algunos p (x#xs) = (p x ∨ algunos p xs)"

text {*
---------------------------------------------------------------------
Ejercicio 3.1. Demostrar o refutar automáticamente
todos (λx. P x ∧ Q x) xs = (todos P xs ∧ todos Q xs)
---------------------------------------------------------------------
*}

{*danrodcha ivamenjim migtermor dancorgar marpoldia1 ferrenseg wilmorort paupeddeg pablucoto anaprarod serrodcal juacabsou rubgonmar josgarsan fraortmoy lucnovdos*}
lemma "todos (λx. P x ∧ Q x) xs = (todos P xs ∧ todos Q xs)"
by (induct xs) auto

text {*
---------------------------------------------------------------------
Ejercicio 3.2. Demostrar o refutar detalladamente
todos (λx. P x ∧ Q x) xs = (todos P xs ∧ todos Q xs)
---------------------------------------------------------------------
*}

{*danrodcha*}
lemma "todos (λx. P x ∧ Q x) xs = (todos P xs ∧ todos Q xs)"
proof (induct xs)
show "?R []" by simp
next
fix a xs
assume HI: "?R xs"
have "todos (λx. P x ∧ Q x) (a#xs) = (P a ∧ Q a ∧ todos (λx. P x ∧ Q x) xs)" by simp
also have "… = (P a ∧ Q a ∧ (todos P xs ∧ todos Q xs))" using HI by simp
also have "… = ((P a ∧ todos P xs) ∧ (Q a ∧ todos Q xs))" by blast
also have "… = (todos P (a#xs) ∧ todos Q (a#xs))" by simp
finally show "?R (a#xs)" by simp
qed

(* ivamenjim wilmorort serrodcal josgarsan*)

lemma "todos (λx. P x ∧ Q x) xs = (todos P xs ∧ todos Q xs)"
proof (induct xs)
show "todos (λx. P x ∧ Q x) [] = (todos P [] ∧ todos Q [])" by simp
next
fix a xs
assume HI: "todos (λx. P x ∧ Q x) xs = (todos P xs ∧ todos Q xs)"
have "todos (λx. P x ∧ Q x) (a # xs) = (P a ∧ Q a ∧ todos (λx. P x ∧ Q x) xs)" by simp
also have "... = (P a ∧ Q a ∧ todos P xs ∧ todos Q xs)" using HI by simp
finally show "todos (λx. P x ∧ Q x) (a # xs) = (todos P (a # xs) ∧ todos Q (a # xs))" by auto
qed

(* dancorgar paupeddeg *)
lemma "todos (λx. P x ∧ Q x) xs = (todos P xs ∧ todos Q xs)"
proof (induct xs)
show "todos (λx. P x ∧ Q x) [] = (todos P [] ∧ todos Q [])" by simp
next
fix y xs
assume HI: "todos (λx. P x ∧ Q x) xs = (todos P xs ∧ todos Q xs)"
have "todos (λx. P x ∧ Q x) (y#xs) = ((P y ∧ Q y) ∧ (todos (λx. P x ∧ Q x) xs))" by simp
also have "… = ((P y ∧ Q y) ∧ (todos P xs ∧ todos Q xs))" using HI by simp
also have "… = ((P y ∧ todos P xs) ∧ (Q y ∧ todos Q xs))" by blast
also have "… = ((todos P (y#xs)) ∧ (todos Q (y#xs)))" by simp
finally show "todos (λx. P x ∧ Q x) (y#xs) = (todos P (y#xs) ∧ todos Q (y#xs))" by simp
qed

(* migtermor *)
lemma "todos (λx. P x ∧ Q x) xs = (todos P xs ∧ todos Q xs)"
proof (induct xs)
show "todos (λx. P x ∧ Q x) [] = (todos P [] ∧ todos Q [])" by simp
next
fix x xs
assume HI: "todos (λx. P x ∧ Q x) xs = (todos P xs ∧ todos Q xs)"
have "(todos (λx. P x ∧ Q x) (x#xs)) = (( P x ∧ Q x) ∧ (todos (λx. P x ∧ Q x) xs))"
by (simp only: todos.simps(2))
also have "… = ((P x ∧ Q x) ∧ (todos P xs ∧ todos Q xs))" using HI by simp
also have "… = (((P x)∧(todos P xs)) ∧ ((Q x) ∧ (todos Q xs)))" by arith
also have "((P x ∧ Q x) ∧ (todos P xs ∧ todos Q xs)) = (((P x)∧(todos P xs)) ∧ ((Q x) ∧ (todos Q xs)))"
by arith (* Este paso es exactamente el mismo que el anterior, pero sin cualquiera de los dos no funciona el "finally show" *)
have "… = (((P x)∧(todos P xs))∧((Q x)∧(todos Q xs)))" by simp
have "… = ((todos P (x#xs))∧(todos Q (x#xs)))" by simp
finally show "(todos (λx. P x ∧ Q x) (x#xs)) = ((todos P (x#xs)) ∧ (todos Q (x#xs)))" by simp
qed

(* marpoldia1 *)
lemma "todos (λx. P x ∧ Q x) xs = (todos P xs ∧ todos Q xs)"
proof (induct xs)
show "todos (λx. P x ∧ Q x) [] = (todos P [] ∧ todos Q [])" by simp
next
fix a xs
assume HI: "todos (λx. P x ∧ Q x) xs = (todos P xs ∧ todos Q xs)"
have "todos (λx. P x ∧ Q x) (a # xs) = ((P a ∧ Q a) ∧ (todos P xs ∧ todos Q xs))" using HI by simp
also have "... = ((P a ∧ todos P xs) ∧ (Q a ∧ todos Q xs))" by blast
also have "... = (todos P (a#xs) ∧ todos Q (a#xs)) " by simp
finally show "todos (λx. P x ∧ Q x) (a#xs) = (todos P (a#xs) ∧ todos Q (a#xs))" by simp
qed

(* ferrenseg *)

lemma "todos (λx. P x ∧ Q x) xs = (todos P xs ∧ todos Q xs)"
proof (induct xs)
show "todos (λx. P x ∧ Q x) [] = (todos P [] ∧ todos Q [])" by simp
next
fix n xs
assume HI: "todos (λx. P x ∧ Q x) xs = (todos P xs ∧ todos Q xs)"
have "todos (λx. P x ∧ Q x) (n # xs) = (P n ∧ Q n ∧ todos P xs ∧ todos Q xs)" using HI by simp
also have "… = P n ∧ Q n ∧ todos P xs ∧ todos Q xs" by blast
also have "⋯ = todos P (n # xs) ∧ todos Q (n # xs)" by simp
finally show "todos (λx. P x ∧ Q x) (n # xs) = todos P (n # xs) ∧ todos Q (n # xs)" by simp
qed

(* lucnovdos *)

lemma "todos (λx. P x ∧ Q x) xs = (todos P xs ∧ todos Q xs)"
proof (induct xs)
show "todos (λx. P x ∧ Q x) [] = (todos P [] ∧ todos Q [])" by simp
next
fix n xs
assume HI: "todos (λx. P x ∧ Q x) xs = (todos P xs ∧ todos Q xs)"
have "todos (λx. P x ∧ Q x) (n # xs) = ((P n ∧ Q n) ∧ (todos P xs ∧ todos Q xs))" using HI by simp
also have "... = ((P n ∧ todos P xs) ∧ (Q n ∧ todos Q xs))" by arith
also have "... = ((todos P(n#xs)) ∧ (todos Q(n#xs)))" by simp

finally show "todos (λx. P x ∧ Q x) (n#xs) = (todos P (n#xs) ∧ todos Q (n#xs))" by simp
qed

{* wilmorort pablucoto crigomgom anaprarod juacabsou rubgonmar fraortmoy *}

lemma "todos (λx. P x ∧ Q x) xs = (todos P xs ∧ todos Q xs)"
proof (induct xs)
show "todos (λx. P x ∧ Q x) [] = (todos P [] ∧ todos Q [])" by simp
next
fix a xs
assume HI: "todos (λx. P x ∧ Q x) xs = (todos P xs ∧ todos Q xs)"
have "todos (λx. P x ∧ Q x) (a # xs) = (P a ∧ Q a ∧ todos (λx. P x ∧ Q x) xs)" by simp
also have "... = (P a ∧ Q a ∧ todos P xs ∧ todos Q xs)" using HI by simp
also have "... = (P a ∧ todos P xs ∧ Q a ∧ todos Q xs)" by arith
also have "... = (todos P (a # xs) ∧ todos Q (a # xs))" by simp (* Este paso se puede obviar*)
finally show "todos (λx. P x ∧ Q x) (a # xs) = (todos P (a # xs) ∧ todos Q (a # xs))" by simp
qed

text {*
---------------------------------------------------------------------
Ejercicio 4.1. Demostrar o refutar automáticamente
todos P (x @ y) = (todos P x ∧ todos P y)
---------------------------------------------------------------------
*}
{*danrodcha ivamenjim marpoldia1 migtermor ferrenseg wilmorort paupeddeg crigomgom anaprarod serrodcal juacabsou rubgonmar pablucoto fraortmoy josgarsan lucnovdos*}
lemma "todos P (x @ y) = (todos P x ∧ todos P y)"
by (induct x) auto

{* anaprarod dancorgar *}
lemma "todos P (x @ y) = (todos P x ∧ todos P y)"
apply (induct x)
apply auto
done

text {*
---------------------------------------------------------------------
Ejercicio 4.2. Demostrar o refutar detalladamente
todos P (x @ y) = (todos P x ∧ todos P y)
---------------------------------------------------------------------
*}

(* ivamenjim ferrenseg wilmorort dancorgar fraortmoy josgarsan lucnovdos*)

lemma todos_append:
"todos P (x @ y) = (todos P x ∧ todos P y)"
proof (induct x)
show "todos P ([] @ y) = (todos P [] ∧ todos P y)" by simp
next
fix a x
assume HI: "todos P (x @ y) = (todos P x ∧ todos P y)"
have "todos P ((a#x) @ y) = (P a ∧ (todos P (x @ y)))" by simp
also have "... = (P a ∧ (todos P x ∧ todos P y))" using HI by simp
finally show "todos P ((a # x) @ y) = (todos P (a # x) ∧ todos P y)" by simp
qed

(* migtermor serrodcal *)

lemma todos_append1:
"todos P (x @ y) = (todos P x ∧ todos P y)" (is "?P x")
proof (induct x)
show "?P []" by simp
next
fix a x
assume HI: "?P x"
have "todos P ((a#x) @ y) = (P a ∧ (todos P (x @ y)))" by simp
also have "… = (P a ∧ (todos P x ∧ todos P y))" using HI by simp
finally show "?P (a#x)" by simp
qed

(* marpoldia1 *)
lemma todos_append:
"todos P (x @ y) = (todos P x ∧ todos P y)"
proof (induct x)
show "todos P ([] @ y) = (todos P [] ∧ todos P y)" by simp
next
fix a x
assume HI: "todos P (x @ y) = (todos P x ∧ todos P y)"
have "todos P ((a # x) @ y) = todos P (a#(x@y))" by simp
also have "... = (P a ∧ todos P (x @ y))" by simp
also have "... = (P a ∧ (todos P x ∧ todos P y))" using HI by simp
finally show "todos P ((a # x) @ y) = (todos P (a # x) ∧ todos P y)" by simp
qed

(* paupeddeg *)
lemma todos_append3:
"todos P (x @ y) = (todos P x ∧ todos P y)"
proof (induct x)
show "todos P ([] @ y) = (todos P [] ∧ todos P y)" by simp
next
fix a x
assume HI: "todos P (x @ y) = (todos P x ∧ todos P y)"
have "todos P ((a # x) @ y) = todos P (a # (x @ y))" by simp
also have "... = (todos P [a] ∧ todos P (x @ y)) " by simp
also have "... = (todos P [a] ∧ (todos P x ∧ todos P y))" using HI by simp
finally show "todos P ((a # x) @ y) = (todos P (a # x) ∧ todos P y)" by simp
qed

(* crigomgom juacabsou anaprarod*)

lemma todos_append4:
"todos P (x @ y) = (todos P x ∧ todos P y)"
proof (induct x)
show "todos P ([] @ y) = (todos P [] ∧ todos P y)" by simp
next
fix x xs
assume HI:"todos P (xs @ y) = (todos P xs ∧ todos P y)"
have "todos P ((x # xs) @ y) = (P x ∧ todos P (xs @ y ))" by simp
also have "... = (P x ∧ (todos P xs ∧ todos P y)) " using HI by simp
also have "... = ((P x ∧ todos P xs) ∧ todos P y)" by simp
finally show "todos P ((x # xs) @ y) = (todos P (x # xs) ∧ todos P y)" by simp
qed

(* pablucoto *)
lemma todos_append5:
"todos P (x @ y) = (todos P x ∧ todos P y)"
proof (induct x )
show "todos P ([] @ y) = (todos P [] ∧ todos P y)" by simp
next
fix a x
assume HI:"todos P (x @ y) = (todos P x ∧ todos P y) "
have "todos P ((a # x) @ y) = ( P a ∧ todos P (x @ y))" by simp
also have " ... =( P a ∧ (todos P x ∧ todos P y))" using HI by simp
also have "... = (todos P (a#x) ∧ todos P y)" by simp
finally show "todos P ((a # x) @ y) = (todos P (a # x) ∧ todos P y)" by auto
qed

{*danrodcha*}
lemma todos_append6:
"todos P (x @ y) = (todos P x ∧ todos P y)" (is "?Q x")
proof (induct x)
show "?Q []" by simp
next
fix x a assume HI: "?Q x"
have "todos P ((a # x) @ y) = todos P (a # x @ y)" by simp
also have "… = ((P a) ∧ todos P (x @ y))" by simp
also have "… = ((P a) ∧ (todos P x ∧ todos P y))" using HI by simp
also have "… = (todos P (a # x) ∧ todos P y)" by simp
finally show "?Q (a # x)" by simp
qed

text {*
---------------------------------------------------------------------
Ejercicio 5.1. Demostrar o refutar automáticamente
todos P (rev xs) = todos P xs
---------------------------------------------------------------------
*}

lemma "todos P (rev xs) = todos P xs"
oops

(* migtermor ivamenjim marpoldia1 serrodcal anaprarod paupeddeg *)
lemma "todos P (rev xs) = todos P xs"
apply (induct xs)
apply simp
apply (simp add: todos_append)
apply auto
done

(* ferrenseg crigomgom rubgonmar fraortmoy josgarsan danrodcha lucnovdos*)

lemma "todos P (rev xs) = todos P xs"
by (induct xs) (auto simp add: todos_append)

(* wilmorort *)
lemma "todos P (rev xs) = todos P xs"
by (induct xs,simp,simp add: todos_append,auto)

(* juacabsou *)
lemma "todos P (rev xs) = todos P xs"
apply (induct xs,simp,simp add: todos_append,auto) done

(* dancorgar *)
lemma "todos P (rev xs) = todos P xs"
apply (induct xs)
apply (simp add: todos_append)
apply (simp add: todos_append)
apply auto
done

(* pablucoto * )
lemma "todos P (rev xs) = todos P xs"
by (induct xs) (auto, simp_all add: todos_append)
lemma "todos P (rev xs) = todos P xs"
by (induct xs) ( simp_all add: todos_append, auto)

text {*
---------------------------------------------------------------------
Ejercicio 5.2. Demostrar o refutar detalladamente
todos P (rev xs) = todos P xs
---------------------------------------------------------------------
*}

lemma "todos P (rev xs) = todos P xs"
oops

(* migtermor *)

lemma auxiliar:
"rev (a#xs) = rev xs @ [a]"
by auto

lemma "todos P (rev xs) = todos P xs" (is "?Q xs")
proof (induct xs)
show "?Q []" by (simp only: rev.simps(1))
next
fix a xs
assume HI: "?Q xs"
have "todos P (rev (a#xs)) = (todos P (rev xs @ [a]))" by (simp add: auxiliar)
have "… = ((todos P (rev xs)) ∧ (todos P [a]))" by (simp add: todos_append)
have "… = (todos P (rev xs) ∧ P a)" by simp
also have Aux: "… = (todos P xs ∧ P a)" using HI by simp
have Aux1: "… = (P a ∧ todos P xs)" by arith
have "(todos P (rev xs) ∧ P a) = (P a ∧ todos P xs)" using Aux Aux1 by simp
finally show "todos P (rev (a#xs)) = todos P (a#xs)" by (simp add: todos_append)
qed

(* marpoldia1 ferrenseg crigomgom serrodcal juacabsou rubgonmar josgarsan pablucoto*)
lemma "todos P (rev xs) = todos P xs"
proof (induct xs)
show "todos P (rev []) = todos P []" by simp
next
fix a xs
assume HI: "todos P (rev xs) = todos P xs"
have "todos P (rev (a # xs)) = (todos P ((rev xs)@[a]))" by simp
also have "... = (todos P (rev xs) ∧ todos P [a])" by (simp add:todos_append)
also have "... = (todos P xs ∧ P a)" using HI by simp
also have "... = (P a ∧ todos P xs)" by arith
also have "... = (todos P (a#xs))" by simp
finally show "todos P (rev (a # xs)) = (todos P (a#xs))" by simp
qed

(* ivamenjim *)

lemma "todos P (rev xs) = todos P xs" (is "?P xs")
proof (induct xs)
show "?P []" by simp
next
fix a xs
assume HI: "?P xs"
have "todos P (rev (a # xs)) = (todos P ((rev xs)@[a]))" by simp
also have "... = ((todos P (rev xs)) ∧ todos P [a])" by (simp add: todos_append)
also have "... = (todos P xs ∧ todos P [a])" using HI by simp
also have "... = (todos P [a] ∧ todos P xs)" by arith
finally show "todos P (rev (a # xs)) = todos P (a # xs)" by simp
qed

(* fraortmoy *)

lemma "todos P (rev xs) = todos P xs"
proof (induct xs)
show "todos P (rev []) = todos P []" by simp
next
fix a xs
assume H1: "todos P (rev xs) = todos P xs"
have " todos P (rev (a # xs)) = todos P (rev xs @ [a])" by simp
also have "… = (todos P (rev xs) ∧ todos P [a])" by (simp add:todos_append)
also have "… = (todos P xs ∧ todos P [a])" using H1 by simp
also have "… = (todos P [a] ∧ todos P xs)" by arith
finally show "todos P (rev (a # xs)) = todos P (a#xs)" by simp
qed

(* paupeddeg dancorgar *)
lemma "todos P (rev xs) = todos P xs"
proof (induct xs)
show " todos P (rev []) = todos P []" by simp
next
fix a xs
assume HI: "todos P (rev xs) = todos P xs "
have "todos P (rev (a # xs)) = todos P (rev(xs) @ [a])" by simp
also have "... = (todos P (rev xs) ∧ todos P [a]) " by (simp add: todos_append)
also have "... = (todos P xs ∧ todos P [a])" using HI by simp
also have "... = (todos P [a] ∧ todos P xs)" by arith
also have "... = todos P ([a] @ xs)" by (simp add: todos_append)
finally show "todos P (rev (a # xs)) = todos P (a # xs)" by simp
qed

(* wilmorort *)
lemma "todos P (rev xs) = todos P xs" (is "?P xs")
proof (induct xs)
show "?P []" by simp
next
fix a xs
assume HI: "?P xs"
have "todos P (rev (a#xs)) = todos P ((rev xs) @ [a])" by simp
also have "... = (todos P(rev xs) ∧ todos P [a])" by (simp add: todos_append)
also have "... = (todos P xs ∧ todos P [a])" using HI by simp
also have "... = (todos P [a] ∧ todos P xs)" by (simp add: HOL.conj_commute)
also have "... = todos P([a]@(xs))" by (simp)
finally show "todos P (rev (a#xs))= todos P (a # xs)" by simp
qed

(* anaprarod *)
(* es igual que las anteriores pero con el final también con patrones *)

lemma "todos P (rev xs) = todos P xs" (is "?P xs")
proof (induct xs)
show "?P []" by auto
next
fix a xs
assume HI: "?P xs"
have "todos P (rev (a#xs)) = todos P (rev xs @ [a])" by simp
also have "… = (todos P (rev xs) ∧ todos P [a])" by (simp add: todos_append)
also have "… = (todos P xs ∧ todos P [a])" using HI by simp
also have "… = (todos P [a] ∧ todos P xs)" by arith
also have "… = todos P (a#xs)" by simp
finally show "?P (a # xs)" by simp
qed

{*danrodcha*}
lemma "todos P (rev xs) = todos P xs" (is "?Q xs")
proof (induct xs)
show "?Q []" by simp
next
fix a xs assume HI: "?Q xs"
have "todos P (rev (a#xs)) = todos P ((rev xs) @ [a])" by simp
also have "… = (todos P (rev xs) ∧ todos P [a])" by (simp add:todos_append)
also have "… = (todos P xs ∧ todos P [a])" using HI by simp
also have "… = (todos P [a] ∧ todos P xs)" by (simp add: HOL.conj_commute)
also have "… = todos P (a # xs)" by simp
finally show "?Q (a # xs)" by simp
qed

text {*
---------------------------------------------------------------------
Ejercicio 6. Demostrar o refutar:
algunos (λx. P x ∧ Q x) xs = (algunos P xs ∧ algunos Q xs)
---------------------------------------------------------------------
*}

lemma "algunos (λx. P x ∧ Q x) xs = (algunos P xs ∧ algunos Q xs)"
oops

(* migtermor ivamenjim ferrenseg paupeddeg crigomgom serrodcal wilmorort juacabsou rubgonmar anaprarod marpoldia1 fraortmoy josgarsan danrodcha pablucoto*)
lemma "algunos (λx. P x ∧ Q x) xs = (algunos P xs ∧ algunos Q xs)"
quickcheck
oops
(* Quickcheck encuentra el siguiente contraejemplo: P={a1}, Q={a2}, xs={a1,a2}. En este ejemplo:
· "algunos (λx. P x ∧ Q x) xs = False"
· "(algunos P xs ∧ algunos Q xs) = True" *)

(* dancorgar *)
lemma "algunos (λx. P x ∧ Q x) xs = (algunos P xs ∧ algunos Q xs)"
proof -
assume H1: "xs = [a, b]"
assume H2: "P a = True"
assume H3: "Q a = False"
assume H4: "P b = False"
assume H5: "Q b = True"
have F1: "(algunos P xs ∧ algunos Q xs) = True" using H1 H2 H3 H4 H5 by simp
have F2: "algunos (λx. P x ∧ Q x) xs = False" using H1 H2 H3 H4 H5 by simp
have "algunos (λx. P x ∧ Q x) xs ≠ (algunos P xs ∧ algunos Q xs)" using F1 F2 by simp
oops

text {*
---------------------------------------------------------------------
Ejercicio 7.1. Demostrar o refutar automáticamente
algunos P (map f xs) = algunos (P ∘ f) xs
---------------------------------------------------------------------
*}

(* ivamenjim migtermor marpoldia1 crigomgom rubgonmar wilmorort anaprarod
fraortmoy juacabsou paupeddeg josgarsan danrodcha pablucoto *)

lemma "algunos P (map f xs) = algunos (P o f) xs"
by (induct xs) auto

(* ferrenseg *)

lemma "algunos P (map f xs) = algunos (P o f) xs"
by (induct xs) simp_all

(* anaprarod dancorgar *)
lemma "algunos P (map f xs) = algunos (P o f) xs"
apply (induct xs)
apply auto
done

text {*
---------------------------------------------------------------------
Ejercicio 7.2. Demostrar o refutar detalladamente
algunos P (map f xs) = algunos (P ∘ f) xs
---------------------------------------------------------------------
*}

lemma "algunos P (map f xs) = algunos (P o f) xs"
oops

(* migtermor *)

lemma AUX: "algunos (λa. P (f a)) xs = algunos (P o f) xs"
by (induct xs) auto

lemma "algunos P (map f xs) = algunos (P o f) xs" (is "?Q xs")
proof (induct xs)
show "?Q []" by simp
next
fix x xs
assume HI: "?Q xs"
have "algunos P (map f (x#xs)) = (algunos P ((f x)#(map f xs)))" by simp
also have "… = ((P (f x)) ∨ (algunos P (map f xs)))" by (simp only: algunos.simps(2))
also have "… = ((P (f x)) ∨ (algunos (P o f) xs))"
proof (cases)
assume C1: "(P (f x))"
have Aux: "((P (f x)) ∨ (algunos P (map f xs))) = True" using C1 by simp
have Aux1: "… = ((P (f x)) ∨ (algunos (P o f) xs))" using C1 by simp
then show "((P (f x)) ∨ (algunos P (map f xs))) = ((P (f x)) ∨ (algunos (P o f) xs))"
using Aux Aux1 by simp
next
assume C2: "¬(P (f x))"
have Aux2: "((P (f x)) ∨ (algunos P (map f xs))) = (algunos P (map f xs))" using C2 by simp
have Aux3: "… = (algunos (P o f) xs)" using HI by (simp add: AUX)
also have "… = ((P (f x)) ∨ (algunos (P o f) xs))" using C2 by simp
then show "((P (f x)) ∨ (algunos P (map f xs))) = ((P (f x)) ∨ (algunos (P o f) xs))"
using Aux2 Aux3 by simp
qed
also have "… = (((P o f) x) ∨ (algunos (P o f) xs))" by simp
also have "… = (algunos (P o f) (x#xs))" by simp
finally show "algunos P (map f (x#xs)) = (algunos (P o f) (x#xs))" by simp
qed

(* ferrenseg *)

lemma "algunos P (map f xs) = algunos (P ∘ f) xs"
proof (induct xs)
show "algunos P (map f []) = algunos (P ∘ f) []" by simp
next
fix x xs
assume HI: "algunos P (map f xs) = algunos (P ∘ f) xs"
show "algunos P (map f (x # xs)) = algunos (P ∘ f) (x # xs)"
proof -
have "algunos P (map f (x # xs)) = algunos P ((f x) # (map f xs))" by simp
also have "… = ((P (f x)) ∨ (algunos P (map f xs)))" by simp
also have "… = (((P ∘ f) x) ∨ (algunos (P ∘ f) xs))" using HI by simp
also have "… = algunos (P ∘ f) (x # xs)" by simp
finally show ?thesis
qed
qed

(* ivamenjim *)

lemma "algunos P (map f xs) = algunos (P o f) xs"
proof (induct xs)
show "algunos P (map f []) = algunos (P ∘ f) []" by simp
next
fix a xs
assume HI: "algunos P (map f xs) = algunos (P ∘ f) xs"
have "algunos P (map f (a # xs)) = algunos P ((f a)#map f xs)" by simp
also have "... = (algunos P (map f [a]) ∨ algunos P (map f xs))" by simp
also have "... = (algunos P (map f [a]) ∨ algunos (P ∘ f) xs)" using HI by simp
finally show "algunos P (map f (a # xs)) = algunos (P ∘ f) (a # xs)" by simp
qed

(* crigomgom rubgonmar anaprarod marpoldia1 juacabsou danrodcha paupeddeg *)
lemma "algunos P (map f xs) = algunos (P o f) xs"
proof (induct xs)
show "algunos P (map f []) = algunos (P ∘ f) []" by simp
next
fix x xs
assume HI: "algunos P (map f xs) = algunos (P ∘ f) xs"
have "algunos P (map f (x # xs)) = algunos P ((f x) # (map f xs))" by simp
also have "... = (P (f x) ∨ algunos P (map f xs))" by simp
also have "... = (P (f x) ∨ algunos (P ∘ f) xs)" using HI by simp
also have "... = ((P ∘ f) x ∨ algunos (P ∘ f) xs)" by simp
finally show "algunos P (map f (x # xs)) = algunos (P ∘ f) (x # xs)" by simp
qed

(* wilmorort pablucoto*)
lemma "algunos P (map f xs) = algunos (P o f) xs" (is "?P xs" )
proof (induct xs)
show "?P []" by simp
next
fix a xs
assume HI: "algunos P (map f xs) = algunos (P ∘ f) xs"
have "algunos P (map f (a # xs)) = (P (f a) ∨ algunos P (map f xs))" by simp
also have "... = (P (f a) ∨ algunos (P ∘ f) xs )" using HI by simp
also have "... = ((P ∘ f) a ∨ algunos (P ∘ f)xs)" by simp
finally show " algunos P (map f (a # xs)) = algunos (P ∘ f) (a # xs)" by simp
qed

(* fraortmoy *)

lemma "algunos P (map f xs) = algunos (P o f) xs"
proof (induct xs)
show "algunos P (map f []) = algunos (P ∘ f) []" by simp
next
fix a xs
assume H1: "algunos P (map f xs) = algunos (P ∘ f) xs"
have "algunos P (map f (a # xs)) = (algunos P (map f [a]) ∨ algunos P (map f xs))" by simp
also have "… = (algunos (P ∘ f) [a] ∨ algunos (P ∘ f) xs )" using H1 by simp
also have "… = algunos (P ∘ f) (a#xs)" by simp
finally show "algunos P (map f (a # xs)) = algunos (P ∘ f) (a#xs)" by simp
qed

{*danrodcha*}
lemma "algunos P (map f xs) = algunos (P o f) xs" (is "?Q xs")
proof (induct xs)
show "?Q []" by simp
next
fix a xs assume HI: "?Q xs"
have " algunos P (map f (a # xs)) = algunos P ((f a)#(map f xs))" by simp
also have "… = (P (f a) ∨ algunos P (map f xs))" by simp
also have "… = (P (f a) ∨ algunos (P ∘ f) xs)" using HI by blast
also have "… = ((P ∘ f) a ∨ algunos (P ∘ f) xs)" by simp
also have "… = algunos (P ∘ f) (a # xs)" by simp
finally show "?Q (a # xs)" by blast
qed

text {*
---------------------------------------------------------------------
Ejercicio 8.1. Demostrar o refutar automáticamente
algunos P (xs @ ys) = (algunos P xs ∨ algunos P ys)
---------------------------------------------------------------------
*}

(* ivamenjim marpoldia1 paupeddeg crigomgom rubgonmar wilmorort fraortmoy danrodcha pablucoto*)

lemma "algunos P (xs @ ys) = (algunos P xs ∨ algunos P ys)"
by (induct xs) auto

(* ferrenseg *)
lemma "algunos P (xs @ ys) = (algunos P xs ∨ algunos P ys)"
by (induct xs) simp_all

text {*
---------------------------------------------------------------------
Ejercicio 8.2. Demostrar o refutar detalladamente
algunos P (xs @ ys) = (algunos P xs ∨ algunos P ys)
---------------------------------------------------------------------
*}

(* ivamenjim ferrenseg marpoldia1 wilmorort*)

lemma algunos_append:
"algunos P (xs @ ys) = (algunos P xs ∨ algunos P ys)"
proof (induct xs)
show "algunos P ([] @ ys) = (algunos P [] ∨ algunos P ys)" by simp
next
fix a xs
assume HI: "algunos P (xs @ ys) = (algunos P xs ∨ algunos P ys)"
have "algunos P ((a # xs) @ ys) = (P a ∨ (algunos P (xs @ ys)))" by simp
also have "... = (P a ∨ (algunos P xs ∨ algunos P ys))" using HI by simp
finally show "algunos P ((a # xs) @ ys) = (algunos P (a # xs) ∨ algunos P ys)" by simp
qed

(* migtermor crigomgom *)

lemma algunos_append2:
"algunos P (xs @ ys) = (algunos P xs ∨ algunos P ys)" (is "?Q xs")
proof (induct xs)
show "?Q []" by simp
next
fix x xs
assume HI: "?Q xs"
have "algunos P ((x#xs) @ ys) = algunos P (x#(xs @ ys))" by simp
also have "… = ((P x) ∨ (algunos P (xs @ ys)))" by simp
also have "… = ((P x) ∨ (algunos P xs) ∨ (algunos P ys))" using HI by simp
also have "… = ((algunos P (x#xs)) ∨ (algunos P ys))" by simp
finally show "algunos P ((x#xs) @ ys) = (algunos P (x#xs) ∨ algunos P ys)" by simp
qed

(* paupeddeg*)
lemma algunos_append3:
"algunos P (xs @ ys) = (algunos P xs ∨ algunos P ys)"
proof (induct xs)
show "algunos P ([] @ ys) = (algunos P [] ∨ algunos P ys)" by simp
next
fix a xs
assume HI: "algunos P (xs @ ys) = (algunos P xs ∨ algunos P ys)"
have "algunos P ((a # xs) @ ys) = (algunos P [a] ∨ ( algunos P (xs @ ys)))" by simp
also have "... = (algunos P [a] ∨ (algunos P xs ∨ algunos P ys))" using HI by simp
finally show "algunos P ((a # xs) @ ys) = (algunos P (a # xs) ∨ algunos P ys)" by simp
qed

(* fraortmoy *)
lemma algunos_append4:
"algunos P (xs @ ys) = (algunos P xs ∨ algunos P ys)"
proof (induct xs)
show " algunos P ([] @ ys) = (algunos P [] ∨ algunos P ys)" by simp
next
fix a xs
assume H1: "algunos P (xs @ ys) = (algunos P xs ∨ algunos P ys)"
have "algunos P ((a # xs) @ ys) = (algunos P [a] ∨ algunos P (xs @ ys))" by simp
also have "… = (algunos P [a] ∨ algunos P xs ∨ algunos P ys)" using H1 by simp
also have "… = ((algunos P [a] ∨ algunos P xs) ∨ algunos P ys)" by simp
also have "… = (algunos P (a#xs) ∨ algunos P ys)" by simp
finally show "algunos P ((a # xs) @ ys) = (algunos P (a#xs) ∨ algunos P ys)" by simp
qed

{* danrodcha pablucoto*}
lemma algunos_append5:
"algunos P (xs @ ys) = (algunos P xs ∨ algunos P ys)" (is "?Q xs")
proof (induct xs)
show "?Q []" by simp
next
fix a xs assume HI: "?Q xs"
have "algunos P ((a#xs) @ ys) = algunos P (a#(xs @ ys))" by simp
also have "… = (algunos P [a] ∨ algunos P (xs @ ys))" by simp
also have "… = (algunos P [a] ∨ algunos P xs ∨ algunos P ys)" using HI by simp
also have "… = (algunos P (a # xs) ∨ algunos P ys)" by simp
finally show "?Q (a # xs)" by simp
qed

text {*
---------------------------------------------------------------------
Ejercicio 9.1. Demostrar o refutar automáticamente
algunos P (rev xs) = algunos P xs
---------------------------------------------------------------------
*}

(* ivamenjim migtermor marpoldia1 rubgonmar paupeddeg *)

lemma "algunos P (rev xs) = algunos P xs"
apply (induct xs)
apply simp
apply (simp add: algunos_append)
apply auto
done

(* ferrenseg crigomgom danrodcha pablucoto*)
lemma "algunos P (rev xs) = algunos P xs"
by (induct xs) (auto simp add: algunos_append)

(* wilmorort *)
lemma "algunos P (rev xs) = algunos P xs"
by (induct xs,simp,simp add: algunos_append,auto)

text {*
---------------------------------------------------------------------
Ejercicio 9.2. Demostrar o refutar detalladamente
algunos P (rev xs) = algunos P xs
---------------------------------------------------------------------
*}

lemma "algunos P (rev xs) = algunos P xs"
oops

(* migtermor *)
lemma auxiliar1:
"rev (a#xs) = rev xs @ [a]"
by auto

lemma "algunos P (rev xs) = algunos P xs" (is "?Q xs")
proof (induct xs)
show "?Q []" by simp
next
fix x xs
assume HI: "?Q xs"
have "algunos P (rev (x#xs)) = (algunos P (rev xs @ [x]))" using auxiliar1 by simp
also have "… = ((algunos P (rev xs)) ∨ (algunos P [x]))" by (simp add: algunos_append)
also have "… = ((P x) ∨ (algunos P (rev xs)))" by simp arith
also have "… = ((P x) ∨ (algunos P xs))" using HI by simp
finally show "algunos P (rev (x#xs)) = (algunos P (x#xs))" by simp
qed

(* ferrenseg *)

lemma "algunos P (rev xs) = algunos P xs"
proof (induct xs)
show "algunos P (rev []) = algunos P []" by simp
next
fix x xs
assume HI: "algunos P (rev xs) = algunos P xs"
show "algunos P (rev (x # xs)) = algunos P (x # xs)"
proof -
have "algunos P (rev (x # xs)) = algunos P ((rev xs) @ [x])" by simp
also have "… = (algunos P (rev xs) ∨ algunos P [x])"
by (simp add: algunos_append)
also have "… = (algunos P xs ∨ algunos P [x])" using HI by simp
also have "… = (algunos P xs ∨ P x)" by simp
also have "… = (P x ∨ algunos P xs)" by auto
also have "… = algunos P (x # xs)" by simp
finally show ?thesis
qed
qed

(* ivamenjim marpoldia1*)

lemma "algunos P (rev xs) = algunos P xs" (is "?P xs")
proof (induct xs)
show "?P []" by simp
next
fix a xs
assume HI: "?P xs"
have "algunos P (rev (a # xs)) = (algunos P ((rev xs)@[a]))" by simp
also have "... = ((algunos P (rev xs)) ∨ algunos P [a])" by (simp add: algunos_append)
also have "... = (algunos P xs ∨ algunos P [a])" using HI by simp
also have "... = (algunos P [a] ∨ algunos P xs)" by arith
finally show "algunos P (rev (a # xs)) = algunos P (a # xs)" by simp
qed

(*paupeddeg*)
lemma "algunos P (rev xs) = algunos P xs"
proof (induct xs)
show "algunos P (rev []) = algunos P []" by simp
next
fix a xs
assume HI:"algunos P (rev xs) = algunos P xs"
have "algunos P (rev (a # xs)) = algunos P ((rev xs) @ [a])" by simp
also have "... = ((algunos P (rev xs)) ∨ (algunos P [a]))" by (simp add: algunos_append)
also have "... = ((algunos P xs) ∨ (algunos P [a]))" using HI by simp
also have "... = ((algunos P [a]) ∨ (algunos P xs))" by arith
finally show "algunos P (rev (a # xs)) = algunos P (a # xs)" by simp
qed

(*crigomgom pablucoto*)
lemma "algunos P (rev xs) = algunos P xs"
proof (induct xs)
show "algunos P (rev []) = algunos P []" by simp
next
fix x xs
assume HI: " algunos P (rev xs) = algunos P xs"
have "algunos P (rev (x # xs)) = algunos P ((rev xs) @ [x])" by simp
also have "... = (algunos P (rev xs) ∨ algunos P [x])" by (simp add: algunos_append)
also have "... = (algunos P xs ∨ algunos P [x])" using HI by simp
also have "... = (algunos P xs ∨ P x)" by simp
also have "... = (P x ∨ algunos P xs)" by arith
also have "... = algunos P (x#xs)" by simp
finally show "algunos P (rev (x # xs)) = algunos P (x # xs)" by simp
qed

(*wilmorort*)
lemma "algunos P (rev xs) = algunos P xs" (is "?P xs")
proof (induct xs)
show "?P []" by simp
next
fix a xs
assume HI: "?P xs"
have "algunos P (rev (a#xs)) = algunos P ((rev xs) @ [a])" by simp
also have "... = (algunos P(rev xs) ∨ algunos P [a])" by (simp add: algunos_append)
also have "... = (algunos P xs ∨ algunos P [a])" using HI by simp
also have "... = (algunos P [a] ∨ algunos P xs)" by (simp add: HOL.disj_commute)
also have "... = algunos P([a]@(xs))" by (simp)
finally show "algunos P (rev (a#xs))= algunos P (a # xs)" by simp
qed

{*danrodcha *}
lemma "algunos P (rev xs) = algunos P xs" (is "?Q xs")
proof (induct xs)
show "?Q []" by simp
next
fix a xs assume HI: "?Q xs"
have "algunos P (rev (a#xs)) = algunos P ((rev xs) @ [a])" by simp
also have "… = (algunos P (rev xs) ∨ algunos P [a])" by (simp add:algunos_append)
also have "… = (algunos P xs ∨ algunos P [a])" using HI by simp
also have "… = (algunos P [a] ∨ algunos P xs)" by (simp add: HOL.disj_commute)
also have "… = algunos P (a # xs)" by simp
finally show "?Q (a # xs)" by simp
qed

text {*
---------------------------------------------------------------------
Ejercicio 10. Encontrar un término no trivial Z tal que sea cierta la
siguiente ecuación:
algunos (λx. P x ∨ Q x) xs = Z
y demostrar la equivalencia de forma automática y detallada.
---------------------------------------------------------------------
*}

(* migtermor *)

lemma "algunos (λx. P x ∨ Q x) xs = ((algunos (λx. P x) xs) ∨ (algunos (λx. Q x) xs))"
by (induct xs) auto

lemma "algunos (λx. P x ∨ Q x) xs = ((algunos (λx. P x) xs) ∨ (algunos (λx. Q x) xs))" (is "?R xs")
proof (induct xs)
show "?R []" by simp
next
fix x xs
assume HI: "?R xs"
have "algunos (λx. P x ∨ Q x) (x#xs) = (((λx. P x ∨ Q x) x) ∨ (algunos (λx. P x ∨ Q x) xs))"
by simp
also have "… = ((P x ∨ Q x) ∨ (algunos (λx. P x ∨ Q x) xs))" by simp
also have H1: "… = ((((λx. P x) x) ∨ (algunos (λx. P x) xs)) ∨ (((λx. Q x) x) ∨ (algunos (λx. Q x) xs)))"
using HI by simp arith
have H2: "… = ((algunos (λx. P x) (x#xs)) ∨ (algunos (λx. Q x) (x#xs)))"
by simp
have C: "(algunos (λx. P x ∨ Q x) (x#xs)) =
((algunos (λx. P x) (x#xs)) ∨ (algunos (λx. Q x) (x#xs)))"
using H1 H2 by simp
finally show "(algunos (λx. P x ∨ Q x) (x#xs)) =
((algunos (λx. P x) (x#xs)) ∨ (algunos (λx. Q x) (x#xs)))"
using C by simp
qed

(* ferrenseg *)

lemma "algunos (λx. P x ∨ Q x) xs = (algunos P xs ∨ algunos Q xs)"
proof (induct xs)
show "algunos (λx. P x ∨ Q x) [] = (algunos P [] ∨ algunos Q [])" by simp
next
fix x xs
assume HI: "algunos (λx. (P x ∨ Q x)) xs = (algunos P xs ∨ algunos Q xs)"
show "algunos (λx. P x ∨ Q x) (x # xs) = (algunos P (x # xs) ∨ algunos Q (x # xs))"
proof -
have "algunos (λx. P x ∨ Q x) (x # xs) =
((P x) ∨ (Q x) ∨ algunos (λx. P x ∨ Q x) xs)" by simp
also have "… = ((P x) ∨ (Q x) ∨ algunos P xs ∨ algunos Q xs)"
using HI by simp
also have "… = (((P x) ∨ algunos P xs) ∨ ((Q x) ∨ algunos Q xs))" by auto
also have "… = (algunos P (x # xs) ∨ algunos Q (x # xs))" by simp
finally show ?thesis
qed
qed

(* ivamenjim marpoldia1 paupeddeg *)

lemma "algunos (λx. P x ∨ Q x) xs = (algunos P xs ∨ algunos Q xs)"
by (induct xs) auto

lemma "algunos (λx. P x ∨ Q x) xs = (algunos P xs ∨ algunos Q xs)"
proof (induct xs)
show "algunos (λx. P x ∨ Q x) [] = (algunos P [] ∨ algunos Q [])" by simp
next
fix a xs
assume HI: "algunos (λx. P x ∨ Q x) xs = (algunos P xs ∨ algunos Q xs)"
have "algunos (λx. P x ∨ Q x) (a # xs) = (P a ∨ Q a ∨ algunos (λx. P x ∨ Q x) xs)" by simp
also have "... = (P a ∨ Q a ∨ algunos P xs ∨ algunos Q xs)" using HI by simp
finally show "algunos (λx. P x ∨ Q x) (a # xs) = (algunos P (a # xs) ∨ algunos Q (a # xs))" by auto
qed

(* wilmorort danrodcha*)
lemma "algunos (λx. P x ∨ Q x) xs = (¬todos(λx. ¬P x)xs ∨ ¬todos(λx. ¬Q x)xs)"
by (induct xs) auto

(* wilmorort *)
lemma "algunos (λx. P x ∨ Q x) xs = (¬todos(λx. ¬P x)xs ∨ ¬todos(λx. ¬Q x)xs)"
(is "?P xs")
proof (induct xs)
show "?P []" by simp
next
fix a xs
assume HI: "?P xs"
have "algunos (λx. P x ∨ Q x) (a # xs) = (P a ∨ Q a ∨ algunos (λx. P x ∨ Q x) xs)" by simp
also have "... = (P a ∨ Q a ∨ (¬ todos (λx. ¬ P x) xs ∨ ¬ todos (λx. ¬ Q x) xs))" using HI by simp
also have "... = (P a ∨ ¬ todos (λx. ¬ P x) xs ∨ Q a ∨ ¬ todos (λx. ¬ Q x) xs )" by arith
finally show "algunos (λx. P x ∨ Q x) (a # xs) = (~ todos (λx. ¬ P x) (a # xs) ∨ ~ todos (λx. ¬ Q x) (a # xs))" by simp
qed

{* danrodcha *}
lemma "algunos (λx. P x ∨ Q x) xs = (algunos P xs ∨ algunos Q xs)" (is "?R xs")
proof (induct xs)
show "?R []" by simp
next
fix a xs assume HI: "?R xs"
have 1:" (Q a ∨ algunos P xs) = (algunos P xs ∨ Q a)" by (simp add: HOL.disj_commute)
have "algunos (λx. P x ∨ Q x) (a # xs) =
(algunos P [a] ∨ algunos Q [a] ∨ algunos (λx. P x ∨ Q x) xs)" by simp
also have "… = (P a ∨ (Q a ∨ algunos P xs) ∨ algunos Q xs)" using HI by simp
also have "… = (P a ∨ algunos P xs ∨ Q a ∨ algunos Q xs)" using 1 by simp
also have "… = (algunos P (a # xs) ∨ algunos Q (a # xs))" by simp
finally show "?R (a # xs)" by simp
qed

text {*
---------------------------------------------------------------------
Ejercicio 11.1. Demostrar o refutar automáticamente
algunos P xs = (¬ todos (λx. (¬ P x)) xs)
---------------------------------------------------------------------
*}

(* ivamenjim wilmorort migtermor marpoldia1 crigomgom rubgonmar paupeddeg danrodcha pablucoto*)

lemma "algunos P xs = (¬ todos (λx. (¬ P x)) xs)"
by (induct xs) auto

(* ferrenseg *)

lemma "algunos P xs = (¬ todos (λx. (¬ P x)) xs)"
by (induct xs) simp_all

text {*
---------------------------------------------------------------------
Ejercicio 11.2. Demostrar o refutar datalladamente
algunos P xs = (¬ todos (λx. (¬ P x)) xs)
---------------------------------------------------------------------
*}

(* ivamenjim migtermor marpoldia1 crigomgom paupeddeg*)

lemma "algunos P xs = (¬ todos (λx. (¬ P x)) xs)"
proof (induct xs)
show "algunos P [] = (¬ todos (λx. ¬ P x) [])" by simp
next
fix a xs
assume HI: "algunos P xs = (¬ todos (λx. ¬ P x) xs)"
have "algunos P (a # xs) = (P a ∨ (algunos P xs))" by simp
also have "... = (P a ∨ (¬ todos (λx. ¬ P x) xs))" using HI by simp
finally show "algunos P (a # xs) = (¬ todos (λx. ¬ P x) (a # xs))" by simp
qed

(* ferrenseg *)

lemma "algunos P xs = (¬ todos (λx. (¬ P x)) xs)"
proof (induct xs)
show "algunos P [] = (¬ todos (λx. (¬ P x)) [])" by simp
next
fix x xs
assume HI: "algunos P xs = (¬ todos (λx. (¬ P x)) xs)"
show "algunos P (x # xs) = (¬ todos (λx. (¬ P x)) (x # xs))"
proof -
have "algunos P (x # xs) = ((P x) ∨ algunos P xs)" by simp
also have "… = ((P x) ∨ ¬ todos (λx. (¬ P x)) xs)" using HI by simp
also have "… = (¬ (¬ (P x) ∧ todos (λx. (¬ P x)) xs))" by simp
also have "… = (¬ todos (λx. (¬ P x)) (x # xs))" by simp
finally show ?thesis
qed
qed

{* danrodcha *}
lemma "algunos P xs = (¬ todos (λx. (¬ P x)) xs)" (is "?Q xs")
proof (induct xs)
show "?Q []" by simp
next
fix a xs assume HI: "?Q xs"
have "algunos P (a # xs) = (P a ∨ algunos P xs)" by simp
also have "… = (P a ∨ (¬ todos (λx. (¬ P x)) xs))" using HI by simp
also have "… = (¬(¬(P a) ∧ todos(λx. (¬ P x)) xs))" by simp
also have "… = (¬ todos(λx. (¬ P x)) (a # xs))" by simp
finally show "?Q (a # xs)" by simp
qed

text {*
---------------------------------------------------------------------
Ejercicio 12. Definir la funcion primitiva recursiva
estaEn :: 'a ⇒ 'a list ⇒ bool
tal que (estaEn x xs) se verifica si el elemento x está en la lista
xs. Por ejemplo,
estaEn (2::nat) [3,2,4] = True
estaEn (1::nat) [3,2,4] = False
---------------------------------------------------------------------
*}

(* ivamenjim ferrenseg migtermor serrodcal crigomgom rubgonmar marpoldia1 paupeddeg danrodcha*)

fun estaEn :: "'a ⇒ 'a list ⇒ bool" where
"estaEn x [] = False"
| "estaEn x (a#xs) = ((a = x) ∨ (estaEn x xs))"

value "estaEn (2::nat) [3,2,4] = True"
value "estaEn (1::nat) [3,2,4] = False"

text {*
---------------------------------------------------------------------
Ejercicio 13. Expresar la relación existente entre estaEn y algunos.
Demostrar dicha relación de forma automática y detallada.
---------------------------------------------------------------------
*}

(* migtermor crigomgom *)

lemma "estaEn x xs = (algunos (λa. a=x) xs)"
by (induct xs) auto

lemma auxiliar13:
"(x=a) = ((λa. a=x) a)"
by auto

lemma "estaEn x xs = (algunos (λa. a=x) xs)" (is "?P xs")
proof (induct xs)
show "?P []" by simp
next
fix a xs
assume HI: "?P xs"
have "estaEn x (a#xs) = ((a=x) ∨ (estaEn x xs))" by simp
also have H: "… = ((a=x) ∨ (algunos (λa. a=x) xs))" using HI by simp
also have "… = (((λa. a=x) a) ∨ (algunos (λa. a=x) xs))" using auxiliar13 by simp
also have "… = (algunos (λa. a=x) (a#xs))" by simp
have C: "estaEn x (a#xs) = (algunos (λa. a=x) (a#xs))" using H by simp
finally show "estaEn x (a#xs) = (algunos (λa. a=x) (a#xs))" using C by simp
qed

(* ferrenseg *)

lemma estaEn_algunos:
"estaEn y xs = algunos (λx. x=y) xs"
by (induct xs) auto

lemma estaEn_algunos_2:
"estaEn y xs = algunos (λx. x=y) xs"
proof (induct xs)
show "estaEn y [] = algunos (λx. x=y) []" by simp
next
fix x xs
assume HI: "estaEn y xs = algunos (λx. x=y) xs"
show "estaEn y (x # xs) = algunos (λx. x=y) (x # xs)"
proof -
have "estaEn y (x # xs) = (y=x ∨ estaEn y xs)" by simp
also have "… = (y=x ∨ algunos (λx. x=y) xs)" using HI by simp
also have "… = (x=y ∨ algunos (λx. x=y) xs)" by auto
also have "… = algunos (λx. x=y) (x # xs)" by simp
finally show ?thesis
qed
qed

(*crigomgom paupeddeg*)

lemma "estaEn a xs = algunos (λx. x = a) xs"
by (induct xs) auto

lemma "estaEn x xs = (algunos (λa. a=x) xs)"
proof (induct xs)
show " estaEn x [] = algunos (λa. a = x) []" by simp
next
fix a xs
assume HI: "estaEn x xs = algunos (λa. a = x) xs"
have "estaEn x (a # xs) = (a = x ∨ estaEn x xs)" by simp
also have "... = (a = x ∨ algunos (λa. a = x) xs)" using HI by simp
finally show " estaEn x (a # xs) = algunos (λa. a = x) (a # xs)" by simp
qed

(* ivamenjim marpoldia1 *)

lemma "estaEn x xs = algunos (λy. x=y) xs"
by (induct xs) auto

lemma "estaEn x xs = algunos (λy. x=y) xs"
proof (induct xs)
show "estaEn x [] = algunos (op = x) []" by simp
next
fix a xs
assume HI: "estaEn x xs = algunos (op = x) xs"
have "estaEn x (a # xs) = ((a = x) ∨ (estaEn x xs))" by simp
also have "... = (a = x ∨ algunos (op = x) xs)" using HI by simp
finally show " estaEn x (a # xs) = algunos (op = x) (a # xs)" by auto
qed

{* danrodcha *}
lemma "estaEn a xs = algunos (op = a) xs"
by (induct xs) auto

{* danrodcha *}
lemma "estaEn a xs = algunos (op = a) xs"
proof (induct xs)
show "estaEn a [] = algunos (op = a) []" by simp
next
fix x xs
assume HI: "estaEn a xs = algunos (op = a) xs"
have "estaEn a (x # xs) = ((x = a) ∨ (estaEn a xs))" by simp
also have "… = (x = a ∨ algunos (op = a) xs)" using HI by simp
also have "… = ((op = a) x ∨ algunos (op = a) xs)" by (simp add:HOL.eq_commute)
also have "… = algunos (op = a) (x # xs)" by simp
finally show "estaEn a (x # xs) = algunos (op = a) (x # xs)" by simp
qed

end
</source>

Relación 4

2016-11-23T00:31:12Z

Pablucoto:

<source lang="isar">
chapter {* R4: Cuantificadores sobre listas *}

theory R4_Cuantificadores_sobre_listas
imports Main
begin

text {*
---------------------------------------------------------------------
Ejercicio 1. Definir la función
todos :: ('a ⇒ bool) ⇒ 'a list ⇒ bool
tal que (todos p xs) se verifica si todos los elementos de la lista
xs cumplen la propiedad p. Por ejemplo, se verifica
todos (λx. 1<length x) [[2,1,4],[1,3]]
¬todos (λx. 1<length x) [[2,1,4],[3]]

Nota: La función todos es equivalente a la predefinida list_all.
---------------------------------------------------------------------
*}

{*danrodcha*}
fun todos :: "('a ⇒ bool) ⇒ 'a list ⇒ bool" where
"todos p xs = "todos p [] = True"
| "todos p (x#xs) = (p x ∧ todos p xs)""

(* ivamenjim migtermor dancorgar wilmorort marpoldia1 ferrenseg paupeddeg pablucoto crigomgom anaprarod serrodcal juacabsou rubgonmar josgarsan fraortmoy lucnovdos*)
fun todos :: "('a ⇒ bool) ⇒ 'a list ⇒ bool" where
"todos p [] = True"
| "todos p (x#xs) = (p x ∧ todos p xs)"

text {*
---------------------------------------------------------------------
Ejercicio 2. Definir la función
algunos :: ('a ⇒ bool) ⇒ 'a list ⇒ bool
tal que (algunos p xs) se verifica si algunos elementos de la lista
xs cumplen la propiedad p. Por ejemplo, se verifica
algunos (λx. 1<length x) [[2,1,4],[3]]
¬algunos (λx. 1<length x) [[],[3]]"

Nota: La función algunos es equivalente a la predefinida list_ex.
---------------------------------------------------------------------
*}

{*danrodcha ivamenjim migtermor dancorgar marpoldia1 ferrenseg wilmorort paupeddeg pablucoto crigomgom anaprarod serrodcal juacabsou rubgonmar josgarsan fraortmoy lucnovdos*}
fun algunos :: "('a ⇒ bool) ⇒ 'a list ⇒ bool" where
"algunos p [] = False"
| "algunos p (x#xs) = (p x ∨ algunos p xs)"

text {*
---------------------------------------------------------------------
Ejercicio 3.1. Demostrar o refutar automáticamente
todos (λx. P x ∧ Q x) xs = (todos P xs ∧ todos Q xs)
---------------------------------------------------------------------
*}

{*danrodcha ivamenjim migtermor dancorgar marpoldia1 ferrenseg wilmorort paupeddeg pablucoto anaprarod serrodcal juacabsou rubgonmar josgarsan fraortmoy lucnovdos*}
lemma "todos (λx. P x ∧ Q x) xs = (todos P xs ∧ todos Q xs)"
by (induct xs) auto

text {*
---------------------------------------------------------------------
Ejercicio 3.2. Demostrar o refutar detalladamente
todos (λx. P x ∧ Q x) xs = (todos P xs ∧ todos Q xs)
---------------------------------------------------------------------
*}

{*danrodcha*}
lemma "todos (λx. P x ∧ Q x) xs = (todos P xs ∧ todos Q xs)"
proof (induct xs)
show "?R []" by simp
next
fix a xs
assume HI: "?R xs"
have "todos (λx. P x ∧ Q x) (a#xs) = (P a ∧ Q a ∧ todos (λx. P x ∧ Q x) xs)" by simp
also have "… = (P a ∧ Q a ∧ (todos P xs ∧ todos Q xs))" using HI by simp
also have "… = ((P a ∧ todos P xs) ∧ (Q a ∧ todos Q xs))" by blast
also have "… = (todos P (a#xs) ∧ todos Q (a#xs))" by simp
finally show "?R (a#xs)" by simp
qed

(* ivamenjim wilmorort serrodcal josgarsan*)

lemma "todos (λx. P x ∧ Q x) xs = (todos P xs ∧ todos Q xs)"
proof (induct xs)
show "todos (λx. P x ∧ Q x) [] = (todos P [] ∧ todos Q [])" by simp
next
fix a xs
assume HI: "todos (λx. P x ∧ Q x) xs = (todos P xs ∧ todos Q xs)"
have "todos (λx. P x ∧ Q x) (a # xs) = (P a ∧ Q a ∧ todos (λx. P x ∧ Q x) xs)" by simp
also have "... = (P a ∧ Q a ∧ todos P xs ∧ todos Q xs)" using HI by simp
finally show "todos (λx. P x ∧ Q x) (a # xs) = (todos P (a # xs) ∧ todos Q (a # xs))" by auto
qed

(* dancorgar paupeddeg *)
lemma "todos (λx. P x ∧ Q x) xs = (todos P xs ∧ todos Q xs)"
proof (induct xs)
show "todos (λx. P x ∧ Q x) [] = (todos P [] ∧ todos Q [])" by simp
next
fix y xs
assume HI: "todos (λx. P x ∧ Q x) xs = (todos P xs ∧ todos Q xs)"
have "todos (λx. P x ∧ Q x) (y#xs) = ((P y ∧ Q y) ∧ (todos (λx. P x ∧ Q x) xs))" by simp
also have "… = ((P y ∧ Q y) ∧ (todos P xs ∧ todos Q xs))" using HI by simp
also have "… = ((P y ∧ todos P xs) ∧ (Q y ∧ todos Q xs))" by blast
also have "… = ((todos P (y#xs)) ∧ (todos Q (y#xs)))" by simp
finally show "todos (λx. P x ∧ Q x) (y#xs) = (todos P (y#xs) ∧ todos Q (y#xs))" by simp
qed

(* migtermor *)
lemma "todos (λx. P x ∧ Q x) xs = (todos P xs ∧ todos Q xs)"
proof (induct xs)
show "todos (λx. P x ∧ Q x) [] = (todos P [] ∧ todos Q [])" by simp
next
fix x xs
assume HI: "todos (λx. P x ∧ Q x) xs = (todos P xs ∧ todos Q xs)"
have "(todos (λx. P x ∧ Q x) (x#xs)) = (( P x ∧ Q x) ∧ (todos (λx. P x ∧ Q x) xs))"
by (simp only: todos.simps(2))
also have "… = ((P x ∧ Q x) ∧ (todos P xs ∧ todos Q xs))" using HI by simp
also have "… = (((P x)∧(todos P xs)) ∧ ((Q x) ∧ (todos Q xs)))" by arith
also have "((P x ∧ Q x) ∧ (todos P xs ∧ todos Q xs)) = (((P x)∧(todos P xs)) ∧ ((Q x) ∧ (todos Q xs)))"
by arith (* Este paso es exactamente el mismo que el anterior, pero sin cualquiera de los dos no funciona el "finally show" *)
have "… = (((P x)∧(todos P xs))∧((Q x)∧(todos Q xs)))" by simp
have "… = ((todos P (x#xs))∧(todos Q (x#xs)))" by simp
finally show "(todos (λx. P x ∧ Q x) (x#xs)) = ((todos P (x#xs)) ∧ (todos Q (x#xs)))" by simp
qed

(* marpoldia1 *)
lemma "todos (λx. P x ∧ Q x) xs = (todos P xs ∧ todos Q xs)"
proof (induct xs)
show "todos (λx. P x ∧ Q x) [] = (todos P [] ∧ todos Q [])" by simp
next
fix a xs
assume HI: "todos (λx. P x ∧ Q x) xs = (todos P xs ∧ todos Q xs)"
have "todos (λx. P x ∧ Q x) (a # xs) = ((P a ∧ Q a) ∧ (todos P xs ∧ todos Q xs))" using HI by simp
also have "... = ((P a ∧ todos P xs) ∧ (Q a ∧ todos Q xs))" by blast
also have "... = (todos P (a#xs) ∧ todos Q (a#xs)) " by simp
finally show "todos (λx. P x ∧ Q x) (a#xs) = (todos P (a#xs) ∧ todos Q (a#xs))" by simp
qed

(* ferrenseg *)

lemma "todos (λx. P x ∧ Q x) xs = (todos P xs ∧ todos Q xs)"
proof (induct xs)
show "todos (λx. P x ∧ Q x) [] = (todos P [] ∧ todos Q [])" by simp
next
fix n xs
assume HI: "todos (λx. P x ∧ Q x) xs = (todos P xs ∧ todos Q xs)"
have "todos (λx. P x ∧ Q x) (n # xs) = (P n ∧ Q n ∧ todos P xs ∧ todos Q xs)" using HI by simp
also have "… = P n ∧ Q n ∧ todos P xs ∧ todos Q xs" by blast
also have "⋯ = todos P (n # xs) ∧ todos Q (n # xs)" by simp
finally show "todos (λx. P x ∧ Q x) (n # xs) = todos P (n # xs) ∧ todos Q (n # xs)" by simp
qed

(* lucnovdos *)

lemma "todos (λx. P x ∧ Q x) xs = (todos P xs ∧ todos Q xs)"
proof (induct xs)
show "todos (λx. P x ∧ Q x) [] = (todos P [] ∧ todos Q [])" by simp
next
fix n xs
assume HI: "todos (λx. P x ∧ Q x) xs = (todos P xs ∧ todos Q xs)"
have "todos (λx. P x ∧ Q x) (n # xs) = ((P n ∧ Q n) ∧ (todos P xs ∧ todos Q xs))" using HI by simp
also have "... = ((P n ∧ todos P xs) ∧ (Q n ∧ todos Q xs))" by arith
also have "... = ((todos P(n#xs)) ∧ (todos Q(n#xs)))" by simp

finally show "todos (λx. P x ∧ Q x) (n#xs) = (todos P (n#xs) ∧ todos Q (n#xs))" by simp
qed

{* wilmorort pablucoto crigomgom anaprarod juacabsou rubgonmar fraortmoy *}

lemma "todos (λx. P x ∧ Q x) xs = (todos P xs ∧ todos Q xs)"
proof (induct xs)
show "todos (λx. P x ∧ Q x) [] = (todos P [] ∧ todos Q [])" by simp
next
fix a xs
assume HI: "todos (λx. P x ∧ Q x) xs = (todos P xs ∧ todos Q xs)"
have "todos (λx. P x ∧ Q x) (a # xs) = (P a ∧ Q a ∧ todos (λx. P x ∧ Q x) xs)" by simp
also have "... = (P a ∧ Q a ∧ todos P xs ∧ todos Q xs)" using HI by simp
also have "... = (P a ∧ todos P xs ∧ Q a ∧ todos Q xs)" by arith
also have "... = (todos P (a # xs) ∧ todos Q (a # xs))" by simp (* Este paso se puede obviar*)
finally show "todos (λx. P x ∧ Q x) (a # xs) = (todos P (a # xs) ∧ todos Q (a # xs))" by simp
qed

text {*
---------------------------------------------------------------------
Ejercicio 4.1. Demostrar o refutar automáticamente
todos P (x @ y) = (todos P x ∧ todos P y)
---------------------------------------------------------------------
*}
{*danrodcha ivamenjim marpoldia1 migtermor ferrenseg wilmorort paupeddeg crigomgom anaprarod serrodcal juacabsou rubgonmar pablucoto fraortmoy josgarsan lucnovdos*}
lemma "todos P (x @ y) = (todos P x ∧ todos P y)"
by (induct x) auto

{* anaprarod dancorgar *}
lemma "todos P (x @ y) = (todos P x ∧ todos P y)"
apply (induct x)
apply auto
done

text {*
---------------------------------------------------------------------
Ejercicio 4.2. Demostrar o refutar detalladamente
todos P (x @ y) = (todos P x ∧ todos P y)
---------------------------------------------------------------------
*}

(* ivamenjim ferrenseg wilmorort dancorgar fraortmoy josgarsan lucnovdos*)

lemma todos_append:
"todos P (x @ y) = (todos P x ∧ todos P y)"
proof (induct x)
show "todos P ([] @ y) = (todos P [] ∧ todos P y)" by simp
next
fix a x
assume HI: "todos P (x @ y) = (todos P x ∧ todos P y)"
have "todos P ((a#x) @ y) = (P a ∧ (todos P (x @ y)))" by simp
also have "... = (P a ∧ (todos P x ∧ todos P y))" using HI by simp
finally show "todos P ((a # x) @ y) = (todos P (a # x) ∧ todos P y)" by simp
qed

(* migtermor serrodcal *)

lemma todos_append1:
"todos P (x @ y) = (todos P x ∧ todos P y)" (is "?P x")
proof (induct x)
show "?P []" by simp
next
fix a x
assume HI: "?P x"
have "todos P ((a#x) @ y) = (P a ∧ (todos P (x @ y)))" by simp
also have "… = (P a ∧ (todos P x ∧ todos P y))" using HI by simp
finally show "?P (a#x)" by simp
qed

(* marpoldia1 *)
lemma todos_append:
"todos P (x @ y) = (todos P x ∧ todos P y)"
proof (induct x)
show "todos P ([] @ y) = (todos P [] ∧ todos P y)" by simp
next
fix a x
assume HI: "todos P (x @ y) = (todos P x ∧ todos P y)"
have "todos P ((a # x) @ y) = todos P (a#(x@y))" by simp
also have "... = (P a ∧ todos P (x @ y))" by simp
also have "... = (P a ∧ (todos P x ∧ todos P y))" using HI by simp
finally show "todos P ((a # x) @ y) = (todos P (a # x) ∧ todos P y)" by simp
qed

(* paupeddeg *)
lemma todos_append3:
"todos P (x @ y) = (todos P x ∧ todos P y)"
proof (induct x)
show "todos P ([] @ y) = (todos P [] ∧ todos P y)" by simp
next
fix a x
assume HI: "todos P (x @ y) = (todos P x ∧ todos P y)"
have "todos P ((a # x) @ y) = todos P (a # (x @ y))" by simp
also have "... = (todos P [a] ∧ todos P (x @ y)) " by simp
also have "... = (todos P [a] ∧ (todos P x ∧ todos P y))" using HI by simp
finally show "todos P ((a # x) @ y) = (todos P (a # x) ∧ todos P y)" by simp
qed

(* crigomgom juacabsou anaprarod*)

lemma todos_append4:
"todos P (x @ y) = (todos P x ∧ todos P y)"
proof (induct x)
show "todos P ([] @ y) = (todos P [] ∧ todos P y)" by simp
next
fix x xs
assume HI:"todos P (xs @ y) = (todos P xs ∧ todos P y)"
have "todos P ((x # xs) @ y) = (P x ∧ todos P (xs @ y ))" by simp
also have "... = (P x ∧ (todos P xs ∧ todos P y)) " using HI by simp
also have "... = ((P x ∧ todos P xs) ∧ todos P y)" by simp
finally show "todos P ((x # xs) @ y) = (todos P (x # xs) ∧ todos P y)" by simp
qed

(* pablucoto *)
lemma todos_append5:
"todos P (x @ y) = (todos P x ∧ todos P y)"
proof (induct x )
show "todos P ([] @ y) = (todos P [] ∧ todos P y)" by simp
next
fix a x
assume HI:"todos P (x @ y) = (todos P x ∧ todos P y) "
have "todos P ((a # x) @ y) = ( P a ∧ todos P (x @ y))" by simp
also have " ... =( P a ∧ (todos P x ∧ todos P y))" using HI by simp
also have "... = (todos P (a#x) ∧ todos P y)" by simp
finally show "todos P ((a # x) @ y) = (todos P (a # x) ∧ todos P y)" by auto
qed

{*danrodcha*}
lemma todos_append6:
"todos P (x @ y) = (todos P x ∧ todos P y)" (is "?Q x")
proof (induct x)
show "?Q []" by simp
next
fix x a assume HI: "?Q x"
have "todos P ((a # x) @ y) = todos P (a # x @ y)" by simp
also have "… = ((P a) ∧ todos P (x @ y))" by simp
also have "… = ((P a) ∧ (todos P x ∧ todos P y))" using HI by simp
also have "… = (todos P (a # x) ∧ todos P y)" by simp
finally show "?Q (a # x)" by simp
qed

text {*
---------------------------------------------------------------------
Ejercicio 5.1. Demostrar o refutar automáticamente
todos P (rev xs) = todos P xs
---------------------------------------------------------------------
*}

lemma "todos P (rev xs) = todos P xs"
oops

(* migtermor ivamenjim marpoldia1 serrodcal anaprarod paupeddeg *)
lemma "todos P (rev xs) = todos P xs"
apply (induct xs)
apply simp
apply (simp add: todos_append)
apply auto
done

(* ferrenseg crigomgom rubgonmar fraortmoy josgarsan danrodcha lucnovdos*)

lemma "todos P (rev xs) = todos P xs"
by (induct xs) (auto simp add: todos_append)

(* wilmorort *)
lemma "todos P (rev xs) = todos P xs"
by (induct xs,simp,simp add: todos_append,auto)

(* juacabsou *)
lemma "todos P (rev xs) = todos P xs"
apply (induct xs,simp,simp add: todos_append,auto) done

(* dancorgar *)
lemma "todos P (rev xs) = todos P xs"
apply (induct xs)
apply (simp add: todos_append)
apply (simp add: todos_append)
apply auto
done

(* pablucoto * )
lemma "todos P (rev xs) = todos P xs"
by (induct xs) (auto, simp_all add: todos_append)
lemma "todos P (rev xs) = todos P xs"
by (induct xs) ( simp_all add: todos_append, auto)

text {*
---------------------------------------------------------------------
Ejercicio 5.2. Demostrar o refutar detalladamente
todos P (rev xs) = todos P xs
---------------------------------------------------------------------
*}

lemma "todos P (rev xs) = todos P xs"
oops

(* migtermor *)

lemma auxiliar:
"rev (a#xs) = rev xs @ [a]"
by auto

lemma "todos P (rev xs) = todos P xs" (is "?Q xs")
proof (induct xs)
show "?Q []" by (simp only: rev.simps(1))
next
fix a xs
assume HI: "?Q xs"
have "todos P (rev (a#xs)) = (todos P (rev xs @ [a]))" by (simp add: auxiliar)
have "… = ((todos P (rev xs)) ∧ (todos P [a]))" by (simp add: todos_append)
have "… = (todos P (rev xs) ∧ P a)" by simp
also have Aux: "… = (todos P xs ∧ P a)" using HI by simp
have Aux1: "… = (P a ∧ todos P xs)" by arith
have "(todos P (rev xs) ∧ P a) = (P a ∧ todos P xs)" using Aux Aux1 by simp
finally show "todos P (rev (a#xs)) = todos P (a#xs)" by (simp add: todos_append)
qed

(* marpoldia1 ferrenseg crigomgom serrodcal juacabsou rubgonmar josgarsan pablucoto*)
lemma "todos P (rev xs) = todos P xs"
proof (induct xs)
show "todos P (rev []) = todos P []" by simp
next
fix a xs
assume HI: "todos P (rev xs) = todos P xs"
have "todos P (rev (a # xs)) = (todos P ((rev xs)@[a]))" by simp
also have "... = (todos P (rev xs) ∧ todos P [a])" by (simp add:todos_append)
also have "... = (todos P xs ∧ P a)" using HI by simp
also have "... = (P a ∧ todos P xs)" by arith
also have "... = (todos P (a#xs))" by simp
finally show "todos P (rev (a # xs)) = (todos P (a#xs))" by simp
qed

(* ivamenjim *)

lemma "todos P (rev xs) = todos P xs" (is "?P xs")
proof (induct xs)
show "?P []" by simp
next
fix a xs
assume HI: "?P xs"
have "todos P (rev (a # xs)) = (todos P ((rev xs)@[a]))" by simp
also have "... = ((todos P (rev xs)) ∧ todos P [a])" by (simp add: todos_append)
also have "... = (todos P xs ∧ todos P [a])" using HI by simp
also have "... = (todos P [a] ∧ todos P xs)" by arith
finally show "todos P (rev (a # xs)) = todos P (a # xs)" by simp
qed

(* fraortmoy *)

lemma "todos P (rev xs) = todos P xs"
proof (induct xs)
show "todos P (rev []) = todos P []" by simp
next
fix a xs
assume H1: "todos P (rev xs) = todos P xs"
have " todos P (rev (a # xs)) = todos P (rev xs @ [a])" by simp
also have "… = (todos P (rev xs) ∧ todos P [a])" by (simp add:todos_append)
also have "… = (todos P xs ∧ todos P [a])" using H1 by simp
also have "… = (todos P [a] ∧ todos P xs)" by arith
finally show "todos P (rev (a # xs)) = todos P (a#xs)" by simp
qed

(* paupeddeg dancorgar *)
lemma "todos P (rev xs) = todos P xs"
proof (induct xs)
show " todos P (rev []) = todos P []" by simp
next
fix a xs
assume HI: "todos P (rev xs) = todos P xs "
have "todos P (rev (a # xs)) = todos P (rev(xs) @ [a])" by simp
also have "... = (todos P (rev xs) ∧ todos P [a]) " by (simp add: todos_append)
also have "... = (todos P xs ∧ todos P [a])" using HI by simp
also have "... = (todos P [a] ∧ todos P xs)" by arith
also have "... = todos P ([a] @ xs)" by (simp add: todos_append)
finally show "todos P (rev (a # xs)) = todos P (a # xs)" by simp
qed

(* wilmorort *)
lemma "todos P (rev xs) = todos P xs" (is "?P xs")
proof (induct xs)
show "?P []" by simp
next
fix a xs
assume HI: "?P xs"
have "todos P (rev (a#xs)) = todos P ((rev xs) @ [a])" by simp
also have "... = (todos P(rev xs) ∧ todos P [a])" by (simp add: todos_append)
also have "... = (todos P xs ∧ todos P [a])" using HI by simp
also have "... = (todos P [a] ∧ todos P xs)" by (simp add: HOL.conj_commute)
also have "... = todos P([a]@(xs))" by (simp)
finally show "todos P (rev (a#xs))= todos P (a # xs)" by simp
qed

(* anaprarod *)
(* es igual que las anteriores pero con el final también con patrones *)

lemma "todos P (rev xs) = todos P xs" (is "?P xs")
proof (induct xs)
show "?P []" by auto
next
fix a xs
assume HI: "?P xs"
have "todos P (rev (a#xs)) = todos P (rev xs @ [a])" by simp
also have "… = (todos P (rev xs) ∧ todos P [a])" by (simp add: todos_append)
also have "… = (todos P xs ∧ todos P [a])" using HI by simp
also have "… = (todos P [a] ∧ todos P xs)" by arith
also have "… = todos P (a#xs)" by simp
finally show "?P (a # xs)" by simp
qed

{*danrodcha*}
lemma "todos P (rev xs) = todos P xs" (is "?Q xs")
proof (induct xs)
show "?Q []" by simp
next
fix a xs assume HI: "?Q xs"
have "todos P (rev (a#xs)) = todos P ((rev xs) @ [a])" by simp
also have "… = (todos P (rev xs) ∧ todos P [a])" by (simp add:todos_append)
also have "… = (todos P xs ∧ todos P [a])" using HI by simp
also have "… = (todos P [a] ∧ todos P xs)" by (simp add: HOL.conj_commute)
also have "… = todos P (a # xs)" by simp
finally show "?Q (a # xs)" by simp
qed

text {*
---------------------------------------------------------------------
Ejercicio 6. Demostrar o refutar:
algunos (λx. P x ∧ Q x) xs = (algunos P xs ∧ algunos Q xs)
---------------------------------------------------------------------
*}

lemma "algunos (λx. P x ∧ Q x) xs = (algunos P xs ∧ algunos Q xs)"
oops

(* migtermor ivamenjim ferrenseg paupeddeg crigomgom serrodcal wilmorort juacabsou rubgonmar anaprarod marpoldia1 fraortmoy josgarsan danrodcha pablucoto*)
lemma "algunos (λx. P x ∧ Q x) xs = (algunos P xs ∧ algunos Q xs)"
quickcheck
oops
(* Quickcheck encuentra el siguiente contraejemplo: P={a1}, Q={a2}, xs={a1,a2}. En este ejemplo:
· "algunos (λx. P x ∧ Q x) xs = False"
· "(algunos P xs ∧ algunos Q xs) = True" *)

text {*
---------------------------------------------------------------------
Ejercicio 7.1. Demostrar o refutar automáticamente
algunos P (map f xs) = algunos (P ∘ f) xs
---------------------------------------------------------------------
*}

(* ivamenjim migtermor marpoldia1 crigomgom rubgonmar wilmorort anaprarod
fraortmoy juacabsou paupeddeg josgarsan danrodcha pablucoto *)

lemma "algunos P (map f xs) = algunos (P o f) xs"
by (induct xs) auto

(* ferrenseg *)

lemma "algunos P (map f xs) = algunos (P o f) xs"
by (induct xs) simp_all

(* anaprarod *)
lemma "algunos P (map f xs) = algunos (P o f) xs"
apply (induct xs)
apply auto
done

text {*
---------------------------------------------------------------------
Ejercicio 7.2. Demostrar o refutar datalladamente
algunos P (map f xs) = algunos (P ∘ f) xs
---------------------------------------------------------------------
*}

lemma "algunos P (map f xs) = algunos (P o f) xs"
oops

(* migtermor *)

lemma AUX: "algunos (λa. P (f a)) xs = algunos (P o f) xs"
by (induct xs) auto

lemma "algunos P (map f xs) = algunos (P o f) xs" (is "?Q xs")
proof (induct xs)
show "?Q []" by simp
next
fix x xs
assume HI: "?Q xs"
have "algunos P (map f (x#xs)) = (algunos P ((f x)#(map f xs)))" by simp
also have "… = ((P (f x)) ∨ (algunos P (map f xs)))" by (simp only: algunos.simps(2))
also have "… = ((P (f x)) ∨ (algunos (P o f) xs))"
proof (cases)
assume C1: "(P (f x))"
have Aux: "((P (f x)) ∨ (algunos P (map f xs))) = True" using C1 by simp
have Aux1: "… = ((P (f x)) ∨ (algunos (P o f) xs))" using C1 by simp
then show "((P (f x)) ∨ (algunos P (map f xs))) = ((P (f x)) ∨ (algunos (P o f) xs))"
using Aux Aux1 by simp
next
assume C2: "¬(P (f x))"
have Aux2: "((P (f x)) ∨ (algunos P (map f xs))) = (algunos P (map f xs))" using C2 by simp
have Aux3: "… = (algunos (P o f) xs)" using HI by (simp add: AUX)
also have "… = ((P (f x)) ∨ (algunos (P o f) xs))" using C2 by simp
then show "((P (f x)) ∨ (algunos P (map f xs))) = ((P (f x)) ∨ (algunos (P o f) xs))"
using Aux2 Aux3 by simp
qed
also have "… = (((P o f) x) ∨ (algunos (P o f) xs))" by simp
also have "… = (algunos (P o f) (x#xs))" by simp
finally show "algunos P (map f (x#xs)) = (algunos (P o f) (x#xs))" by simp
qed

(* ferrenseg *)

lemma "algunos P (map f xs) = algunos (P ∘ f) xs"
proof (induct xs)
show "algunos P (map f []) = algunos (P ∘ f) []" by simp
next
fix x xs
assume HI: "algunos P (map f xs) = algunos (P ∘ f) xs"
show "algunos P (map f (x # xs)) = algunos (P ∘ f) (x # xs)"
proof -
have "algunos P (map f (x # xs)) = algunos P ((f x) # (map f xs))" by simp
also have "… = ((P (f x)) ∨ (algunos P (map f xs)))" by simp
also have "… = (((P ∘ f) x) ∨ (algunos (P ∘ f) xs))" using HI by simp
also have "… = algunos (P ∘ f) (x # xs)" by simp
finally show ?thesis
qed
qed

(* ivamenjim *)

lemma "algunos P (map f xs) = algunos (P o f) xs"
proof (induct xs)
show "algunos P (map f []) = algunos (P ∘ f) []" by simp
next
fix a xs
assume HI: "algunos P (map f xs) = algunos (P ∘ f) xs"
have "algunos P (map f (a # xs)) = algunos P ((f a)#map f xs)" by simp
also have "... = (algunos P (map f [a]) ∨ algunos P (map f xs))" by simp
also have "... = (algunos P (map f [a]) ∨ algunos (P ∘ f) xs)" using HI by simp
finally show "algunos P (map f (a # xs)) = algunos (P ∘ f) (a # xs)" by simp
qed

(* crigomgom rubgonmar anaprarod marpoldia1 juacabsou danrodcha paupeddeg *)
lemma "algunos P (map f xs) = algunos (P o f) xs"
proof (induct xs)
show "algunos P (map f []) = algunos (P ∘ f) []" by simp
next
fix x xs
assume HI: "algunos P (map f xs) = algunos (P ∘ f) xs"
have "algunos P (map f (x # xs)) = algunos P ((f x) # (map f xs))" by simp
also have "... = (P (f x) ∨ algunos P (map f xs))" by simp
also have "... = (P (f x) ∨ algunos (P ∘ f) xs)" using HI by simp
also have "... = ((P ∘ f) x ∨ algunos (P ∘ f) xs)" by simp
finally show "algunos P (map f (x # xs)) = algunos (P ∘ f) (x # xs)" by simp
qed

(* wilmorort pablucoto*)
lemma "algunos P (map f xs) = algunos (P o f) xs" (is "?P xs" )
proof (induct xs)
show "?P []" by simp
next
fix a xs
assume HI: "algunos P (map f xs) = algunos (P ∘ f) xs"
have "algunos P (map f (a # xs)) = (P (f a) ∨ algunos P (map f xs))" by simp
also have "... = (P (f a) ∨ algunos (P ∘ f) xs )" using HI by simp
also have "... = ((P ∘ f) a ∨ algunos (P ∘ f)xs)" by simp
finally show " algunos P (map f (a # xs)) = algunos (P ∘ f) (a # xs)" by simp
qed

(* fraortmoy *)

lemma "algunos P (map f xs) = algunos (P o f) xs"
proof (induct xs)
show "algunos P (map f []) = algunos (P ∘ f) []" by simp
next
fix a xs
assume H1: "algunos P (map f xs) = algunos (P ∘ f) xs"
have "algunos P (map f (a # xs)) = (algunos P (map f [a]) ∨ algunos P (map f xs))" by simp
also have "… = (algunos (P ∘ f) [a] ∨ algunos (P ∘ f) xs )" using H1 by simp
also have "… = algunos (P ∘ f) (a#xs)" by simp
finally show "algunos P (map f (a # xs)) = algunos (P ∘ f) (a#xs)" by simp
qed

{*danrodcha*}
lemma "algunos P (map f xs) = algunos (P o f) xs" (is "?Q xs")
proof (induct xs)
show "?Q []" by simp
next
fix a xs assume HI: "?Q xs"
have " algunos P (map f (a # xs)) = algunos P ((f a)#(map f xs))" by simp
also have "… = (P (f a) ∨ algunos P (map f xs))" by simp
also have "… = (P (f a) ∨ algunos (P ∘ f) xs)" using HI by blast
also have "… = ((P ∘ f) a ∨ algunos (P ∘ f) xs)" by simp
also have "… = algunos (P ∘ f) (a # xs)" by simp
finally show "?Q (a # xs)" by blast
qed

text {*
---------------------------------------------------------------------
Ejercicio 8.1. Demostrar o refutar automáticamente
algunos P (xs @ ys) = (algunos P xs ∨ algunos P ys)
---------------------------------------------------------------------
*}

(* ivamenjim marpoldia1 paupeddeg crigomgom rubgonmar wilmorort fraortmoy danrodcha pablucoto*)

lemma "algunos P (xs @ ys) = (algunos P xs ∨ algunos P ys)"
by (induct xs) auto

(* ferrenseg *)
lemma "algunos P (xs @ ys) = (algunos P xs ∨ algunos P ys)"
by (induct xs) simp_all

text {*
---------------------------------------------------------------------
Ejercicio 8.2. Demostrar o refutar detalladamente
algunos P (xs @ ys) = (algunos P xs ∨ algunos P ys)
---------------------------------------------------------------------
*}

(* ivamenjim ferrenseg marpoldia1 wilmorort*)

lemma algunos_append:
"algunos P (xs @ ys) = (algunos P xs ∨ algunos P ys)"
proof (induct xs)
show "algunos P ([] @ ys) = (algunos P [] ∨ algunos P ys)" by simp
next
fix a xs
assume HI: "algunos P (xs @ ys) = (algunos P xs ∨ algunos P ys)"
have "algunos P ((a # xs) @ ys) = (P a ∨ (algunos P (xs @ ys)))" by simp
also have "... = (P a ∨ (algunos P xs ∨ algunos P ys))" using HI by simp
finally show "algunos P ((a # xs) @ ys) = (algunos P (a # xs) ∨ algunos P ys)" by simp
qed

(* migtermor crigomgom *)

lemma algunos_append2:
"algunos P (xs @ ys) = (algunos P xs ∨ algunos P ys)" (is "?Q xs")
proof (induct xs)
show "?Q []" by simp
next
fix x xs
assume HI: "?Q xs"
have "algunos P ((x#xs) @ ys) = algunos P (x#(xs @ ys))" by simp
also have "… = ((P x) ∨ (algunos P (xs @ ys)))" by simp
also have "… = ((P x) ∨ (algunos P xs) ∨ (algunos P ys))" using HI by simp
also have "… = ((algunos P (x#xs)) ∨ (algunos P ys))" by simp
finally show "algunos P ((x#xs) @ ys) = (algunos P (x#xs) ∨ algunos P ys)" by simp
qed

(* paupeddeg*)
lemma algunos_append3:
"algunos P (xs @ ys) = (algunos P xs ∨ algunos P ys)"
proof (induct xs)
show "algunos P ([] @ ys) = (algunos P [] ∨ algunos P ys)" by simp
next
fix a xs
assume HI: "algunos P (xs @ ys) = (algunos P xs ∨ algunos P ys)"
have "algunos P ((a # xs) @ ys) = (algunos P [a] ∨ ( algunos P (xs @ ys)))" by simp
also have "... = (algunos P [a] ∨ (algunos P xs ∨ algunos P ys))" using HI by simp
finally show "algunos P ((a # xs) @ ys) = (algunos P (a # xs) ∨ algunos P ys)" by simp
qed

(* fraortmoy *)
lemma algunos_append4:
"algunos P (xs @ ys) = (algunos P xs ∨ algunos P ys)"
proof (induct xs)
show " algunos P ([] @ ys) = (algunos P [] ∨ algunos P ys)" by simp
next
fix a xs
assume H1: "algunos P (xs @ ys) = (algunos P xs ∨ algunos P ys)"
have "algunos P ((a # xs) @ ys) = (algunos P [a] ∨ algunos P (xs @ ys))" by simp
also have "… = (algunos P [a] ∨ algunos P xs ∨ algunos P ys)" using H1 by simp
also have "… = ((algunos P [a] ∨ algunos P xs) ∨ algunos P ys)" by simp
also have "… = (algunos P (a#xs) ∨ algunos P ys)" by simp
finally show "algunos P ((a # xs) @ ys) = (algunos P (a#xs) ∨ algunos P ys)" by simp
qed

{* danrodcha pablucoto*}
lemma algunos_append5:
"algunos P (xs @ ys) = (algunos P xs ∨ algunos P ys)" (is "?Q xs")
proof (induct xs)
show "?Q []" by simp
next
fix a xs assume HI: "?Q xs"
have "algunos P ((a#xs) @ ys) = algunos P (a#(xs @ ys))" by simp
also have "… = (algunos P [a] ∨ algunos P (xs @ ys))" by simp
also have "… = (algunos P [a] ∨ algunos P xs ∨ algunos P ys)" using HI by simp
also have "… = (algunos P (a # xs) ∨ algunos P ys)" by simp
finally show "?Q (a # xs)" by simp
qed

text {*
---------------------------------------------------------------------
Ejercicio 9.1. Demostrar o refutar automáticamente
algunos P (rev xs) = algunos P xs
---------------------------------------------------------------------
*}

(* ivamenjim migtermor marpoldia1 rubgonmar paupeddeg *)

lemma "algunos P (rev xs) = algunos P xs"
apply (induct xs)
apply simp
apply (simp add: algunos_append)
apply auto
done

(* ferrenseg crigomgom danrodcha pablucoto*)
lemma "algunos P (rev xs) = algunos P xs"
by (induct xs) (auto simp add: algunos_append)

(* wilmorort *)
lemma "algunos P (rev xs) = algunos P xs"
by (induct xs,simp,simp add: algunos_append,auto)

text {*
---------------------------------------------------------------------
Ejercicio 9.2. Demostrar o refutar detalladamente
algunos P (rev xs) = algunos P xs
---------------------------------------------------------------------
*}

lemma "algunos P (rev xs) = algunos P xs"
oops

(* migtermor *)
lemma auxiliar1:
"rev (a#xs) = rev xs @ [a]"
by auto

lemma "algunos P (rev xs) = algunos P xs" (is "?Q xs")
proof (induct xs)
show "?Q []" by simp
next
fix x xs
assume HI: "?Q xs"
have "algunos P (rev (x#xs)) = (algunos P (rev xs @ [x]))" using auxiliar1 by simp
also have "… = ((algunos P (rev xs)) ∨ (algunos P [x]))" by (simp add: algunos_append)
also have "… = ((P x) ∨ (algunos P (rev xs)))" by simp arith
also have "… = ((P x) ∨ (algunos P xs))" using HI by simp
finally show "algunos P (rev (x#xs)) = (algunos P (x#xs))" by simp
qed

(* ferrenseg *)

lemma "algunos P (rev xs) = algunos P xs"
proof (induct xs)
show "algunos P (rev []) = algunos P []" by simp
next
fix x xs
assume HI: "algunos P (rev xs) = algunos P xs"
show "algunos P (rev (x # xs)) = algunos P (x # xs)"
proof -
have "algunos P (rev (x # xs)) = algunos P ((rev xs) @ [x])" by simp
also have "… = (algunos P (rev xs) ∨ algunos P [x])"
by (simp add: algunos_append)
also have "… = (algunos P xs ∨ algunos P [x])" using HI by simp
also have "… = (algunos P xs ∨ P x)" by simp
also have "… = (P x ∨ algunos P xs)" by auto
also have "… = algunos P (x # xs)" by simp
finally show ?thesis
qed
qed

(* ivamenjim marpoldia1*)

lemma "algunos P (rev xs) = algunos P xs" (is "?P xs")
proof (induct xs)
show "?P []" by simp
next
fix a xs
assume HI: "?P xs"
have "algunos P (rev (a # xs)) = (algunos P ((rev xs)@[a]))" by simp
also have "... = ((algunos P (rev xs)) ∨ algunos P [a])" by (simp add: algunos_append)
also have "... = (algunos P xs ∨ algunos P [a])" using HI by simp
also have "... = (algunos P [a] ∨ algunos P xs)" by arith
finally show "algunos P (rev (a # xs)) = algunos P (a # xs)" by simp
qed

(*paupeddeg*)
lemma "algunos P (rev xs) = algunos P xs"
proof (induct xs)
show "algunos P (rev []) = algunos P []" by simp
next
fix a xs
assume HI:"algunos P (rev xs) = algunos P xs"
have "algunos P (rev (a # xs)) = algunos P ((rev xs) @ [a])" by simp
also have "... = ((algunos P (rev xs)) ∨ (algunos P [a]))" by (simp add: algunos_append)
also have "... = ((algunos P xs) ∨ (algunos P [a]))" using HI by simp
also have "... = ((algunos P [a]) ∨ (algunos P xs))" by arith
finally show "algunos P (rev (a # xs)) = algunos P (a # xs)" by simp
qed

(*crigomgom *)
lemma "algunos P (rev xs) = algunos P xs"
proof (induct xs)
show "algunos P (rev []) = algunos P []" by simp
next
fix x xs
assume HI: " algunos P (rev xs) = algunos P xs"
have "algunos P (rev (x # xs)) = algunos P ((rev xs) @ [x])" by simp
also have "... = (algunos P (rev xs) ∨ algunos P [x])" by (simp add: algunos_append)
also have "... = (algunos P xs ∨ algunos P [x])" using HI by simp
also have "... = (algunos P xs ∨ P x)" by simp
also have "... = (P x ∨ algunos P xs)" by arith
also have "... = algunos P (x#xs)" by simp
finally show "algunos P (rev (x # xs)) = algunos P (x # xs)" by simp
qed

(*wilmorort*)
lemma "algunos P (rev xs) = algunos P xs" (is "?P xs")
proof (induct xs)
show "?P []" by simp
next
fix a xs
assume HI: "?P xs"
have "algunos P (rev (a#xs)) = algunos P ((rev xs) @ [a])" by simp
also have "... = (algunos P(rev xs) ∨ algunos P [a])" by (simp add: algunos_append)
also have "... = (algunos P xs ∨ algunos P [a])" using HI by simp
also have "... = (algunos P [a] ∨ algunos P xs)" by (simp add: HOL.disj_commute)
also have "... = algunos P([a]@(xs))" by (simp)
finally show "algunos P (rev (a#xs))= algunos P (a # xs)" by simp
qed

{*danrodcha *}
lemma "algunos P (rev xs) = algunos P xs" (is "?Q xs")
proof (induct xs)
show "?Q []" by simp
next
fix a xs assume HI: "?Q xs"
have "algunos P (rev (a#xs)) = algunos P ((rev xs) @ [a])" by simp
also have "… = (algunos P (rev xs) ∨ algunos P [a])" by (simp add:algunos_append)
also have "… = (algunos P xs ∨ algunos P [a])" using HI by simp
also have "… = (algunos P [a] ∨ algunos P xs)" by (simp add: HOL.disj_commute)
also have "… = algunos P (a # xs)" by simp
finally show "?Q (a # xs)" by simp
qed

text {*
---------------------------------------------------------------------
Ejercicio 10. Encontrar un término no trivial Z tal que sea cierta la
siguiente ecuación:
algunos (λx. P x ∨ Q x) xs = Z
y demostrar la equivalencia de forma automática y detallada.
---------------------------------------------------------------------
*}

(* migtermor *)

lemma "algunos (λx. P x ∨ Q x) xs = ((algunos (λx. P x) xs) ∨ (algunos (λx. Q x) xs))"
by (induct xs) auto

lemma "algunos (λx. P x ∨ Q x) xs = ((algunos (λx. P x) xs) ∨ (algunos (λx. Q x) xs))" (is "?R xs")
proof (induct xs)
show "?R []" by simp
next
fix x xs
assume HI: "?R xs"
have "algunos (λx. P x ∨ Q x) (x#xs) = (((λx. P x ∨ Q x) x) ∨ (algunos (λx. P x ∨ Q x) xs))"
by simp
also have "… = ((P x ∨ Q x) ∨ (algunos (λx. P x ∨ Q x) xs))" by simp
also have H1: "… = ((((λx. P x) x) ∨ (algunos (λx. P x) xs)) ∨ (((λx. Q x) x) ∨ (algunos (λx. Q x) xs)))"
using HI by simp arith
have H2: "… = ((algunos (λx. P x) (x#xs)) ∨ (algunos (λx. Q x) (x#xs)))"
by simp
have C: "(algunos (λx. P x ∨ Q x) (x#xs)) =
((algunos (λx. P x) (x#xs)) ∨ (algunos (λx. Q x) (x#xs)))"
using H1 H2 by simp
finally show "(algunos (λx. P x ∨ Q x) (x#xs)) =
((algunos (λx. P x) (x#xs)) ∨ (algunos (λx. Q x) (x#xs)))"
using C by simp
qed

(* ferrenseg *)

lemma "algunos (λx. P x ∨ Q x) xs = (algunos P xs ∨ algunos Q xs)"
proof (induct xs)
show "algunos (λx. P x ∨ Q x) [] = (algunos P [] ∨ algunos Q [])" by simp
next
fix x xs
assume HI: "algunos (λx. (P x ∨ Q x)) xs = (algunos P xs ∨ algunos Q xs)"
show "algunos (λx. P x ∨ Q x) (x # xs) = (algunos P (x # xs) ∨ algunos Q (x # xs))"
proof -
have "algunos (λx. P x ∨ Q x) (x # xs) =
((P x) ∨ (Q x) ∨ algunos (λx. P x ∨ Q x) xs)" by simp
also have "… = ((P x) ∨ (Q x) ∨ algunos P xs ∨ algunos Q xs)"
using HI by simp
also have "… = (((P x) ∨ algunos P xs) ∨ ((Q x) ∨ algunos Q xs))" by auto
also have "… = (algunos P (x # xs) ∨ algunos Q (x # xs))" by simp
finally show ?thesis
qed
qed

(* ivamenjim marpoldia1 paupeddeg *)

lemma "algunos (λx. P x ∨ Q x) xs = (algunos P xs ∨ algunos Q xs)"
by (induct xs) auto

lemma "algunos (λx. P x ∨ Q x) xs = (algunos P xs ∨ algunos Q xs)"
proof (induct xs)
show "algunos (λx. P x ∨ Q x) [] = (algunos P [] ∨ algunos Q [])" by simp
next
fix a xs
assume HI: "algunos (λx. P x ∨ Q x) xs = (algunos P xs ∨ algunos Q xs)"
have "algunos (λx. P x ∨ Q x) (a # xs) = (P a ∨ Q a ∨ algunos (λx. P x ∨ Q x) xs)" by simp
also have "... = (P a ∨ Q a ∨ algunos P xs ∨ algunos Q xs)" using HI by simp
finally show "algunos (λx. P x ∨ Q x) (a # xs) = (algunos P (a # xs) ∨ algunos Q (a # xs))" by auto
qed

(* wilmorort danrodcha*)
lemma "algunos (λx. P x ∨ Q x) xs = (¬todos(λx. ¬P x)xs ∨ ¬todos(λx. ¬Q x)xs)"
by (induct xs) auto

(* wilmorort *)
lemma "algunos (λx. P x ∨ Q x) xs = (¬todos(λx. ¬P x)xs ∨ ¬todos(λx. ¬Q x)xs)"
(is "?P xs")
proof (induct xs)
show "?P []" by simp
next
fix a xs
assume HI: "?P xs"
have "algunos (λx. P x ∨ Q x) (a # xs) = (P a ∨ Q a ∨ algunos (λx. P x ∨ Q x) xs)" by simp
also have "... = (P a ∨ Q a ∨ (¬ todos (λx. ¬ P x) xs ∨ ¬ todos (λx. ¬ Q x) xs))" using HI by simp
also have "... = (P a ∨ ¬ todos (λx. ¬ P x) xs ∨ Q a ∨ ¬ todos (λx. ¬ Q x) xs )" by arith
finally show "algunos (λx. P x ∨ Q x) (a # xs) = (~ todos (λx. ¬ P x) (a # xs) ∨ ~ todos (λx. ¬ Q x) (a # xs))" by simp
qed

{* danrodcha *}
lemma "algunos (λx. P x ∨ Q x) xs = (algunos P xs ∨ algunos Q xs)" (is "?R xs")
proof (induct xs)
show "?R []" by simp
next
fix a xs assume HI: "?R xs"
have 1:" (Q a ∨ algunos P xs) = (algunos P xs ∨ Q a)" by (simp add: HOL.disj_commute)
have "algunos (λx. P x ∨ Q x) (a # xs) =
(algunos P [a] ∨ algunos Q [a] ∨ algunos (λx. P x ∨ Q x) xs)" by simp
also have "… = (P a ∨ (Q a ∨ algunos P xs) ∨ algunos Q xs)" using HI by simp
also have "… = (P a ∨ algunos P xs ∨ Q a ∨ algunos Q xs)" using 1 by simp
also have "… = (algunos P (a # xs) ∨ algunos Q (a # xs))" by simp
finally show "?R (a # xs)" by simp
qed

text {*
---------------------------------------------------------------------
Ejercicio 11.1. Demostrar o refutar automáticamente
algunos P xs = (¬ todos (λx. (¬ P x)) xs)
---------------------------------------------------------------------
*}

(* ivamenjim wilmorort migtermor marpoldia1 crigomgom rubgonmar paupeddeg danrodcha*)

lemma "algunos P xs = (¬ todos (λx. (¬ P x)) xs)"
by (induct xs) auto

(* ferrenseg *)

lemma "algunos P xs = (¬ todos (λx. (¬ P x)) xs)"
by (induct xs) simp_all

text {*
---------------------------------------------------------------------
Ejercicio 11.2. Demostrar o refutar datalladamente
algunos P xs = (¬ todos (λx. (¬ P x)) xs)
---------------------------------------------------------------------
*}

(* ivamenjim migtermor marpoldia1 crigomgom paupeddeg*)

lemma "algunos P xs = (¬ todos (λx. (¬ P x)) xs)"
proof (induct xs)
show "algunos P [] = (¬ todos (λx. ¬ P x) [])" by simp
next
fix a xs
assume HI: "algunos P xs = (¬ todos (λx. ¬ P x) xs)"
have "algunos P (a # xs) = (P a ∨ (algunos P xs))" by simp
also have "... = (P a ∨ (¬ todos (λx. ¬ P x) xs))" using HI by simp
finally show "algunos P (a # xs) = (¬ todos (λx. ¬ P x) (a # xs))" by simp
qed

(* ferrenseg *)

lemma "algunos P xs = (¬ todos (λx. (¬ P x)) xs)"
proof (induct xs)
show "algunos P [] = (¬ todos (λx. (¬ P x)) [])" by simp
next
fix x xs
assume HI: "algunos P xs = (¬ todos (λx. (¬ P x)) xs)"
show "algunos P (x # xs) = (¬ todos (λx. (¬ P x)) (x # xs))"
proof -
have "algunos P (x # xs) = ((P x) ∨ algunos P xs)" by simp
also have "… = ((P x) ∨ ¬ todos (λx. (¬ P x)) xs)" using HI by simp
also have "… = (¬ (¬ (P x) ∧ todos (λx. (¬ P x)) xs))" by simp
also have "… = (¬ todos (λx. (¬ P x)) (x # xs))" by simp
finally show ?thesis
qed
qed

{* danrodcha *}
lemma "algunos P xs = (¬ todos (λx. (¬ P x)) xs)" (is "?Q xs")
proof (induct xs)
show "?Q []" by simp
next
fix a xs assume HI: "?Q xs"
have "algunos P (a # xs) = (P a ∨ algunos P xs)" by simp
also have "… = (P a ∨ (¬ todos (λx. (¬ P x)) xs))" using HI by simp
also have "… = (¬(¬(P a) ∧ todos(λx. (¬ P x)) xs))" by simp
also have "… = (¬ todos(λx. (¬ P x)) (a # xs))" by simp
finally show "?Q (a # xs)" by simp
qed

text {*
---------------------------------------------------------------------
Ejercicio 12. Definir la funcion primitiva recursiva
estaEn :: 'a ⇒ 'a list ⇒ bool
tal que (estaEn x xs) se verifica si el elemento x está en la lista
xs. Por ejemplo,
estaEn (2::nat) [3,2,4] = True
estaEn (1::nat) [3,2,4] = False
---------------------------------------------------------------------
*}

(* ivamenjim ferrenseg migtermor serrodcal crigomgom rubgonmar marpoldia1 paupeddeg danrodcha*)

fun estaEn :: "'a ⇒ 'a list ⇒ bool" where
"estaEn x [] = False"
| "estaEn x (a#xs) = ((a = x) ∨ (estaEn x xs))"

value "estaEn (2::nat) [3,2,4] = True"
value "estaEn (1::nat) [3,2,4] = False"

text {*
---------------------------------------------------------------------
Ejercicio 13. Expresar la relación existente entre estaEn y algunos.
Demostrar dicha relación de forma automática y detallada.
---------------------------------------------------------------------
*}

(* migtermor crigomgom *)

lemma "estaEn x xs = (algunos (λa. a=x) xs)"
by (induct xs) auto

lemma auxiliar13:
"(x=a) = ((λa. a=x) a)"
by auto

lemma "estaEn x xs = (algunos (λa. a=x) xs)" (is "?P xs")
proof (induct xs)
show "?P []" by simp
next
fix a xs
assume HI: "?P xs"
have "estaEn x (a#xs) = ((a=x) ∨ (estaEn x xs))" by simp
also have H: "… = ((a=x) ∨ (algunos (λa. a=x) xs))" using HI by simp
also have "… = (((λa. a=x) a) ∨ (algunos (λa. a=x) xs))" using auxiliar13 by simp
also have "… = (algunos (λa. a=x) (a#xs))" by simp
have C: "estaEn x (a#xs) = (algunos (λa. a=x) (a#xs))" using H by simp
finally show "estaEn x (a#xs) = (algunos (λa. a=x) (a#xs))" using C by simp
qed

(* ferrenseg *)

lemma estaEn_algunos:
"estaEn y xs = algunos (λx. x=y) xs"
by (induct xs) auto

lemma estaEn_algunos_2:
"estaEn y xs = algunos (λx. x=y) xs"
proof (induct xs)
show "estaEn y [] = algunos (λx. x=y) []" by simp
next
fix x xs
assume HI: "estaEn y xs = algunos (λx. x=y) xs"
show "estaEn y (x # xs) = algunos (λx. x=y) (x # xs)"
proof -
have "estaEn y (x # xs) = (y=x ∨ estaEn y xs)" by simp
also have "… = (y=x ∨ algunos (λx. x=y) xs)" using HI by simp
also have "… = (x=y ∨ algunos (λx. x=y) xs)" by auto
also have "… = algunos (λx. x=y) (x # xs)" by simp
finally show ?thesis
qed
qed

(*crigomgom paupeddeg*)

lemma "estaEn a xs = algunos (λx. x = a) xs"
by (induct xs) auto

lemma "estaEn x xs = (algunos (λa. a=x) xs)"
proof (induct xs)
show " estaEn x [] = algunos (λa. a = x) []" by simp
next
fix a xs
assume HI: "estaEn x xs = algunos (λa. a = x) xs"
have "estaEn x (a # xs) = (a = x ∨ estaEn x xs)" by simp
also have "... = (a = x ∨ algunos (λa. a = x) xs)" using HI by simp
finally show " estaEn x (a # xs) = algunos (λa. a = x) (a # xs)" by simp
qed

(* ivamenjim marpoldia1 *)

lemma "estaEn x xs = algunos (λy. x=y) xs"
by (induct xs) auto

lemma "estaEn x xs = algunos (λy. x=y) xs"
proof (induct xs)
show "estaEn x [] = algunos (op = x) []" by simp
next
fix a xs
assume HI: "estaEn x xs = algunos (op = x) xs"
have "estaEn x (a # xs) = ((a = x) ∨ (estaEn x xs))" by simp
also have "... = (a = x ∨ algunos (op = x) xs)" using HI by simp
finally show " estaEn x (a # xs) = algunos (op = x) (a # xs)" by auto
qed

{* danrodcha *}
lemma "estaEn a xs = algunos (op = a) xs"
by (induct xs) auto

{* danrodcha *}
lemma "estaEn a xs = algunos (op = a) xs"
proof (induct xs)
show "estaEn a [] = algunos (op = a) []" by simp
next
fix x xs
assume HI: "estaEn a xs = algunos (op = a) xs"
have "estaEn a (x # xs) = ((x = a) ∨ (estaEn a xs))" by simp
also have "… = (x = a ∨ algunos (op = a) xs)" using HI by simp
also have "… = ((op = a) x ∨ algunos (op = a) xs)" by (simp add:HOL.eq_commute)
also have "… = algunos (op = a) (x # xs)" by simp
finally show "estaEn a (x # xs) = algunos (op = a) (x # xs)" by simp
qed

end
</source>

Relación 4

2016-11-23T00:16:19Z

Pablucoto:

<source lang="isar">
chapter {* R4: Cuantificadores sobre listas *}

theory R4_Cuantificadores_sobre_listas
imports Main
begin

text {*
---------------------------------------------------------------------
Ejercicio 1. Definir la función
todos :: ('a ⇒ bool) ⇒ 'a list ⇒ bool
tal que (todos p xs) se verifica si todos los elementos de la lista
xs cumplen la propiedad p. Por ejemplo, se verifica
todos (λx. 1<length x) [[2,1,4],[1,3]]
¬todos (λx. 1<length x) [[2,1,4],[3]]

Nota: La función todos es equivalente a la predefinida list_all.
---------------------------------------------------------------------
*}

{*danrodcha*}
fun todos :: "('a ⇒ bool) ⇒ 'a list ⇒ bool" where
"todos p xs = "todos p [] = True"
| "todos p (x#xs) = (p x ∧ todos p xs)""

(* ivamenjim migtermor dancorgar wilmorort marpoldia1 ferrenseg paupeddeg pablucoto crigomgom anaprarod serrodcal juacabsou rubgonmar josgarsan fraortmoy lucnovdos*)
fun todos :: "('a ⇒ bool) ⇒ 'a list ⇒ bool" where
"todos p [] = True"
| "todos p (x#xs) = (p x ∧ todos p xs)"

text {*
---------------------------------------------------------------------
Ejercicio 2. Definir la función
algunos :: ('a ⇒ bool) ⇒ 'a list ⇒ bool
tal que (algunos p xs) se verifica si algunos elementos de la lista
xs cumplen la propiedad p. Por ejemplo, se verifica
algunos (λx. 1<length x) [[2,1,4],[3]]
¬algunos (λx. 1<length x) [[],[3]]"

Nota: La función algunos es equivalente a la predefinida list_ex.
---------------------------------------------------------------------
*}

{*danrodcha ivamenjim migtermor dancorgar marpoldia1 ferrenseg wilmorort paupeddeg pablucoto crigomgom anaprarod serrodcal juacabsou rubgonmar josgarsan fraortmoy lucnovdos*}
fun algunos :: "('a ⇒ bool) ⇒ 'a list ⇒ bool" where
"algunos p [] = False"
| "algunos p (x#xs) = (p x ∨ algunos p xs)"

text {*
---------------------------------------------------------------------
Ejercicio 3.1. Demostrar o refutar automáticamente
todos (λx. P x ∧ Q x) xs = (todos P xs ∧ todos Q xs)
---------------------------------------------------------------------
*}

{*danrodcha ivamenjim migtermor dancorgar marpoldia1 ferrenseg wilmorort paupeddeg pablucoto anaprarod serrodcal juacabsou rubgonmar josgarsan fraortmoy lucnovdos*}
lemma "todos (λx. P x ∧ Q x) xs = (todos P xs ∧ todos Q xs)"
by (induct xs) auto

text {*
---------------------------------------------------------------------
Ejercicio 3.2. Demostrar o refutar detalladamente
todos (λx. P x ∧ Q x) xs = (todos P xs ∧ todos Q xs)
---------------------------------------------------------------------
*}

{*danrodcha*}
lemma "todos (λx. P x ∧ Q x) xs = (todos P xs ∧ todos Q xs)"
proof (induct xs)
show "?R []" by simp
next
fix a xs
assume HI: "?R xs"
have "todos (λx. P x ∧ Q x) (a#xs) = (P a ∧ Q a ∧ todos (λx. P x ∧ Q x) xs)" by simp
also have "… = (P a ∧ Q a ∧ (todos P xs ∧ todos Q xs))" using HI by simp
also have "… = ((P a ∧ todos P xs) ∧ (Q a ∧ todos Q xs))" by blast
also have "… = (todos P (a#xs) ∧ todos Q (a#xs))" by simp
finally show "?R (a#xs)" by simp
qed

(* ivamenjim wilmorort serrodcal josgarsan*)

lemma "todos (λx. P x ∧ Q x) xs = (todos P xs ∧ todos Q xs)"
proof (induct xs)
show "todos (λx. P x ∧ Q x) [] = (todos P [] ∧ todos Q [])" by simp
next
fix a xs
assume HI: "todos (λx. P x ∧ Q x) xs = (todos P xs ∧ todos Q xs)"
have "todos (λx. P x ∧ Q x) (a # xs) = (P a ∧ Q a ∧ todos (λx. P x ∧ Q x) xs)" by simp
also have "... = (P a ∧ Q a ∧ todos P xs ∧ todos Q xs)" using HI by simp
finally show "todos (λx. P x ∧ Q x) (a # xs) = (todos P (a # xs) ∧ todos Q (a # xs))" by auto
qed

(* dancorgar paupeddeg *)
lemma "todos (λx. P x ∧ Q x) xs = (todos P xs ∧ todos Q xs)"
proof (induct xs)
show "todos (λx. P x ∧ Q x) [] = (todos P [] ∧ todos Q [])" by simp
next
fix y xs
assume HI: "todos (λx. P x ∧ Q x) xs = (todos P xs ∧ todos Q xs)"
have "todos (λx. P x ∧ Q x) (y#xs) = ((P y ∧ Q y) ∧ (todos (λx. P x ∧ Q x) xs))" by simp
also have "… = ((P y ∧ Q y) ∧ (todos P xs ∧ todos Q xs))" using HI by simp
also have "… = ((P y ∧ todos P xs) ∧ (Q y ∧ todos Q xs))" by blast
also have "… = ((todos P (y#xs)) ∧ (todos Q (y#xs)))" by simp
finally show "todos (λx. P x ∧ Q x) (y#xs) = (todos P (y#xs) ∧ todos Q (y#xs))" by simp
qed

(* migtermor *)
lemma "todos (λx. P x ∧ Q x) xs = (todos P xs ∧ todos Q xs)"
proof (induct xs)
show "todos (λx. P x ∧ Q x) [] = (todos P [] ∧ todos Q [])" by simp
next
fix x xs
assume HI: "todos (λx. P x ∧ Q x) xs = (todos P xs ∧ todos Q xs)"
have "(todos (λx. P x ∧ Q x) (x#xs)) = (( P x ∧ Q x) ∧ (todos (λx. P x ∧ Q x) xs))"
by (simp only: todos.simps(2))
also have "… = ((P x ∧ Q x) ∧ (todos P xs ∧ todos Q xs))" using HI by simp
also have "… = (((P x)∧(todos P xs)) ∧ ((Q x) ∧ (todos Q xs)))" by arith
also have "((P x ∧ Q x) ∧ (todos P xs ∧ todos Q xs)) = (((P x)∧(todos P xs)) ∧ ((Q x) ∧ (todos Q xs)))"
by arith (* Este paso es exactamente el mismo que el anterior, pero sin cualquiera de los dos no funciona el "finally show" *)
have "… = (((P x)∧(todos P xs))∧((Q x)∧(todos Q xs)))" by simp
have "… = ((todos P (x#xs))∧(todos Q (x#xs)))" by simp
finally show "(todos (λx. P x ∧ Q x) (x#xs)) = ((todos P (x#xs)) ∧ (todos Q (x#xs)))" by simp
qed

(* marpoldia1 *)
lemma "todos (λx. P x ∧ Q x) xs = (todos P xs ∧ todos Q xs)"
proof (induct xs)
show "todos (λx. P x ∧ Q x) [] = (todos P [] ∧ todos Q [])" by simp
next
fix a xs
assume HI: "todos (λx. P x ∧ Q x) xs = (todos P xs ∧ todos Q xs)"
have "todos (λx. P x ∧ Q x) (a # xs) = ((P a ∧ Q a) ∧ (todos P xs ∧ todos Q xs))" using HI by simp
also have "... = ((P a ∧ todos P xs) ∧ (Q a ∧ todos Q xs))" by blast
also have "... = (todos P (a#xs) ∧ todos Q (a#xs)) " by simp
finally show "todos (λx. P x ∧ Q x) (a#xs) = (todos P (a#xs) ∧ todos Q (a#xs))" by simp
qed

(* ferrenseg *)

lemma "todos (λx. P x ∧ Q x) xs = (todos P xs ∧ todos Q xs)"
proof (induct xs)
show "todos (λx. P x ∧ Q x) [] = (todos P [] ∧ todos Q [])" by simp
next
fix n xs
assume HI: "todos (λx. P x ∧ Q x) xs = (todos P xs ∧ todos Q xs)"
have "todos (λx. P x ∧ Q x) (n # xs) = (P n ∧ Q n ∧ todos P xs ∧ todos Q xs)" using HI by simp
also have "… = P n ∧ Q n ∧ todos P xs ∧ todos Q xs" by blast
also have "⋯ = todos P (n # xs) ∧ todos Q (n # xs)" by simp
finally show "todos (λx. P x ∧ Q x) (n # xs) = todos P (n # xs) ∧ todos Q (n # xs)" by simp
qed

(* lucnovdos *)

lemma "todos (λx. P x ∧ Q x) xs = (todos P xs ∧ todos Q xs)"
proof (induct xs)
show "todos (λx. P x ∧ Q x) [] = (todos P [] ∧ todos Q [])" by simp
next
fix n xs
assume HI: "todos (λx. P x ∧ Q x) xs = (todos P xs ∧ todos Q xs)"
have "todos (λx. P x ∧ Q x) (n # xs) = ((P n ∧ Q n) ∧ (todos P xs ∧ todos Q xs))" using HI by simp
also have "... = ((P n ∧ todos P xs) ∧ (Q n ∧ todos Q xs))" by arith
also have "... = ((todos P(n#xs)) ∧ (todos Q(n#xs)))" by simp

finally show "todos (λx. P x ∧ Q x) (n#xs) = (todos P (n#xs) ∧ todos Q (n#xs))" by simp
qed

{* wilmorort pablucoto crigomgom anaprarod juacabsou rubgonmar fraortmoy *}

lemma "todos (λx. P x ∧ Q x) xs = (todos P xs ∧ todos Q xs)"
proof (induct xs)
show "todos (λx. P x ∧ Q x) [] = (todos P [] ∧ todos Q [])" by simp
next
fix a xs
assume HI: "todos (λx. P x ∧ Q x) xs = (todos P xs ∧ todos Q xs)"
have "todos (λx. P x ∧ Q x) (a # xs) = (P a ∧ Q a ∧ todos (λx. P x ∧ Q x) xs)" by simp
also have "... = (P a ∧ Q a ∧ todos P xs ∧ todos Q xs)" using HI by simp
also have "... = (P a ∧ todos P xs ∧ Q a ∧ todos Q xs)" by arith
also have "... = (todos P (a # xs) ∧ todos Q (a # xs))" by simp (* Este paso se puede obviar*)
finally show "todos (λx. P x ∧ Q x) (a # xs) = (todos P (a # xs) ∧ todos Q (a # xs))" by simp
qed

text {*
---------------------------------------------------------------------
Ejercicio 4.1. Demostrar o refutar automáticamente
todos P (x @ y) = (todos P x ∧ todos P y)
---------------------------------------------------------------------
*}
{*danrodcha ivamenjim marpoldia1 migtermor ferrenseg wilmorort paupeddeg crigomgom anaprarod serrodcal juacabsou rubgonmar pablucoto fraortmoy josgarsan lucnovdos*}
lemma "todos P (x @ y) = (todos P x ∧ todos P y)"
by (induct x) auto

{* anaprarod dancorgar *}
lemma "todos P (x @ y) = (todos P x ∧ todos P y)"
apply (induct x)
apply auto
done

text {*
---------------------------------------------------------------------
Ejercicio 4.2. Demostrar o refutar detalladamente
todos P (x @ y) = (todos P x ∧ todos P y)
---------------------------------------------------------------------
*}

(* ivamenjim ferrenseg wilmorort dancorgar fraortmoy josgarsan lucnovdos*)

lemma todos_append:
"todos P (x @ y) = (todos P x ∧ todos P y)"
proof (induct x)
show "todos P ([] @ y) = (todos P [] ∧ todos P y)" by simp
next
fix a x
assume HI: "todos P (x @ y) = (todos P x ∧ todos P y)"
have "todos P ((a#x) @ y) = (P a ∧ (todos P (x @ y)))" by simp
also have "... = (P a ∧ (todos P x ∧ todos P y))" using HI by simp
finally show "todos P ((a # x) @ y) = (todos P (a # x) ∧ todos P y)" by simp
qed

(* migtermor serrodcal *)

lemma todos_append1:
"todos P (x @ y) = (todos P x ∧ todos P y)" (is "?P x")
proof (induct x)
show "?P []" by simp
next
fix a x
assume HI: "?P x"
have "todos P ((a#x) @ y) = (P a ∧ (todos P (x @ y)))" by simp
also have "… = (P a ∧ (todos P x ∧ todos P y))" using HI by simp
finally show "?P (a#x)" by simp
qed

(* marpoldia1 *)
lemma todos_append:
"todos P (x @ y) = (todos P x ∧ todos P y)"
proof (induct x)
show "todos P ([] @ y) = (todos P [] ∧ todos P y)" by simp
next
fix a x
assume HI: "todos P (x @ y) = (todos P x ∧ todos P y)"
have "todos P ((a # x) @ y) = todos P (a#(x@y))" by simp
also have "... = (P a ∧ todos P (x @ y))" by simp
also have "... = (P a ∧ (todos P x ∧ todos P y))" using HI by simp
finally show "todos P ((a # x) @ y) = (todos P (a # x) ∧ todos P y)" by simp
qed

(* paupeddeg *)
lemma todos_append3:
"todos P (x @ y) = (todos P x ∧ todos P y)"
proof (induct x)
show "todos P ([] @ y) = (todos P [] ∧ todos P y)" by simp
next
fix a x
assume HI: "todos P (x @ y) = (todos P x ∧ todos P y)"
have "todos P ((a # x) @ y) = todos P (a # (x @ y))" by simp
also have "... = (todos P [a] ∧ todos P (x @ y)) " by simp
also have "... = (todos P [a] ∧ (todos P x ∧ todos P y))" using HI by simp
finally show "todos P ((a # x) @ y) = (todos P (a # x) ∧ todos P y)" by simp
qed

(* crigomgom juacabsou anaprarod*)

lemma todos_append4:
"todos P (x @ y) = (todos P x ∧ todos P y)"
proof (induct x)
show "todos P ([] @ y) = (todos P [] ∧ todos P y)" by simp
next
fix x xs
assume HI:"todos P (xs @ y) = (todos P xs ∧ todos P y)"
have "todos P ((x # xs) @ y) = (P x ∧ todos P (xs @ y ))" by simp
also have "... = (P x ∧ (todos P xs ∧ todos P y)) " using HI by simp
also have "... = ((P x ∧ todos P xs) ∧ todos P y)" by simp
finally show "todos P ((x # xs) @ y) = (todos P (x # xs) ∧ todos P y)" by simp
qed

(* pablucoto *)
lemma todos_append5:
"todos P (x @ y) = (todos P x ∧ todos P y)"
proof (induct x )
show "todos P ([] @ y) = (todos P [] ∧ todos P y)" by simp
next
fix a x
assume HI:"todos P (x @ y) = (todos P x ∧ todos P y) "
have "todos P ((a # x) @ y) = ( P a ∧ todos P (x @ y))" by simp
also have " ... =( P a ∧ (todos P x ∧ todos P y))" using HI by simp
also have "... = (todos P (a#x) ∧ todos P y)" by simp
finally show "todos P ((a # x) @ y) = (todos P (a # x) ∧ todos P y)" by auto
qed

{*danrodcha*}
lemma todos_append6:
"todos P (x @ y) = (todos P x ∧ todos P y)" (is "?Q x")
proof (induct x)
show "?Q []" by simp
next
fix x a assume HI: "?Q x"
have "todos P ((a # x) @ y) = todos P (a # x @ y)" by simp
also have "… = ((P a) ∧ todos P (x @ y))" by simp
also have "… = ((P a) ∧ (todos P x ∧ todos P y))" using HI by simp
also have "… = (todos P (a # x) ∧ todos P y)" by simp
finally show "?Q (a # x)" by simp
qed

text {*
---------------------------------------------------------------------
Ejercicio 5.1. Demostrar o refutar automáticamente
todos P (rev xs) = todos P xs
---------------------------------------------------------------------
*}

lemma "todos P (rev xs) = todos P xs"
oops

(* migtermor ivamenjim marpoldia1 serrodcal anaprarod paupeddeg *)
lemma "todos P (rev xs) = todos P xs"
apply (induct xs)
apply simp
apply (simp add: todos_append)
apply auto
done

(* ferrenseg crigomgom rubgonmar fraortmoy josgarsan danrodcha lucnovdos*)

lemma "todos P (rev xs) = todos P xs"
by (induct xs) (auto simp add: todos_append)

(* wilmorort *)
lemma "todos P (rev xs) = todos P xs"
by (induct xs,simp,simp add: todos_append,auto)

(* juacabsou *)
lemma "todos P (rev xs) = todos P xs"
apply (induct xs,simp,simp add: todos_append,auto) done

(* dancorgar *)
lemma "todos P (rev xs) = todos P xs"
apply (induct xs)
apply (simp add: todos_append)
apply (simp add: todos_append)
apply auto
done

(* pablucoto * )
lemma "todos P (rev xs) = todos P xs"
by (induct xs) (auto, simp_all add: todos_append)
lemma "todos P (rev xs) = todos P xs"
by (induct xs) ( simp_all add: todos_append, auto)

text {*
---------------------------------------------------------------------
Ejercicio 5.2. Demostrar o refutar detalladamente
todos P (rev xs) = todos P xs
---------------------------------------------------------------------
*}

lemma "todos P (rev xs) = todos P xs"
oops

(* migtermor *)

lemma auxiliar:
"rev (a#xs) = rev xs @ [a]"
by auto

lemma "todos P (rev xs) = todos P xs" (is "?Q xs")
proof (induct xs)
show "?Q []" by (simp only: rev.simps(1))
next
fix a xs
assume HI: "?Q xs"
have "todos P (rev (a#xs)) = (todos P (rev xs @ [a]))" by (simp add: auxiliar)
have "… = ((todos P (rev xs)) ∧ (todos P [a]))" by (simp add: todos_append)
have "… = (todos P (rev xs) ∧ P a)" by simp
also have Aux: "… = (todos P xs ∧ P a)" using HI by simp
have Aux1: "… = (P a ∧ todos P xs)" by arith
have "(todos P (rev xs) ∧ P a) = (P a ∧ todos P xs)" using Aux Aux1 by simp
finally show "todos P (rev (a#xs)) = todos P (a#xs)" by (simp add: todos_append)
qed

(* marpoldia1 ferrenseg crigomgom serrodcal juacabsou rubgonmar josgarsan pablucoto*)
lemma "todos P (rev xs) = todos P xs"
proof (induct xs)
show "todos P (rev []) = todos P []" by simp
next
fix a xs
assume HI: "todos P (rev xs) = todos P xs"
have "todos P (rev (a # xs)) = (todos P ((rev xs)@[a]))" by simp
also have "... = (todos P (rev xs) ∧ todos P [a])" by (simp add:todos_append)
also have "... = (todos P xs ∧ P a)" using HI by simp
also have "... = (P a ∧ todos P xs)" by arith
also have "... = (todos P (a#xs))" by simp
finally show "todos P (rev (a # xs)) = (todos P (a#xs))" by simp
qed

(* ivamenjim *)

lemma "todos P (rev xs) = todos P xs" (is "?P xs")
proof (induct xs)
show "?P []" by simp
next
fix a xs
assume HI: "?P xs"
have "todos P (rev (a # xs)) = (todos P ((rev xs)@[a]))" by simp
also have "... = ((todos P (rev xs)) ∧ todos P [a])" by (simp add: todos_append)
also have "... = (todos P xs ∧ todos P [a])" using HI by simp
also have "... = (todos P [a] ∧ todos P xs)" by arith
finally show "todos P (rev (a # xs)) = todos P (a # xs)" by simp
qed

(* fraortmoy *)

lemma "todos P (rev xs) = todos P xs"
proof (induct xs)
show "todos P (rev []) = todos P []" by simp
next
fix a xs
assume H1: "todos P (rev xs) = todos P xs"
have " todos P (rev (a # xs)) = todos P (rev xs @ [a])" by simp
also have "… = (todos P (rev xs) ∧ todos P [a])" by (simp add:todos_append)
also have "… = (todos P xs ∧ todos P [a])" using H1 by simp
also have "… = (todos P [a] ∧ todos P xs)" by arith
finally show "todos P (rev (a # xs)) = todos P (a#xs)" by simp
qed

(* paupeddeg dancorgar *)
lemma "todos P (rev xs) = todos P xs"
proof (induct xs)
show " todos P (rev []) = todos P []" by simp
next
fix a xs
assume HI: "todos P (rev xs) = todos P xs "
have "todos P (rev (a # xs)) = todos P (rev(xs) @ [a])" by simp
also have "... = (todos P (rev xs) ∧ todos P [a]) " by (simp add: todos_append)
also have "... = (todos P xs ∧ todos P [a])" using HI by simp
also have "... = (todos P [a] ∧ todos P xs)" by arith
also have "... = todos P ([a] @ xs)" by (simp add: todos_append)
finally show "todos P (rev (a # xs)) = todos P (a # xs)" by simp
qed

(* wilmorort *)
lemma "todos P (rev xs) = todos P xs" (is "?P xs")
proof (induct xs)
show "?P []" by simp
next
fix a xs
assume HI: "?P xs"
have "todos P (rev (a#xs)) = todos P ((rev xs) @ [a])" by simp
also have "... = (todos P(rev xs) ∧ todos P [a])" by (simp add: todos_append)
also have "... = (todos P xs ∧ todos P [a])" using HI by simp
also have "... = (todos P [a] ∧ todos P xs)" by (simp add: HOL.conj_commute)
also have "... = todos P([a]@(xs))" by (simp)
finally show "todos P (rev (a#xs))= todos P (a # xs)" by simp
qed

(* anaprarod *)
(* es igual que las anteriores pero con el final también con patrones *)

lemma "todos P (rev xs) = todos P xs" (is "?P xs")
proof (induct xs)
show "?P []" by auto
next
fix a xs
assume HI: "?P xs"
have "todos P (rev (a#xs)) = todos P (rev xs @ [a])" by simp
also have "… = (todos P (rev xs) ∧ todos P [a])" by (simp add: todos_append)
also have "… = (todos P xs ∧ todos P [a])" using HI by simp
also have "… = (todos P [a] ∧ todos P xs)" by arith
also have "… = todos P (a#xs)" by simp
finally show "?P (a # xs)" by simp
qed

{*danrodcha*}
lemma "todos P (rev xs) = todos P xs" (is "?Q xs")
proof (induct xs)
show "?Q []" by simp
next
fix a xs assume HI: "?Q xs"
have "todos P (rev (a#xs)) = todos P ((rev xs) @ [a])" by simp
also have "… = (todos P (rev xs) ∧ todos P [a])" by (simp add:todos_append)
also have "… = (todos P xs ∧ todos P [a])" using HI by simp
also have "… = (todos P [a] ∧ todos P xs)" by (simp add: HOL.conj_commute)
also have "… = todos P (a # xs)" by simp
finally show "?Q (a # xs)" by simp
qed

text {*
---------------------------------------------------------------------
Ejercicio 6. Demostrar o refutar:
algunos (λx. P x ∧ Q x) xs = (algunos P xs ∧ algunos Q xs)
---------------------------------------------------------------------
*}

lemma "algunos (λx. P x ∧ Q x) xs = (algunos P xs ∧ algunos Q xs)"
oops

(* migtermor ivamenjim ferrenseg paupeddeg crigomgom serrodcal wilmorort juacabsou rubgonmar anaprarod marpoldia1 fraortmoy josgarsan danrodcha pablucoto*)
lemma "algunos (λx. P x ∧ Q x) xs = (algunos P xs ∧ algunos Q xs)"
quickcheck
oops
(* Quickcheck encuentra el siguiente contraejemplo: P={a1}, Q={a2}, xs={a1,a2}. En este ejemplo:
· "algunos (λx. P x ∧ Q x) xs = False"
· "(algunos P xs ∧ algunos Q xs) = True" *)

text {*
---------------------------------------------------------------------
Ejercicio 7.1. Demostrar o refutar automáticamente
algunos P (map f xs) = algunos (P ∘ f) xs
---------------------------------------------------------------------
*}

(* ivamenjim migtermor marpoldia1 crigomgom rubgonmar wilmorort anaprarod
fraortmoy juacabsou paupeddeg josgarsan danrodcha pablucoto *)

lemma "algunos P (map f xs) = algunos (P o f) xs"
by (induct xs) auto

(* ferrenseg *)

lemma "algunos P (map f xs) = algunos (P o f) xs"
by (induct xs) simp_all

(* anaprarod *)
lemma "algunos P (map f xs) = algunos (P o f) xs"
apply (induct xs)
apply auto
done

text {*
---------------------------------------------------------------------
Ejercicio 7.2. Demostrar o refutar datalladamente
algunos P (map f xs) = algunos (P ∘ f) xs
---------------------------------------------------------------------
*}

lemma "algunos P (map f xs) = algunos (P o f) xs"
oops

(* migtermor *)

lemma AUX: "algunos (λa. P (f a)) xs = algunos (P o f) xs"
by (induct xs) auto

lemma "algunos P (map f xs) = algunos (P o f) xs" (is "?Q xs")
proof (induct xs)
show "?Q []" by simp
next
fix x xs
assume HI: "?Q xs"
have "algunos P (map f (x#xs)) = (algunos P ((f x)#(map f xs)))" by simp
also have "… = ((P (f x)) ∨ (algunos P (map f xs)))" by (simp only: algunos.simps(2))
also have "… = ((P (f x)) ∨ (algunos (P o f) xs))"
proof (cases)
assume C1: "(P (f x))"
have Aux: "((P (f x)) ∨ (algunos P (map f xs))) = True" using C1 by simp
have Aux1: "… = ((P (f x)) ∨ (algunos (P o f) xs))" using C1 by simp
then show "((P (f x)) ∨ (algunos P (map f xs))) = ((P (f x)) ∨ (algunos (P o f) xs))"
using Aux Aux1 by simp
next
assume C2: "¬(P (f x))"
have Aux2: "((P (f x)) ∨ (algunos P (map f xs))) = (algunos P (map f xs))" using C2 by simp
have Aux3: "… = (algunos (P o f) xs)" using HI by (simp add: AUX)
also have "… = ((P (f x)) ∨ (algunos (P o f) xs))" using C2 by simp
then show "((P (f x)) ∨ (algunos P (map f xs))) = ((P (f x)) ∨ (algunos (P o f) xs))"
using Aux2 Aux3 by simp
qed
also have "… = (((P o f) x) ∨ (algunos (P o f) xs))" by simp
also have "… = (algunos (P o f) (x#xs))" by simp
finally show "algunos P (map f (x#xs)) = (algunos (P o f) (x#xs))" by simp
qed

(* ferrenseg *)

lemma "algunos P (map f xs) = algunos (P ∘ f) xs"
proof (induct xs)
show "algunos P (map f []) = algunos (P ∘ f) []" by simp
next
fix x xs
assume HI: "algunos P (map f xs) = algunos (P ∘ f) xs"
show "algunos P (map f (x # xs)) = algunos (P ∘ f) (x # xs)"
proof -
have "algunos P (map f (x # xs)) = algunos P ((f x) # (map f xs))" by simp
also have "… = ((P (f x)) ∨ (algunos P (map f xs)))" by simp
also have "… = (((P ∘ f) x) ∨ (algunos (P ∘ f) xs))" using HI by simp
also have "… = algunos (P ∘ f) (x # xs)" by simp
finally show ?thesis
qed
qed

(* ivamenjim *)

lemma "algunos P (map f xs) = algunos (P o f) xs"
proof (induct xs)
show "algunos P (map f []) = algunos (P ∘ f) []" by simp
next
fix a xs
assume HI: "algunos P (map f xs) = algunos (P ∘ f) xs"
have "algunos P (map f (a # xs)) = algunos P ((f a)#map f xs)" by simp
also have "... = (algunos P (map f [a]) ∨ algunos P (map f xs))" by simp
also have "... = (algunos P (map f [a]) ∨ algunos (P ∘ f) xs)" using HI by simp
finally show "algunos P (map f (a # xs)) = algunos (P ∘ f) (a # xs)" by simp
qed

(* crigomgom rubgonmar anaprarod marpoldia1 juacabsou danrodcha paupeddeg *)
lemma "algunos P (map f xs) = algunos (P o f) xs"
proof (induct xs)
show "algunos P (map f []) = algunos (P ∘ f) []" by simp
next
fix x xs
assume HI: "algunos P (map f xs) = algunos (P ∘ f) xs"
have "algunos P (map f (x # xs)) = algunos P ((f x) # (map f xs))" by simp
also have "... = (P (f x) ∨ algunos P (map f xs))" by simp
also have "... = (P (f x) ∨ algunos (P ∘ f) xs)" using HI by simp
also have "... = ((P ∘ f) x ∨ algunos (P ∘ f) xs)" by simp
finally show "algunos P (map f (x # xs)) = algunos (P ∘ f) (x # xs)" by simp
qed

(* wilmorort pablucoto*)
lemma "algunos P (map f xs) = algunos (P o f) xs" (is "?P xs" )
proof (induct xs)
show "?P []" by simp
next
fix a xs
assume HI: "algunos P (map f xs) = algunos (P ∘ f) xs"
have "algunos P (map f (a # xs)) = (P (f a) ∨ algunos P (map f xs))" by simp
also have "... = (P (f a) ∨ algunos (P ∘ f) xs )" using HI by simp
also have "... = ((P ∘ f) a ∨ algunos (P ∘ f)xs)" by simp
finally show " algunos P (map f (a # xs)) = algunos (P ∘ f) (a # xs)" by simp
qed

(* fraortmoy *)

lemma "algunos P (map f xs) = algunos (P o f) xs"
proof (induct xs)
show "algunos P (map f []) = algunos (P ∘ f) []" by simp
next
fix a xs
assume H1: "algunos P (map f xs) = algunos (P ∘ f) xs"
have "algunos P (map f (a # xs)) = (algunos P (map f [a]) ∨ algunos P (map f xs))" by simp
also have "… = (algunos (P ∘ f) [a] ∨ algunos (P ∘ f) xs )" using H1 by simp
also have "… = algunos (P ∘ f) (a#xs)" by simp
finally show "algunos P (map f (a # xs)) = algunos (P ∘ f) (a#xs)" by simp
qed

{*danrodcha*}
lemma "algunos P (map f xs) = algunos (P o f) xs" (is "?Q xs")
proof (induct xs)
show "?Q []" by simp
next
fix a xs assume HI: "?Q xs"
have " algunos P (map f (a # xs)) = algunos P ((f a)#(map f xs))" by simp
also have "… = (P (f a) ∨ algunos P (map f xs))" by simp
also have "… = (P (f a) ∨ algunos (P ∘ f) xs)" using HI by blast
also have "… = ((P ∘ f) a ∨ algunos (P ∘ f) xs)" by simp
also have "… = algunos (P ∘ f) (a # xs)" by simp
finally show "?Q (a # xs)" by blast
qed

text {*
---------------------------------------------------------------------
Ejercicio 8.1. Demostrar o refutar automáticamente
algunos P (xs @ ys) = (algunos P xs ∨ algunos P ys)
---------------------------------------------------------------------
*}

(* ivamenjim marpoldia1 paupeddeg crigomgom rubgonmar wilmorort fraortmoy danrodcha pablucoto*)

lemma "algunos P (xs @ ys) = (algunos P xs ∨ algunos P ys)"
by (induct xs) auto

(* ferrenseg *)
lemma "algunos P (xs @ ys) = (algunos P xs ∨ algunos P ys)"
by (induct xs) simp_all

text {*
---------------------------------------------------------------------
Ejercicio 8.2. Demostrar o refutar detalladamente
algunos P (xs @ ys) = (algunos P xs ∨ algunos P ys)
---------------------------------------------------------------------
*}

(* ivamenjim ferrenseg marpoldia1 wilmorort*)

lemma algunos_append:
"algunos P (xs @ ys) = (algunos P xs ∨ algunos P ys)"
proof (induct xs)
show "algunos P ([] @ ys) = (algunos P [] ∨ algunos P ys)" by simp
next
fix a xs
assume HI: "algunos P (xs @ ys) = (algunos P xs ∨ algunos P ys)"
have "algunos P ((a # xs) @ ys) = (P a ∨ (algunos P (xs @ ys)))" by simp
also have "... = (P a ∨ (algunos P xs ∨ algunos P ys))" using HI by simp
finally show "algunos P ((a # xs) @ ys) = (algunos P (a # xs) ∨ algunos P ys)" by simp
qed

(* migtermor crigomgom *)

lemma algunos_append2:
"algunos P (xs @ ys) = (algunos P xs ∨ algunos P ys)" (is "?Q xs")
proof (induct xs)
show "?Q []" by simp
next
fix x xs
assume HI: "?Q xs"
have "algunos P ((x#xs) @ ys) = algunos P (x#(xs @ ys))" by simp
also have "… = ((P x) ∨ (algunos P (xs @ ys)))" by simp
also have "… = ((P x) ∨ (algunos P xs) ∨ (algunos P ys))" using HI by simp
also have "… = ((algunos P (x#xs)) ∨ (algunos P ys))" by simp
finally show "algunos P ((x#xs) @ ys) = (algunos P (x#xs) ∨ algunos P ys)" by simp
qed

(* paupeddeg*)
lemma algunos_append3:
"algunos P (xs @ ys) = (algunos P xs ∨ algunos P ys)"
proof (induct xs)
show "algunos P ([] @ ys) = (algunos P [] ∨ algunos P ys)" by simp
next
fix a xs
assume HI: "algunos P (xs @ ys) = (algunos P xs ∨ algunos P ys)"
have "algunos P ((a # xs) @ ys) = (algunos P [a] ∨ ( algunos P (xs @ ys)))" by simp
also have "... = (algunos P [a] ∨ (algunos P xs ∨ algunos P ys))" using HI by simp
finally show "algunos P ((a # xs) @ ys) = (algunos P (a # xs) ∨ algunos P ys)" by simp
qed

(* fraortmoy *)
lemma algunos_append4:
"algunos P (xs @ ys) = (algunos P xs ∨ algunos P ys)"
proof (induct xs)
show " algunos P ([] @ ys) = (algunos P [] ∨ algunos P ys)" by simp
next
fix a xs
assume H1: "algunos P (xs @ ys) = (algunos P xs ∨ algunos P ys)"
have "algunos P ((a # xs) @ ys) = (algunos P [a] ∨ algunos P (xs @ ys))" by simp
also have "… = (algunos P [a] ∨ algunos P xs ∨ algunos P ys)" using H1 by simp
also have "… = ((algunos P [a] ∨ algunos P xs) ∨ algunos P ys)" by simp
also have "… = (algunos P (a#xs) ∨ algunos P ys)" by simp
finally show "algunos P ((a # xs) @ ys) = (algunos P (a#xs) ∨ algunos P ys)" by simp
qed

{* danrodcha *}
lemma algunos_append5:
"algunos P (xs @ ys) = (algunos P xs ∨ algunos P ys)" (is "?Q xs")
proof (induct xs)
show "?Q []" by simp
next
fix a xs assume HI: "?Q xs"
have "algunos P ((a#xs) @ ys) = algunos P (a#(xs @ ys))" by simp
also have "… = (algunos P [a] ∨ algunos P (xs @ ys))" by simp
also have "… = (algunos P [a] ∨ algunos P xs ∨ algunos P ys)" using HI by simp
also have "… = (algunos P (a # xs) ∨ algunos P ys)" by simp
finally show "?Q (a # xs)" by simp
qed

text {*
---------------------------------------------------------------------
Ejercicio 9.1. Demostrar o refutar automáticamente
algunos P (rev xs) = algunos P xs
---------------------------------------------------------------------
*}

(* ivamenjim migtermor marpoldia1 rubgonmar paupeddeg *)

lemma "algunos P (rev xs) = algunos P xs"
apply (induct xs)
apply simp
apply (simp add: algunos_append)
apply auto
done

(* ferrenseg crigomgom danrodcha*)
lemma "algunos P (rev xs) = algunos P xs"
by (induct xs) (auto simp add: algunos_append)

(* wilmorort *)
lemma "algunos P (rev xs) = algunos P xs"
by (induct xs,simp,simp add: algunos_append,auto)

text {*
---------------------------------------------------------------------
Ejercicio 9.2. Demostrar o refutar detalladamente
algunos P (rev xs) = algunos P xs
---------------------------------------------------------------------
*}

lemma "algunos P (rev xs) = algunos P xs"
oops

(* migtermor *)
lemma auxiliar1:
"rev (a#xs) = rev xs @ [a]"
by auto

lemma "algunos P (rev xs) = algunos P xs" (is "?Q xs")
proof (induct xs)
show "?Q []" by simp
next
fix x xs
assume HI: "?Q xs"
have "algunos P (rev (x#xs)) = (algunos P (rev xs @ [x]))" using auxiliar1 by simp
also have "… = ((algunos P (rev xs)) ∨ (algunos P [x]))" by (simp add: algunos_append)
also have "… = ((P x) ∨ (algunos P (rev xs)))" by simp arith
also have "… = ((P x) ∨ (algunos P xs))" using HI by simp
finally show "algunos P (rev (x#xs)) = (algunos P (x#xs))" by simp
qed

(* ferrenseg *)

lemma "algunos P (rev xs) = algunos P xs"
proof (induct xs)
show "algunos P (rev []) = algunos P []" by simp
next
fix x xs
assume HI: "algunos P (rev xs) = algunos P xs"
show "algunos P (rev (x # xs)) = algunos P (x # xs)"
proof -
have "algunos P (rev (x # xs)) = algunos P ((rev xs) @ [x])" by simp
also have "… = (algunos P (rev xs) ∨ algunos P [x])"
by (simp add: algunos_append)
also have "… = (algunos P xs ∨ algunos P [x])" using HI by simp
also have "… = (algunos P xs ∨ P x)" by simp
also have "… = (P x ∨ algunos P xs)" by auto
also have "… = algunos P (x # xs)" by simp
finally show ?thesis
qed
qed

(* ivamenjim marpoldia1*)

lemma "algunos P (rev xs) = algunos P xs" (is "?P xs")
proof (induct xs)
show "?P []" by simp
next
fix a xs
assume HI: "?P xs"
have "algunos P (rev (a # xs)) = (algunos P ((rev xs)@[a]))" by simp
also have "... = ((algunos P (rev xs)) ∨ algunos P [a])" by (simp add: algunos_append)
also have "... = (algunos P xs ∨ algunos P [a])" using HI by simp
also have "... = (algunos P [a] ∨ algunos P xs)" by arith
finally show "algunos P (rev (a # xs)) = algunos P (a # xs)" by simp
qed

(*paupeddeg*)
lemma "algunos P (rev xs) = algunos P xs"
proof (induct xs)
show "algunos P (rev []) = algunos P []" by simp
next
fix a xs
assume HI:"algunos P (rev xs) = algunos P xs"
have "algunos P (rev (a # xs)) = algunos P ((rev xs) @ [a])" by simp
also have "... = ((algunos P (rev xs)) ∨ (algunos P [a]))" by (simp add: algunos_append)
also have "... = ((algunos P xs) ∨ (algunos P [a]))" using HI by simp
also have "... = ((algunos P [a]) ∨ (algunos P xs))" by arith
finally show "algunos P (rev (a # xs)) = algunos P (a # xs)" by simp
qed

(*crigomgom *)
lemma "algunos P (rev xs) = algunos P xs"
proof (induct xs)
show "algunos P (rev []) = algunos P []" by simp
next
fix x xs
assume HI: " algunos P (rev xs) = algunos P xs"
have "algunos P (rev (x # xs)) = algunos P ((rev xs) @ [x])" by simp
also have "... = (algunos P (rev xs) ∨ algunos P [x])" by (simp add: algunos_append)
also have "... = (algunos P xs ∨ algunos P [x])" using HI by simp
also have "... = (algunos P xs ∨ P x)" by simp
also have "... = (P x ∨ algunos P xs)" by arith
also have "... = algunos P (x#xs)" by simp
finally show "algunos P (rev (x # xs)) = algunos P (x # xs)" by simp
qed

(*wilmorort*)
lemma "algunos P (rev xs) = algunos P xs" (is "?P xs")
proof (induct xs)
show "?P []" by simp
next
fix a xs
assume HI: "?P xs"
have "algunos P (rev (a#xs)) = algunos P ((rev xs) @ [a])" by simp
also have "... = (algunos P(rev xs) ∨ algunos P [a])" by (simp add: algunos_append)
also have "... = (algunos P xs ∨ algunos P [a])" using HI by simp
also have "... = (algunos P [a] ∨ algunos P xs)" by (simp add: HOL.disj_commute)
also have "... = algunos P([a]@(xs))" by (simp)
finally show "algunos P (rev (a#xs))= algunos P (a # xs)" by simp
qed

{*danrodcha *}
lemma "algunos P (rev xs) = algunos P xs" (is "?Q xs")
proof (induct xs)
show "?Q []" by simp
next
fix a xs assume HI: "?Q xs"
have "algunos P (rev (a#xs)) = algunos P ((rev xs) @ [a])" by simp
also have "… = (algunos P (rev xs) ∨ algunos P [a])" by (simp add:algunos_append)
also have "… = (algunos P xs ∨ algunos P [a])" using HI by simp
also have "… = (algunos P [a] ∨ algunos P xs)" by (simp add: HOL.disj_commute)
also have "… = algunos P (a # xs)" by simp
finally show "?Q (a # xs)" by simp
qed

text {*
---------------------------------------------------------------------
Ejercicio 10. Encontrar un término no trivial Z tal que sea cierta la
siguiente ecuación:
algunos (λx. P x ∨ Q x) xs = Z
y demostrar la equivalencia de forma automática y detallada.
---------------------------------------------------------------------
*}

(* migtermor *)

lemma "algunos (λx. P x ∨ Q x) xs = ((algunos (λx. P x) xs) ∨ (algunos (λx. Q x) xs))"
by (induct xs) auto

lemma "algunos (λx. P x ∨ Q x) xs = ((algunos (λx. P x) xs) ∨ (algunos (λx. Q x) xs))" (is "?R xs")
proof (induct xs)
show "?R []" by simp
next
fix x xs
assume HI: "?R xs"
have "algunos (λx. P x ∨ Q x) (x#xs) = (((λx. P x ∨ Q x) x) ∨ (algunos (λx. P x ∨ Q x) xs))"
by simp
also have "… = ((P x ∨ Q x) ∨ (algunos (λx. P x ∨ Q x) xs))" by simp
also have H1: "… = ((((λx. P x) x) ∨ (algunos (λx. P x) xs)) ∨ (((λx. Q x) x) ∨ (algunos (λx. Q x) xs)))"
using HI by simp arith
have H2: "… = ((algunos (λx. P x) (x#xs)) ∨ (algunos (λx. Q x) (x#xs)))"
by simp
have C: "(algunos (λx. P x ∨ Q x) (x#xs)) =
((algunos (λx. P x) (x#xs)) ∨ (algunos (λx. Q x) (x#xs)))"
using H1 H2 by simp
finally show "(algunos (λx. P x ∨ Q x) (x#xs)) =
((algunos (λx. P x) (x#xs)) ∨ (algunos (λx. Q x) (x#xs)))"
using C by simp
qed

(* ferrenseg *)

lemma "algunos (λx. P x ∨ Q x) xs = (algunos P xs ∨ algunos Q xs)"
proof (induct xs)
show "algunos (λx. P x ∨ Q x) [] = (algunos P [] ∨ algunos Q [])" by simp
next
fix x xs
assume HI: "algunos (λx. (P x ∨ Q x)) xs = (algunos P xs ∨ algunos Q xs)"
show "algunos (λx. P x ∨ Q x) (x # xs) = (algunos P (x # xs) ∨ algunos Q (x # xs))"
proof -
have "algunos (λx. P x ∨ Q x) (x # xs) =
((P x) ∨ (Q x) ∨ algunos (λx. P x ∨ Q x) xs)" by simp
also have "… = ((P x) ∨ (Q x) ∨ algunos P xs ∨ algunos Q xs)"
using HI by simp
also have "… = (((P x) ∨ algunos P xs) ∨ ((Q x) ∨ algunos Q xs))" by auto
also have "… = (algunos P (x # xs) ∨ algunos Q (x # xs))" by simp
finally show ?thesis
qed
qed

(* ivamenjim marpoldia1 paupeddeg *)

lemma "algunos (λx. P x ∨ Q x) xs = (algunos P xs ∨ algunos Q xs)"
by (induct xs) auto

lemma "algunos (λx. P x ∨ Q x) xs = (algunos P xs ∨ algunos Q xs)"
proof (induct xs)
show "algunos (λx. P x ∨ Q x) [] = (algunos P [] ∨ algunos Q [])" by simp
next
fix a xs
assume HI: "algunos (λx. P x ∨ Q x) xs = (algunos P xs ∨ algunos Q xs)"
have "algunos (λx. P x ∨ Q x) (a # xs) = (P a ∨ Q a ∨ algunos (λx. P x ∨ Q x) xs)" by simp
also have "... = (P a ∨ Q a ∨ algunos P xs ∨ algunos Q xs)" using HI by simp
finally show "algunos (λx. P x ∨ Q x) (a # xs) = (algunos P (a # xs) ∨ algunos Q (a # xs))" by auto
qed

(* wilmorort danrodcha*)
lemma "algunos (λx. P x ∨ Q x) xs = (¬todos(λx. ¬P x)xs ∨ ¬todos(λx. ¬Q x)xs)"
by (induct xs) auto

(* wilmorort *)
lemma "algunos (λx. P x ∨ Q x) xs = (¬todos(λx. ¬P x)xs ∨ ¬todos(λx. ¬Q x)xs)"
(is "?P xs")
proof (induct xs)
show "?P []" by simp
next
fix a xs
assume HI: "?P xs"
have "algunos (λx. P x ∨ Q x) (a # xs) = (P a ∨ Q a ∨ algunos (λx. P x ∨ Q x) xs)" by simp
also have "... = (P a ∨ Q a ∨ (¬ todos (λx. ¬ P x) xs ∨ ¬ todos (λx. ¬ Q x) xs))" using HI by simp
also have "... = (P a ∨ ¬ todos (λx. ¬ P x) xs ∨ Q a ∨ ¬ todos (λx. ¬ Q x) xs )" by arith
finally show "algunos (λx. P x ∨ Q x) (a # xs) = (~ todos (λx. ¬ P x) (a # xs) ∨ ~ todos (λx. ¬ Q x) (a # xs))" by simp
qed

{* danrodcha *}
lemma "algunos (λx. P x ∨ Q x) xs = (algunos P xs ∨ algunos Q xs)" (is "?R xs")
proof (induct xs)
show "?R []" by simp
next
fix a xs assume HI: "?R xs"
have 1:" (Q a ∨ algunos P xs) = (algunos P xs ∨ Q a)" by (simp add: HOL.disj_commute)
have "algunos (λx. P x ∨ Q x) (a # xs) =
(algunos P [a] ∨ algunos Q [a] ∨ algunos (λx. P x ∨ Q x) xs)" by simp
also have "… = (P a ∨ (Q a ∨ algunos P xs) ∨ algunos Q xs)" using HI by simp
also have "… = (P a ∨ algunos P xs ∨ Q a ∨ algunos Q xs)" using 1 by simp
also have "… = (algunos P (a # xs) ∨ algunos Q (a # xs))" by simp
finally show "?R (a # xs)" by simp
qed

text {*
---------------------------------------------------------------------
Ejercicio 11.1. Demostrar o refutar automáticamente
algunos P xs = (¬ todos (λx. (¬ P x)) xs)
---------------------------------------------------------------------
*}

(* ivamenjim wilmorort migtermor marpoldia1 crigomgom rubgonmar paupeddeg danrodcha*)

lemma "algunos P xs = (¬ todos (λx. (¬ P x)) xs)"
by (induct xs) auto

(* ferrenseg *)

lemma "algunos P xs = (¬ todos (λx. (¬ P x)) xs)"
by (induct xs) simp_all

text {*
---------------------------------------------------------------------
Ejercicio 11.2. Demostrar o refutar datalladamente
algunos P xs = (¬ todos (λx. (¬ P x)) xs)
---------------------------------------------------------------------
*}

(* ivamenjim migtermor marpoldia1 crigomgom paupeddeg*)

lemma "algunos P xs = (¬ todos (λx. (¬ P x)) xs)"
proof (induct xs)
show "algunos P [] = (¬ todos (λx. ¬ P x) [])" by simp
next
fix a xs
assume HI: "algunos P xs = (¬ todos (λx. ¬ P x) xs)"
have "algunos P (a # xs) = (P a ∨ (algunos P xs))" by simp
also have "... = (P a ∨ (¬ todos (λx. ¬ P x) xs))" using HI by simp
finally show "algunos P (a # xs) = (¬ todos (λx. ¬ P x) (a # xs))" by simp
qed

(* ferrenseg *)

lemma "algunos P xs = (¬ todos (λx. (¬ P x)) xs)"
proof (induct xs)
show "algunos P [] = (¬ todos (λx. (¬ P x)) [])" by simp
next
fix x xs
assume HI: "algunos P xs = (¬ todos (λx. (¬ P x)) xs)"
show "algunos P (x # xs) = (¬ todos (λx. (¬ P x)) (x # xs))"
proof -
have "algunos P (x # xs) = ((P x) ∨ algunos P xs)" by simp
also have "… = ((P x) ∨ ¬ todos (λx. (¬ P x)) xs)" using HI by simp
also have "… = (¬ (¬ (P x) ∧ todos (λx. (¬ P x)) xs))" by simp
also have "… = (¬ todos (λx. (¬ P x)) (x # xs))" by simp
finally show ?thesis
qed
qed

{* danrodcha *}
lemma "algunos P xs = (¬ todos (λx. (¬ P x)) xs)" (is "?Q xs")
proof (induct xs)
show "?Q []" by simp
next
fix a xs assume HI: "?Q xs"
have "algunos P (a # xs) = (P a ∨ algunos P xs)" by simp
also have "… = (P a ∨ (¬ todos (λx. (¬ P x)) xs))" using HI by simp
also have "… = (¬(¬(P a) ∧ todos(λx. (¬ P x)) xs))" by simp
also have "… = (¬ todos(λx. (¬ P x)) (a # xs))" by simp
finally show "?Q (a # xs)" by simp
qed

text {*
---------------------------------------------------------------------
Ejercicio 12. Definir la funcion primitiva recursiva
estaEn :: 'a ⇒ 'a list ⇒ bool
tal que (estaEn x xs) se verifica si el elemento x está en la lista
xs. Por ejemplo,
estaEn (2::nat) [3,2,4] = True
estaEn (1::nat) [3,2,4] = False
---------------------------------------------------------------------
*}

(* ivamenjim ferrenseg migtermor serrodcal crigomgom rubgonmar marpoldia1 paupeddeg danrodcha*)

fun estaEn :: "'a ⇒ 'a list ⇒ bool" where
"estaEn x [] = False"
| "estaEn x (a#xs) = ((a = x) ∨ (estaEn x xs))"

value "estaEn (2::nat) [3,2,4] = True"
value "estaEn (1::nat) [3,2,4] = False"

text {*
---------------------------------------------------------------------
Ejercicio 13. Expresar la relación existente entre estaEn y algunos.
Demostrar dicha relación de forma automática y detallada.
---------------------------------------------------------------------
*}

(* migtermor crigomgom *)

lemma "estaEn x xs = (algunos (λa. a=x) xs)"
by (induct xs) auto

lemma auxiliar13:
"(x=a) = ((λa. a=x) a)"
by auto

lemma "estaEn x xs = (algunos (λa. a=x) xs)" (is "?P xs")
proof (induct xs)
show "?P []" by simp
next
fix a xs
assume HI: "?P xs"
have "estaEn x (a#xs) = ((a=x) ∨ (estaEn x xs))" by simp
also have H: "… = ((a=x) ∨ (algunos (λa. a=x) xs))" using HI by simp
also have "… = (((λa. a=x) a) ∨ (algunos (λa. a=x) xs))" using auxiliar13 by simp
also have "… = (algunos (λa. a=x) (a#xs))" by simp
have C: "estaEn x (a#xs) = (algunos (λa. a=x) (a#xs))" using H by simp
finally show "estaEn x (a#xs) = (algunos (λa. a=x) (a#xs))" using C by simp
qed

(* ferrenseg *)

lemma estaEn_algunos:
"estaEn y xs = algunos (λx. x=y) xs"
by (induct xs) auto

lemma estaEn_algunos_2:
"estaEn y xs = algunos (λx. x=y) xs"
proof (induct xs)
show "estaEn y [] = algunos (λx. x=y) []" by simp
next
fix x xs
assume HI: "estaEn y xs = algunos (λx. x=y) xs"
show "estaEn y (x # xs) = algunos (λx. x=y) (x # xs)"
proof -
have "estaEn y (x # xs) = (y=x ∨ estaEn y xs)" by simp
also have "… = (y=x ∨ algunos (λx. x=y) xs)" using HI by simp
also have "… = (x=y ∨ algunos (λx. x=y) xs)" by auto
also have "… = algunos (λx. x=y) (x # xs)" by simp
finally show ?thesis
qed
qed

(*crigomgom paupeddeg*)

lemma "estaEn a xs = algunos (λx. x = a) xs"
by (induct xs) auto

lemma "estaEn x xs = (algunos (λa. a=x) xs)"
proof (induct xs)
show " estaEn x [] = algunos (λa. a = x) []" by simp
next
fix a xs
assume HI: "estaEn x xs = algunos (λa. a = x) xs"
have "estaEn x (a # xs) = (a = x ∨ estaEn x xs)" by simp
also have "... = (a = x ∨ algunos (λa. a = x) xs)" using HI by simp
finally show " estaEn x (a # xs) = algunos (λa. a = x) (a # xs)" by simp
qed

(* ivamenjim marpoldia1 *)

lemma "estaEn x xs = algunos (λy. x=y) xs"
by (induct xs) auto

lemma "estaEn x xs = algunos (λy. x=y) xs"
proof (induct xs)
show "estaEn x [] = algunos (op = x) []" by simp
next
fix a xs
assume HI: "estaEn x xs = algunos (op = x) xs"
have "estaEn x (a # xs) = ((a = x) ∨ (estaEn x xs))" by simp
also have "... = (a = x ∨ algunos (op = x) xs)" using HI by simp
finally show " estaEn x (a # xs) = algunos (op = x) (a # xs)" by auto
qed

{* danrodcha *}
lemma "estaEn a xs = algunos (op = a) xs"
by (induct xs) auto

{* danrodcha *}
lemma "estaEn a xs = algunos (op = a) xs"
proof (induct xs)
show "estaEn a [] = algunos (op = a) []" by simp
next
fix x xs
assume HI: "estaEn a xs = algunos (op = a) xs"
have "estaEn a (x # xs) = ((x = a) ∨ (estaEn a xs))" by simp
also have "… = (x = a ∨ algunos (op = a) xs)" using HI by simp
also have "… = ((op = a) x ∨ algunos (op = a) xs)" by (simp add:HOL.eq_commute)
also have "… = algunos (op = a) (x # xs)" by simp
finally show "estaEn a (x # xs) = algunos (op = a) (x # xs)" by simp
qed

end
</source>

Relación 4

2016-11-22T23:02:21Z

Pablucoto:

<source lang="isar">
chapter {* R4: Cuantificadores sobre listas *}

theory R4_Cuantificadores_sobre_listas
imports Main
begin

text {*
---------------------------------------------------------------------
Ejercicio 1. Definir la función
todos :: ('a ⇒ bool) ⇒ 'a list ⇒ bool
tal que (todos p xs) se verifica si todos los elementos de la lista
xs cumplen la propiedad p. Por ejemplo, se verifica
todos (λx. 1<length x) [[2,1,4],[1,3]]
¬todos (λx. 1<length x) [[2,1,4],[3]]

Nota: La función todos es equivalente a la predefinida list_all.
---------------------------------------------------------------------
*}

{*danrodcha*}
fun todos :: "('a ⇒ bool) ⇒ 'a list ⇒ bool" where
"todos p xs = "todos p [] = True"
| "todos p (x#xs) = (p x ∧ todos p xs)""

(* ivamenjim migtermor dancorgar wilmorort marpoldia1 ferrenseg paupeddeg pablucoto crigomgom anaprarod serrodcal juacabsou rubgonmar josgarsan fraortmoy lucnovdos*)
fun todos :: "('a ⇒ bool) ⇒ 'a list ⇒ bool" where
"todos p [] = True"
| "todos p (x#xs) = (p x ∧ todos p xs)"

text {*
---------------------------------------------------------------------
Ejercicio 2. Definir la función
algunos :: ('a ⇒ bool) ⇒ 'a list ⇒ bool
tal que (algunos p xs) se verifica si algunos elementos de la lista
xs cumplen la propiedad p. Por ejemplo, se verifica
algunos (λx. 1<length x) [[2,1,4],[3]]
¬algunos (λx. 1<length x) [[],[3]]"

Nota: La función algunos es equivalente a la predefinida list_ex.
---------------------------------------------------------------------
*}

{*danrodcha ivamenjim migtermor dancorgar marpoldia1 ferrenseg wilmorort paupeddeg pablucoto crigomgom anaprarod serrodcal juacabsou rubgonmar josgarsan fraortmoy lucnovdos*}
fun algunos :: "('a ⇒ bool) ⇒ 'a list ⇒ bool" where
"algunos p [] = False"
| "algunos p (x#xs) = (p x ∨ algunos p xs)"

text {*
---------------------------------------------------------------------
Ejercicio 3.1. Demostrar o refutar automáticamente
todos (λx. P x ∧ Q x) xs = (todos P xs ∧ todos Q xs)
---------------------------------------------------------------------
*}

{*danrodcha ivamenjim migtermor dancorgar marpoldia1 ferrenseg wilmorort paupeddeg pablucoto anaprarod serrodcal juacabsou rubgonmar josgarsan fraortmoy lucnovdos*}
lemma "todos (λx. P x ∧ Q x) xs = (todos P xs ∧ todos Q xs)"
by (induct xs) auto

text {*
---------------------------------------------------------------------
Ejercicio 3.2. Demostrar o refutar detalladamente
todos (λx. P x ∧ Q x) xs = (todos P xs ∧ todos Q xs)
---------------------------------------------------------------------
*}

{*danrodcha*}
lemma "todos (λx. P x ∧ Q x) xs = (todos P xs ∧ todos Q xs)"
proof (induct xs)
show "?R []" by simp
next
fix a xs
assume HI: "?R xs"
have "todos (λx. P x ∧ Q x) (a#xs) = (P a ∧ Q a ∧ todos (λx. P x ∧ Q x) xs)" by simp
also have "… = (P a ∧ Q a ∧ (todos P xs ∧ todos Q xs))" using HI by simp
also have "… = ((P a ∧ todos P xs) ∧ (Q a ∧ todos Q xs))" by blast
also have "… = (todos P (a#xs) ∧ todos Q (a#xs))" by simp
finally show "?R (a#xs)" by simp
qed

(* ivamenjim wilmorort serrodcal josgarsan*)

lemma "todos (λx. P x ∧ Q x) xs = (todos P xs ∧ todos Q xs)"
proof (induct xs)
show "todos (λx. P x ∧ Q x) [] = (todos P [] ∧ todos Q [])" by simp
next
fix a xs
assume HI: "todos (λx. P x ∧ Q x) xs = (todos P xs ∧ todos Q xs)"
have "todos (λx. P x ∧ Q x) (a # xs) = (P a ∧ Q a ∧ todos (λx. P x ∧ Q x) xs)" by simp
also have "... = (P a ∧ Q a ∧ todos P xs ∧ todos Q xs)" using HI by simp
finally show "todos (λx. P x ∧ Q x) (a # xs) = (todos P (a # xs) ∧ todos Q (a # xs))" by auto
qed

(* dancorgar paupeddeg *)
lemma "todos (λx. P x ∧ Q x) xs = (todos P xs ∧ todos Q xs)"
proof (induct xs)
show "todos (λx. P x ∧ Q x) [] = (todos P [] ∧ todos Q [])" by simp
next
fix y xs
assume HI: "todos (λx. P x ∧ Q x) xs = (todos P xs ∧ todos Q xs)"
have "todos (λx. P x ∧ Q x) (y#xs) = ((P y ∧ Q y) ∧ (todos (λx. P x ∧ Q x) xs))" by simp
also have "… = ((P y ∧ Q y) ∧ (todos P xs ∧ todos Q xs))" using HI by simp
also have "… = ((P y ∧ todos P xs) ∧ (Q y ∧ todos Q xs))" by blast
also have "… = ((todos P (y#xs)) ∧ (todos Q (y#xs)))" by simp
finally show "todos (λx. P x ∧ Q x) (y#xs) = (todos P (y#xs) ∧ todos Q (y#xs))" by simp
qed

(* migtermor *)
lemma "todos (λx. P x ∧ Q x) xs = (todos P xs ∧ todos Q xs)"
proof (induct xs)
show "todos (λx. P x ∧ Q x) [] = (todos P [] ∧ todos Q [])" by simp
next
fix x xs
assume HI: "todos (λx. P x ∧ Q x) xs = (todos P xs ∧ todos Q xs)"
have "(todos (λx. P x ∧ Q x) (x#xs)) = (( P x ∧ Q x) ∧ (todos (λx. P x ∧ Q x) xs))"
by (simp only: todos.simps(2))
also have "… = ((P x ∧ Q x) ∧ (todos P xs ∧ todos Q xs))" using HI by simp
also have "… = (((P x)∧(todos P xs)) ∧ ((Q x) ∧ (todos Q xs)))" by arith
also have "((P x ∧ Q x) ∧ (todos P xs ∧ todos Q xs)) = (((P x)∧(todos P xs)) ∧ ((Q x) ∧ (todos Q xs)))"
by arith (* Este paso es exactamente el mismo que el anterior, pero sin cualquiera de los dos no funciona el "finally show" *)
have "… = (((P x)∧(todos P xs))∧((Q x)∧(todos Q xs)))" by simp
have "… = ((todos P (x#xs))∧(todos Q (x#xs)))" by simp
finally show "(todos (λx. P x ∧ Q x) (x#xs)) = ((todos P (x#xs)) ∧ (todos Q (x#xs)))" by simp
qed

(* marpoldia1 *)
lemma "todos (λx. P x ∧ Q x) xs = (todos P xs ∧ todos Q xs)"
proof (induct xs)
show "todos (λx. P x ∧ Q x) [] = (todos P [] ∧ todos Q [])" by simp
next
fix a xs
assume HI: "todos (λx. P x ∧ Q x) xs = (todos P xs ∧ todos Q xs)"
have "todos (λx. P x ∧ Q x) (a # xs) = ((P a ∧ Q a) ∧ (todos P xs ∧ todos Q xs))" using HI by simp
also have "... = ((P a ∧ todos P xs) ∧ (Q a ∧ todos Q xs))" by blast
also have "... = (todos P (a#xs) ∧ todos Q (a#xs)) " by simp
finally show "todos (λx. P x ∧ Q x) (a#xs) = (todos P (a#xs) ∧ todos Q (a#xs))" by simp
qed

(* ferrenseg *)

lemma "todos (λx. P x ∧ Q x) xs = (todos P xs ∧ todos Q xs)"
proof (induct xs)
show "todos (λx. P x ∧ Q x) [] = (todos P [] ∧ todos Q [])" by simp
next
fix n xs
assume HI: "todos (λx. P x ∧ Q x) xs = (todos P xs ∧ todos Q xs)"
have "todos (λx. P x ∧ Q x) (n # xs) = (P n ∧ Q n ∧ todos P xs ∧ todos Q xs)" using HI by simp
also have "… = P n ∧ Q n ∧ todos P xs ∧ todos Q xs" by blast
also have "⋯ = todos P (n # xs) ∧ todos Q (n # xs)" by simp
finally show "todos (λx. P x ∧ Q x) (n # xs) = todos P (n # xs) ∧ todos Q (n # xs)" by simp
qed

(* lucnovdos *)

lemma "todos (λx. P x ∧ Q x) xs = (todos P xs ∧ todos Q xs)"
proof (induct xs)
show "todos (λx. P x ∧ Q x) [] = (todos P [] ∧ todos Q [])" by simp
next
fix n xs
assume HI: "todos (λx. P x ∧ Q x) xs = (todos P xs ∧ todos Q xs)"
have "todos (λx. P x ∧ Q x) (n # xs) = ((P n ∧ Q n) ∧ (todos P xs ∧ todos Q xs))" using HI by simp
also have "... = ((P n ∧ todos P xs) ∧ (Q n ∧ todos Q xs))" by arith
also have "... = ((todos P(n#xs)) ∧ (todos Q(n#xs)))" by simp

finally show "todos (λx. P x ∧ Q x) (n#xs) = (todos P (n#xs) ∧ todos Q (n#xs))" by simp
qed

{* wilmorort pablucoto crigomgom anaprarod juacabsou rubgonmar fraortmoy *}

lemma "todos (λx. P x ∧ Q x) xs = (todos P xs ∧ todos Q xs)"
proof (induct xs)
show "todos (λx. P x ∧ Q x) [] = (todos P [] ∧ todos Q [])" by simp
next
fix a xs
assume HI: "todos (λx. P x ∧ Q x) xs = (todos P xs ∧ todos Q xs)"
have "todos (λx. P x ∧ Q x) (a # xs) = (P a ∧ Q a ∧ todos (λx. P x ∧ Q x) xs)" by simp
also have "... = (P a ∧ Q a ∧ todos P xs ∧ todos Q xs)" using HI by simp
also have "... = (P a ∧ todos P xs ∧ Q a ∧ todos Q xs)" by arith
also have "... = (todos P (a # xs) ∧ todos Q (a # xs))" by simp (* Este paso se puede obviar*)
finally show "todos (λx. P x ∧ Q x) (a # xs) = (todos P (a # xs) ∧ todos Q (a # xs))" by simp
qed

text {*
---------------------------------------------------------------------
Ejercicio 4.1. Demostrar o refutar automáticamente
todos P (x @ y) = (todos P x ∧ todos P y)
---------------------------------------------------------------------
*}
{*danrodcha ivamenjim marpoldia1 migtermor ferrenseg wilmorort paupeddeg crigomgom anaprarod serrodcal juacabsou rubgonmar pablucoto fraortmoy josgarsan lucnovdos*}
lemma "todos P (x @ y) = (todos P x ∧ todos P y)"
by (induct x) auto

{* anaprarod dancorgar *}
lemma "todos P (x @ y) = (todos P x ∧ todos P y)"
apply (induct x)
apply auto
done

text {*
---------------------------------------------------------------------
Ejercicio 4.2. Demostrar o refutar detalladamente
todos P (x @ y) = (todos P x ∧ todos P y)
---------------------------------------------------------------------
*}

(* ivamenjim ferrenseg wilmorort dancorgar fraortmoy josgarsan*)

lemma todos_append:
"todos P (x @ y) = (todos P x ∧ todos P y)"
proof (induct x)
show "todos P ([] @ y) = (todos P [] ∧ todos P y)" by simp
next
fix a x
assume HI: "todos P (x @ y) = (todos P x ∧ todos P y)"
have "todos P ((a#x) @ y) = (P a ∧ (todos P (x @ y)))" by simp
also have "... = (P a ∧ (todos P x ∧ todos P y))" using HI by simp
finally show "todos P ((a # x) @ y) = (todos P (a # x) ∧ todos P y)" by simp
qed

(* migtermor serrodcal *)

lemma todos_append1:
"todos P (x @ y) = (todos P x ∧ todos P y)" (is "?P x")
proof (induct x)
show "?P []" by simp
next
fix a x
assume HI: "?P x"
have "todos P ((a#x) @ y) = (P a ∧ (todos P (x @ y)))" by simp
also have "… = (P a ∧ (todos P x ∧ todos P y))" using HI by simp
finally show "?P (a#x)" by simp
qed

(* marpoldia1 *)
lemma todos_append:
"todos P (x @ y) = (todos P x ∧ todos P y)"
proof (induct x)
show "todos P ([] @ y) = (todos P [] ∧ todos P y)" by simp
next
fix a x
assume HI: "todos P (x @ y) = (todos P x ∧ todos P y)"
have "todos P ((a # x) @ y) = todos P (a#(x@y))" by simp
also have "... = (P a ∧ todos P (x @ y))" by simp
also have "... = (P a ∧ (todos P x ∧ todos P y))" using HI by simp
finally show "todos P ((a # x) @ y) = (todos P (a # x) ∧ todos P y)" by simp
qed

(* paupeddeg *)
lemma todos_append3:
"todos P (x @ y) = (todos P x ∧ todos P y)"
proof (induct x)
show "todos P ([] @ y) = (todos P [] ∧ todos P y)" by simp
next
fix a x
assume HI: "todos P (x @ y) = (todos P x ∧ todos P y)"
have "todos P ((a # x) @ y) = todos P (a # (x @ y))" by simp
also have "... = (todos P [a] ∧ todos P (x @ y)) " by simp
also have "... = (todos P [a] ∧ (todos P x ∧ todos P y))" using HI by simp
finally show "todos P ((a # x) @ y) = (todos P (a # x) ∧ todos P y)" by simp
qed

(* crigomgom juacabsou anaprarod*)

lemma todos_append4:
"todos P (x @ y) = (todos P x ∧ todos P y)"
proof (induct x)
show "todos P ([] @ y) = (todos P [] ∧ todos P y)" by simp
next
fix x xs
assume HI:"todos P (xs @ y) = (todos P xs ∧ todos P y)"
have "todos P ((x # xs) @ y) = (P x ∧ todos P (xs @ y ))" by simp
also have "... = (P x ∧ (todos P xs ∧ todos P y)) " using HI by simp
also have "... = ((P x ∧ todos P xs) ∧ todos P y)" by simp
finally show "todos P ((x # xs) @ y) = (todos P (x # xs) ∧ todos P y)" by simp
qed

(* pablucoto *)
lemma todos_append5:
"todos P (x @ y) = (todos P x ∧ todos P y)"
proof (induct x )
show "todos P ([] @ y) = (todos P [] ∧ todos P y)" by simp
next
fix a x
assume HI:"todos P (x @ y) = (todos P x ∧ todos P y) "
have "todos P ((a # x) @ y) = ( P a ∧ todos P (x @ y))" by simp
also have " ... =( P a ∧ (todos P x ∧ todos P y))" using HI by simp
also have "... = (todos P (a#x) ∧ todos P y)" by simp
finally show "todos P ((a # x) @ y) = (todos P (a # x) ∧ todos P y)" by auto
qed

{*danrodcha*}
lemma todos_append6:
"todos P (x @ y) = (todos P x ∧ todos P y)" (is "?Q x")
proof (induct x)
show "?Q []" by simp
next
fix x a assume HI: "?Q x"
have "todos P ((a # x) @ y) = todos P (a # x @ y)" by simp
also have "… = ((P a) ∧ todos P (x @ y))" by simp
also have "… = ((P a) ∧ (todos P x ∧ todos P y))" using HI by simp
also have "… = (todos P (a # x) ∧ todos P y)" by simp
finally show "?Q (a # x)" by simp
qed

text {*
---------------------------------------------------------------------
Ejercicio 5.1. Demostrar o refutar automáticamente
todos P (rev xs) = todos P xs
---------------------------------------------------------------------
*}

lemma "todos P (rev xs) = todos P xs"
oops

(* migtermor ivamenjim marpoldia1 serrodcal anaprarod paupeddeg *)
lemma "todos P (rev xs) = todos P xs"
apply (induct xs)
apply simp
apply (simp add: todos_append)
apply auto
done

(* ferrenseg crigomgom rubgonmar fraortmoy josgarsan danrodcha*)

lemma "todos P (rev xs) = todos P xs"
by (induct xs) (auto simp add: todos_append)

(* wilmorort *)
lemma "todos P (rev xs) = todos P xs"
by (induct xs,simp,simp add: todos_append,auto)

(* juacabsou *)
lemma "todos P (rev xs) = todos P xs"
apply (induct xs,simp,simp add: todos_append,auto) done

(* dancorgar *)
lemma "todos P (rev xs) = todos P xs"
apply (induct xs)
apply (simp add: todos_append)
apply (simp add: todos_append)
apply auto
done

(* pablucoto * )
lemma "todos P (rev xs) = todos P xs"
by (induct xs) (auto, simp_all add: todos_append)
lemma "todos P (rev xs) = todos P xs"
by (induct xs) ( simp_all add: todos_append, auto)

text {*
---------------------------------------------------------------------
Ejercicio 5.2. Demostrar o refutar detalladamente
todos P (rev xs) = todos P xs
---------------------------------------------------------------------
*}

lemma "todos P (rev xs) = todos P xs"
oops

(* migtermor *)

lemma auxiliar:
"rev (a#xs) = rev xs @ [a]"
by auto

lemma "todos P (rev xs) = todos P xs" (is "?Q xs")
proof (induct xs)
show "?Q []" by (simp only: rev.simps(1))
next
fix a xs
assume HI: "?Q xs"
have "todos P (rev (a#xs)) = (todos P (rev xs @ [a]))" by (simp add: auxiliar)
have "… = ((todos P (rev xs)) ∧ (todos P [a]))" by (simp add: todos_append)
have "… = (todos P (rev xs) ∧ P a)" by simp
also have Aux: "… = (todos P xs ∧ P a)" using HI by simp
have Aux1: "… = (P a ∧ todos P xs)" by arith
have "(todos P (rev xs) ∧ P a) = (P a ∧ todos P xs)" using Aux Aux1 by simp
finally show "todos P (rev (a#xs)) = todos P (a#xs)" by (simp add: todos_append)
qed

(* marpoldia1 ferrenseg crigomgom serrodcal juacabsou rubgonmar josgarsan pablucoto*)
lemma "todos P (rev xs) = todos P xs"
proof (induct xs)
show "todos P (rev []) = todos P []" by simp
next
fix a xs
assume HI: "todos P (rev xs) = todos P xs"
have "todos P (rev (a # xs)) = (todos P ((rev xs)@[a]))" by simp
also have "... = (todos P (rev xs) ∧ todos P [a])" by (simp add:todos_append)
also have "... = (todos P xs ∧ P a)" using HI by simp
also have "... = (P a ∧ todos P xs)" by arith
also have "... = (todos P (a#xs))" by simp
finally show "todos P (rev (a # xs)) = (todos P (a#xs))" by simp
qed

(* ivamenjim *)

lemma "todos P (rev xs) = todos P xs" (is "?P xs")
proof (induct xs)
show "?P []" by simp
next
fix a xs
assume HI: "?P xs"
have "todos P (rev (a # xs)) = (todos P ((rev xs)@[a]))" by simp
also have "... = ((todos P (rev xs)) ∧ todos P [a])" by (simp add: todos_append)
also have "... = (todos P xs ∧ todos P [a])" using HI by simp
also have "... = (todos P [a] ∧ todos P xs)" by arith
finally show "todos P (rev (a # xs)) = todos P (a # xs)" by simp
qed

(* fraortmoy *)

lemma "todos P (rev xs) = todos P xs"
proof (induct xs)
show "todos P (rev []) = todos P []" by simp
next
fix a xs
assume H1: "todos P (rev xs) = todos P xs"
have " todos P (rev (a # xs)) = todos P (rev xs @ [a])" by simp
also have "… = (todos P (rev xs) ∧ todos P [a])" by (simp add:todos_append)
also have "… = (todos P xs ∧ todos P [a])" using H1 by simp
also have "… = (todos P [a] ∧ todos P xs)" by arith
finally show "todos P (rev (a # xs)) = todos P (a#xs)" by simp
qed

(* paupeddeg dancorgar *)
lemma "todos P (rev xs) = todos P xs"
proof (induct xs)
show " todos P (rev []) = todos P []" by simp
next
fix a xs
assume HI: "todos P (rev xs) = todos P xs "
have "todos P (rev (a # xs)) = todos P (rev(xs) @ [a])" by simp
also have "... = (todos P (rev xs) ∧ todos P [a]) " by (simp add: todos_append)
also have "... = (todos P xs ∧ todos P [a])" using HI by simp
also have "... = (todos P [a] ∧ todos P xs)" by arith
also have "... = todos P ([a] @ xs)" by (simp add: todos_append)
finally show "todos P (rev (a # xs)) = todos P (a # xs)" by simp
qed

(* wilmorort *)
lemma "todos P (rev xs) = todos P xs" (is "?P xs")
proof (induct xs)
show "?P []" by simp
next
fix a xs
assume HI: "?P xs"
have "todos P (rev (a#xs)) = todos P ((rev xs) @ [a])" by simp
also have "... = (todos P(rev xs) ∧ todos P [a])" by (simp add: todos_append)
also have "... = (todos P xs ∧ todos P [a])" using HI by simp
also have "... = (todos P [a] ∧ todos P xs)" by (simp add: HOL.conj_commute)
also have "... = todos P([a]@(xs))" by (simp)
finally show "todos P (rev (a#xs))= todos P (a # xs)" by simp
qed

(* anaprarod *)
(* es igual que las anteriores pero con el final también con patrones *)

lemma "todos P (rev xs) = todos P xs" (is "?P xs")
proof (induct xs)
show "?P []" by auto
next
fix a xs
assume HI: "?P xs"
have "todos P (rev (a#xs)) = todos P (rev xs @ [a])" by simp
also have "… = (todos P (rev xs) ∧ todos P [a])" by (simp add: todos_append)
also have "… = (todos P xs ∧ todos P [a])" using HI by simp
also have "… = (todos P [a] ∧ todos P xs)" by arith
also have "… = todos P (a#xs)" by simp
finally show "?P (a # xs)" by simp
qed

{*danrodcha*}
lemma "todos P (rev xs) = todos P xs" (is "?Q xs")
proof (induct xs)
show "?Q []" by simp
next
fix a xs assume HI: "?Q xs"
have "todos P (rev (a#xs)) = todos P ((rev xs) @ [a])" by simp
also have "… = (todos P (rev xs) ∧ todos P [a])" by (simp add:todos_append)
also have "… = (todos P xs ∧ todos P [a])" using HI by simp
also have "… = (todos P [a] ∧ todos P xs)" by (simp add: HOL.conj_commute)
also have "… = todos P (a # xs)" by simp
finally show "?Q (a # xs)" by simp
qed

text {*
---------------------------------------------------------------------
Ejercicio 6. Demostrar o refutar:
algunos (λx. P x ∧ Q x) xs = (algunos P xs ∧ algunos Q xs)
---------------------------------------------------------------------
*}

lemma "algunos (λx. P x ∧ Q x) xs = (algunos P xs ∧ algunos Q xs)"
oops

(* migtermor ivamenjim ferrenseg paupeddeg crigomgom serrodcal wilmorort juacabsou rubgonmar anaprarod marpoldia1 fraortmoy josgarsan danrodcha pablucoto*)
lemma "algunos (λx. P x ∧ Q x) xs = (algunos P xs ∧ algunos Q xs)"
quickcheck
oops
(* Quickcheck encuentra el siguiente contraejemplo: P={a1}, Q={a2}, xs={a1,a2}. En este ejemplo:
· "algunos (λx. P x ∧ Q x) xs = False"
· "(algunos P xs ∧ algunos Q xs) = True" *)

text {*
---------------------------------------------------------------------
Ejercicio 7.1. Demostrar o refutar automáticamente
algunos P (map f xs) = algunos (P ∘ f) xs
---------------------------------------------------------------------
*}

(* ivamenjim migtermor marpoldia1 crigomgom rubgonmar wilmorort anaprarod
fraortmoy juacabsou paupeddeg josgarsan danrodcha*)

lemma "algunos P (map f xs) = algunos (P o f) xs"
by (induct xs) auto

(* ferrenseg *)

lemma "algunos P (map f xs) = algunos (P o f) xs"
by (induct xs) simp_all

(* anaprarod *)
lemma "algunos P (map f xs) = algunos (P o f) xs"
apply (induct xs)
apply auto
done

text {*
---------------------------------------------------------------------
Ejercicio 7.2. Demostrar o refutar datalladamente
algunos P (map f xs) = algunos (P ∘ f) xs
---------------------------------------------------------------------
*}

lemma "algunos P (map f xs) = algunos (P o f) xs"
oops

(* migtermor *)

lemma AUX: "algunos (λa. P (f a)) xs = algunos (P o f) xs"
by (induct xs) auto

lemma "algunos P (map f xs) = algunos (P o f) xs" (is "?Q xs")
proof (induct xs)
show "?Q []" by simp
next
fix x xs
assume HI: "?Q xs"
have "algunos P (map f (x#xs)) = (algunos P ((f x)#(map f xs)))" by simp
also have "… = ((P (f x)) ∨ (algunos P (map f xs)))" by (simp only: algunos.simps(2))
also have "… = ((P (f x)) ∨ (algunos (P o f) xs))"
proof (cases)
assume C1: "(P (f x))"
have Aux: "((P (f x)) ∨ (algunos P (map f xs))) = True" using C1 by simp
have Aux1: "… = ((P (f x)) ∨ (algunos (P o f) xs))" using C1 by simp
then show "((P (f x)) ∨ (algunos P (map f xs))) = ((P (f x)) ∨ (algunos (P o f) xs))"
using Aux Aux1 by simp
next
assume C2: "¬(P (f x))"
have Aux2: "((P (f x)) ∨ (algunos P (map f xs))) = (algunos P (map f xs))" using C2 by simp
have Aux3: "… = (algunos (P o f) xs)" using HI by (simp add: AUX)
also have "… = ((P (f x)) ∨ (algunos (P o f) xs))" using C2 by simp
then show "((P (f x)) ∨ (algunos P (map f xs))) = ((P (f x)) ∨ (algunos (P o f) xs))"
using Aux2 Aux3 by simp
qed
also have "… = (((P o f) x) ∨ (algunos (P o f) xs))" by simp
also have "… = (algunos (P o f) (x#xs))" by simp
finally show "algunos P (map f (x#xs)) = (algunos (P o f) (x#xs))" by simp
qed

(* ferrenseg *)

lemma "algunos P (map f xs) = algunos (P ∘ f) xs"
proof (induct xs)
show "algunos P (map f []) = algunos (P ∘ f) []" by simp
next
fix x xs
assume HI: "algunos P (map f xs) = algunos (P ∘ f) xs"
show "algunos P (map f (x # xs)) = algunos (P ∘ f) (x # xs)"
proof -
have "algunos P (map f (x # xs)) = algunos P ((f x) # (map f xs))" by simp
also have "… = ((P (f x)) ∨ (algunos P (map f xs)))" by simp
also have "… = (((P ∘ f) x) ∨ (algunos (P ∘ f) xs))" using HI by simp
also have "… = algunos (P ∘ f) (x # xs)" by simp
finally show ?thesis
qed
qed

(* ivamenjim *)

lemma "algunos P (map f xs) = algunos (P o f) xs"
proof (induct xs)
show "algunos P (map f []) = algunos (P ∘ f) []" by simp
next
fix a xs
assume HI: "algunos P (map f xs) = algunos (P ∘ f) xs"
have "algunos P (map f (a # xs)) = algunos P ((f a)#map f xs)" by simp
also have "... = (algunos P (map f [a]) ∨ algunos P (map f xs))" by simp
also have "... = (algunos P (map f [a]) ∨ algunos (P ∘ f) xs)" using HI by simp
finally show "algunos P (map f (a # xs)) = algunos (P ∘ f) (a # xs)" by simp
qed

(* crigomgom rubgonmar anaprarod marpoldia1 juacabsou danrodcha paupeddeg*)
lemma "algunos P (map f xs) = algunos (P o f) xs"
proof (induct xs)
show "algunos P (map f []) = algunos (P ∘ f) []" by simp
next
fix x xs
assume HI: "algunos P (map f xs) = algunos (P ∘ f) xs"
have "algunos P (map f (x # xs)) = algunos P ((f x) # (map f xs))" by simp
also have "... = (P (f x) ∨ algunos P (map f xs))" by simp
also have "... = (P (f x) ∨ algunos (P ∘ f) xs)" using HI by simp
also have "... = ((P ∘ f) x ∨ algunos (P ∘ f) xs)" by simp
finally show "algunos P (map f (x # xs)) = algunos (P ∘ f) (x # xs)" by simp
qed

(* wilmorort *)
lemma "algunos P (map f xs) = algunos (P o f) xs" (is "?P xs" )
proof (induct xs)
show "?P []" by simp
next
fix a xs
assume HI: "algunos P (map f xs) = algunos (P ∘ f) xs"
have "algunos P (map f (a # xs)) = (P (f a) ∨ algunos P (map f xs))" by simp
also have "... = (P (f a) ∨ algunos (P ∘ f) xs )" using HI by simp
also have "... = ((P ∘ f) a ∨ algunos (P ∘ f)xs)" by simp
finally show " algunos P (map f (a # xs)) = algunos (P ∘ f) (a # xs)" by simp
qed

(* fraortmoy *)

lemma "algunos P (map f xs) = algunos (P o f) xs"
proof (induct xs)
show "algunos P (map f []) = algunos (P ∘ f) []" by simp
next
fix a xs
assume H1: "algunos P (map f xs) = algunos (P ∘ f) xs"
have "algunos P (map f (a # xs)) = (algunos P (map f [a]) ∨ algunos P (map f xs))" by simp
also have "… = (algunos (P ∘ f) [a] ∨ algunos (P ∘ f) xs )" using H1 by simp
also have "… = algunos (P ∘ f) (a#xs)" by simp
finally show "algunos P (map f (a # xs)) = algunos (P ∘ f) (a#xs)" by simp
qed

{*danrodcha*}
lemma "algunos P (map f xs) = algunos (P o f) xs" (is "?Q xs")
proof (induct xs)
show "?Q []" by simp
next
fix a xs assume HI: "?Q xs"
have " algunos P (map f (a # xs)) = algunos P ((f a)#(map f xs))" by simp
also have "… = (P (f a) ∨ algunos P (map f xs))" by simp
also have "… = (P (f a) ∨ algunos (P ∘ f) xs)" using HI by blast
also have "… = ((P ∘ f) a ∨ algunos (P ∘ f) xs)" by simp
also have "… = algunos (P ∘ f) (a # xs)" by simp
finally show "?Q (a # xs)" by blast
qed

text {*
---------------------------------------------------------------------
Ejercicio 8.1. Demostrar o refutar automáticamente
algunos P (xs @ ys) = (algunos P xs ∨ algunos P ys)
---------------------------------------------------------------------
*}

(* ivamenjim marpoldia1 paupeddeg crigomgom rubgonmar wilmorort fraortmoy danrodcha *)

lemma "algunos P (xs @ ys) = (algunos P xs ∨ algunos P ys)"
by (induct xs) auto

(* ferrenseg *)
lemma "algunos P (xs @ ys) = (algunos P xs ∨ algunos P ys)"
by (induct xs) simp_all

text {*
---------------------------------------------------------------------
Ejercicio 8.2. Demostrar o refutar detalladamente
algunos P (xs @ ys) = (algunos P xs ∨ algunos P ys)
---------------------------------------------------------------------
*}

(* ivamenjim ferrenseg marpoldia1 wilmorort*)

lemma algunos_append:
"algunos P (xs @ ys) = (algunos P xs ∨ algunos P ys)"
proof (induct xs)
show "algunos P ([] @ ys) = (algunos P [] ∨ algunos P ys)" by simp
next
fix a xs
assume HI: "algunos P (xs @ ys) = (algunos P xs ∨ algunos P ys)"
have "algunos P ((a # xs) @ ys) = (P a ∨ (algunos P (xs @ ys)))" by simp
also have "... = (P a ∨ (algunos P xs ∨ algunos P ys))" using HI by simp
finally show "algunos P ((a # xs) @ ys) = (algunos P (a # xs) ∨ algunos P ys)" by simp
qed

(* migtermor crigomgom *)

lemma algunos_append2:
"algunos P (xs @ ys) = (algunos P xs ∨ algunos P ys)" (is "?Q xs")
proof (induct xs)
show "?Q []" by simp
next
fix x xs
assume HI: "?Q xs"
have "algunos P ((x#xs) @ ys) = algunos P (x#(xs @ ys))" by simp
also have "… = ((P x) ∨ (algunos P (xs @ ys)))" by simp
also have "… = ((P x) ∨ (algunos P xs) ∨ (algunos P ys))" using HI by simp
also have "… = ((algunos P (x#xs)) ∨ (algunos P ys))" by simp
finally show "algunos P ((x#xs) @ ys) = (algunos P (x#xs) ∨ algunos P ys)" by simp
qed

(* paupeddeg*)
lemma algunos_append3:
"algunos P (xs @ ys) = (algunos P xs ∨ algunos P ys)"
proof (induct xs)
show "algunos P ([] @ ys) = (algunos P [] ∨ algunos P ys)" by simp
next
fix a xs
assume HI: "algunos P (xs @ ys) = (algunos P xs ∨ algunos P ys)"
have "algunos P ((a # xs) @ ys) = (algunos P [a] ∨ ( algunos P (xs @ ys)))" by simp
also have "... = (algunos P [a] ∨ (algunos P xs ∨ algunos P ys))" using HI by simp
finally show "algunos P ((a # xs) @ ys) = (algunos P (a # xs) ∨ algunos P ys)" by simp
qed

(* fraortmoy *)
lemma algunos_append4:
"algunos P (xs @ ys) = (algunos P xs ∨ algunos P ys)"
proof (induct xs)
show " algunos P ([] @ ys) = (algunos P [] ∨ algunos P ys)" by simp
next
fix a xs
assume H1: "algunos P (xs @ ys) = (algunos P xs ∨ algunos P ys)"
have "algunos P ((a # xs) @ ys) = (algunos P [a] ∨ algunos P (xs @ ys))" by simp
also have "… = (algunos P [a] ∨ algunos P xs ∨ algunos P ys)" using H1 by simp
also have "… = ((algunos P [a] ∨ algunos P xs) ∨ algunos P ys)" by simp
also have "… = (algunos P (a#xs) ∨ algunos P ys)" by simp
finally show "algunos P ((a # xs) @ ys) = (algunos P (a#xs) ∨ algunos P ys)" by simp
qed

{* danrodcha *}
lemma algunos_append5:
"algunos P (xs @ ys) = (algunos P xs ∨ algunos P ys)" (is "?Q xs")
proof (induct xs)
show "?Q []" by simp
next
fix a xs assume HI: "?Q xs"
have "algunos P ((a#xs) @ ys) = algunos P (a#(xs @ ys))" by simp
also have "… = (algunos P [a] ∨ algunos P (xs @ ys))" by simp
also have "… = (algunos P [a] ∨ algunos P xs ∨ algunos P ys)" using HI by simp
also have "… = (algunos P (a # xs) ∨ algunos P ys)" by simp
finally show "?Q (a # xs)" by simp
qed

text {*
---------------------------------------------------------------------
Ejercicio 9.1. Demostrar o refutar automáticamente
algunos P (rev xs) = algunos P xs
---------------------------------------------------------------------
*}

(* ivamenjim migtermor marpoldia1 rubgonmar paupeddeg *)

lemma "algunos P (rev xs) = algunos P xs"
apply (induct xs)
apply simp
apply (simp add: algunos_append)
apply auto
done

(* ferrenseg crigomgom danrodcha*)
lemma "algunos P (rev xs) = algunos P xs"
by (induct xs) (auto simp add: algunos_append)

(* wilmorort *)
lemma "algunos P (rev xs) = algunos P xs"
by (induct xs,simp,simp add: algunos_append,auto)

text {*
---------------------------------------------------------------------
Ejercicio 9.2. Demostrar o refutar detalladamente
algunos P (rev xs) = algunos P xs
---------------------------------------------------------------------
*}

lemma "algunos P (rev xs) = algunos P xs"
oops

(* migtermor *)
lemma auxiliar1:
"rev (a#xs) = rev xs @ [a]"
by auto

lemma "algunos P (rev xs) = algunos P xs" (is "?Q xs")
proof (induct xs)
show "?Q []" by simp
next
fix x xs
assume HI: "?Q xs"
have "algunos P (rev (x#xs)) = (algunos P (rev xs @ [x]))" using auxiliar1 by simp
also have "… = ((algunos P (rev xs)) ∨ (algunos P [x]))" by (simp add: algunos_append)
also have "… = ((P x) ∨ (algunos P (rev xs)))" by simp arith
also have "… = ((P x) ∨ (algunos P xs))" using HI by simp
finally show "algunos P (rev (x#xs)) = (algunos P (x#xs))" by simp
qed

(* ferrenseg *)

lemma "algunos P (rev xs) = algunos P xs"
proof (induct xs)
show "algunos P (rev []) = algunos P []" by simp
next
fix x xs
assume HI: "algunos P (rev xs) = algunos P xs"
show "algunos P (rev (x # xs)) = algunos P (x # xs)"
proof -
have "algunos P (rev (x # xs)) = algunos P ((rev xs) @ [x])" by simp
also have "… = (algunos P (rev xs) ∨ algunos P [x])"
by (simp add: algunos_append)
also have "… = (algunos P xs ∨ algunos P [x])" using HI by simp
also have "… = (algunos P xs ∨ P x)" by simp
also have "… = (P x ∨ algunos P xs)" by auto
also have "… = algunos P (x # xs)" by simp
finally show ?thesis
qed
qed

(* ivamenjim marpoldia1*)

lemma "algunos P (rev xs) = algunos P xs" (is "?P xs")
proof (induct xs)
show "?P []" by simp
next
fix a xs
assume HI: "?P xs"
have "algunos P (rev (a # xs)) = (algunos P ((rev xs)@[a]))" by simp
also have "... = ((algunos P (rev xs)) ∨ algunos P [a])" by (simp add: algunos_append)
also have "... = (algunos P xs ∨ algunos P [a])" using HI by simp
also have "... = (algunos P [a] ∨ algunos P xs)" by arith
finally show "algunos P (rev (a # xs)) = algunos P (a # xs)" by simp
qed

(*paupeddeg*)
lemma "algunos P (rev xs) = algunos P xs"
proof (induct xs)
show "algunos P (rev []) = algunos P []" by simp
next
fix a xs
assume HI:"algunos P (rev xs) = algunos P xs"
have "algunos P (rev (a # xs)) = algunos P ((rev xs) @ [a])" by simp
also have "... = ((algunos P (rev xs)) ∨ (algunos P [a]))" by (simp add: algunos_append)
also have "... = ((algunos P xs) ∨ (algunos P [a]))" using HI by simp
also have "... = ((algunos P [a]) ∨ (algunos P xs))" by arith
finally show "algunos P (rev (a # xs)) = algunos P (a # xs)" by simp
qed

(*crigomgom *)
lemma "algunos P (rev xs) = algunos P xs"
proof (induct xs)
show "algunos P (rev []) = algunos P []" by simp
next
fix x xs
assume HI: " algunos P (rev xs) = algunos P xs"
have "algunos P (rev (x # xs)) = algunos P ((rev xs) @ [x])" by simp
also have "... = (algunos P (rev xs) ∨ algunos P [x])" by (simp add: algunos_append)
also have "... = (algunos P xs ∨ algunos P [x])" using HI by simp
also have "... = (algunos P xs ∨ P x)" by simp
also have "... = (P x ∨ algunos P xs)" by arith
also have "... = algunos P (x#xs)" by simp
finally show "algunos P (rev (x # xs)) = algunos P (x # xs)" by simp
qed

(*wilmorort*)
lemma "algunos P (rev xs) = algunos P xs" (is "?P xs")
proof (induct xs)
show "?P []" by simp
next
fix a xs
assume HI: "?P xs"
have "algunos P (rev (a#xs)) = algunos P ((rev xs) @ [a])" by simp
also have "... = (algunos P(rev xs) ∨ algunos P [a])" by (simp add: algunos_append)
also have "... = (algunos P xs ∨ algunos P [a])" using HI by simp
also have "... = (algunos P [a] ∨ algunos P xs)" by (simp add: HOL.disj_commute)
also have "... = algunos P([a]@(xs))" by (simp)
finally show "algunos P (rev (a#xs))= algunos P (a # xs)" by simp
qed

{*danrodcha *}
lemma "algunos P (rev xs) = algunos P xs" (is "?Q xs")
proof (induct xs)
show "?Q []" by simp
next
fix a xs assume HI: "?Q xs"
have "algunos P (rev (a#xs)) = algunos P ((rev xs) @ [a])" by simp
also have "… = (algunos P (rev xs) ∨ algunos P [a])" by (simp add:algunos_append)
also have "… = (algunos P xs ∨ algunos P [a])" using HI by simp
also have "… = (algunos P [a] ∨ algunos P xs)" by (simp add: HOL.disj_commute)
also have "… = algunos P (a # xs)" by simp
finally show "?Q (a # xs)" by simp
qed

text {*
---------------------------------------------------------------------
Ejercicio 10. Encontrar un término no trivial Z tal que sea cierta la
siguiente ecuación:
algunos (λx. P x ∨ Q x) xs = Z
y demostrar la equivalencia de forma automática y detallada.
---------------------------------------------------------------------
*}

(* migtermor *)

lemma "algunos (λx. P x ∨ Q x) xs = ((algunos (λx. P x) xs) ∨ (algunos (λx. Q x) xs))"
by (induct xs) auto

lemma "algunos (λx. P x ∨ Q x) xs = ((algunos (λx. P x) xs) ∨ (algunos (λx. Q x) xs))" (is "?R xs")
proof (induct xs)
show "?R []" by simp
next
fix x xs
assume HI: "?R xs"
have "algunos (λx. P x ∨ Q x) (x#xs) = (((λx. P x ∨ Q x) x) ∨ (algunos (λx. P x ∨ Q x) xs))"
by simp
also have "… = ((P x ∨ Q x) ∨ (algunos (λx. P x ∨ Q x) xs))" by simp
also have H1: "… = ((((λx. P x) x) ∨ (algunos (λx. P x) xs)) ∨ (((λx. Q x) x) ∨ (algunos (λx. Q x) xs)))"
using HI by simp arith
have H2: "… = ((algunos (λx. P x) (x#xs)) ∨ (algunos (λx. Q x) (x#xs)))"
by simp
have C: "(algunos (λx. P x ∨ Q x) (x#xs)) =
((algunos (λx. P x) (x#xs)) ∨ (algunos (λx. Q x) (x#xs)))"
using H1 H2 by simp
finally show "(algunos (λx. P x ∨ Q x) (x#xs)) =
((algunos (λx. P x) (x#xs)) ∨ (algunos (λx. Q x) (x#xs)))"
using C by simp
qed

(* ferrenseg *)

lemma "algunos (λx. P x ∨ Q x) xs = (algunos P xs ∨ algunos Q xs)"
proof (induct xs)
show "algunos (λx. P x ∨ Q x) [] = (algunos P [] ∨ algunos Q [])" by simp
next
fix x xs
assume HI: "algunos (λx. (P x ∨ Q x)) xs = (algunos P xs ∨ algunos Q xs)"
show "algunos (λx. P x ∨ Q x) (x # xs) = (algunos P (x # xs) ∨ algunos Q (x # xs))"
proof -
have "algunos (λx. P x ∨ Q x) (x # xs) =
((P x) ∨ (Q x) ∨ algunos (λx. P x ∨ Q x) xs)" by simp
also have "… = ((P x) ∨ (Q x) ∨ algunos P xs ∨ algunos Q xs)"
using HI by simp
also have "… = (((P x) ∨ algunos P xs) ∨ ((Q x) ∨ algunos Q xs))" by auto
also have "… = (algunos P (x # xs) ∨ algunos Q (x # xs))" by simp
finally show ?thesis
qed
qed

(* ivamenjim marpoldia1 *)

lemma "algunos (λx. P x ∨ Q x) xs = (algunos P xs ∨ algunos Q xs)"
by (induct xs) auto

lemma "algunos (λx. P x ∨ Q x) xs = (algunos P xs ∨ algunos Q xs)"
proof (induct xs)
show "algunos (λx. P x ∨ Q x) [] = (algunos P [] ∨ algunos Q [])" by simp
next
fix a xs
assume HI: "algunos (λx. P x ∨ Q x) xs = (algunos P xs ∨ algunos Q xs)"
have "algunos (λx. P x ∨ Q x) (a # xs) = (P a ∨ Q a ∨ algunos (λx. P x ∨ Q x) xs)" by simp
also have "... = (P a ∨ Q a ∨ algunos P xs ∨ algunos Q xs)" using HI by simp
finally show "algunos (λx. P x ∨ Q x) (a # xs) = (algunos P (a # xs) ∨ algunos Q (a # xs))" by auto
qed

(* wilmorort danrodcha*)
lemma "algunos (λx. P x ∨ Q x) xs = (¬todos(λx. ¬P x)xs ∨ ¬todos(λx. ¬Q x)xs)"
by (induct xs) auto

(* wilmorort *)
lemma "algunos (λx. P x ∨ Q x) xs = (¬todos(λx. ¬P x)xs ∨ ¬todos(λx. ¬Q x)xs)"
(is "?P xs")
proof (induct xs)
show "?P []" by simp
next
fix a xs
assume HI: "?P xs"
have "algunos (λx. P x ∨ Q x) (a # xs) = (P a ∨ Q a ∨ algunos (λx. P x ∨ Q x) xs)" by simp
also have "... = (P a ∨ Q a ∨ (¬ todos (λx. ¬ P x) xs ∨ ¬ todos (λx. ¬ Q x) xs))" using HI by simp
also have "... = (P a ∨ ¬ todos (λx. ¬ P x) xs ∨ Q a ∨ ¬ todos (λx. ¬ Q x) xs )" by arith
finally show "algunos (λx. P x ∨ Q x) (a # xs) = (~ todos (λx. ¬ P x) (a # xs) ∨ ~ todos (λx. ¬ Q x) (a # xs))" by simp
qed

{* danrodcha *}
lemma "algunos (λx. P x ∨ Q x) xs = (algunos P xs ∨ algunos Q xs)" (is "?R xs")
proof (induct xs)
show "?R []" by simp
next
fix a xs assume HI: "?R xs"
have 1:" (Q a ∨ algunos P xs) = (algunos P xs ∨ Q a)" by (simp add: HOL.disj_commute)
have "algunos (λx. P x ∨ Q x) (a # xs) =
(algunos P [a] ∨ algunos Q [a] ∨ algunos (λx. P x ∨ Q x) xs)" by simp
also have "… = (P a ∨ (Q a ∨ algunos P xs) ∨ algunos Q xs)" using HI by simp
also have "… = (P a ∨ algunos P xs ∨ Q a ∨ algunos Q xs)" using 1 by simp
also have "… = (algunos P (a # xs) ∨ algunos Q (a # xs))" by simp
finally show "?R (a # xs)" by simp
qed

text {*
---------------------------------------------------------------------
Ejercicio 11.1. Demostrar o refutar automáticamente
algunos P xs = (¬ todos (λx. (¬ P x)) xs)
---------------------------------------------------------------------
*}

(* ivamenjim wilmorort migtermor marpoldia1 crigomgom rubgonmar paupeddeg danrodcha*)

lemma "algunos P xs = (¬ todos (λx. (¬ P x)) xs)"
by (induct xs) auto

(* ferrenseg *)

lemma "algunos P xs = (¬ todos (λx. (¬ P x)) xs)"
by (induct xs) simp_all

text {*
---------------------------------------------------------------------
Ejercicio 11.2. Demostrar o refutar datalladamente
algunos P xs = (¬ todos (λx. (¬ P x)) xs)
---------------------------------------------------------------------
*}

(* ivamenjim migtermor marpoldia1 crigomgom paupeddeg*)

lemma "algunos P xs = (¬ todos (λx. (¬ P x)) xs)"
proof (induct xs)
show "algunos P [] = (¬ todos (λx. ¬ P x) [])" by simp
next
fix a xs
assume HI: "algunos P xs = (¬ todos (λx. ¬ P x) xs)"
have "algunos P (a # xs) = (P a ∨ (algunos P xs))" by simp
also have "... = (P a ∨ (¬ todos (λx. ¬ P x) xs))" using HI by simp
finally show "algunos P (a # xs) = (¬ todos (λx. ¬ P x) (a # xs))" by simp
qed

(* ferrenseg *)

lemma "algunos P xs = (¬ todos (λx. (¬ P x)) xs)"
proof (induct xs)
show "algunos P [] = (¬ todos (λx. (¬ P x)) [])" by simp
next
fix x xs
assume HI: "algunos P xs = (¬ todos (λx. (¬ P x)) xs)"
show "algunos P (x # xs) = (¬ todos (λx. (¬ P x)) (x # xs))"
proof -
have "algunos P (x # xs) = ((P x) ∨ algunos P xs)" by simp
also have "… = ((P x) ∨ ¬ todos (λx. (¬ P x)) xs)" using HI by simp
also have "… = (¬ (¬ (P x) ∧ todos (λx. (¬ P x)) xs))" by simp
also have "… = (¬ todos (λx. (¬ P x)) (x # xs))" by simp
finally show ?thesis
qed
qed

{* danrodcha *}
lemma "algunos P xs = (¬ todos (λx. (¬ P x)) xs)" (is "?Q xs")
proof (induct xs)
show "?Q []" by simp
next
fix a xs assume HI: "?Q xs"
have "algunos P (a # xs) = (P a ∨ algunos P xs)" by simp
also have "… = (P a ∨ (¬ todos (λx. (¬ P x)) xs))" using HI by simp
also have "… = (¬(¬(P a) ∧ todos(λx. (¬ P x)) xs))" by simp
also have "… = (¬ todos(λx. (¬ P x)) (a # xs))" by simp
finally show "?Q (a # xs)" by simp
qed

text {*
---------------------------------------------------------------------
Ejercicio 12. Definir la funcion primitiva recursiva
estaEn :: 'a ⇒ 'a list ⇒ bool
tal que (estaEn x xs) se verifica si el elemento x está en la lista
xs. Por ejemplo,
estaEn (2::nat) [3,2,4] = True
estaEn (1::nat) [3,2,4] = False
---------------------------------------------------------------------
*}

(* ivamenjim ferrenseg migtermor serrodcal crigomgom rubgonmar marpoldia1 paupeddeg danrodcha*)

fun estaEn :: "'a ⇒ 'a list ⇒ bool" where
"estaEn x [] = False"
| "estaEn x (a#xs) = ((a = x) ∨ (estaEn x xs))"

value "estaEn (2::nat) [3,2,4] = True"
value "estaEn (1::nat) [3,2,4] = False"

text {*
---------------------------------------------------------------------
Ejercicio 13. Expresar la relación existente entre estaEn y algunos.
Demostrar dicha relación de forma automática y detallada.
---------------------------------------------------------------------
*}

(* migtermor crigomgom *)

lemma "estaEn x xs = (algunos (λa. a=x) xs)"
by (induct xs) auto

lemma auxiliar13:
"(x=a) = ((λa. a=x) a)"
by auto

lemma "estaEn x xs = (algunos (λa. a=x) xs)" (is "?P xs")
proof (induct xs)
show "?P []" by simp
next
fix a xs
assume HI: "?P xs"
have "estaEn x (a#xs) = ((a=x) ∨ (estaEn x xs))" by simp
also have H: "… = ((a=x) ∨ (algunos (λa. a=x) xs))" using HI by simp
also have "… = (((λa. a=x) a) ∨ (algunos (λa. a=x) xs))" using auxiliar13 by simp
also have "… = (algunos (λa. a=x) (a#xs))" by simp
have C: "estaEn x (a#xs) = (algunos (λa. a=x) (a#xs))" using H by simp
finally show "estaEn x (a#xs) = (algunos (λa. a=x) (a#xs))" using C by simp
qed

(* ferrenseg *)

lemma estaEn_algunos:
"estaEn y xs = algunos (λx. x=y) xs"
by (induct xs) auto

lemma estaEn_algunos_2:
"estaEn y xs = algunos (λx. x=y) xs"
proof (induct xs)
show "estaEn y [] = algunos (λx. x=y) []" by simp
next
fix x xs
assume HI: "estaEn y xs = algunos (λx. x=y) xs"
show "estaEn y (x # xs) = algunos (λx. x=y) (x # xs)"
proof -
have "estaEn y (x # xs) = (y=x ∨ estaEn y xs)" by simp
also have "… = (y=x ∨ algunos (λx. x=y) xs)" using HI by simp
also have "… = (x=y ∨ algunos (λx. x=y) xs)" by auto
also have "… = algunos (λx. x=y) (x # xs)" by simp
finally show ?thesis
qed
qed

(*crigomgom paupeddeg*)

lemma "estaEn a xs = algunos (λx. x = a) xs"
by (induct xs) auto

lemma "estaEn x xs = (algunos (λa. a=x) xs)"
proof (induct xs)
show " estaEn x [] = algunos (λa. a = x) []" by simp
next
fix a xs
assume HI: "estaEn x xs = algunos (λa. a = x) xs"
have "estaEn x (a # xs) = (a = x ∨ estaEn x xs)" by simp
also have "... = (a = x ∨ algunos (λa. a = x) xs)" using HI by simp
finally show " estaEn x (a # xs) = algunos (λa. a = x) (a # xs)" by simp
qed

(* ivamenjim marpoldia1 *)

lemma "estaEn x xs = algunos (λy. x=y) xs"
by (induct xs) auto

lemma "estaEn x xs = algunos (λy. x=y) xs"
proof (induct xs)
show "estaEn x [] = algunos (op = x) []" by simp
next
fix a xs
assume HI: "estaEn x xs = algunos (op = x) xs"
have "estaEn x (a # xs) = ((a = x) ∨ (estaEn x xs))" by simp
also have "... = (a = x ∨ algunos (op = x) xs)" using HI by simp
finally show " estaEn x (a # xs) = algunos (op = x) (a # xs)" by auto
qed

{* danrodcha *}
lemma "estaEn a xs = algunos (op = a) xs"
by (induct xs) auto

{* danrodcha *}
lemma "estaEn a xs = algunos (op = a) xs"
proof (induct xs)
show "estaEn a [] = algunos (op = a) []" by simp
next
fix x xs
assume HI: "estaEn a xs = algunos (op = a) xs"
have "estaEn a (x # xs) = ((x = a) ∨ (estaEn a xs))" by simp
also have "… = (x = a ∨ algunos (op = a) xs)" using HI by simp
also have "… = ((op = a) x ∨ algunos (op = a) xs)" by (simp add:HOL.eq_commute)
also have "… = algunos (op = a) (x # xs)" by simp
finally show "estaEn a (x # xs) = algunos (op = a) (x # xs)" by simp
qed

end
</source>

Relación 4

2016-11-22T21:31:19Z

Pablucoto:

<source lang="isar">
chapter {* R4: Cuantificadores sobre listas *}

theory R4_Cuantificadores_sobre_listas
imports Main
begin

text {*
---------------------------------------------------------------------
Ejercicio 1. Definir la función
todos :: ('a ⇒ bool) ⇒ 'a list ⇒ bool
tal que (todos p xs) se verifica si todos los elementos de la lista
xs cumplen la propiedad p. Por ejemplo, se verifica
todos (λx. 1<length x) [[2,1,4],[1,3]]
¬todos (λx. 1<length x) [[2,1,4],[3]]

Nota: La función todos es equivalente a la predefinida list_all.
---------------------------------------------------------------------
*}

{*danrodcha*}
fun todos :: "('a ⇒ bool) ⇒ 'a list ⇒ bool" where
"todos p xs = "todos p [] = True"
| "todos p (x#xs) = (p x ∧ todos p xs)""

(* ivamenjim migtermor dancorgar wilmorort marpoldia1 ferrenseg paupeddeg pablucoto crigomgom anaprarod serrodcal juacabsou rubgonmar josgarsan fraortmoy lucnovdos*)
fun todos :: "('a ⇒ bool) ⇒ 'a list ⇒ bool" where
"todos p [] = True"
| "todos p (x#xs) = (p x ∧ todos p xs)"

text {*
---------------------------------------------------------------------
Ejercicio 2. Definir la función
algunos :: ('a ⇒ bool) ⇒ 'a list ⇒ bool
tal que (algunos p xs) se verifica si algunos elementos de la lista
xs cumplen la propiedad p. Por ejemplo, se verifica
algunos (λx. 1<length x) [[2,1,4],[3]]
¬algunos (λx. 1<length x) [[],[3]]"

Nota: La función algunos es equivalente a la predefinida list_ex.
---------------------------------------------------------------------
*}

{*danrodcha ivamenjim migtermor dancorgar marpoldia1 ferrenseg wilmorort paupeddeg pablucoto crigomgom anaprarod serrodcal juacabsou rubgonmar josgarsan fraortmoy lucnovdos*}
fun algunos :: "('a ⇒ bool) ⇒ 'a list ⇒ bool" where
"algunos p [] = False"
| "algunos p (x#xs) = (p x ∨ algunos p xs)"

text {*
---------------------------------------------------------------------
Ejercicio 3.1. Demostrar o refutar automáticamente
todos (λx. P x ∧ Q x) xs = (todos P xs ∧ todos Q xs)
---------------------------------------------------------------------
*}

{*danrodcha ivamenjim migtermor dancorgar marpoldia1 ferrenseg wilmorort paupeddeg pablucoto anaprarod serrodcal juacabsou rubgonmar josgarsan fraortmoy lucnovdos*}
lemma "todos (λx. P x ∧ Q x) xs = (todos P xs ∧ todos Q xs)"
by (induct xs) auto

text {*
---------------------------------------------------------------------
Ejercicio 3.2. Demostrar o refutar detalladamente
todos (λx. P x ∧ Q x) xs = (todos P xs ∧ todos Q xs)
---------------------------------------------------------------------
*}

{*danrodcha*}
lemma "todos (λx. P x ∧ Q x) xs = (todos P xs ∧ todos Q xs)"
proof (induct xs)
show "?R []" by simp
next
fix a xs
assume HI: "?R xs"
have "todos (λx. P x ∧ Q x) (a#xs) = (P a ∧ Q a ∧ todos (λx. P x ∧ Q x) xs)" by simp
also have "… = (P a ∧ Q a ∧ (todos P xs ∧ todos Q xs))" using HI by simp
also have "… = ((P a ∧ todos P xs) ∧ (Q a ∧ todos Q xs))" by blast
also have "… = (todos P (a#xs) ∧ todos Q (a#xs))" by simp
finally show "?R (a#xs)" by simp
qed

(* ivamenjim wilmorort serrodcal josgarsan*)

lemma "todos (λx. P x ∧ Q x) xs = (todos P xs ∧ todos Q xs)"
proof (induct xs)
show "todos (λx. P x ∧ Q x) [] = (todos P [] ∧ todos Q [])" by simp
next
fix a xs
assume HI: "todos (λx. P x ∧ Q x) xs = (todos P xs ∧ todos Q xs)"
have "todos (λx. P x ∧ Q x) (a # xs) = (P a ∧ Q a ∧ todos (λx. P x ∧ Q x) xs)" by simp
also have "... = (P a ∧ Q a ∧ todos P xs ∧ todos Q xs)" using HI by simp
finally show "todos (λx. P x ∧ Q x) (a # xs) = (todos P (a # xs) ∧ todos Q (a # xs))" by auto
qed

(* dancorgar paupeddeg *)
lemma "todos (λx. P x ∧ Q x) xs = (todos P xs ∧ todos Q xs)"
proof (induct xs)
show "todos (λx. P x ∧ Q x) [] = (todos P [] ∧ todos Q [])" by simp
next
fix y xs
assume HI: "todos (λx. P x ∧ Q x) xs = (todos P xs ∧ todos Q xs)"
have "todos (λx. P x ∧ Q x) (y#xs) = ((P y ∧ Q y) ∧ (todos (λx. P x ∧ Q x) xs))" by simp
also have "… = ((P y ∧ Q y) ∧ (todos P xs ∧ todos Q xs))" using HI by simp
also have "… = ((P y ∧ todos P xs) ∧ (Q y ∧ todos Q xs))" by blast
also have "… = ((todos P (y#xs)) ∧ (todos Q (y#xs)))" by simp
finally show "todos (λx. P x ∧ Q x) (y#xs) = (todos P (y#xs) ∧ todos Q (y#xs))" by simp
qed

(* migtermor *)
lemma "todos (λx. P x ∧ Q x) xs = (todos P xs ∧ todos Q xs)"
proof (induct xs)
show "todos (λx. P x ∧ Q x) [] = (todos P [] ∧ todos Q [])" by simp
next
fix x xs
assume HI: "todos (λx. P x ∧ Q x) xs = (todos P xs ∧ todos Q xs)"
have "(todos (λx. P x ∧ Q x) (x#xs)) = (( P x ∧ Q x) ∧ (todos (λx. P x ∧ Q x) xs))"
by (simp only: todos.simps(2))
also have "… = ((P x ∧ Q x) ∧ (todos P xs ∧ todos Q xs))" using HI by simp
also have "… = (((P x)∧(todos P xs)) ∧ ((Q x) ∧ (todos Q xs)))" by arith
also have "((P x ∧ Q x) ∧ (todos P xs ∧ todos Q xs)) = (((P x)∧(todos P xs)) ∧ ((Q x) ∧ (todos Q xs)))"
by arith (* Este paso es exactamente el mismo que el anterior, pero sin cualquiera de los dos no funciona el "finally show" *)
have "… = (((P x)∧(todos P xs))∧((Q x)∧(todos Q xs)))" by simp
have "… = ((todos P (x#xs))∧(todos Q (x#xs)))" by simp
finally show "(todos (λx. P x ∧ Q x) (x#xs)) = ((todos P (x#xs)) ∧ (todos Q (x#xs)))" by simp
qed

(* marpoldia1 *)
lemma "todos (λx. P x ∧ Q x) xs = (todos P xs ∧ todos Q xs)"
proof (induct xs)
show "todos (λx. P x ∧ Q x) [] = (todos P [] ∧ todos Q [])" by simp
next
fix a xs
assume HI: "todos (λx. P x ∧ Q x) xs = (todos P xs ∧ todos Q xs)"
have "todos (λx. P x ∧ Q x) (a # xs) = ((P a ∧ Q a) ∧ (todos P xs ∧ todos Q xs))" using HI by simp
also have "... = ((P a ∧ todos P xs) ∧ (Q a ∧ todos Q xs))" by blast
also have "... = (todos P (a#xs) ∧ todos Q (a#xs)) " by simp
finally show "todos (λx. P x ∧ Q x) (a#xs) = (todos P (a#xs) ∧ todos Q (a#xs))" by simp
qed

(* ferrenseg *)

lemma "todos (λx. P x ∧ Q x) xs = (todos P xs ∧ todos Q xs)"
proof (induct xs)
show "todos (λx. P x ∧ Q x) [] = (todos P [] ∧ todos Q [])" by simp
next
fix n xs
assume HI: "todos (λx. P x ∧ Q x) xs = (todos P xs ∧ todos Q xs)"
have "todos (λx. P x ∧ Q x) (n # xs) = (P n ∧ Q n ∧ todos P xs ∧ todos Q xs)" using HI by simp
also have "… = P n ∧ Q n ∧ todos P xs ∧ todos Q xs" by blast
also have "⋯ = todos P (n # xs) ∧ todos Q (n # xs)" by simp
finally show "todos (λx. P x ∧ Q x) (n # xs) = todos P (n # xs) ∧ todos Q (n # xs)" by simp
qed

{* wilmorort pablucoto crigomgom anaprarod juacabsou rubgonmar fraortmoy *}

lemma "todos (λx. P x ∧ Q x) xs = (todos P xs ∧ todos Q xs)"
proof (induct xs)
show "todos (λx. P x ∧ Q x) [] = (todos P [] ∧ todos Q [])" by simp
next
fix a xs
assume HI: "todos (λx. P x ∧ Q x) xs = (todos P xs ∧ todos Q xs)"
have "todos (λx. P x ∧ Q x) (a # xs) = (P a ∧ Q a ∧ todos (λx. P x ∧ Q x) xs)" by simp
also have "... = (P a ∧ Q a ∧ todos P xs ∧ todos Q xs)" using HI by simp
also have "... = (P a ∧ todos P xs ∧ Q a ∧ todos Q xs)" by arith
also have "... = (todos P (a # xs) ∧ todos Q (a # xs))" by simp (* Este paso se puede obviar*)
finally show "todos (λx. P x ∧ Q x) (a # xs) = (todos P (a # xs) ∧ todos Q (a # xs))" by simp
qed

text {*
---------------------------------------------------------------------
Ejercicio 4.1. Demostrar o refutar automáticamente
todos P (x @ y) = (todos P x ∧ todos P y)
---------------------------------------------------------------------
*}
{*danrodcha ivamenjim marpoldia1 migtermor ferrenseg wilmorort paupeddeg crigomgom anaprarod serrodcal juacabsou rubgonmar pablucoto fraortmoy josgarsan*}
lemma "todos P (x @ y) = (todos P x ∧ todos P y)"
by (induct x) auto

{* anaprarod dancorgar *}
lemma "todos P (x @ y) = (todos P x ∧ todos P y)"
apply (induct x)
apply auto
done

text {*
---------------------------------------------------------------------
Ejercicio 4.2. Demostrar o refutar detalladamente
todos P (x @ y) = (todos P x ∧ todos P y)
---------------------------------------------------------------------
*}

(* ivamenjim ferrenseg wilmorort dancorgar fraortmoy josgarsan*)

lemma todos_append:
"todos P (x @ y) = (todos P x ∧ todos P y)"
proof (induct x)
show "todos P ([] @ y) = (todos P [] ∧ todos P y)" by simp
next
fix a x
assume HI: "todos P (x @ y) = (todos P x ∧ todos P y)"
have "todos P ((a#x) @ y) = (P a ∧ (todos P (x @ y)))" by simp
also have "... = (P a ∧ (todos P x ∧ todos P y))" using HI by simp
finally show "todos P ((a # x) @ y) = (todos P (a # x) ∧ todos P y)" by simp
qed

(* migtermor serrodcal *)

lemma todos_append1:
"todos P (x @ y) = (todos P x ∧ todos P y)" (is "?P x")
proof (induct x)
show "?P []" by simp
next
fix a x
assume HI: "?P x"
have "todos P ((a#x) @ y) = (P a ∧ (todos P (x @ y)))" by simp
also have "… = (P a ∧ (todos P x ∧ todos P y))" using HI by simp
finally show "?P (a#x)" by simp
qed

(* marpoldia1 *)
lemma todos_append:
"todos P (x @ y) = (todos P x ∧ todos P y)"
proof (induct x)
show "todos P ([] @ y) = (todos P [] ∧ todos P y)" by simp
next
fix a x
assume HI: "todos P (x @ y) = (todos P x ∧ todos P y)"
have "todos P ((a # x) @ y) = todos P (a#(x@y))" by simp
also have "... = (P a ∧ todos P (x @ y))" by simp
also have "... = (P a ∧ (todos P x ∧ todos P y))" using HI by simp
finally show "todos P ((a # x) @ y) = (todos P (a # x) ∧ todos P y)" by simp
qed

(* paupeddeg *)
lemma todos_append3:
"todos P (x @ y) = (todos P x ∧ todos P y)"
proof (induct x)
show "todos P ([] @ y) = (todos P [] ∧ todos P y)" by simp
next
fix a x
assume HI: "todos P (x @ y) = (todos P x ∧ todos P y)"
have "todos P ((a # x) @ y) = todos P (a # (x @ y))" by simp
also have "... = (todos P [a] ∧ todos P (x @ y)) " by simp
also have "... = (todos P [a] ∧ (todos P x ∧ todos P y))" using HI by simp
finally show "todos P ((a # x) @ y) = (todos P (a # x) ∧ todos P y)" by simp
qed

(* crigomgom juacabsou anaprarod*)

lemma todos_append4:
"todos P (x @ y) = (todos P x ∧ todos P y)"
proof (induct x)
show "todos P ([] @ y) = (todos P [] ∧ todos P y)" by simp
next
fix x xs
assume HI:"todos P (xs @ y) = (todos P xs ∧ todos P y)"
have "todos P ((x # xs) @ y) = (P x ∧ todos P (xs @ y ))" by simp
also have "... = (P x ∧ (todos P xs ∧ todos P y)) " using HI by simp
also have "... = ((P x ∧ todos P xs) ∧ todos P y)" by simp
finally show "todos P ((x # xs) @ y) = (todos P (x # xs) ∧ todos P y)" by simp
qed

(* pablucoto *)
lemma todos_append5:
"todos P (x @ y) = (todos P x ∧ todos P y)"
proof (induct x )
show "todos P ([] @ y) = (todos P [] ∧ todos P y)" by simp
next
fix a x
assume HI:"todos P (x @ y) = (todos P x ∧ todos P y) "
have "todos P ((a # x) @ y) = ( P a ∧ todos P (x @ y))" by simp
also have " ... =( P a ∧ (todos P x ∧ todos P y))" using HI by simp
also have "... = (todos P (a#x) ∧ todos P y)" by simp
finally show "todos P ((a # x) @ y) = (todos P (a # x) ∧ todos P y)" by auto
qed

text {*
---------------------------------------------------------------------
Ejercicio 5.1. Demostrar o refutar automáticamente
todos P (rev xs) = todos P xs
---------------------------------------------------------------------
*}

lemma "todos P (rev xs) = todos P xs"
oops

(* migtermor ivamenjim marpoldia1 serrodcal anaprarod paupeddeg *)
lemma "todos P (rev xs) = todos P xs"
apply (induct xs)
apply simp
apply (simp add: todos_append)
apply auto
done

(* ferrenseg crigomgom rubgonmar fraortmoy josgarsan*)

lemma "todos P (rev xs) = todos P xs"
by (induct xs) (auto simp add: todos_append)

(* wilmorort *)
lemma "todos P (rev xs) = todos P xs"
by (induct xs,simp,simp add: todos_append,auto)

(* juacabsou *)
lemma "todos P (rev xs) = todos P xs"
apply (induct xs,simp,simp add: todos_append,auto) done

(* dancorgar *)
lemma "todos P (rev xs) = todos P xs"
apply (induct xs)
apply (simp add: todos_append)
apply (simp add: todos_append)
apply auto
done

(* pablucoto * )
lemma "todos P (rev xs) = todos P xs"
by (induct xs) (auto, simp_all add: todos_append)
lemma "todos P (rev xs) = todos P xs"
by (induct xs) ( simp_all add: todos_append, auto)

text {*
---------------------------------------------------------------------
Ejercicio 5.2. Demostrar o refutar detalladamente
todos P (rev xs) = todos P xs
---------------------------------------------------------------------
*}

lemma "todos P (rev xs) = todos P xs"
oops

(* migtermor *)

lemma auxiliar:
"rev (a#xs) = rev xs @ [a]"
by auto

lemma "todos P (rev xs) = todos P xs" (is "?Q xs")
proof (induct xs)
show "?Q []" by (simp only: rev.simps(1))
next
fix a xs
assume HI: "?Q xs"
have "todos P (rev (a#xs)) = (todos P (rev xs @ [a]))" by (simp add: auxiliar)
have "… = ((todos P (rev xs)) ∧ (todos P [a]))" by (simp add: todos_append)
have "… = (todos P (rev xs) ∧ P a)" by simp
also have Aux: "… = (todos P xs ∧ P a)" using HI by simp
have Aux1: "… = (P a ∧ todos P xs)" by arith
have "(todos P (rev xs) ∧ P a) = (P a ∧ todos P xs)" using Aux Aux1 by simp
finally show "todos P (rev (a#xs)) = todos P (a#xs)" by (simp add: todos_append)
qed

(* marpoldia1 ferrenseg crigomgom serrodcal juacabsou rubgonmar josgarsan pablucoto*)
lemma "todos P (rev xs) = todos P xs"
proof (induct xs)
show "todos P (rev []) = todos P []" by simp
next
fix a xs
assume HI: "todos P (rev xs) = todos P xs"
have "todos P (rev (a # xs)) = (todos P ((rev xs)@[a]))" by simp
also have "... = (todos P (rev xs) ∧ todos P [a])" by (simp add:todos_append)
also have "... = (todos P xs ∧ P a)" using HI by simp
also have "... = (P a ∧ todos P xs)" by arith
also have "... = (todos P (a#xs))" by simp
finally show "todos P (rev (a # xs)) = (todos P (a#xs))" by simp
qed

(* ivamenjim *)

lemma "todos P (rev xs) = todos P xs" (is "?P xs")
proof (induct xs)
show "?P []" by simp
next
fix a xs
assume HI: "?P xs"
have "todos P (rev (a # xs)) = (todos P ((rev xs)@[a]))" by simp
also have "... = ((todos P (rev xs)) ∧ todos P [a])" by (simp add: todos_append)
also have "... = (todos P xs ∧ todos P [a])" using HI by simp
also have "... = (todos P [a] ∧ todos P xs)" by arith
finally show "todos P (rev (a # xs)) = todos P (a # xs)" by simp
qed

(* fraortmoy *)

lemma "todos P (rev xs) = todos P xs"
proof (induct xs)
show "todos P (rev []) = todos P []" by simp
next
fix a xs
assume H1: "todos P (rev xs) = todos P xs"
have " todos P (rev (a # xs)) = todos P (rev xs @ [a])" by simp
also have "… = (todos P (rev xs) ∧ todos P [a])" by (simp add:todos_append)
also have "… = (todos P xs ∧ todos P [a])" using H1 by simp
also have "… = (todos P [a] ∧ todos P xs)" by arith
finally show "todos P (rev (a # xs)) = todos P (a#xs)" by simp
qed

(* paupeddeg dancorgar *)
lemma "todos P (rev xs) = todos P xs"
proof (induct xs)
show " todos P (rev []) = todos P []" by simp
next
fix a xs
assume HI: "todos P (rev xs) = todos P xs "
have "todos P (rev (a # xs)) = todos P (rev(xs) @ [a])" by simp
also have "... = (todos P (rev xs) ∧ todos P [a]) " by (simp add: todos_append)
also have "... = (todos P xs ∧ todos P [a])" using HI by simp
also have "... = (todos P [a] ∧ todos P xs)" by arith
also have "... = todos P ([a] @ xs)" by (simp add: todos_append)
finally show "todos P (rev (a # xs)) = todos P (a # xs)" by simp
qed

(* wilmorort *)
lemma "todos P (rev xs) = todos P xs" (is "?P xs")
proof (induct xs)
show "?P []" by simp
next
fix a xs
assume HI: "?P xs"
have "todos P (rev (a#xs)) = todos P ((rev xs) @ [a])" by simp
also have "... = (todos P(rev xs) ∧ todos P [a])" by (simp add: todos_append)
also have "... = (todos P xs ∧ todos P [a])" using HI by simp
also have "... = (todos P [a] ∧ todos P xs)" by (simp add: HOL.conj_commute)
also have "... = todos P([a]@(xs))" by (simp)
finally show "todos P (rev (a#xs))= todos P (a # xs)" by simp
qed

(* anaprarod *)
(* es igual que las anteriores pero con el final también con patrones *)

lemma "todos P (rev xs) = todos P xs" (is "?P xs")
proof (induct xs)
show "?P []" by auto
next
fix a xs
assume HI: "?P xs"
have "todos P (rev (a#xs)) = todos P (rev xs @ [a])" by simp
also have "… = (todos P (rev xs) ∧ todos P [a])" by (simp add: todos_append)
also have "… = (todos P xs ∧ todos P [a])" using HI by simp
also have "… = (todos P [a] ∧ todos P xs)" by arith
also have "… = todos P (a#xs)" by simp
finally show "?P (a # xs)" by simp
qed

text {*
---------------------------------------------------------------------
Ejercicio 6. Demostrar o refutar:
algunos (λx. P x ∧ Q x) xs = (algunos P xs ∧ algunos Q xs)
---------------------------------------------------------------------
*}

lemma "algunos (λx. P x ∧ Q x) xs = (algunos P xs ∧ algunos Q xs)"
oops

(* migtermor ivamenjim ferrenseg paupeddeg crigomgom serrodcal wilmorort juacabsou rubgonmar anaprarod marpoldia1 fraortmoy josgarsan*)
lemma "algunos (λx. P x ∧ Q x) xs = (algunos P xs ∧ algunos Q xs)"
quickcheck
oops
(* Quickcheck encuentra el siguiente contraejemplo: P={a1}, Q={a2}, xs={a1,a2}. En este ejemplo:
· "algunos (λx. P x ∧ Q x) xs = False"
· "(algunos P xs ∧ algunos Q xs) = True" *)

text {*
---------------------------------------------------------------------
Ejercicio 7.1. Demostrar o refutar automáticamente
algunos P (map f xs) = algunos (P ∘ f) xs
---------------------------------------------------------------------
*}

(* ivamenjim migtermor marpoldia1 crigomgom rubgonmar wilmorort anaprarod fraortmoy juacabsou paupeddeg josgarsan*)

lemma "algunos P (map f xs) = algunos (P o f) xs"
by (induct xs) auto

(* ferrenseg *)

lemma "algunos P (map f xs) = algunos (P o f) xs"
by (induct xs) simp_all

(* anaprarod *)
lemma "algunos P (map f xs) = algunos (P o f) xs"
apply (induct xs)
apply auto
done

text {*
---------------------------------------------------------------------
Ejercicio 7.2. Demostrar o refutar datalladamente
algunos P (map f xs) = algunos (P ∘ f) xs
---------------------------------------------------------------------
*}

lemma "algunos P (map f xs) = algunos (P o f) xs"
oops

(* migtermor *)

lemma AUX: "algunos (λa. P (f a)) xs = algunos (P o f) xs"
by (induct xs) auto

lemma "algunos P (map f xs) = algunos (P o f) xs" (is "?Q xs")
proof (induct xs)
show "?Q []" by simp
next
fix x xs
assume HI: "?Q xs"
have "algunos P (map f (x#xs)) = (algunos P ((f x)#(map f xs)))" by simp
also have "… = ((P (f x)) ∨ (algunos P (map f xs)))" by (simp only: algunos.simps(2))
also have "… = ((P (f x)) ∨ (algunos (P o f) xs))"
proof (cases)
assume C1: "(P (f x))"
have Aux: "((P (f x)) ∨ (algunos P (map f xs))) = True" using C1 by simp
have Aux1: "… = ((P (f x)) ∨ (algunos (P o f) xs))" using C1 by simp
then show "((P (f x)) ∨ (algunos P (map f xs))) = ((P (f x)) ∨ (algunos (P o f) xs))"
using Aux Aux1 by simp
next
assume C2: "¬(P (f x))"
have Aux2: "((P (f x)) ∨ (algunos P (map f xs))) = (algunos P (map f xs))" using C2 by simp
have Aux3: "… = (algunos (P o f) xs)" using HI by (simp add: AUX)
also have "… = ((P (f x)) ∨ (algunos (P o f) xs))" using C2 by simp
then show "((P (f x)) ∨ (algunos P (map f xs))) = ((P (f x)) ∨ (algunos (P o f) xs))"
using Aux2 Aux3 by simp
qed
also have "… = (((P o f) x) ∨ (algunos (P o f) xs))" by simp
also have "… = (algunos (P o f) (x#xs))" by simp
finally show "algunos P (map f (x#xs)) = (algunos (P o f) (x#xs))" by simp
qed

(* ferrenseg *)

lemma "algunos P (map f xs) = algunos (P ∘ f) xs"
proof (induct xs)
show "algunos P (map f []) = algunos (P ∘ f) []" by simp
next
fix x xs
assume HI: "algunos P (map f xs) = algunos (P ∘ f) xs"
show "algunos P (map f (x # xs)) = algunos (P ∘ f) (x # xs)"
proof -
have "algunos P (map f (x # xs)) = algunos P ((f x) # (map f xs))" by simp
also have "… = ((P (f x)) ∨ (algunos P (map f xs)))" by simp
also have "… = (((P ∘ f) x) ∨ (algunos (P ∘ f) xs))" using HI by simp
also have "… = algunos (P ∘ f) (x # xs)" by simp
finally show ?thesis
qed
qed

(* ivamenjim *)

lemma "algunos P (map f xs) = algunos (P o f) xs"
proof (induct xs)
show "algunos P (map f []) = algunos (P ∘ f) []" by simp
next
fix a xs
assume HI: "algunos P (map f xs) = algunos (P ∘ f) xs"
have "algunos P (map f (a # xs)) = algunos P ((f a)#map f xs)" by simp
also have "... = (algunos P (map f [a]) ∨ algunos P (map f xs))" by simp
also have "... = (algunos P (map f [a]) ∨ algunos (P ∘ f) xs)" using HI by simp
finally show "algunos P (map f (a # xs)) = algunos (P ∘ f) (a # xs)" by simp
qed

(* crigomgom rubgonmar anaprarod marpoldia1 juacabsou *)
lemma "algunos P (map f xs) = algunos (P o f) xs"
proof (induct xs)
show "algunos P (map f []) = algunos (P ∘ f) []" by simp
next
fix x xs
assume HI: "algunos P (map f xs) = algunos (P ∘ f) xs"
have "algunos P (map f (x # xs)) = algunos P ((f x) # (map f xs))" by simp
also have "... = (P (f x) ∨ algunos P (map f xs))" by simp
also have "... = (P (f x) ∨ algunos (P ∘ f) xs)" using HI by simp
also have "... = ((P ∘ f) x ∨ algunos (P ∘ f) xs)" by simp
finally show "algunos P (map f (x # xs)) = algunos (P ∘ f) (x # xs)" by simp
qed

(* wilmorort *)
lemma "algunos P (map f xs) = algunos (P o f) xs" (is "?P xs" )
proof (induct xs)
show "?P []" by simp
next
fix a xs
assume HI: "algunos P (map f xs) = algunos (P ∘ f) xs"
have "algunos P (map f (a # xs)) = (P (f a) ∨ algunos P (map f xs))" by simp
also have "... = (P (f a) ∨ algunos (P ∘ f) xs )" using HI by simp
also have "... = ((P ∘ f) a ∨ algunos (P ∘ f)xs)" by simp
finally show " algunos P (map f (a # xs)) = algunos (P ∘ f) (a # xs)" by simp
qed

(* fraortmoy *)

lemma "algunos P (map f xs) = algunos (P o f) xs"
proof (induct xs)
show "algunos P (map f []) = algunos (P ∘ f) []" by simp
next
fix a xs
assume H1: "algunos P (map f xs) = algunos (P ∘ f) xs"
have "algunos P (map f (a # xs)) = (algunos P (map f [a]) ∨ algunos P (map f xs))" by simp
also have "… = (algunos (P ∘ f) [a] ∨ algunos (P ∘ f) xs )" using H1 by simp
also have "… = algunos (P ∘ f) (a#xs)" by simp
finally show "algunos P (map f (a # xs)) = algunos (P ∘ f) (a#xs)" by simp
qed

text {*
---------------------------------------------------------------------
Ejercicio 8.1. Demostrar o refutar automáticamente
algunos P (xs @ ys) = (algunos P xs ∨ algunos P ys)
---------------------------------------------------------------------
*}

(* ivamenjim marpoldia1 paupeddeg crigomgom rubgonmar wilmorort fraortmoy *)

lemma "algunos P (xs @ ys) = (algunos P xs ∨ algunos P ys)"
by (induct xs) auto

(* ferrenseg *)
lemma "algunos P (xs @ ys) = (algunos P xs ∨ algunos P ys)"
by (induct xs) simp_all

text {*
---------------------------------------------------------------------
Ejercicio 8.2. Demostrar o refutar detalladamente
algunos P (xs @ ys) = (algunos P xs ∨ algunos P ys)
---------------------------------------------------------------------
*}

(* ivamenjim ferrenseg marpoldia1 wilmorort*)

lemma algunos_append:
"algunos P (xs @ ys) = (algunos P xs ∨ algunos P ys)"
proof (induct xs)
show "algunos P ([] @ ys) = (algunos P [] ∨ algunos P ys)" by simp
next
fix a xs
assume HI: "algunos P (xs @ ys) = (algunos P xs ∨ algunos P ys)"
have "algunos P ((a # xs) @ ys) = (P a ∨ (algunos P (xs @ ys)))" by simp
also have "... = (P a ∨ (algunos P xs ∨ algunos P ys))" using HI by simp
finally show "algunos P ((a # xs) @ ys) = (algunos P (a # xs) ∨ algunos P ys)" by simp
qed

(* migtermor crigomgom *)

lemma algunos_append:
"algunos P (xs @ ys) = (algunos P xs ∨ algunos P ys)" (is "?Q xs")
proof (induct xs)
show "?Q []" by simp
next
fix x xs
assume HI: "?Q xs"
have "algunos P ((x#xs) @ ys) = algunos P (x#(xs @ ys))" by simp
also have "… = ((P x) ∨ (algunos P (xs @ ys)))" by simp
also have "… = ((P x) ∨ (algunos P xs) ∨ (algunos P ys))" using HI by simp
also have "… = ((algunos P (x#xs)) ∨ (algunos P ys))" by simp
finally show "algunos P ((x#xs) @ ys) = (algunos P (x#xs) ∨ algunos P ys)" by simp
qed

(* paupeddeg*)
lemma algunos_append:
"algunos P (xs @ ys) = (algunos P xs ∨ algunos P ys)"
proof (induct xs)
show "algunos P ([] @ ys) = (algunos P [] ∨ algunos P ys)" by simp
next
fix a xs
assume HI: "algunos P (xs @ ys) = (algunos P xs ∨ algunos P ys)"
have "algunos P ((a # xs) @ ys) = (algunos P [a] ∨ ( algunos P (xs @ ys)))" by simp
also have "... = (algunos P [a] ∨ (algunos P xs ∨ algunos P ys))" using HI by simp
finally show "algunos P ((a # xs) @ ys) = (algunos P (a # xs) ∨ algunos P ys)" by simp
qed

(* fraortmoy *)
lemma algunos_append:
"algunos P (xs @ ys) = (algunos P xs ∨ algunos P ys)"
proof (induct xs)
show " algunos P ([] @ ys) = (algunos P [] ∨ algunos P ys)" by simp
next
fix a xs
assume H1: "algunos P (xs @ ys) = (algunos P xs ∨ algunos P ys)"
have "algunos P ((a # xs) @ ys) = (algunos P [a] ∨ algunos P (xs @ ys))" by simp
also have "… = (algunos P [a] ∨ algunos P xs ∨ algunos P ys)" using H1 by simp
also have "… = ((algunos P [a] ∨ algunos P xs) ∨ algunos P ys)" by simp
also have "… = (algunos P (a#xs) ∨ algunos P ys)" by simp
finally show "algunos P ((a # xs) @ ys) = (algunos P (a#xs) ∨ algunos P ys)" by simp
qed

text {*
---------------------------------------------------------------------
Ejercicio 9.1. Demostrar o refutar automáticamente
algunos P (rev xs) = algunos P xs
---------------------------------------------------------------------
*}

(* ivamenjim migtermor marpoldia1 rubgonmar paupeddeg *)

lemma "algunos P (rev xs) = algunos P xs"
apply (induct xs)
apply simp
apply (simp add: algunos_append)
apply auto
done

(* ferrenseg crigomgom*)
lemma "algunos P (rev xs) = algunos P xs"
by (induct xs) (auto simp add: algunos_append)

(* wilmorort *)
lemma "algunos P (rev xs) = algunos P xs"
by (induct xs,simp,simp add: algunos_append,auto)

text {*
---------------------------------------------------------------------
Ejercicio 9.2. Demostrar o refutar detalladamente
algunos P (rev xs) = algunos P xs
---------------------------------------------------------------------
*}

lemma "algunos P (rev xs) = algunos P xs"
oops

(* migtermor *)
lemma auxiliar1:
"rev (a#xs) = rev xs @ [a]"
by auto

lemma "algunos P (rev xs) = algunos P xs" (is "?Q xs")
proof (induct xs)
show "?Q []" by simp
next
fix x xs
assume HI: "?Q xs"
have "algunos P (rev (x#xs)) = (algunos P (rev xs @ [x]))" using auxiliar1 by simp
also have "… = ((algunos P (rev xs)) ∨ (algunos P [x]))" by (simp add: algunos_append)
also have "… = ((P x) ∨ (algunos P (rev xs)))" by simp arith
also have "… = ((P x) ∨ (algunos P xs))" using HI by simp
finally show "algunos P (rev (x#xs)) = (algunos P (x#xs))" by simp
qed

(* ferrenseg *)

lemma "algunos P (rev xs) = algunos P xs"
proof (induct xs)
show "algunos P (rev []) = algunos P []" by simp
next
fix x xs
assume HI: "algunos P (rev xs) = algunos P xs"
show "algunos P (rev (x # xs)) = algunos P (x # xs)"
proof -
have "algunos P (rev (x # xs)) = algunos P ((rev xs) @ [x])" by simp
also have "… = (algunos P (rev xs) ∨ algunos P [x])"
by (simp add: algunos_append)
also have "… = (algunos P xs ∨ algunos P [x])" using HI by simp
also have "… = (algunos P xs ∨ P x)" by simp
also have "… = (P x ∨ algunos P xs)" by auto
also have "… = algunos P (x # xs)" by simp
finally show ?thesis
qed
qed

(* ivamenjim marpoldia1*)

lemma "algunos P (rev xs) = algunos P xs" (is "?P xs")
proof (induct xs)
show "?P []" by simp
next
fix a xs
assume HI: "?P xs"
have "algunos P (rev (a # xs)) = (algunos P ((rev xs)@[a]))" by simp
also have "... = ((algunos P (rev xs)) ∨ algunos P [a])" by (simp add: algunos_append)
also have "... = (algunos P xs ∨ algunos P [a])" using HI by simp
also have "... = (algunos P [a] ∨ algunos P xs)" by arith
finally show "algunos P (rev (a # xs)) = algunos P (a # xs)" by simp
qed

(*paupeddeg*)
lemma "algunos P (rev xs) = algunos P xs"
proof (induct xs)
show "algunos P (rev []) = algunos P []" by simp
next
fix a xs
assume HI:"algunos P (rev xs) = algunos P xs"
have "algunos P (rev (a # xs)) = algunos P ((rev xs) @ [a])" by simp
also have "... = ((algunos P (rev xs)) ∨ (algunos P [a]))" by (simp add: algunos_append)
also have "... = ((algunos P xs) ∨ (algunos P [a]))" using HI by simp
also have "... = ((algunos P [a]) ∨ (algunos P xs))" by arith
finally show "algunos P (rev (a # xs)) = algunos P (a # xs)" by simp
qed

(*crigomgom *)
lemma "algunos P (rev xs) = algunos P xs"
proof (induct xs)
show "algunos P (rev []) = algunos P []" by simp
next
fix x xs
assume HI: " algunos P (rev xs) = algunos P xs"
have "algunos P (rev (x # xs)) = algunos P ((rev xs) @ [x])" by simp
also have "... = (algunos P (rev xs) ∨ algunos P [x])" by (simp add: algunos_append)
also have "... = (algunos P xs ∨ algunos P [x])" using HI by simp
also have "... = (algunos P xs ∨ P x)" by simp
also have "... = (P x ∨ algunos P xs)" by arith
also have "... = algunos P (x#xs)" by simp
finally show "algunos P (rev (x # xs)) = algunos P (x # xs)" by simp
qed

(*wilmorort*)
lemma "algunos P (rev xs) = algunos P xs" (is "?P xs")
proof (induct xs)
show "?P []" by simp
next
fix a xs
assume HI: "?P xs"
have "algunos P (rev (a#xs)) = algunos P ((rev xs) @ [a])" by simp
also have "... = (algunos P(rev xs) ∨ algunos P [a])" by (simp add: algunos_append)
also have "... = (algunos P xs ∨ algunos P [a])" using HI by simp
also have "... = (algunos P [a] ∨ algunos P xs)" by (simp add: HOL.disj_commute)
also have "... = algunos P([a]@(xs))" by (simp)
finally show "algunos P (rev (a#xs))= algunos P (a # xs)" by simp
qed

text {*
---------------------------------------------------------------------
Ejercicio 10. Encontrar un término no trivial Z tal que sea cierta la
siguiente ecuación:
algunos (λx. P x ∨ Q x) xs = Z
y demostrar la equivalencia de forma automática y detallada.
---------------------------------------------------------------------
*}

(* migtermor *)

lemma "algunos (λx. P x ∨ Q x) xs = ((algunos (λx. P x) xs) ∨ (algunos (λx. Q x) xs))"
by (induct xs) auto

lemma "algunos (λx. P x ∨ Q x) xs = ((algunos (λx. P x) xs) ∨ (algunos (λx. Q x) xs))" (is "?R xs")
proof (induct xs)
show "?R []" by simp
next
fix x xs
assume HI: "?R xs"
have "algunos (λx. P x ∨ Q x) (x#xs) = (((λx. P x ∨ Q x) x) ∨ (algunos (λx. P x ∨ Q x) xs))"
by simp
also have "… = ((P x ∨ Q x) ∨ (algunos (λx. P x ∨ Q x) xs))" by simp
also have H1: "… = ((((λx. P x) x) ∨ (algunos (λx. P x) xs)) ∨ (((λx. Q x) x) ∨ (algunos (λx. Q x) xs)))"
using HI by simp arith
have H2: "… = ((algunos (λx. P x) (x#xs)) ∨ (algunos (λx. Q x) (x#xs)))"
by simp
have C: "(algunos (λx. P x ∨ Q x) (x#xs)) =
((algunos (λx. P x) (x#xs)) ∨ (algunos (λx. Q x) (x#xs)))"
using H1 H2 by simp
finally show "(algunos (λx. P x ∨ Q x) (x#xs)) =
((algunos (λx. P x) (x#xs)) ∨ (algunos (λx. Q x) (x#xs)))"
using C by simp
qed

(* ferrenseg *)

lemma "algunos (λx. P x ∨ Q x) xs = (algunos P xs ∨ algunos Q xs)"
proof (induct xs)
show "algunos (λx. P x ∨ Q x) [] = (algunos P [] ∨ algunos Q [])" by simp
next
fix x xs
assume HI: "algunos (λx. (P x ∨ Q x)) xs = (algunos P xs ∨ algunos Q xs)"
show "algunos (λx. P x ∨ Q x) (x # xs) = (algunos P (x # xs) ∨ algunos Q (x # xs))"
proof -
have "algunos (λx. P x ∨ Q x) (x # xs) =
((P x) ∨ (Q x) ∨ algunos (λx. P x ∨ Q x) xs)" by simp
also have "… = ((P x) ∨ (Q x) ∨ algunos P xs ∨ algunos Q xs)"
using HI by simp
also have "… = (((P x) ∨ algunos P xs) ∨ ((Q x) ∨ algunos Q xs))" by auto
also have "… = (algunos P (x # xs) ∨ algunos Q (x # xs))" by simp
finally show ?thesis
qed
qed

(* ivamenjim marpoldia1 *)

lemma "algunos (λx. P x ∨ Q x) xs = (algunos P xs ∨ algunos Q xs)"
by (induct xs) auto

lemma "algunos (λx. P x ∨ Q x) xs = (algunos P xs ∨ algunos Q xs)"
proof (induct xs)
show "algunos (λx. P x ∨ Q x) [] = (algunos P [] ∨ algunos Q [])" by simp
next
fix a xs
assume HI: "algunos (λx. P x ∨ Q x) xs = (algunos P xs ∨ algunos Q xs)"
have "algunos (λx. P x ∨ Q x) (a # xs) = (P a ∨ Q a ∨ algunos (λx. P x ∨ Q x) xs)" by simp
also have "... = (P a ∨ Q a ∨ algunos P xs ∨ algunos Q xs)" using HI by simp
finally show "algunos (λx. P x ∨ Q x) (a # xs) = (algunos P (a # xs) ∨ algunos Q (a # xs))" by auto
qed

(* wilmorort *)
lemma "algunos (λx. P x ∨ Q x) xs = (¬todos(λx. ¬P x)xs ∨ ¬todos(λx. ¬Q x)xs)"
by (induct xs) auto

lemma "algunos (λx. P x ∨ Q x) xs = (¬todos(λx. ¬P x)xs ∨ ¬todos(λx. ¬Q x)xs)"
(is "?P xs")
proof (induct xs)
show "?P []" by simp
next
fix a xs
assume HI: "?P xs"
have "algunos (λx. P x ∨ Q x) (a # xs) = (P a ∨ Q a ∨ algunos (λx. P x ∨ Q x) xs)" by simp
also have "... = (P a ∨ Q a ∨ (¬ todos (λx. ¬ P x) xs ∨ ¬ todos (λx. ¬ Q x) xs))" using HI by simp
also have "... = (P a ∨ ¬ todos (λx. ¬ P x) xs ∨ Q a ∨ ¬ todos (λx. ¬ Q x) xs )" by arith
finally show "algunos (λx. P x ∨ Q x) (a # xs) = (~ todos (λx. ¬ P x) (a # xs) ∨ ~ todos (λx. ¬ Q x) (a # xs))" by simp
qed

text {*
---------------------------------------------------------------------
Ejercicio 11.1. Demostrar o refutar automáticamente
algunos P xs = (¬ todos (λx. (¬ P x)) xs)
---------------------------------------------------------------------
*}

(* ivamenjim wilmorort migtermor marpoldia1 crigomgom rubgonmar paupeddeg*)

lemma "algunos P xs = (¬ todos (λx. (¬ P x)) xs)"
by (induct xs) auto

(* ferrenseg *)

lemma "algunos P xs = (¬ todos (λx. (¬ P x)) xs)"
by (induct xs) simp_all

text {*
---------------------------------------------------------------------
Ejercicio 11.2. Demostrar o refutar datalladamente
algunos P xs = (¬ todos (λx. (¬ P x)) xs)
---------------------------------------------------------------------
*}

(* ivamenjim migtermor marpoldia1 crigomgom paupeddeg*)

lemma "algunos P xs = (¬ todos (λx. (¬ P x)) xs)"
proof (induct xs)
show "algunos P [] = (¬ todos (λx. ¬ P x) [])" by simp
next
fix a xs
assume HI: "algunos P xs = (¬ todos (λx. ¬ P x) xs)"
have "algunos P (a # xs) = (P a ∨ (algunos P xs))" by simp
also have "... = (P a ∨ (¬ todos (λx. ¬ P x) xs))" using HI by simp
finally show "algunos P (a # xs) = (¬ todos (λx. ¬ P x) (a # xs))" by simp
qed

(* ferrenseg *)

lemma "algunos P xs = (¬ todos (λx. (¬ P x)) xs)"
proof (induct xs)
show "algunos P [] = (¬ todos (λx. (¬ P x)) [])" by simp
next
fix x xs
assume HI: "algunos P xs = (¬ todos (λx. (¬ P x)) xs)"
show "algunos P (x # xs) = (¬ todos (λx. (¬ P x)) (x # xs))"
proof -
have "algunos P (x # xs) = ((P x) ∨ algunos P xs)" by simp
also have "… = ((P x) ∨ ¬ todos (λx. (¬ P x)) xs)" using HI by simp
also have "… = (¬ (¬ (P x) ∧ todos (λx. (¬ P x)) xs))" by simp
also have "… = (¬ todos (λx. (¬ P x)) (x # xs))" by simp
finally show ?thesis
qed
qed

text {*
---------------------------------------------------------------------
Ejercicio 12. Definir la funcion primitiva recursiva
estaEn :: 'a ⇒ 'a list ⇒ bool
tal que (estaEn x xs) se verifica si el elemento x está en la lista
xs. Por ejemplo,
estaEn (2::nat) [3,2,4] = True
estaEn (1::nat) [3,2,4] = False
---------------------------------------------------------------------
*}

(* ivamenjim ferrenseg migtermor serrodcal crigomgom rubgonmar marpoldia1 paupeddeg*)

fun estaEn :: "'a ⇒ 'a list ⇒ bool" where
"estaEn x [] = False"
| "estaEn x (a#xs) = ((a = x) ∨ (estaEn x xs))"

value "estaEn (2::nat) [3,2,4] = True"
value "estaEn (1::nat) [3,2,4] = False"

text {*
---------------------------------------------------------------------
Ejercicio 13. Expresar la relación existente entre estaEn y algunos.
Demostrar dicha relación de forma automática y detallada.
---------------------------------------------------------------------
*}

(* migtermor crigomgom*)

lemma "estaEn x xs = (algunos (λa. a=x) xs)"
by (induct xs) auto

lemma auxiliar13:
"(x=a) = ((λa. a=x) a)"
by auto

lemma "estaEn x xs = (algunos (λa. a=x) xs)" (is "?P xs")
proof (induct xs)
show "?P []" by simp
next
fix a xs
assume HI: "?P xs"
have "estaEn x (a#xs) = ((a=x) ∨ (estaEn x xs))" by simp
also have H: "… = ((a=x) ∨ (algunos (λa. a=x) xs))" using HI by simp
also have "… = (((λa. a=x) a) ∨ (algunos (λa. a=x) xs))" using auxiliar13 by simp
also have "… = (algunos (λa. a=x) (a#xs))" by simp
have C: "estaEn x (a#xs) = (algunos (λa. a=x) (a#xs))" using H by simp
finally show "estaEn x (a#xs) = (algunos (λa. a=x) (a#xs))" using C by simp
qed

(* ferrenseg *)

lemma estaEn_algunos:
"estaEn y xs = algunos (λx. x=y) xs"
by (induct xs) auto

lemma estaEn_algunos_2:
"estaEn y xs = algunos (λx. x=y) xs"
proof (induct xs)
show "estaEn y [] = algunos (λx. x=y) []" by simp
next
fix x xs
assume HI: "estaEn y xs = algunos (λx. x=y) xs"
show "estaEn y (x # xs) = algunos (λx. x=y) (x # xs)"
proof -
have "estaEn y (x # xs) = (y=x ∨ estaEn y xs)" by simp
also have "… = (y=x ∨ algunos (λx. x=y) xs)" using HI by simp
also have "… = (x=y ∨ algunos (λx. x=y) xs)" by auto
also have "… = algunos (λx. x=y) (x # xs)" by simp
finally show ?thesis
qed
qed

(*crigomgom paupeddeg*)

lemma "estaEn a xs = algunos (λx. x = a) xs"
by (induct xs) auto

lemma "estaEn x xs = (algunos (λa. a=x) xs)"
proof (induct xs)
show " estaEn x [] = algunos (λa. a = x) []" by simp
next
fix a xs
assume HI: "estaEn x xs = algunos (λa. a = x) xs"
have "estaEn x (a # xs) = (a = x ∨ estaEn x xs)" by simp
also have "... = (a = x ∨ algunos (λa. a = x) xs)" using HI by simp
finally show " estaEn x (a # xs) = algunos (λa. a = x) (a # xs)" by simp
qed

(* ivamenjim marpoldia1 *)

lemma "estaEn x xs = algunos (λy. x=y) xs"
by (induct xs) auto

lemma "estaEn x xs = algunos (λy. x=y) xs"
proof (induct xs)
show "estaEn x [] = algunos (op = x) []" by simp
next
fix a xs
assume HI: "estaEn x xs = algunos (op = x) xs"
have "estaEn x (a # xs) = ((a = x) ∨ (estaEn x xs))" by simp
also have "... = (a = x ∨ algunos (op = x) xs)" using HI by simp
finally show " estaEn x (a # xs) = algunos (op = x) (a # xs)" by auto
qed

end
</source>

Relación 4

2016-11-22T04:49:30Z

Pablucoto:

<source lang="isar">
chapter {* R4: Cuantificadores sobre listas *}

theory R4_Cuantificadores_sobre_listas
imports Main
begin

text {*
---------------------------------------------------------------------
Ejercicio 1. Definir la función
todos :: ('a ⇒ bool) ⇒ 'a list ⇒ bool
tal que (todos p xs) se verifica si todos los elementos de la lista
xs cumplen la propiedad p. Por ejemplo, se verifica
todos (λx. 1<length x) [[2,1,4],[1,3]]
¬todos (λx. 1<length x) [[2,1,4],[3]]

Nota: La función todos es equivalente a la predefinida list_all.
---------------------------------------------------------------------
*}

{*danrodcha*}
fun todos :: "('a ⇒ bool) ⇒ 'a list ⇒ bool" where
"todos p xs = "todos p [] = True"
| "todos p (x#xs) = (p x ∧ todos p xs)""

(* ivamenjim migtermor dancorgar wilmorort marpoldia1 ferrenseg paupeddeg pablucoto crigomgom anaprarod serrodcal juacabsou rubgonmar *)
fun todos :: "('a ⇒ bool) ⇒ 'a list ⇒ bool" where
"todos p [] = True"
| "todos p (x#xs) = (p x ∧ todos p xs)"

text {*
---------------------------------------------------------------------
Ejercicio 2. Definir la función
algunos :: ('a ⇒ bool) ⇒ 'a list ⇒ bool
tal que (algunos p xs) se verifica si algunos elementos de la lista
xs cumplen la propiedad p. Por ejemplo, se verifica
algunos (λx. 1<length x) [[2,1,4],[3]]
¬algunos (λx. 1<length x) [[],[3]]"

Nota: La función algunos es equivalente a la predefinida list_ex.
---------------------------------------------------------------------
*}

{*danrodcha ivamenjim migtermor dancorgar marpoldia1 ferrenseg wilmorort paupeddeg pablucoto crigomgom anaprarod serrodcal juacabsou rubgonmar *}
fun algunos :: "('a ⇒ bool) ⇒ 'a list ⇒ bool" where
"algunos p [] = False"
| "algunos p (x#xs) = (p x ∨ algunos p xs)"

text {*
---------------------------------------------------------------------
Ejercicio 3.1. Demostrar o refutar automáticamente
todos (λx. P x ∧ Q x) xs = (todos P xs ∧ todos Q xs)
---------------------------------------------------------------------
*}

{*danrodcha ivamenjim migtermor dancorgar marpoldia1 ferrenseg wilmorort paupeddeg pablucoto anaprarod serrodcal juacabsou rubgonmar *}
lemma "todos (λx. P x ∧ Q x) xs = (todos P xs ∧ todos Q xs)"
by (induct xs) auto

text {*
---------------------------------------------------------------------
Ejercicio 3.2. Demostrar o refutar detalladamente
todos (λx. P x ∧ Q x) xs = (todos P xs ∧ todos Q xs)
---------------------------------------------------------------------
*}

{*danrodcha*}
lemma "todos (λx. P x ∧ Q x) xs = (todos P xs ∧ todos Q xs)"
proof (induct xs)
show "?R []" by simp
next
fix a xs
assume HI: "?R xs"
have "todos (λx. P x ∧ Q x) (a#xs) = (P a ∧ Q a ∧ todos (λx. P x ∧ Q x) xs)" by simp
also have "… = (P a ∧ Q a ∧ (todos P xs ∧ todos Q xs))" using HI by simp
also have "… = ((P a ∧ todos P xs) ∧ (Q a ∧ todos Q xs))" by blast
also have "… = (todos P (a#xs) ∧ todos Q (a#xs))" by simp
finally show "?R (a#xs)" by simp
qed

(* ivamenjim wilmorort serrodcal *)

lemma "todos (λx. P x ∧ Q x) xs = (todos P xs ∧ todos Q xs)"
proof (induct xs)
show "todos (λx. P x ∧ Q x) [] = (todos P [] ∧ todos Q [])" by simp
next
fix a xs
assume HI: "todos (λx. P x ∧ Q x) xs = (todos P xs ∧ todos Q xs)"
have "todos (λx. P x ∧ Q x) (a # xs) = (P a ∧ Q a ∧ todos (λx. P x ∧ Q x) xs)" by simp
also have "... = (P a ∧ Q a ∧ todos P xs ∧ todos Q xs)" using HI by simp
finally show "todos (λx. P x ∧ Q x) (a # xs) = (todos P (a # xs) ∧ todos Q (a # xs))" by auto
qed

(* dancorgar paupeddeg *)
lemma "todos (λx. P x ∧ Q x) xs = (todos P xs ∧ todos Q xs)"
proof (induct xs)
show "todos (λx. P x ∧ Q x) [] = (todos P [] ∧ todos Q [])" by simp
next
fix y xs
assume HI: "todos (λx. P x ∧ Q x) xs = (todos P xs ∧ todos Q xs)"
have "todos (λx. P x ∧ Q x) (y#xs) = ((P y ∧ Q y) ∧ (todos (λx. P x ∧ Q x) xs))" by simp
also have "… = ((P y ∧ Q y) ∧ (todos P xs ∧ todos Q xs))" using HI by simp
also have "… = ((P y ∧ todos P xs) ∧ (Q y ∧ todos Q xs))" by blast
also have "… = ((todos P (y#xs)) ∧ (todos Q (y#xs)))" by simp
finally show "todos (λx. P x ∧ Q x) (y#xs) = (todos P (y#xs) ∧ todos Q (y#xs))" by simp
qed

(* migtermor *)
lemma "todos (λx. P x ∧ Q x) xs = (todos P xs ∧ todos Q xs)"
proof (induct xs)
show "todos (λx. P x ∧ Q x) [] = (todos P [] ∧ todos Q [])" by simp
next
fix x xs
assume HI: "todos (λx. P x ∧ Q x) xs = (todos P xs ∧ todos Q xs)"
have "(todos (λx. P x ∧ Q x) (x#xs)) = (( P x ∧ Q x) ∧ (todos (λx. P x ∧ Q x) xs))"
by (simp only: todos.simps(2))
also have "… = ((P x ∧ Q x) ∧ (todos P xs ∧ todos Q xs))" using HI by simp
also have "… = (((P x)∧(todos P xs)) ∧ ((Q x) ∧ (todos Q xs)))" by arith
also have "((P x ∧ Q x) ∧ (todos P xs ∧ todos Q xs)) = (((P x)∧(todos P xs)) ∧ ((Q x) ∧ (todos Q xs)))"
by arith (* Este paso es exactamente el mismo que el anterior, pero sin cualquiera de los dos no funciona el "finally show" *)
have "… = (((P x)∧(todos P xs))∧((Q x)∧(todos Q xs)))" by simp
have "… = ((todos P (x#xs))∧(todos Q (x#xs)))" by simp
finally show "(todos (λx. P x ∧ Q x) (x#xs)) = ((todos P (x#xs)) ∧ (todos Q (x#xs)))" by simp
qed

(* marpoldia1 *)
lemma "todos (λx. P x ∧ Q x) xs = (todos P xs ∧ todos Q xs)"
proof (induct xs)
show "todos (λx. P x ∧ Q x) [] = (todos P [] ∧ todos Q [])" by simp
next
fix a xs
assume HI: "todos (λx. P x ∧ Q x) xs = (todos P xs ∧ todos Q xs)"
have "todos (λx. P x ∧ Q x) (a # xs) = ((P a ∧ Q a) ∧ (todos P xs ∧ todos Q xs))" using HI by simp
also have "... = ((P a ∧ todos P xs) ∧ (Q a ∧ todos Q xs))" by blast
also have "... = (todos P (a#xs) ∧ todos Q (a#xs)) " by simp
finally show "todos (λx. P x ∧ Q x) (a#xs) = (todos P (a#xs) ∧ todos Q (a#xs))" by simp
qed

(* ferrenseg *)

lemma "todos (λx. P x ∧ Q x) xs = (todos P xs ∧ todos Q xs)"
proof (induct xs)
show "todos (λx. P x ∧ Q x) [] = (todos P [] ∧ todos Q [])" by simp
next
fix n xs
assume HI: "todos (λx. P x ∧ Q x) xs = (todos P xs ∧ todos Q xs)"
have "todos (λx. P x ∧ Q x) (n # xs) = (P n ∧ Q n ∧ todos P xs ∧ todos Q xs)" using HI by simp
also have "… = P n ∧ Q n ∧ todos P xs ∧ todos Q xs" by blast
also have "⋯ = todos P (n # xs) ∧ todos Q (n # xs)" by simp
finally show "todos (λx. P x ∧ Q x) (n # xs) = todos P (n # xs) ∧ todos Q (n # xs)" by simp
qed

{* wilmorort pablucoto crigomgom anaprarod juacabsou*}

lemma "todos (λx. P x ∧ Q x) xs = (todos P xs ∧ todos Q xs)"
proof (induct xs)
show "todos (λx. P x ∧ Q x) [] = (todos P [] ∧ todos Q [])" by simp
next
fix a xs
assume HI: "todos (λx. P x ∧ Q x) xs = (todos P xs ∧ todos Q xs)"
have "todos (λx. P x ∧ Q x) (a # xs) = (P a ∧ Q a ∧ todos (λx. P x ∧ Q x) xs)" by simp
also have "... = (P a ∧ Q a ∧ todos P xs ∧ todos Q xs)" using HI by simp
also have "... = (P a ∧ todos P xs ∧ Q a ∧ todos Q xs)" by arith
also have "... = (todos P (a # xs) ∧ todos Q (a # xs))" by simp (* Este paso se puede obviar*)
finally show "todos (λx. P x ∧ Q x) (a # xs) = (todos P (a # xs) ∧ todos Q (a # xs))" by simp
qed

text {*
---------------------------------------------------------------------
Ejercicio 4.1. Demostrar o refutar automáticamente
todos P (x @ y) = (todos P x ∧ todos P y)
---------------------------------------------------------------------
*}
{*danrodcha ivamenjim marpoldia1 migtermor ferrenseg wilmorort paupeddeg crigomgom anaprarod serrodcal juacabsou rubgonmar pablucoto*}
lemma "todos P (x @ y) = (todos P x ∧ todos P y)"
by (induct x) auto

text {*
---------------------------------------------------------------------
Ejercicio 4.2. Demostrar o refutar detalladamente
todos P (x @ y) = (todos P x ∧ todos P y)
---------------------------------------------------------------------
*}

(* ivamenjim ferrenseg *)

lemma todos_append:
"todos P (x @ y) = (todos P x ∧ todos P y)"
proof (induct x)
show "todos P ([] @ y) = (todos P [] ∧ todos P y)" by simp
next
fix a x
assume HI: "todos P (x @ y) = (todos P x ∧ todos P y)"
have "todos P ((a#x) @ y) = (P a ∧ (todos P (x @ y)))" by simp
also have "... = (P a ∧ (todos P x ∧ todos P y))" using HI by simp
finally show "todos P ((a # x) @ y) = (todos P (a # x) ∧ todos P y)" by simp
qed

(* migtermor serrodcal *)

lemma todos_append1:
"todos P (x @ y) = (todos P x ∧ todos P y)" (is "?P x")
proof (induct x)
show "?P []" by simp
next
fix a x
assume HI: "?P x"
have "todos P ((a#x) @ y) = (P a ∧ (todos P (x @ y)))" by simp
also have "… = (P a ∧ (todos P x ∧ todos P y))" using HI by simp
finally show "?P (a#x)" by simp
qed

(* marpoldia1 *)
lemma todos_append:
"todos P (x @ y) = (todos P x ∧ todos P y)"
proof (induct x)
show "todos P ([] @ y) = (todos P [] ∧ todos P y)" by simp
next
fix a x
assume HI: "todos P (x @ y) = (todos P x ∧ todos P y)"
have "todos P ((a # x) @ y) = todos P (a#(x@y))" by simp
also have "... = (P a ∧ todos P (x @ y))" by simp
also have "... = (P a ∧ (todos P x ∧ todos P y))" using HI by simp
finally show "todos P ((a # x) @ y) = (todos P (a # x) ∧ todos P y)" by simp
qed

(* paupeddeg *)
lemma todos_append3:
"todos P (x @ y) = (todos P x ∧ todos P y)"
proof (induct x)
show "todos P ([] @ y) = (todos P [] ∧ todos P y)" by simp
next
fix a x
assume HI: "todos P (x @ y) = (todos P x ∧ todos P y)"
have "todos P ((a # x) @ y) = todos P (a # (x @ y))" by simp
also have "... = (todos P [a] ∧ todos P (x @ y)) " by simp
also have "... = (todos P [a] ∧ (todos P x ∧ todos P y))" using HI by simp
finally show "todos P ((a # x) @ y) = (todos P (a # x) ∧ todos P y)" by simp
qed

(* crigomgom juacabsou *)

lemma todos_append4:
"todos P (x @ y) = (todos P x ∧ todos P y)"
proof (induct x)
show "todos P ([] @ y) = (todos P [] ∧ todos P y)" by simp
next
fix x xs
assume HI:"todos P (xs @ y) = (todos P xs ∧ todos P y)"
have "todos P ((x # xs) @ y) = (P x ∧ todos P (xs @ y ))" by simp
also have "... = (P x ∧ (todos P xs ∧ todos P y)) " using HI by simp
also have "... = ((P x ∧ todos P xs) ∧ todos P y)" by simp
finally show "todos P ((x # xs) @ y) = (todos P (x # xs) ∧ todos P y)" by simp
qed

(* pablucoto *)
lemma todos_append5:
"todos P (x @ y) = (todos P x ∧ todos P y)"
proof (induct x )
show "todos P ([] @ y) = (todos P [] ∧ todos P y)" by simp
next
fix a x
assume HI:"todos P (x @ y) = (todos P x ∧ todos P y) "
have "todos P ((a # x) @ y) = ( P a ∧ todos P (x @ y))" by simp
also have " ... =( P a ∧ (todos P x ∧ todos P y))" using HI by simp
also have "... = (todos P (a#x) ∧ todos P y)" by simp
finally show "todos P ((a # x) @ y) = (todos P (a # x) ∧ todos P y)" by auto
qed

text {*
---------------------------------------------------------------------
Ejercicio 5.1. Demostrar o refutar automáticamente
todos P (rev xs) = todos P xs
---------------------------------------------------------------------
*}

lemma "todos P (rev xs) = todos P xs"
oops

(* migtermor ivamenjim marpoldia1 serrodcal *)
lemma "todos P (rev xs) = todos P xs"
apply (induct xs)
apply simp
apply (simp add: todos_append)
apply auto
done

(* ferrenseg crigomgom *)

lemma "todos P (rev xs) = todos P xs"
by (induct xs) (auto simp add: todos_append)

(* wilmorort *)
lemma "todos P (rev xs) = todos P xs"
by (induct xs,simp,simp add: todos_append,auto)

(* juacabsou *)
lemma "todos P (rev xs) = todos P xs"
apply (induct xs,simp,simp add: todos_append,auto) done

(* pablucoto * )
lemma "todos P (rev xs) = todos P xs"
by (induct xs) (auto, simp_all add: todos_append)
lemma "todos P (rev xs) = todos P xs"
by (induct xs) ( simp_all add: todos_append, auto)

text {*
---------------------------------------------------------------------
Ejercicio 5.2. Demostrar o refutar detalladamente
todos P (rev xs) = todos P xs
---------------------------------------------------------------------
*}

lemma "todos P (rev xs) = todos P xs"
oops

(* migtermor *)

lemma auxiliar:
"rev (a#xs) = rev xs @ [a]"
by auto

lemma "todos P (rev xs) = todos P xs" (is "?Q xs")
proof (induct xs)
show "?Q []" by (simp only: rev.simps(1))
next
fix a xs
assume HI: "?Q xs"
have "todos P (rev (a#xs)) = (todos P (rev xs @ [a]))" by (simp add: auxiliar)
have "… = ((todos P (rev xs)) ∧ (todos P [a]))" by (simp add: todos_append)
have "… = (todos P (rev xs) ∧ P a)" by simp
also have Aux: "… = (todos P xs ∧ P a)" using HI by simp
have Aux1: "… = (P a ∧ todos P xs)" by arith
have "(todos P (rev xs) ∧ P a) = (P a ∧ todos P xs)" using Aux Aux1 by simp
finally show "todos P (rev (a#xs)) = todos P (a#xs)" by (simp add: todos_append)
qed

(* marpoldia1 ferrenseg crigomgom serrodcal juacabsou *)
lemma "todos P (rev xs) = todos P xs"
proof (induct xs)
show "todos P (rev []) = todos P []" by simp
next
fix a xs
assume HI: "todos P (rev xs) = todos P xs"
have "todos P (rev (a # xs)) = (todos P ((rev xs)@[a]))" by simp
also have "... = (todos P (rev xs) ∧ todos P [a])" by (simp add:todos_append)
also have "... = (todos P xs ∧ P a)" using HI by simp
also have "... = (P a ∧ todos P xs)" by arith
also have "... = (todos P (a#xs))" by simp
finally show "todos P (rev (a # xs)) = (todos P (a#xs))" by simp
qed

(* ivamenjim *)

lemma "todos P (rev xs) = todos P xs" (is "?P xs")
proof (induct xs)
show "?P []" by simp
next
fix a xs
assume HI: "?P xs"
have "todos P (rev (a # xs)) = (todos P ((rev xs)@[a]))" by simp
also have "... = ((todos P (rev xs)) ∧ todos P [a])" by (simp add: todos_append)
also have "... = (todos P xs ∧ todos P [a])" using HI by simp
also have "... = (todos P [a] ∧ todos P xs)" by arith
finally show "todos P (rev (a # xs)) = todos P (a # xs)" by simp
qed

(* paupeddeg*)
lemma "todos P (rev xs) = todos P xs"
proof (induct xs)
show " todos P (rev []) = todos P []" by simp
next
fix a xs
assume HI: "todos P (rev xs) = todos P xs "
have "todos P (rev (a # xs)) = todos P (rev(xs) @ [a])" by simp
also have "... = (todos P (rev xs) ∧ todos P [a]) " by (simp add: todos_append)
also have "... = (todos P xs ∧ todos P [a])" using HI by simp
also have "... = (todos P [a] ∧ todos P xs)" by arith
also have "... = todos P ([a] @ xs)" by (simp add: todos_append)
finally show "todos P (rev (a # xs)) = todos P (a # xs)" by simp
qed

(* wilmorort *)
lemma "todos P (rev xs) = todos P xs" (is "?P xs")
proof (induct xs)
show "?P []" by simp
next
fix a xs
assume HI: "?P xs"
have "todos P (rev (a#xs)) = todos P ((rev xs) @ [a])" by simp
also have "... = (todos P(rev xs) ∧ todos P [a])" by (simp add: todos_append)
also have "... = (todos P xs ∧ todos P [a])" using HI by simp
also have "... = (todos P [a] ∧ todos P xs)" by (simp add: HOL.conj_commute)
also have "... = todos P([a]@(xs))" by (simp)
finally show "todos P (rev (a#xs))= todos P (a # xs)" by simp
qed

text {*
---------------------------------------------------------------------
Ejercicio 6. Demostrar o refutar:
algunos (λx. P x ∧ Q x) xs = (algunos P xs ∧ algunos Q xs)
---------------------------------------------------------------------
*}

lemma "algunos (λx. P x ∧ Q x) xs = (algunos P xs ∧ algunos Q xs)"
oops

(* migtermor ivamenjim ferrenseg paupeddeg crigomgom serrodcal wilmorort juacabsou *)
lemma "algunos (λx. P x ∧ Q x) xs = (algunos P xs ∧ algunos Q xs)"
quickcheck
oops
(* Quickcheck encuentra el siguiente contraejemplo: P={a1}, Q={a2}, xs={a1,a2}. En este ejemplo:
· "algunos (λx. P x ∧ Q x) xs = False"
· "(algunos P xs ∧ algunos Q xs) = True" *)

text {*
---------------------------------------------------------------------
Ejercicio 7.1. Demostrar o refutar automáticamente
algunos P (map f xs) = algunos (P ∘ f) xs
---------------------------------------------------------------------
*}

(* ivamenjim migtermor marpoldia1 crigomgom *)

lemma "algunos P (map f xs) = algunos (P o f) xs"
by (induct xs) auto

(* ferrenseg *)

lemma "algunos P (map f xs) = algunos (P o f) xs"
by (induct xs) simp_all

text {*
---------------------------------------------------------------------
Ejercicio 7.2. Demostrar o refutar datalladamente
algunos P (map f xs) = algunos (P ∘ f) xs
---------------------------------------------------------------------
*}

lemma "algunos P (map f xs) = algunos (P o f) xs"
oops

(* migtermor *)

lemma AUX: "algunos (λa. P (f a)) xs = algunos (P o f) xs"
by (induct xs) auto

lemma "algunos P (map f xs) = algunos (P o f) xs" (is "?Q xs")
proof (induct xs)
show "?Q []" by simp
next
fix x xs
assume HI: "?Q xs"
have "algunos P (map f (x#xs)) = (algunos P ((f x)#(map f xs)))" by simp
also have "… = ((P (f x)) ∨ (algunos P (map f xs)))" by (simp only: algunos.simps(2))
also have "… = ((P (f x)) ∨ (algunos (P o f) xs))"
proof (cases)
assume C1: "(P (f x))"
have Aux: "((P (f x)) ∨ (algunos P (map f xs))) = True" using C1 by simp
have Aux1: "… = ((P (f x)) ∨ (algunos (P o f) xs))" using C1 by simp
then show "((P (f x)) ∨ (algunos P (map f xs))) = ((P (f x)) ∨ (algunos (P o f) xs))"
using Aux Aux1 by simp
next
assume C2: "¬(P (f x))"
have Aux2: "((P (f x)) ∨ (algunos P (map f xs))) = (algunos P (map f xs))" using C2 by simp
have Aux3: "… = (algunos (P o f) xs)" using HI by (simp add: AUX)
also have "… = ((P (f x)) ∨ (algunos (P o f) xs))" using C2 by simp
then show "((P (f x)) ∨ (algunos P (map f xs))) = ((P (f x)) ∨ (algunos (P o f) xs))"
using Aux2 Aux3 by simp
qed
also have "… = (((P o f) x) ∨ (algunos (P o f) xs))" by simp
also have "… = (algunos (P o f) (x#xs))" by simp
finally show "algunos P (map f (x#xs)) = (algunos (P o f) (x#xs))" by simp
qed

(* ferrenseg *)

lemma "algunos P (map f xs) = algunos (P ∘ f) xs"
proof (induct xs)
show "algunos P (map f []) = algunos (P ∘ f) []" by simp
next
fix x xs
assume HI: "algunos P (map f xs) = algunos (P ∘ f) xs"
show "algunos P (map f (x # xs)) = algunos (P ∘ f) (x # xs)"
proof -
have "algunos P (map f (x # xs)) = algunos P ((f x) # (map f xs))" by simp
also have "… = ((P (f x)) ∨ (algunos P (map f xs)))" by simp
also have "… = (((P ∘ f) x) ∨ (algunos (P ∘ f) xs))" using HI by simp
also have "… = algunos (P ∘ f) (x # xs)" by simp
finally show ?thesis
qed
qed

(* ivamenjim *)

lemma "algunos P (map f xs) = algunos (P o f) xs"
proof (induct xs)
show "algunos P (map f []) = algunos (P ∘ f) []" by simp
next
fix a xs
assume HI: "algunos P (map f xs) = algunos (P ∘ f) xs"
have "algunos P (map f (a # xs)) = algunos P ((f a)#map f xs)" by simp
also have "... = (algunos P (map f [a]) ∨ algunos P (map f xs))" by simp
also have "... = (algunos P (map f [a]) ∨ algunos (P ∘ f) xs)" using HI by simp
finally show "algunos P (map f (a # xs)) = algunos (P ∘ f) (a # xs)" by simp
qed

(* crigomgom *)
lemma "algunos P (map f xs) = algunos (P o f) xs"
proof (induct xs)
show "algunos P (map f []) = algunos (P ∘ f) []" by simp
next
fix x xs
assume HI: "algunos P (map f xs) = algunos (P ∘ f) xs"
have "algunos P (map f (x # xs)) = algunos P ((f x) # (map f xs))" by simp
also have "... = (P (f x) ∨ algunos P (map f xs))" by simp
also have "... = (P (f x) ∨ algunos (P ∘ f) xs)" using HI by simp
also have "... = ((P ∘ f) x ∨ algunos (P ∘ f) xs)" by simp
finally show "algunos P (map f (x # xs)) = algunos (P ∘ f) (x # xs)" by simp
qed

text {*
---------------------------------------------------------------------
Ejercicio 8.1. Demostrar o refutar automáticamente
algunos P (xs @ ys) = (algunos P xs ∨ algunos P ys)
---------------------------------------------------------------------
*}

(* ivamenjim marpoldia1 paupeddeg crigomgom*)

lemma "algunos P (xs @ ys) = (algunos P xs ∨ algunos P ys)"
by (induct xs) auto

(* ferrenseg *)
lemma "algunos P (xs @ ys) = (algunos P xs ∨ algunos P ys)"
by (induct xs) simp_all

text {*
---------------------------------------------------------------------
Ejercicio 8.2. Demostrar o refutar detalladamente
algunos P (xs @ ys) = (algunos P xs ∨ algunos P ys)
---------------------------------------------------------------------
*}

(* ivamenjim ferrenseg marpoldia1*)

lemma algunos_append:
"algunos P (xs @ ys) = (algunos P xs ∨ algunos P ys)"
proof (induct xs)
show "algunos P ([] @ ys) = (algunos P [] ∨ algunos P ys)" by simp
next
fix a xs
assume HI: "algunos P (xs @ ys) = (algunos P xs ∨ algunos P ys)"
have "algunos P ((a # xs) @ ys) = (P a ∨ (algunos P (xs @ ys)))" by simp
also have "... = (P a ∨ (algunos P xs ∨ algunos P ys))" using HI by simp
finally show "algunos P ((a # xs) @ ys) = (algunos P (a # xs) ∨ algunos P ys)" by simp
qed

(* migtermor crigomgom *)

lemma algunos_append:
"algunos P (xs @ ys) = (algunos P xs ∨ algunos P ys)" (is "?Q xs")
proof (induct xs)
show "?Q []" by simp
next
fix x xs
assume HI: "?Q xs"
have "algunos P ((x#xs) @ ys) = algunos P (x#(xs @ ys))" by simp
also have "… = ((P x) ∨ (algunos P (xs @ ys)))" by simp
also have "… = ((P x) ∨ (algunos P xs) ∨ (algunos P ys))" using HI by simp
also have "… = ((algunos P (x#xs)) ∨ (algunos P ys))" by simp
finally show "algunos P ((x#xs) @ ys) = (algunos P (x#xs) ∨ algunos P ys)" by simp
qed

(* paupeddeg*)
lemma algunos_append:
"algunos P (xs @ ys) = (algunos P xs ∨ algunos P ys)"
proof (induct xs)
show "algunos P ([] @ ys) = (algunos P [] ∨ algunos P ys)" by simp
next
fix a xs
assume HI: "algunos P (xs @ ys) = (algunos P xs ∨ algunos P ys)"
have "algunos P ((a # xs) @ ys) = (algunos P [a] ∨ ( algunos P (xs @ ys)))" by simp
also have "... = (algunos P [a] ∨ (algunos P xs ∨ algunos P ys))" using HI by simp
finally show "algunos P ((a # xs) @ ys) = (algunos P (a # xs) ∨ algunos P ys)" by simp
qed

text {*
---------------------------------------------------------------------
Ejercicio 9.1. Demostrar o refutar automáticamente
algunos P (rev xs) = algunos P xs
---------------------------------------------------------------------
*}

(* ivamenjim migtermor marpoldia1*)

lemma "algunos P (rev xs) = algunos P xs"
apply (induct xs)
apply simp
apply (simp add: algunos_append)
apply auto
done

(* ferrenseg crigomgom*)
lemma "algunos P (rev xs) = algunos P xs"
by (induct xs) (auto simp add: algunos_append)

text {*
---------------------------------------------------------------------
Ejercicio 9.2. Demostrar o refutar detalladamente
algunos P (rev xs) = algunos P xs
---------------------------------------------------------------------
*}

lemma "algunos P (rev xs) = algunos P xs"
oops

(* migtermor *)
lemma auxiliar1:
"rev (a#xs) = rev xs @ [a]"
by auto

lemma "algunos P (rev xs) = algunos P xs" (is "?Q xs")
proof (induct xs)
show "?Q []" by simp
next
fix x xs
assume HI: "?Q xs"
have "algunos P (rev (x#xs)) = (algunos P (rev xs @ [x]))" using auxiliar1 by simp
also have "… = ((algunos P (rev xs)) ∨ (algunos P [x]))" by (simp add: algunos_append)
also have "… = ((P x) ∨ (algunos P (rev xs)))" by simp arith
also have "… = ((P x) ∨ (algunos P xs))" using HI by simp
finally show "algunos P (rev (x#xs)) = (algunos P (x#xs))" by simp
qed

(* ferrenseg *)

lemma "algunos P (rev xs) = algunos P xs"
proof (induct xs)
show "algunos P (rev []) = algunos P []" by simp
next
fix x xs
assume HI: "algunos P (rev xs) = algunos P xs"
show "algunos P (rev (x # xs)) = algunos P (x # xs)"
proof -
have "algunos P (rev (x # xs)) = algunos P ((rev xs) @ [x])" by simp
also have "… = (algunos P (rev xs) ∨ algunos P [x])"
by (simp add: algunos_append)
also have "… = (algunos P xs ∨ algunos P [x])" using HI by simp
also have "… = (algunos P xs ∨ P x)" by simp
also have "… = (P x ∨ algunos P xs)" by auto
also have "… = algunos P (x # xs)" by simp
finally show ?thesis
qed
qed

(* ivamenjim marpoldia1*)

lemma "algunos P (rev xs) = algunos P xs" (is "?P xs")
proof (induct xs)
show "?P []" by simp
next
fix a xs
assume HI: "?P xs"
have "algunos P (rev (a # xs)) = (algunos P ((rev xs)@[a]))" by simp
also have "... = ((algunos P (rev xs)) ∨ algunos P [a])" by (simp add: algunos_append)
also have "... = (algunos P xs ∨ algunos P [a])" using HI by simp
also have "... = (algunos P [a] ∨ algunos P xs)" by arith
finally show "algunos P (rev (a # xs)) = algunos P (a # xs)" by simp
qed

(*paupeddeg*)
lemma "algunos P (rev xs) = algunos P xs"
proof (induct xs)
show "algunos P (rev []) = algunos P []" by simp
next
fix a xs
assume HI:"algunos P (rev xs) = algunos P xs"
have "algunos P (rev (a # xs)) = algunos P ((rev xs) @ [a])" by simp
also have "... = ((algunos P (rev xs)) ∨ (algunos P [a]))" by (simp add: algunos_append)
also have "... = ((algunos P xs) ∨ (algunos P [a]))" using HI by simp
also have "... = ((algunos P [a]) ∨ (algunos P xs))" by arith
finally show "algunos P (rev (a # xs)) = algunos P (a # xs)" by simp
qed

(*crigomgom *)
lemma "algunos P (rev xs) = algunos P xs"
proof (induct xs)
show "algunos P (rev []) = algunos P []" by simp
next
fix x xs
assume HI: " algunos P (rev xs) = algunos P xs"
have "algunos P (rev (x # xs)) = algunos P ((rev xs) @ [x])" by simp
also have "... = (algunos P (rev xs) ∨ algunos P [x])" by (simp add: algunos_append)
also have "... = (algunos P xs ∨ algunos P [x])" using HI by simp
also have "... = (algunos P xs ∨ P x)" by simp
also have "... = (P x ∨ algunos P xs)" by arith
also have "... = algunos P (x#xs)" by simp
finally show "algunos P (rev (x # xs)) = algunos P (x # xs)" by simp
qed

text {*
---------------------------------------------------------------------
Ejercicio 10. Encontrar un término no trivial Z tal que sea cierta la
siguiente ecuación:
algunos (λx. P x ∨ Q x) xs = Z
y demostrar la equivalencia de forma automática y detallada.
---------------------------------------------------------------------
*}

(* migtermor *)

lemma "algunos (λx. P x ∨ Q x) xs = ((algunos (λx. P x) xs) ∨ (algunos (λx. Q x) xs))"
by (induct xs) auto

lemma "algunos (λx. P x ∨ Q x) xs = ((algunos (λx. P x) xs) ∨ (algunos (λx. Q x) xs))" (is "?R xs")
proof (induct xs)
show "?R []" by simp
next
fix x xs
assume HI: "?R xs"
have "algunos (λx. P x ∨ Q x) (x#xs) = (((λx. P x ∨ Q x) x) ∨ (algunos (λx. P x ∨ Q x) xs))"
by simp
also have "… = ((P x ∨ Q x) ∨ (algunos (λx. P x ∨ Q x) xs))" by simp
also have H1: "… = ((((λx. P x) x) ∨ (algunos (λx. P x) xs)) ∨ (((λx. Q x) x) ∨ (algunos (λx. Q x) xs)))"
using HI by simp arith
have H2: "… = ((algunos (λx. P x) (x#xs)) ∨ (algunos (λx. Q x) (x#xs)))"
by simp
have C: "(algunos (λx. P x ∨ Q x) (x#xs)) =
((algunos (λx. P x) (x#xs)) ∨ (algunos (λx. Q x) (x#xs)))"
using H1 H2 by simp
finally show "(algunos (λx. P x ∨ Q x) (x#xs)) =
((algunos (λx. P x) (x#xs)) ∨ (algunos (λx. Q x) (x#xs)))"
using C by simp
qed

(* ferrenseg *)

lemma "algunos (λx. P x ∨ Q x) xs = (algunos P xs ∨ algunos Q xs)"
proof (induct xs)
show "algunos (λx. P x ∨ Q x) [] = (algunos P [] ∨ algunos Q [])" by simp
next
fix x xs
assume HI: "algunos (λx. (P x ∨ Q x)) xs = (algunos P xs ∨ algunos Q xs)"
show "algunos (λx. P x ∨ Q x) (x # xs) = (algunos P (x # xs) ∨ algunos Q (x # xs))"
proof -
have "algunos (λx. P x ∨ Q x) (x # xs) =
((P x) ∨ (Q x) ∨ algunos (λx. P x ∨ Q x) xs)" by simp
also have "… = ((P x) ∨ (Q x) ∨ algunos P xs ∨ algunos Q xs)"
using HI by simp
also have "… = (((P x) ∨ algunos P xs) ∨ ((Q x) ∨ algunos Q xs))" by auto
also have "… = (algunos P (x # xs) ∨ algunos Q (x # xs))" by simp
finally show ?thesis
qed
qed

(* ivamenjim *)

lemma "algunos (λx. P x ∨ Q x) xs = (algunos P xs ∨ algunos Q xs)"
by (induct xs) auto

lemma "algunos (λx. P x ∨ Q x) xs = (algunos P xs ∨ algunos Q xs)"
proof (induct xs)
show "algunos (λx. P x ∨ Q x) [] = (algunos P [] ∨ algunos Q [])" by simp
next
fix a xs
assume HI: "algunos (λx. P x ∨ Q x) xs = (algunos P xs ∨ algunos Q xs)"
have "algunos (λx. P x ∨ Q x) (a # xs) = (P a ∨ Q a ∨ algunos (λx. P x ∨ Q x) xs)" by simp
also have "... = (P a ∨ Q a ∨ algunos P xs ∨ algunos Q xs)" using HI by simp
finally show "algunos (λx. P x ∨ Q x) (a # xs) = (algunos P (a # xs) ∨ algunos Q (a # xs))" by auto
qed

text {*
---------------------------------------------------------------------
Ejercicio 11.1. Demostrar o refutar automáticamente
algunos P xs = (¬ todos (λx. (¬ P x)) xs)
---------------------------------------------------------------------
*}

(* ivamenjim wilmorort migtermor marpoldia1 crigomgom*)

lemma "algunos P xs = (¬ todos (λx. (¬ P x)) xs)"
by (induct xs) auto

(* ferrenseg *)

lemma "algunos P xs = (¬ todos (λx. (¬ P x)) xs)"
by (induct xs) simp_all

text {*
---------------------------------------------------------------------
Ejercicio 11.2. Demostrar o refutar datalladamente
algunos P xs = (¬ todos (λx. (¬ P x)) xs)
---------------------------------------------------------------------
*}

(* ivamenjim migtermor marpoldia1 crigomgom*)

lemma "algunos P xs = (¬ todos (λx. (¬ P x)) xs)"
proof (induct xs)
show "algunos P [] = (¬ todos (λx. ¬ P x) [])" by simp
next
fix a xs
assume HI: "algunos P xs = (¬ todos (λx. ¬ P x) xs)"
have "algunos P (a # xs) = (P a ∨ (algunos P xs))" by simp
also have "... = (P a ∨ (¬ todos (λx. ¬ P x) xs))" using HI by simp
finally show "algunos P (a # xs) = (¬ todos (λx. ¬ P x) (a # xs))" by simp
qed

(* ferrenseg *)

lemma "algunos P xs = (¬ todos (λx. (¬ P x)) xs)"
proof (induct xs)
show "algunos P [] = (¬ todos (λx. (¬ P x)) [])" by simp
next
fix x xs
assume HI: "algunos P xs = (¬ todos (λx. (¬ P x)) xs)"
show "algunos P (x # xs) = (¬ todos (λx. (¬ P x)) (x # xs))"
proof -
have "algunos P (x # xs) = ((P x) ∨ algunos P xs)" by simp
also have "… = ((P x) ∨ ¬ todos (λx. (¬ P x)) xs)" using HI by simp
also have "… = (¬ (¬ (P x) ∧ todos (λx. (¬ P x)) xs))" by simp
also have "… = (¬ todos (λx. (¬ P x)) (x # xs))" by simp
finally show ?thesis
qed
qed

text {*
---------------------------------------------------------------------
Ejercicio 12. Definir la funcion primitiva recursiva
estaEn :: 'a ⇒ 'a list ⇒ bool
tal que (estaEn x xs) se verifica si el elemento x está en la lista
xs. Por ejemplo,
estaEn (2::nat) [3,2,4] = True
estaEn (1::nat) [3,2,4] = False
---------------------------------------------------------------------
*}

(* ivamenjim ferrenseg migtermor serrodcal crigomgom*)

fun estaEn :: "'a ⇒ 'a list ⇒ bool" where
"estaEn x [] = False"
| "estaEn x (a#xs) = ((a = x) ∨ (estaEn x xs))"

value "estaEn (2::nat) [3,2,4] = True"
value "estaEn (1::nat) [3,2,4] = False"

text {*
---------------------------------------------------------------------
Ejercicio 13. Expresar la relación existente entre estaEn y algunos.
Demostrar dicha relación de forma automática y detallada.
---------------------------------------------------------------------
*}

(* migtermor crigomgom*)

lemma "estaEn x xs = (algunos (λa. a=x) xs)"
by (induct xs) auto

lemma auxiliar13:
"(x=a) = ((λa. a=x) a)"
by auto

lemma "estaEn x xs = (algunos (λa. a=x) xs)" (is "?P xs")
proof (induct xs)
show "?P []" by simp
next
fix a xs
assume HI: "?P xs"
have "estaEn x (a#xs) = ((a=x) ∨ (estaEn x xs))" by simp
also have H: "… = ((a=x) ∨ (algunos (λa. a=x) xs))" using HI by simp
also have "… = (((λa. a=x) a) ∨ (algunos (λa. a=x) xs))" using auxiliar13 by simp
also have "… = (algunos (λa. a=x) (a#xs))" by simp
have C: "estaEn x (a#xs) = (algunos (λa. a=x) (a#xs))" using H by simp
finally show "estaEn x (a#xs) = (algunos (λa. a=x) (a#xs))" using C by simp
qed

(* ferrenseg *)

lemma estaEn_algunos:
"estaEn y xs = algunos (λx. x=y) xs"
by (induct xs) auto

lemma estaEn_algunos_2:
"estaEn y xs = algunos (λx. x=y) xs"
proof (induct xs)
show "estaEn y [] = algunos (λx. x=y) []" by simp
next
fix x xs
assume HI: "estaEn y xs = algunos (λx. x=y) xs"
show "estaEn y (x # xs) = algunos (λx. x=y) (x # xs)"
proof -
have "estaEn y (x # xs) = (y=x ∨ estaEn y xs)" by simp
also have "… = (y=x ∨ algunos (λx. x=y) xs)" using HI by simp
also have "… = (x=y ∨ algunos (λx. x=y) xs)" by auto
also have "… = algunos (λx. x=y) (x # xs)" by simp
finally show ?thesis
qed
qed

(*crigomgom*)
lemma "estaEn x xs = (algunos (λa. a=x) xs)"
proof (induct xs)
show " estaEn x [] = algunos (λa. a = x) []" by simp
next
fix a xs
assume HI: "estaEn x xs = algunos (λa. a = x) xs"
have "estaEn x (a # xs) = (a = x ∨ estaEn x xs)" by simp
also have "... = (a = x ∨ algunos (λa. a = x) xs)" using HI by simp
finally show " estaEn x (a # xs) = algunos (λa. a = x) (a # xs)" by simp
qed

(* ivamenjim *)

lemma "estaEn x xs = algunos (λy. x=y) xs"
by (induct xs) auto

lemma "estaEn x xs = algunos (λy. x=y) xs"
proof (induct xs)
show "estaEn x [] = algunos (op = x) []" by simp
next
fix a xs
assume HI: "estaEn x xs = algunos (op = x) xs"
have "estaEn x (a # xs) = ((a = x) ∨ (estaEn x xs))" by simp
also have "... = (a = x ∨ algunos (op = x) xs)" using HI by simp
finally show " estaEn x (a # xs) = algunos (op = x) (a # xs)" by auto
qed

end
</source>

Relación 4

2016-11-22T04:46:45Z

Pablucoto:

<source lang="isar">
chapter {* R4: Cuantificadores sobre listas *}

theory R4_Cuantificadores_sobre_listas
imports Main
begin

text {*
---------------------------------------------------------------------
Ejercicio 1. Definir la función
todos :: ('a ⇒ bool) ⇒ 'a list ⇒ bool
tal que (todos p xs) se verifica si todos los elementos de la lista
xs cumplen la propiedad p. Por ejemplo, se verifica
todos (λx. 1<length x) [[2,1,4],[1,3]]
¬todos (λx. 1<length x) [[2,1,4],[3]]

Nota: La función todos es equivalente a la predefinida list_all.
---------------------------------------------------------------------
*}

{*danrodcha*}
fun todos :: "('a ⇒ bool) ⇒ 'a list ⇒ bool" where
"todos p xs = "todos p [] = True"
| "todos p (x#xs) = (p x ∧ todos p xs)""

(* ivamenjim migtermor dancorgar wilmorort marpoldia1 ferrenseg paupeddeg pablucoto crigomgom anaprarod serrodcal juacabsou rubgonmar *)
fun todos :: "('a ⇒ bool) ⇒ 'a list ⇒ bool" where
"todos p [] = True"
| "todos p (x#xs) = (p x ∧ todos p xs)"

text {*
---------------------------------------------------------------------
Ejercicio 2. Definir la función
algunos :: ('a ⇒ bool) ⇒ 'a list ⇒ bool
tal que (algunos p xs) se verifica si algunos elementos de la lista
xs cumplen la propiedad p. Por ejemplo, se verifica
algunos (λx. 1<length x) [[2,1,4],[3]]
¬algunos (λx. 1<length x) [[],[3]]"

Nota: La función algunos es equivalente a la predefinida list_ex.
---------------------------------------------------------------------
*}

{*danrodcha ivamenjim migtermor dancorgar marpoldia1 ferrenseg wilmorort paupeddeg pablucoto crigomgom anaprarod serrodcal juacabsou rubgonmar *}
fun algunos :: "('a ⇒ bool) ⇒ 'a list ⇒ bool" where
"algunos p [] = False"
| "algunos p (x#xs) = (p x ∨ algunos p xs)"

text {*
---------------------------------------------------------------------
Ejercicio 3.1. Demostrar o refutar automáticamente
todos (λx. P x ∧ Q x) xs = (todos P xs ∧ todos Q xs)
---------------------------------------------------------------------
*}

{*danrodcha ivamenjim migtermor dancorgar marpoldia1 ferrenseg wilmorort paupeddeg pablucoto anaprarod serrodcal juacabsou rubgonmar *}
lemma "todos (λx. P x ∧ Q x) xs = (todos P xs ∧ todos Q xs)"
by (induct xs) auto

text {*
---------------------------------------------------------------------
Ejercicio 3.2. Demostrar o refutar detalladamente
todos (λx. P x ∧ Q x) xs = (todos P xs ∧ todos Q xs)
---------------------------------------------------------------------
*}

{*danrodcha*}
lemma "todos (λx. P x ∧ Q x) xs = (todos P xs ∧ todos Q xs)"
proof (induct xs)
show "?R []" by simp
next
fix a xs
assume HI: "?R xs"
have "todos (λx. P x ∧ Q x) (a#xs) = (P a ∧ Q a ∧ todos (λx. P x ∧ Q x) xs)" by simp
also have "… = (P a ∧ Q a ∧ (todos P xs ∧ todos Q xs))" using HI by simp
also have "… = ((P a ∧ todos P xs) ∧ (Q a ∧ todos Q xs))" by blast
also have "… = (todos P (a#xs) ∧ todos Q (a#xs))" by simp
finally show "?R (a#xs)" by simp
qed

(* ivamenjim wilmorort serrodcal *)

lemma "todos (λx. P x ∧ Q x) xs = (todos P xs ∧ todos Q xs)"
proof (induct xs)
show "todos (λx. P x ∧ Q x) [] = (todos P [] ∧ todos Q [])" by simp
next
fix a xs
assume HI: "todos (λx. P x ∧ Q x) xs = (todos P xs ∧ todos Q xs)"
have "todos (λx. P x ∧ Q x) (a # xs) = (P a ∧ Q a ∧ todos (λx. P x ∧ Q x) xs)" by simp
also have "... = (P a ∧ Q a ∧ todos P xs ∧ todos Q xs)" using HI by simp
finally show "todos (λx. P x ∧ Q x) (a # xs) = (todos P (a # xs) ∧ todos Q (a # xs))" by auto
qed

(* dancorgar paupeddeg *)
lemma "todos (λx. P x ∧ Q x) xs = (todos P xs ∧ todos Q xs)"
proof (induct xs)
show "todos (λx. P x ∧ Q x) [] = (todos P [] ∧ todos Q [])" by simp
next
fix y xs
assume HI: "todos (λx. P x ∧ Q x) xs = (todos P xs ∧ todos Q xs)"
have "todos (λx. P x ∧ Q x) (y#xs) = ((P y ∧ Q y) ∧ (todos (λx. P x ∧ Q x) xs))" by simp
also have "… = ((P y ∧ Q y) ∧ (todos P xs ∧ todos Q xs))" using HI by simp
also have "… = ((P y ∧ todos P xs) ∧ (Q y ∧ todos Q xs))" by blast
also have "… = ((todos P (y#xs)) ∧ (todos Q (y#xs)))" by simp
finally show "todos (λx. P x ∧ Q x) (y#xs) = (todos P (y#xs) ∧ todos Q (y#xs))" by simp
qed

(* migtermor *)
lemma "todos (λx. P x ∧ Q x) xs = (todos P xs ∧ todos Q xs)"
proof (induct xs)
show "todos (λx. P x ∧ Q x) [] = (todos P [] ∧ todos Q [])" by simp
next
fix x xs
assume HI: "todos (λx. P x ∧ Q x) xs = (todos P xs ∧ todos Q xs)"
have "(todos (λx. P x ∧ Q x) (x#xs)) = (( P x ∧ Q x) ∧ (todos (λx. P x ∧ Q x) xs))"
by (simp only: todos.simps(2))
also have "… = ((P x ∧ Q x) ∧ (todos P xs ∧ todos Q xs))" using HI by simp
also have "… = (((P x)∧(todos P xs)) ∧ ((Q x) ∧ (todos Q xs)))" by arith
also have "((P x ∧ Q x) ∧ (todos P xs ∧ todos Q xs)) = (((P x)∧(todos P xs)) ∧ ((Q x) ∧ (todos Q xs)))"
by arith (* Este paso es exactamente el mismo que el anterior, pero sin cualquiera de los dos no funciona el "finally show" *)
have "… = (((P x)∧(todos P xs))∧((Q x)∧(todos Q xs)))" by simp
have "… = ((todos P (x#xs))∧(todos Q (x#xs)))" by simp
finally show "(todos (λx. P x ∧ Q x) (x#xs)) = ((todos P (x#xs)) ∧ (todos Q (x#xs)))" by simp
qed

(* marpoldia1 *)
lemma "todos (λx. P x ∧ Q x) xs = (todos P xs ∧ todos Q xs)"
proof (induct xs)
show "todos (λx. P x ∧ Q x) [] = (todos P [] ∧ todos Q [])" by simp
next
fix a xs
assume HI: "todos (λx. P x ∧ Q x) xs = (todos P xs ∧ todos Q xs)"
have "todos (λx. P x ∧ Q x) (a # xs) = ((P a ∧ Q a) ∧ (todos P xs ∧ todos Q xs))" using HI by simp
also have "... = ((P a ∧ todos P xs) ∧ (Q a ∧ todos Q xs))" by blast
also have "... = (todos P (a#xs) ∧ todos Q (a#xs)) " by simp
finally show "todos (λx. P x ∧ Q x) (a#xs) = (todos P (a#xs) ∧ todos Q (a#xs))" by simp
qed

(* ferrenseg *)

lemma "todos (λx. P x ∧ Q x) xs = (todos P xs ∧ todos Q xs)"
proof (induct xs)
show "todos (λx. P x ∧ Q x) [] = (todos P [] ∧ todos Q [])" by simp
next
fix n xs
assume HI: "todos (λx. P x ∧ Q x) xs = (todos P xs ∧ todos Q xs)"
have "todos (λx. P x ∧ Q x) (n # xs) = (P n ∧ Q n ∧ todos P xs ∧ todos Q xs)" using HI by simp
also have "… = P n ∧ Q n ∧ todos P xs ∧ todos Q xs" by blast
also have "⋯ = todos P (n # xs) ∧ todos Q (n # xs)" by simp
finally show "todos (λx. P x ∧ Q x) (n # xs) = todos P (n # xs) ∧ todos Q (n # xs)" by simp
qed

{* wilmorort pablucoto crigomgom anaprarod juacabsou*}

lemma "todos (λx. P x ∧ Q x) xs = (todos P xs ∧ todos Q xs)"
proof (induct xs)
show "todos (λx. P x ∧ Q x) [] = (todos P [] ∧ todos Q [])" by simp
next
fix a xs
assume HI: "todos (λx. P x ∧ Q x) xs = (todos P xs ∧ todos Q xs)"
have "todos (λx. P x ∧ Q x) (a # xs) = (P a ∧ Q a ∧ todos (λx. P x ∧ Q x) xs)" by simp
also have "... = (P a ∧ Q a ∧ todos P xs ∧ todos Q xs)" using HI by simp
also have "... = (P a ∧ todos P xs ∧ Q a ∧ todos Q xs)" by arith
also have "... = (todos P (a # xs) ∧ todos Q (a # xs))" by simp (* Este paso se puede obviar*)
finally show "todos (λx. P x ∧ Q x) (a # xs) = (todos P (a # xs) ∧ todos Q (a # xs))" by simp
qed

text {*
---------------------------------------------------------------------
Ejercicio 4.1. Demostrar o refutar automáticamente
todos P (x @ y) = (todos P x ∧ todos P y)
---------------------------------------------------------------------
*}
{*danrodcha ivamenjim marpoldia1 migtermor ferrenseg wilmorort paupeddeg crigomgom anaprarod serrodcal juacabsou rubgonmar pablucoto*}
lemma "todos P (x @ y) = (todos P x ∧ todos P y)"
by (induct x) auto

text {*
---------------------------------------------------------------------
Ejercicio 4.2. Demostrar o refutar detalladamente
todos P (x @ y) = (todos P x ∧ todos P y)
---------------------------------------------------------------------
*}

(* ivamenjim ferrenseg *)

lemma todos_append:
"todos P (x @ y) = (todos P x ∧ todos P y)"
proof (induct x)
show "todos P ([] @ y) = (todos P [] ∧ todos P y)" by simp
next
fix a x
assume HI: "todos P (x @ y) = (todos P x ∧ todos P y)"
have "todos P ((a#x) @ y) = (P a ∧ (todos P (x @ y)))" by simp
also have "... = (P a ∧ (todos P x ∧ todos P y))" using HI by simp
finally show "todos P ((a # x) @ y) = (todos P (a # x) ∧ todos P y)" by simp
qed

(* migtermor serrodcal *)

lemma todos_append1:
"todos P (x @ y) = (todos P x ∧ todos P y)" (is "?P x")
proof (induct x)
show "?P []" by simp
next
fix a x
assume HI: "?P x"
have "todos P ((a#x) @ y) = (P a ∧ (todos P (x @ y)))" by simp
also have "… = (P a ∧ (todos P x ∧ todos P y))" using HI by simp
finally show "?P (a#x)" by simp
qed

(* marpoldia1 *)
lemma todos_append:
"todos P (x @ y) = (todos P x ∧ todos P y)"
proof (induct x)
show "todos P ([] @ y) = (todos P [] ∧ todos P y)" by simp
next
fix a x
assume HI: "todos P (x @ y) = (todos P x ∧ todos P y)"
have "todos P ((a # x) @ y) = todos P (a#(x@y))" by simp
also have "... = (P a ∧ todos P (x @ y))" by simp
also have "... = (P a ∧ (todos P x ∧ todos P y))" using HI by simp
finally show "todos P ((a # x) @ y) = (todos P (a # x) ∧ todos P y)" by simp
qed

(* paupeddeg *)
lemma todos_append3:
"todos P (x @ y) = (todos P x ∧ todos P y)"
proof (induct x)
show "todos P ([] @ y) = (todos P [] ∧ todos P y)" by simp
next
fix a x
assume HI: "todos P (x @ y) = (todos P x ∧ todos P y)"
have "todos P ((a # x) @ y) = todos P (a # (x @ y))" by simp
also have "... = (todos P [a] ∧ todos P (x @ y)) " by simp
also have "... = (todos P [a] ∧ (todos P x ∧ todos P y))" using HI by simp
finally show "todos P ((a # x) @ y) = (todos P (a # x) ∧ todos P y)" by simp
qed

(* crigomgom juacabsou *)

lemma todos_append4:
"todos P (x @ y) = (todos P x ∧ todos P y)"
proof (induct x)
show "todos P ([] @ y) = (todos P [] ∧ todos P y)" by simp
next
fix x xs
assume HI:"todos P (xs @ y) = (todos P xs ∧ todos P y)"
have "todos P ((x # xs) @ y) = (P x ∧ todos P (xs @ y ))" by simp
also have "... = (P x ∧ (todos P xs ∧ todos P y)) " using HI by simp
also have "... = ((P x ∧ todos P xs) ∧ todos P y)" by simp
finally show "todos P ((x # xs) @ y) = (todos P (x # xs) ∧ todos P y)" by simp
qed

(* pablucoto *)
lemma todos_append5:
"todos P (x @ y) = (todos P x ∧ todos P y)"
proof (induct x )
show "todos P ([] @ y) = (todos P [] ∧ todos P y)" by simp
next
fix a x
assume HI:"todos P (x @ y) = (todos P x ∧ todos P y) "
have "todos P ((a # x) @ y) = ( P a ∧ todos P (x @ y))" by simp
also have " ... =( P a ∧ (todos P x ∧ todos P y))" using HI by simp
also have "... = (todos P (a#x) ∧ todos P y)" by simp
finally show "todos P ((a # x) @ y) = (todos P (a # x) ∧ todos P y)" by auto
qed

text {*
---------------------------------------------------------------------
Ejercicio 5.1. Demostrar o refutar automáticamente
todos P (rev xs) = todos P xs
---------------------------------------------------------------------
*}

lemma "todos P (rev xs) = todos P xs"
oops

(* migtermor ivamenjim marpoldia1 serrodcal *)
lemma "todos P (rev xs) = todos P xs"
apply (induct xs)
apply simp
apply (simp add: todos_append)
apply auto
done

(* ferrenseg crigomgom *)

lemma "todos P (rev xs) = todos P xs"
by (induct xs) (auto simp add: todos_append)

(* wilmorort *)
lemma "todos P (rev xs) = todos P xs"
by (induct xs,simp,simp add: todos_append,auto)

(* juacabsou *)
lemma "todos P (rev xs) = todos P xs"
apply (induct xs,simp,simp add: todos_append,auto) done

(* pablucoto * )
lemma "todos P (rev xs) = todos P xs"
by (induct xs) (auto, simp_all add: todos_append)

lemma "todos P (rev xs) = todos P xs"
by (induct xs) ( simp_all add: todos_append, auto)

text {*
---------------------------------------------------------------------
Ejercicio 5.2. Demostrar o refutar detalladamente
todos P (rev xs) = todos P xs
---------------------------------------------------------------------
*}

lemma "todos P (rev xs) = todos P xs"
oops

(* migtermor *)

lemma auxiliar:
"rev (a#xs) = rev xs @ [a]"
by auto

lemma "todos P (rev xs) = todos P xs" (is "?Q xs")
proof (induct xs)
show "?Q []" by (simp only: rev.simps(1))
next
fix a xs
assume HI: "?Q xs"
have "todos P (rev (a#xs)) = (todos P (rev xs @ [a]))" by (simp add: auxiliar)
have "… = ((todos P (rev xs)) ∧ (todos P [a]))" by (simp add: todos_append)
have "… = (todos P (rev xs) ∧ P a)" by simp
also have Aux: "… = (todos P xs ∧ P a)" using HI by simp
have Aux1: "… = (P a ∧ todos P xs)" by arith
have "(todos P (rev xs) ∧ P a) = (P a ∧ todos P xs)" using Aux Aux1 by simp
finally show "todos P (rev (a#xs)) = todos P (a#xs)" by (simp add: todos_append)
qed

(* marpoldia1 ferrenseg crigomgom serrodcal juacabsou *)
lemma "todos P (rev xs) = todos P xs"
proof (induct xs)
show "todos P (rev []) = todos P []" by simp
next
fix a xs
assume HI: "todos P (rev xs) = todos P xs"
have "todos P (rev (a # xs)) = (todos P ((rev xs)@[a]))" by simp
also have "... = (todos P (rev xs) ∧ todos P [a])" by (simp add:todos_append)
also have "... = (todos P xs ∧ P a)" using HI by simp
also have "... = (P a ∧ todos P xs)" by arith
also have "... = (todos P (a#xs))" by simp
finally show "todos P (rev (a # xs)) = (todos P (a#xs))" by simp
qed

(* ivamenjim *)

lemma "todos P (rev xs) = todos P xs" (is "?P xs")
proof (induct xs)
show "?P []" by simp
next
fix a xs
assume HI: "?P xs"
have "todos P (rev (a # xs)) = (todos P ((rev xs)@[a]))" by simp
also have "... = ((todos P (rev xs)) ∧ todos P [a])" by (simp add: todos_append)
also have "... = (todos P xs ∧ todos P [a])" using HI by simp
also have "... = (todos P [a] ∧ todos P xs)" by arith
finally show "todos P (rev (a # xs)) = todos P (a # xs)" by simp
qed

(* paupeddeg*)
lemma "todos P (rev xs) = todos P xs"
proof (induct xs)
show " todos P (rev []) = todos P []" by simp
next
fix a xs
assume HI: "todos P (rev xs) = todos P xs "
have "todos P (rev (a # xs)) = todos P (rev(xs) @ [a])" by simp
also have "... = (todos P (rev xs) ∧ todos P [a]) " by (simp add: todos_append)
also have "... = (todos P xs ∧ todos P [a])" using HI by simp
also have "... = (todos P [a] ∧ todos P xs)" by arith
also have "... = todos P ([a] @ xs)" by (simp add: todos_append)
finally show "todos P (rev (a # xs)) = todos P (a # xs)" by simp
qed

(* wilmorort *)
lemma "todos P (rev xs) = todos P xs" (is "?P xs")
proof (induct xs)
show "?P []" by simp
next
fix a xs
assume HI: "?P xs"
have "todos P (rev (a#xs)) = todos P ((rev xs) @ [a])" by simp
also have "... = (todos P(rev xs) ∧ todos P [a])" by (simp add: todos_append)
also have "... = (todos P xs ∧ todos P [a])" using HI by simp
also have "... = (todos P [a] ∧ todos P xs)" by (simp add: HOL.conj_commute)
also have "... = todos P([a]@(xs))" by (simp)
finally show "todos P (rev (a#xs))= todos P (a # xs)" by simp
qed

text {*
---------------------------------------------------------------------
Ejercicio 6. Demostrar o refutar:
algunos (λx. P x ∧ Q x) xs = (algunos P xs ∧ algunos Q xs)
---------------------------------------------------------------------
*}

lemma "algunos (λx. P x ∧ Q x) xs = (algunos P xs ∧ algunos Q xs)"
oops

(* migtermor ivamenjim ferrenseg paupeddeg crigomgom serrodcal wilmorort juacabsou *)
lemma "algunos (λx. P x ∧ Q x) xs = (algunos P xs ∧ algunos Q xs)"
quickcheck
oops
(* Quickcheck encuentra el siguiente contraejemplo: P={a1}, Q={a2}, xs={a1,a2}. En este ejemplo:
· "algunos (λx. P x ∧ Q x) xs = False"
· "(algunos P xs ∧ algunos Q xs) = True" *)

text {*
---------------------------------------------------------------------
Ejercicio 7.1. Demostrar o refutar automáticamente
algunos P (map f xs) = algunos (P ∘ f) xs
---------------------------------------------------------------------
*}

(* ivamenjim migtermor marpoldia1 crigomgom *)

lemma "algunos P (map f xs) = algunos (P o f) xs"
by (induct xs) auto

(* ferrenseg *)

lemma "algunos P (map f xs) = algunos (P o f) xs"
by (induct xs) simp_all

text {*
---------------------------------------------------------------------
Ejercicio 7.2. Demostrar o refutar datalladamente
algunos P (map f xs) = algunos (P ∘ f) xs
---------------------------------------------------------------------
*}

lemma "algunos P (map f xs) = algunos (P o f) xs"
oops

(* migtermor *)

lemma AUX: "algunos (λa. P (f a)) xs = algunos (P o f) xs"
by (induct xs) auto

lemma "algunos P (map f xs) = algunos (P o f) xs" (is "?Q xs")
proof (induct xs)
show "?Q []" by simp
next
fix x xs
assume HI: "?Q xs"
have "algunos P (map f (x#xs)) = (algunos P ((f x)#(map f xs)))" by simp
also have "… = ((P (f x)) ∨ (algunos P (map f xs)))" by (simp only: algunos.simps(2))
also have "… = ((P (f x)) ∨ (algunos (P o f) xs))"
proof (cases)
assume C1: "(P (f x))"
have Aux: "((P (f x)) ∨ (algunos P (map f xs))) = True" using C1 by simp
have Aux1: "… = ((P (f x)) ∨ (algunos (P o f) xs))" using C1 by simp
then show "((P (f x)) ∨ (algunos P (map f xs))) = ((P (f x)) ∨ (algunos (P o f) xs))"
using Aux Aux1 by simp
next
assume C2: "¬(P (f x))"
have Aux2: "((P (f x)) ∨ (algunos P (map f xs))) = (algunos P (map f xs))" using C2 by simp
have Aux3: "… = (algunos (P o f) xs)" using HI by (simp add: AUX)
also have "… = ((P (f x)) ∨ (algunos (P o f) xs))" using C2 by simp
then show "((P (f x)) ∨ (algunos P (map f xs))) = ((P (f x)) ∨ (algunos (P o f) xs))"
using Aux2 Aux3 by simp
qed
also have "… = (((P o f) x) ∨ (algunos (P o f) xs))" by simp
also have "… = (algunos (P o f) (x#xs))" by simp
finally show "algunos P (map f (x#xs)) = (algunos (P o f) (x#xs))" by simp
qed

(* ferrenseg *)

lemma "algunos P (map f xs) = algunos (P ∘ f) xs"
proof (induct xs)
show "algunos P (map f []) = algunos (P ∘ f) []" by simp
next
fix x xs
assume HI: "algunos P (map f xs) = algunos (P ∘ f) xs"
show "algunos P (map f (x # xs)) = algunos (P ∘ f) (x # xs)"
proof -
have "algunos P (map f (x # xs)) = algunos P ((f x) # (map f xs))" by simp
also have "… = ((P (f x)) ∨ (algunos P (map f xs)))" by simp
also have "… = (((P ∘ f) x) ∨ (algunos (P ∘ f) xs))" using HI by simp
also have "… = algunos (P ∘ f) (x # xs)" by simp
finally show ?thesis
qed
qed

(* ivamenjim *)

lemma "algunos P (map f xs) = algunos (P o f) xs"
proof (induct xs)
show "algunos P (map f []) = algunos (P ∘ f) []" by simp
next
fix a xs
assume HI: "algunos P (map f xs) = algunos (P ∘ f) xs"
have "algunos P (map f (a # xs)) = algunos P ((f a)#map f xs)" by simp
also have "... = (algunos P (map f [a]) ∨ algunos P (map f xs))" by simp
also have "... = (algunos P (map f [a]) ∨ algunos (P ∘ f) xs)" using HI by simp
finally show "algunos P (map f (a # xs)) = algunos (P ∘ f) (a # xs)" by simp
qed

(* crigomgom *)
lemma "algunos P (map f xs) = algunos (P o f) xs"
proof (induct xs)
show "algunos P (map f []) = algunos (P ∘ f) []" by simp
next
fix x xs
assume HI: "algunos P (map f xs) = algunos (P ∘ f) xs"
have "algunos P (map f (x # xs)) = algunos P ((f x) # (map f xs))" by simp
also have "... = (P (f x) ∨ algunos P (map f xs))" by simp
also have "... = (P (f x) ∨ algunos (P ∘ f) xs)" using HI by simp
also have "... = ((P ∘ f) x ∨ algunos (P ∘ f) xs)" by simp
finally show "algunos P (map f (x # xs)) = algunos (P ∘ f) (x # xs)" by simp
qed

text {*
---------------------------------------------------------------------
Ejercicio 8.1. Demostrar o refutar automáticamente
algunos P (xs @ ys) = (algunos P xs ∨ algunos P ys)
---------------------------------------------------------------------
*}

(* ivamenjim marpoldia1 paupeddeg crigomgom*)

lemma "algunos P (xs @ ys) = (algunos P xs ∨ algunos P ys)"
by (induct xs) auto

(* ferrenseg *)
lemma "algunos P (xs @ ys) = (algunos P xs ∨ algunos P ys)"
by (induct xs) simp_all

text {*
---------------------------------------------------------------------
Ejercicio 8.2. Demostrar o refutar detalladamente
algunos P (xs @ ys) = (algunos P xs ∨ algunos P ys)
---------------------------------------------------------------------
*}

(* ivamenjim ferrenseg marpoldia1*)

lemma algunos_append:
"algunos P (xs @ ys) = (algunos P xs ∨ algunos P ys)"
proof (induct xs)
show "algunos P ([] @ ys) = (algunos P [] ∨ algunos P ys)" by simp
next
fix a xs
assume HI: "algunos P (xs @ ys) = (algunos P xs ∨ algunos P ys)"
have "algunos P ((a # xs) @ ys) = (P a ∨ (algunos P (xs @ ys)))" by simp
also have "... = (P a ∨ (algunos P xs ∨ algunos P ys))" using HI by simp
finally show "algunos P ((a # xs) @ ys) = (algunos P (a # xs) ∨ algunos P ys)" by simp
qed

(* migtermor crigomgom *)

lemma algunos_append:
"algunos P (xs @ ys) = (algunos P xs ∨ algunos P ys)" (is "?Q xs")
proof (induct xs)
show "?Q []" by simp
next
fix x xs
assume HI: "?Q xs"
have "algunos P ((x#xs) @ ys) = algunos P (x#(xs @ ys))" by simp
also have "… = ((P x) ∨ (algunos P (xs @ ys)))" by simp
also have "… = ((P x) ∨ (algunos P xs) ∨ (algunos P ys))" using HI by simp
also have "… = ((algunos P (x#xs)) ∨ (algunos P ys))" by simp
finally show "algunos P ((x#xs) @ ys) = (algunos P (x#xs) ∨ algunos P ys)" by simp
qed

(* paupeddeg*)
lemma algunos_append:
"algunos P (xs @ ys) = (algunos P xs ∨ algunos P ys)"
proof (induct xs)
show "algunos P ([] @ ys) = (algunos P [] ∨ algunos P ys)" by simp
next
fix a xs
assume HI: "algunos P (xs @ ys) = (algunos P xs ∨ algunos P ys)"
have "algunos P ((a # xs) @ ys) = (algunos P [a] ∨ ( algunos P (xs @ ys)))" by simp
also have "... = (algunos P [a] ∨ (algunos P xs ∨ algunos P ys))" using HI by simp
finally show "algunos P ((a # xs) @ ys) = (algunos P (a # xs) ∨ algunos P ys)" by simp
qed

text {*
---------------------------------------------------------------------
Ejercicio 9.1. Demostrar o refutar automáticamente
algunos P (rev xs) = algunos P xs
---------------------------------------------------------------------
*}

(* ivamenjim migtermor marpoldia1*)

lemma "algunos P (rev xs) = algunos P xs"
apply (induct xs)
apply simp
apply (simp add: algunos_append)
apply auto
done

(* ferrenseg crigomgom*)
lemma "algunos P (rev xs) = algunos P xs"
by (induct xs) (auto simp add: algunos_append)

text {*
---------------------------------------------------------------------
Ejercicio 9.2. Demostrar o refutar detalladamente
algunos P (rev xs) = algunos P xs
---------------------------------------------------------------------
*}

lemma "algunos P (rev xs) = algunos P xs"
oops

(* migtermor *)
lemma auxiliar1:
"rev (a#xs) = rev xs @ [a]"
by auto

lemma "algunos P (rev xs) = algunos P xs" (is "?Q xs")
proof (induct xs)
show "?Q []" by simp
next
fix x xs
assume HI: "?Q xs"
have "algunos P (rev (x#xs)) = (algunos P (rev xs @ [x]))" using auxiliar1 by simp
also have "… = ((algunos P (rev xs)) ∨ (algunos P [x]))" by (simp add: algunos_append)
also have "… = ((P x) ∨ (algunos P (rev xs)))" by simp arith
also have "… = ((P x) ∨ (algunos P xs))" using HI by simp
finally show "algunos P (rev (x#xs)) = (algunos P (x#xs))" by simp
qed

(* ferrenseg *)

lemma "algunos P (rev xs) = algunos P xs"
proof (induct xs)
show "algunos P (rev []) = algunos P []" by simp
next
fix x xs
assume HI: "algunos P (rev xs) = algunos P xs"
show "algunos P (rev (x # xs)) = algunos P (x # xs)"
proof -
have "algunos P (rev (x # xs)) = algunos P ((rev xs) @ [x])" by simp
also have "… = (algunos P (rev xs) ∨ algunos P [x])"
by (simp add: algunos_append)
also have "… = (algunos P xs ∨ algunos P [x])" using HI by simp
also have "… = (algunos P xs ∨ P x)" by simp
also have "… = (P x ∨ algunos P xs)" by auto
also have "… = algunos P (x # xs)" by simp
finally show ?thesis
qed
qed

(* ivamenjim marpoldia1*)

lemma "algunos P (rev xs) = algunos P xs" (is "?P xs")
proof (induct xs)
show "?P []" by simp
next
fix a xs
assume HI: "?P xs"
have "algunos P (rev (a # xs)) = (algunos P ((rev xs)@[a]))" by simp
also have "... = ((algunos P (rev xs)) ∨ algunos P [a])" by (simp add: algunos_append)
also have "... = (algunos P xs ∨ algunos P [a])" using HI by simp
also have "... = (algunos P [a] ∨ algunos P xs)" by arith
finally show "algunos P (rev (a # xs)) = algunos P (a # xs)" by simp
qed

(*paupeddeg*)
lemma "algunos P (rev xs) = algunos P xs"
proof (induct xs)
show "algunos P (rev []) = algunos P []" by simp
next
fix a xs
assume HI:"algunos P (rev xs) = algunos P xs"
have "algunos P (rev (a # xs)) = algunos P ((rev xs) @ [a])" by simp
also have "... = ((algunos P (rev xs)) ∨ (algunos P [a]))" by (simp add: algunos_append)
also have "... = ((algunos P xs) ∨ (algunos P [a]))" using HI by simp
also have "... = ((algunos P [a]) ∨ (algunos P xs))" by arith
finally show "algunos P (rev (a # xs)) = algunos P (a # xs)" by simp
qed

(*crigomgom *)
lemma "algunos P (rev xs) = algunos P xs"
proof (induct xs)
show "algunos P (rev []) = algunos P []" by simp
next
fix x xs
assume HI: " algunos P (rev xs) = algunos P xs"
have "algunos P (rev (x # xs)) = algunos P ((rev xs) @ [x])" by simp
also have "... = (algunos P (rev xs) ∨ algunos P [x])" by (simp add: algunos_append)
also have "... = (algunos P xs ∨ algunos P [x])" using HI by simp
also have "... = (algunos P xs ∨ P x)" by simp
also have "... = (P x ∨ algunos P xs)" by arith
also have "... = algunos P (x#xs)" by simp
finally show "algunos P (rev (x # xs)) = algunos P (x # xs)" by simp
qed

text {*
---------------------------------------------------------------------
Ejercicio 10. Encontrar un término no trivial Z tal que sea cierta la
siguiente ecuación:
algunos (λx. P x ∨ Q x) xs = Z
y demostrar la equivalencia de forma automática y detallada.
---------------------------------------------------------------------
*}

(* migtermor *)

lemma "algunos (λx. P x ∨ Q x) xs = ((algunos (λx. P x) xs) ∨ (algunos (λx. Q x) xs))"
by (induct xs) auto

lemma "algunos (λx. P x ∨ Q x) xs = ((algunos (λx. P x) xs) ∨ (algunos (λx. Q x) xs))" (is "?R xs")
proof (induct xs)
show "?R []" by simp
next
fix x xs
assume HI: "?R xs"
have "algunos (λx. P x ∨ Q x) (x#xs) = (((λx. P x ∨ Q x) x) ∨ (algunos (λx. P x ∨ Q x) xs))"
by simp
also have "… = ((P x ∨ Q x) ∨ (algunos (λx. P x ∨ Q x) xs))" by simp
also have H1: "… = ((((λx. P x) x) ∨ (algunos (λx. P x) xs)) ∨ (((λx. Q x) x) ∨ (algunos (λx. Q x) xs)))"
using HI by simp arith
have H2: "… = ((algunos (λx. P x) (x#xs)) ∨ (algunos (λx. Q x) (x#xs)))"
by simp
have C: "(algunos (λx. P x ∨ Q x) (x#xs)) =
((algunos (λx. P x) (x#xs)) ∨ (algunos (λx. Q x) (x#xs)))"
using H1 H2 by simp
finally show "(algunos (λx. P x ∨ Q x) (x#xs)) =
((algunos (λx. P x) (x#xs)) ∨ (algunos (λx. Q x) (x#xs)))"
using C by simp
qed

(* ferrenseg *)

lemma "algunos (λx. P x ∨ Q x) xs = (algunos P xs ∨ algunos Q xs)"
proof (induct xs)
show "algunos (λx. P x ∨ Q x) [] = (algunos P [] ∨ algunos Q [])" by simp
next
fix x xs
assume HI: "algunos (λx. (P x ∨ Q x)) xs = (algunos P xs ∨ algunos Q xs)"
show "algunos (λx. P x ∨ Q x) (x # xs) = (algunos P (x # xs) ∨ algunos Q (x # xs))"
proof -
have "algunos (λx. P x ∨ Q x) (x # xs) =
((P x) ∨ (Q x) ∨ algunos (λx. P x ∨ Q x) xs)" by simp
also have "… = ((P x) ∨ (Q x) ∨ algunos P xs ∨ algunos Q xs)"
using HI by simp
also have "… = (((P x) ∨ algunos P xs) ∨ ((Q x) ∨ algunos Q xs))" by auto
also have "… = (algunos P (x # xs) ∨ algunos Q (x # xs))" by simp
finally show ?thesis
qed
qed

(* ivamenjim *)

lemma "algunos (λx. P x ∨ Q x) xs = (algunos P xs ∨ algunos Q xs)"
by (induct xs) auto

lemma "algunos (λx. P x ∨ Q x) xs = (algunos P xs ∨ algunos Q xs)"
proof (induct xs)
show "algunos (λx. P x ∨ Q x) [] = (algunos P [] ∨ algunos Q [])" by simp
next
fix a xs
assume HI: "algunos (λx. P x ∨ Q x) xs = (algunos P xs ∨ algunos Q xs)"
have "algunos (λx. P x ∨ Q x) (a # xs) = (P a ∨ Q a ∨ algunos (λx. P x ∨ Q x) xs)" by simp
also have "... = (P a ∨ Q a ∨ algunos P xs ∨ algunos Q xs)" using HI by simp
finally show "algunos (λx. P x ∨ Q x) (a # xs) = (algunos P (a # xs) ∨ algunos Q (a # xs))" by auto
qed

text {*
---------------------------------------------------------------------
Ejercicio 11.1. Demostrar o refutar automáticamente
algunos P xs = (¬ todos (λx. (¬ P x)) xs)
---------------------------------------------------------------------
*}

(* ivamenjim wilmorort migtermor marpoldia1 crigomgom*)

lemma "algunos P xs = (¬ todos (λx. (¬ P x)) xs)"
by (induct xs) auto

(* ferrenseg *)

lemma "algunos P xs = (¬ todos (λx. (¬ P x)) xs)"
by (induct xs) simp_all

text {*
---------------------------------------------------------------------
Ejercicio 11.2. Demostrar o refutar datalladamente
algunos P xs = (¬ todos (λx. (¬ P x)) xs)
---------------------------------------------------------------------
*}

(* ivamenjim migtermor marpoldia1 crigomgom*)

lemma "algunos P xs = (¬ todos (λx. (¬ P x)) xs)"
proof (induct xs)
show "algunos P [] = (¬ todos (λx. ¬ P x) [])" by simp
next
fix a xs
assume HI: "algunos P xs = (¬ todos (λx. ¬ P x) xs)"
have "algunos P (a # xs) = (P a ∨ (algunos P xs))" by simp
also have "... = (P a ∨ (¬ todos (λx. ¬ P x) xs))" using HI by simp
finally show "algunos P (a # xs) = (¬ todos (λx. ¬ P x) (a # xs))" by simp
qed

(* ferrenseg *)

lemma "algunos P xs = (¬ todos (λx. (¬ P x)) xs)"
proof (induct xs)
show "algunos P [] = (¬ todos (λx. (¬ P x)) [])" by simp
next
fix x xs
assume HI: "algunos P xs = (¬ todos (λx. (¬ P x)) xs)"
show "algunos P (x # xs) = (¬ todos (λx. (¬ P x)) (x # xs))"
proof -
have "algunos P (x # xs) = ((P x) ∨ algunos P xs)" by simp
also have "… = ((P x) ∨ ¬ todos (λx. (¬ P x)) xs)" using HI by simp
also have "… = (¬ (¬ (P x) ∧ todos (λx. (¬ P x)) xs))" by simp
also have "… = (¬ todos (λx. (¬ P x)) (x # xs))" by simp
finally show ?thesis
qed
qed

text {*
---------------------------------------------------------------------
Ejercicio 12. Definir la funcion primitiva recursiva
estaEn :: 'a ⇒ 'a list ⇒ bool
tal que (estaEn x xs) se verifica si el elemento x está en la lista
xs. Por ejemplo,
estaEn (2::nat) [3,2,4] = True
estaEn (1::nat) [3,2,4] = False
---------------------------------------------------------------------
*}

(* ivamenjim ferrenseg migtermor serrodcal crigomgom*)

fun estaEn :: "'a ⇒ 'a list ⇒ bool" where
"estaEn x [] = False"
| "estaEn x (a#xs) = ((a = x) ∨ (estaEn x xs))"

value "estaEn (2::nat) [3,2,4] = True"
value "estaEn (1::nat) [3,2,4] = False"

text {*
---------------------------------------------------------------------
Ejercicio 13. Expresar la relación existente entre estaEn y algunos.
Demostrar dicha relación de forma automática y detallada.
---------------------------------------------------------------------
*}

(* migtermor crigomgom*)

lemma "estaEn x xs = (algunos (λa. a=x) xs)"
by (induct xs) auto

lemma auxiliar13:
"(x=a) = ((λa. a=x) a)"
by auto

lemma "estaEn x xs = (algunos (λa. a=x) xs)" (is "?P xs")
proof (induct xs)
show "?P []" by simp
next
fix a xs
assume HI: "?P xs"
have "estaEn x (a#xs) = ((a=x) ∨ (estaEn x xs))" by simp
also have H: "… = ((a=x) ∨ (algunos (λa. a=x) xs))" using HI by simp
also have "… = (((λa. a=x) a) ∨ (algunos (λa. a=x) xs))" using auxiliar13 by simp
also have "… = (algunos (λa. a=x) (a#xs))" by simp
have C: "estaEn x (a#xs) = (algunos (λa. a=x) (a#xs))" using H by simp
finally show "estaEn x (a#xs) = (algunos (λa. a=x) (a#xs))" using C by simp
qed

(* ferrenseg *)

lemma estaEn_algunos:
"estaEn y xs = algunos (λx. x=y) xs"
by (induct xs) auto

lemma estaEn_algunos_2:
"estaEn y xs = algunos (λx. x=y) xs"
proof (induct xs)
show "estaEn y [] = algunos (λx. x=y) []" by simp
next
fix x xs
assume HI: "estaEn y xs = algunos (λx. x=y) xs"
show "estaEn y (x # xs) = algunos (λx. x=y) (x # xs)"
proof -
have "estaEn y (x # xs) = (y=x ∨ estaEn y xs)" by simp
also have "… = (y=x ∨ algunos (λx. x=y) xs)" using HI by simp
also have "… = (x=y ∨ algunos (λx. x=y) xs)" by auto
also have "… = algunos (λx. x=y) (x # xs)" by simp
finally show ?thesis
qed
qed

(*crigomgom*)
lemma "estaEn x xs = (algunos (λa. a=x) xs)"
proof (induct xs)
show " estaEn x [] = algunos (λa. a = x) []" by simp
next
fix a xs
assume HI: "estaEn x xs = algunos (λa. a = x) xs"
have "estaEn x (a # xs) = (a = x ∨ estaEn x xs)" by simp
also have "... = (a = x ∨ algunos (λa. a = x) xs)" using HI by simp
finally show " estaEn x (a # xs) = algunos (λa. a = x) (a # xs)" by simp
qed

(* ivamenjim *)

lemma "estaEn x xs = algunos (λy. x=y) xs"
by (induct xs) auto

lemma "estaEn x xs = algunos (λy. x=y) xs"
proof (induct xs)
show "estaEn x [] = algunos (op = x) []" by simp
next
fix a xs
assume HI: "estaEn x xs = algunos (op = x) xs"
have "estaEn x (a # xs) = ((a = x) ∨ (estaEn x xs))" by simp
also have "... = (a = x ∨ algunos (op = x) xs)" using HI by simp
finally show " estaEn x (a # xs) = algunos (op = x) (a # xs)" by auto
qed

end
</source>

Relación 4

2016-11-22T04:34:48Z

Pablucoto: