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

Ejercicios

2018-07-16T11:13:09Z

WikiSysop:

En esta sección se publicarán las relaciones de ejercicios. Las soluciones se escriben de forma colaborativa por los alumnos del curso y no deben tomarse como definitivas.

* '''Relación 1''': Programación funcional en Isabelle/HOL. ([[R1 |Enunciado]]).
* '''Relación 2''': Razonamiento automático sobre programas en Isabelle/HOL. ([[R2 |Enunciado]]).
* '''Relación 3''': Razonamiento estructurado sobre programas en Isabelle/HOL. ([[R3 |Enunciado]]).
* '''Relación 4''': Cuantificadores sobre listas. ([[R4 |Enunciado]]).
* '''Relación 5''': Eliminación de duplicados. ([[R5 |Enunciado]]).
* '''Relación 6''': Recorridos de árboles. ([[R6 |Enunciado]]).
* '''Relación 7''': Árboles binarios completos. ([[R7 |Enunciado]]).
* '''Relación 8''': Deducción natural proposicional en Isabelle/HOL. ([[R8 |Enunciado]]).
* '''Relación 9''': Deducción natural LPO en Isabelle/HOL. ([[R9 |Enunciado]]).
* '''Relación 10''': Formalización y argumentación en Isabelle/HOL. ([[R10 |Enunciado]]).

Tema 6a: Verificación de la ordenación por inserción

2018-07-16T11:11:24Z

WikiSysop: Texto reemplazado: «isar» por «isabelle»

<source lang="isabelle">
chapter {* T6a: Verificación de la ordenación por inserción *}

theory T6a_Verificacion_de_la_ordenacion_por_insercion
imports Main
begin

text {*
En este de tema se define el algoritmo de ordenación de listas
por inserción y se demuestra que es correcto. *}

text {*
---------------------------------------------------------------------
Ejercicio 1. Definir la función
inserta :: int ⇒ int list ⇒ int list
tal que (inserta a xs) es la lista obtenida insertando a delante del
primer elemento de xs que es mayor o igual que a. Por ejemplo,
inserta 3 [2,5,1,7] = [2,3,5,1,7]
------------------------------------------------------------------ *}

fun inserta :: "int ⇒ int list ⇒ int list" where
"inserta a [] = [a]"
| "inserta a (x#xs) = (if a ≤ x
then a # x # xs
else x # inserta a xs)"

value "inserta 3 [2,5,1,7] = [2,3,5,1,7]"

text {*
---------------------------------------------------------------------
Ejercicio 2. Definir la función
ordena :: int list ⇒ int list
tal que (ordena xs) es la lista obtenida ordenando xs por inserción.
Por ejemplo,
ordena [3,2,5,3] = [2,3,3,5]
------------------------------------------------------------------ *}

fun ordena :: "int list ⇒ int list" where
"ordena [] = []"
| "ordena (x#xs) = inserta x (ordena xs)"

value "ordena [3,2,5,3] = [2,3,3,5]"

text {*
---------------------------------------------------------------------
Ejercicio 3. Definir la función
menor :: int ⇒ int list ⇒ bool
tal que (menor a xs) se verifica si a es menor o igual que todos los
elementos de xs.Por ejemplo,
menor 2 [3,2,5] = True
menor 2 [3,0,5] = False
------------------------------------------------------------------ *}

fun menor :: "int ⇒ int list ⇒ bool" where
"menor a [] = True"
| "menor a (x#xs) = (a ≤ x ∧ menor a xs)"

value "menor 2 [3,2,5] = True"
value "menor 2 [3,0,5] = False"

text {*
---------------------------------------------------------------------
Ejercicio 4. Definir la función
ordenada :: int list ⇒ bool
tal que (ordenada xs) se verifica si xs es una lista ordenada de
manera creciente. Por ejemplo,
ordenada [2,3,3,5] = True
ordenada [2,4,3,5] = False
------------------------------------------------------------------ *}

fun ordenada :: "int list ⇒ bool" where
"ordenada [] = True"
| "ordenada (x#xs) = (menor x xs & ordenada xs)"

value "ordenada [2,3,3,5] = True"
value "ordenada [2,4,3,5] = False"

text {*
---------------------------------------------------------------------
Ejercicio 5. Demostrar que si y es una cota inferior de zs y x ≤ y,
entonces x es una cota inferior de zs.
------------------------------------------------------------------ *}

-- "La demostración automática es"
lemma menor_menor:
assumes "x ≤ y"
shows "menor y zs ⟶ menor x zs"
using assms
by (induct zs) auto

-- "La demostración estructurada es"
lemma menor_menor_2:
assumes "x ≤ y"
shows "menor y zs ⟶ menor x zs"
proof (induct zs)
show "menor y [] ⟶ menor x []" by simp
next
fix z zs
assume HI: "menor y zs ⟶ menor x zs"
show "menor y (z # zs) ⟶ menor x (z # zs)"
proof
assume sup: "menor y (z # zs)"
show "menor x (z # zs)"
proof (simp only: menor.simps(2))
show "x ≤ z ∧ menor x zs"
proof
have "x ≤ y" using assms .
also have "y ≤ z" using sup by simp
finally show "x ≤ z" .
next
have "menor y zs" using sup by simp
with HI show "menor x zs" by simp
qed
qed
qed
qed

text {*
---------------------------------------------------------------------
Ejercicio 6. Demostrar el siguiente teorema de corrección: x es una
cota inferior de la lista obtenida insertando y en zs syss x ≤ y y x
es una cota inferior de zs.
------------------------------------------------------------------ *}

-- "La demostración automática es"
lemma menor_inserta:
"menor x (inserta y zs) = (x ≤ y ∧ menor x zs)"
by (induct zs) auto

-- "La demostración estructurada es"
lemma menor_inserta_2:
"menor x (inserta y zs) = (x ≤ y ∧ menor x zs)"
proof (induct zs)
show "menor x (inserta y []) = (x ≤ y ∧ menor x [])" by simp
next
fix z zs
assume HI: "menor x (inserta y zs) = (x ≤ y ∧ menor x zs)"
show "menor x (inserta y (z#zs)) = (x ≤ y ∧ menor x (z#zs))"
proof (cases "y ≤ z")
assume "y ≤ z"
hence "menor x (inserta y (z#zs)) = menor x (y#z#zs)" by simp
also have "… = (x ≤ y ∧ menor x (z#zs))" by simp
finally show ?thesis by simp
next
assume "¬(y ≤ z)"
hence "menor x (inserta y (z#zs)) =
menor x (z # inserta y zs)" by simp
also have "… = (x ≤ z ∧ menor x (inserta y zs))" by simp
also have "… = (x ≤ z ∧ x ≤ y ∧ menor x zs)" using HI by simp
also have "… = (x ≤ y ∧ menor x (z#zs))" by auto
finally show ?thesis by simp
qed
qed

text {*
---------------------------------------------------------------------
Ejercicio 6. Demostrar que al insertar un elemento la lista obtenida
está ordenada syss lo estaba la original.
------------------------------------------------------------------ *}

-- "La demostración automática es"
lemma ordenada_inserta:
"ordenada (inserta a xs) = ordenada xs"
by (induct xs) (auto simp add: menor_menor menor_inserta)

-- "La demostración estructurada es"
lemma ordenada_inserta_2:
"ordenada (inserta a xs) = ordenada xs"
proof (induct xs)
show "ordenada (inserta a []) = ordenada []" by simp
next
fix x xs
assume HI: "ordenada (inserta a xs) = ordenada xs"
show "ordenada (inserta a (x # xs)) = ordenada (x # xs)"
proof (cases "a ≤ x")
assume "a ≤ x"
hence "ordenada (inserta a (x # xs)) =
ordenada (a # x # xs)" by simp
also have "… = (menor a (x#xs) ∧ ordenada (x # xs))" by simp
also have "… = ordenada (x # xs)"
using `a ≤ x` by (auto simp add: menor_menor)
finally show "ordenada (inserta a (x # xs)) = ordenada (x # xs)"
by simp
next
assume "¬(a ≤ x)"
hence "ordenada (inserta a (x # xs)) =
ordenada (x # inserta a xs)" by simp
also have "… = (menor x (inserta a xs) ∧ ordenada (inserta a xs))"
by simp
also have "… = (menor x (inserta a xs) ∧ ordenada xs)"
using HI by simp
also have "… = (menor x xs ∧ ordenada xs)"
using `¬(a ≤ x)` by (simp add: menor_inserta)
also have "… = ordenada (x # xs)" by simp
finally show "ordenada (inserta a (x # xs)) = ordenada (x # xs)"
by simp
qed
qed

text {*
---------------------------------------------------------------------
Ejercicio 7. Demostrar que, para toda lista xs, (ordena xs) está
ordenada.
------------------------------------------------------------------ *}

-- "La demostración automática es"
theorem ordenada_ordena:
"ordenada (ordena xs)"
by (induct xs) (auto simp add: ordenada_inserta)

-- "La demostración estructurada es"
theorem ordenada_ordena_2:
"ordenada (ordena xs)"
proof (induct xs)
show "ordenada (ordena [])" by simp
next
fix x xs
assume "ordenada (ordena xs)"
then have "ordenada (inserta x (ordena xs))"
by (simp add: ordenada_inserta)
then show "ordenada (ordena (x # xs))" by simp
qed

text {*
---------------------------------------------------------------------
Nota. El teorema anterior no garantiza que ordena sea correcta, ya que
puede que (ordena xs) no tenga los mismos elementos que xs. Por
ejemplo, si se define (ordena xs) como [] se tiene que (ordena xs)
está ordenada pero no es una ordenación de xs.

Para garantizarlo, definimos la función cuenta.
------------------------------------------------------------------ *}

text {*
---------------------------------------------------------------------
Ejercicio 8. Definir la función
cuenta :: int list ⇒ int ⇒ nat
tal que (cuenta xs y) es el número de veces que aparece el elemento y
en la lista xs. Por ejemplo,
cuenta [1,3,4,3,5] 3 = 2
------------------------------------------------------------------ *}

fun cuenta :: "int list ⇒ int ⇒ nat" where
"cuenta [] y = 0"
| "cuenta (x#xs) y = (if x=y
then Suc (cuenta xs y)
else cuenta xs y)"

value "cuenta [1,3,4,3,5] 3 = 2"

text {*
---------------------------------------------------------------------
Ejercicio 9. Demostrar que el número de veces que aparece y en
(inserta x xs) es
* uno más el número de veces que aparece en xs, si y = x;
* el número de veces que aparece en xs, si y ≠ x;
------------------------------------------------------------------ *}

-- "La demostración automática es"
lemma cuenta_inserta:
"cuenta (inserta x xs) y =
(if x=y then Suc (cuenta xs y) else cuenta xs y)"
by (induct xs) auto

text {*
---------------------------------------------------------------------
Ejercicio 10. Demostrar que el número de veces que aparece y en
(ordena xs) es el número de veces que aparece en xs.
------------------------------------------------------------------ *}

-- "La demostración automática es"
theorem cuenta_ordena:
"cuenta (ordena xs) y = cuenta xs y"
by (induct xs) (auto simp add: cuenta_inserta)

-- "La demostración estructurada es"
theorem cuenta_ordena_2:
"cuenta (ordena xs) y = cuenta xs y"
proof (induct xs)
show "cuenta (ordena []) y = cuenta [] y" by simp
next
fix x xs
assume HI: "cuenta (ordena xs) y = cuenta xs y"
show "cuenta (ordena (x # xs)) y = cuenta (x # xs) y"
proof (cases "x = y")
assume "x = y"
have "cuenta (ordena (x # xs)) y = cuenta (inserta x (ordena xs)) y"
by simp
also have "… = Suc (cuenta (ordena xs) y)" using `x = y`
by (simp add: cuenta_inserta)
also have "… = Suc (cuenta xs y)" using HI by simp
also have "… = cuenta (x # xs) y" using `x = y` by simp
finally show "cuenta (ordena (x # xs)) y = cuenta (x # xs) y" by simp
next
assume "x ≠ y"
have "cuenta (ordena (x # xs)) y = cuenta (inserta x (ordena xs)) y"
by simp
also have "… = cuenta (ordena xs) y" using `x ≠ y`
by (simp add: cuenta_inserta)
also have "… = cuenta xs y" using HI by simp
also have "… = cuenta (x # xs) y" using `x ≠ y` by simp
finally show "cuenta (ordena (x # xs)) y = cuenta (x # xs) y"
by simp
qed
qed

end
</source>

Tema 6b: Verificación de la ordenación por mezcla

2018-07-16T11:11:24Z

WikiSysop: Texto reemplazado: «isar» por «isabelle»

<source lang="isabelle">
chapter {* T6b: Verificación de la ordenación por mezcla *}

theory T6b_Verificacion_de_la_ordenacion_por_mezcla_sol
imports Main
begin

text {*
En esta relación de ejercicios se define el algoritmo de ordenación de
listas por mezcla y se demuestra que es correcto.
*}

section {* Ordenación de listas *}

text {*
---------------------------------------------------------------------
Ejercicio 1. Definir la función
menor :: int ⇒ int list ⇒ bool
tal que (menor a xs) se verifica si a es menor o igual que todos los
elementos de xs.Por ejemplo,
menor 2 [3,2,5] = True
menor 2 [3,0,5] = False
------------------------------------------------------------------ *}

fun menor :: "int ⇒ int list ⇒ bool" where
"menor a [] = True"
| "menor a (x#xs) = (a ≤ x ∧ menor a xs)"

value "menor 2 [3,2,5] = True"
value "menor 2 [3,0,5] = False"

text {*
---------------------------------------------------------------------
Ejercicio 2. Definir la función
ordenada :: int list ⇒ bool
tal que (ordenada xs) se verifica si xs es una lista ordenada de
manera creciente. Por ejemplo,
ordenada [2,3,3,5] = True
ordenada [2,4,3,5] = False
------------------------------------------------------------------ *}

fun ordenada :: "int list ⇒ bool" where
"ordenada [] = True"
| "ordenada (x#xs) = (menor x xs & ordenada xs)"

value "ordenada [2,3,3,5] = True"
value "ordenada [2,4,3,5] = False"

text {*
---------------------------------------------------------------------
Ejercicio 3. Definir la función
cuenta :: int list => int => nat
tal que (cuenta xs y) es el número de veces que aparece el elemento y
en la lista xs. Por ejemplo,
cuenta [1,3,4,3,5] 3 = 2
------------------------------------------------------------------ *}

fun cuenta :: "int list => int => nat" where
"cuenta [] y = 0"
| "cuenta (x#xs) y = (if x=y
then Suc(cuenta xs y)
else cuenta xs y)"

value "cuenta [1,3,4,3,5] 3 = 2"

section {* Ordenación por mezcla *}

text {*
---------------------------------------------------------------------
Ejercicio 4. Definir la función
mezcla :: int list ⇒ int list ⇒ int list
tal que (mezcla xs ys) es la lista obtenida mezclando las listas
ordenadas xs e ys. Por ejemplo,
mezcla [1,2,5] [3,5,7] = [1,2,3,5,5,7]
------------------------------------------------------------------ *}

fun mezcla :: "int list ⇒ int list ⇒ int list" where
"mezcla [] ys = ys"
| "mezcla xs [] = xs"
| "mezcla (x # xs) (y # ys) = (if x ≤ y
then x # mezcla xs (y # ys)
else y # mezcla (x # xs) ys)"

value "mezcla [1,2,5] [3,5,7] = [1,2,3,5,5,7]"

text {*
---------------------------------------------------------------------
Ejercicio 5. Definir la función
ordenaM :: int list ⇒ int list
tal que (ordenaM xs) es la lista obtenida ordenando la lista xs
mediante mezclas; es decir, la divide en dos mitades, las ordena y las
mezcla. Por ejemplo,
ordenaM [3,2,5,2] = [2,2,3,5]
------------------------------------------------------------------ *}

fun ordenaM :: "int list ⇒ int list" where
"ordenaM [] = []"
| "ordenaM [x] = [x]"
| "ordenaM xs =
(let mitad = length xs div 2 in
mezcla (ordenaM (take mitad xs))
(ordenaM (drop mitad xs)))"

value "ordenaM [3,2,5,2] = [2,2,3,5]"

text {*
---------------------------------------------------------------------
Ejercicio 6. Sea x ≤ y. Si y es menor o igual que todos los elementos
de xs, entonces x es menor o igual que todos los elementos de xs
------------------------------------------------------------------ *}

lemma menor_menor:
"x ≤ y ⟹ menor y xs ⟶ menor x xs"
by (induct xs) auto

text {*
---------------------------------------------------------------------
Ejercicio 7. Demostrar que el número de veces que aparece n en la
mezcla de dos listas es igual a la suma del número de apariciones en
cada una de las listas
------------------------------------------------------------------ *}

lemma cuenta_mezcla:
"cuenta (mezcla xs ys) n = cuenta xs n + cuenta ys n"
by (induct xs ys rule: mezcla.induct) auto

text {*
---------------------------------------------------------------------
Ejercicio 8. Demostrar que si x es menor que todos los elementos de
ys y de zs, entonces también lo es de su mezcla.
------------------------------------------------------------------ *}

lemma menor_mezcla:
assumes "menor x ys"
"menor x zs"
shows "menor x (mezcla ys zs)"
using assms
by (induct ys zs rule: mezcla.induct) simp_all

text {*
---------------------------------------------------------------------
Ejercicio 9. Demostrar que la mezcla de dos listas ordenadas es una
lista ordenada.
Indicación: Usar los siguientes lemas
· linorder_not_le: (¬ x ≤ y) = (y < x)
· order_less_le: (x < y) = (x ≤ y ∧ x ≠ y)
------------------------------------------------------------------ *}

lemma ordenada_mezcla:
assumes "ordenada xs"
"ordenada ys"
shows "ordenada (mezcla xs ys)"
using assms
by (induct xs ys rule: mezcla.induct)
(auto simp add: menor_mezcla
menor_menor
linorder_not_le
order_less_le)

text {*
---------------------------------------------------------------------
Ejercicio 10. Demostrar que si x es mayor que 1, entonces el mínimo de
x y su mitad es menor que x.
------------------------------------------------------------------ *}

lemma min_mitad:
"1 < x ⟹ min x (x div 2::int) < x"
by simp

text {*
---------------------------------------------------------------------
Ejercicio 11. Demostrar que si x es mayor que 1, entonces x menos su
mitad es menor que x.
------------------------------------------------------------------ *}

lemma menos_mitad:
"1 < x ⟹ x - x div (2::int) < x"
by arith

text {*
---------------------------------------------------------------------
Ejercicio 11. Demostrar que (ordenaM xs) está ordenada.
------------------------------------------------------------------ *}

theorem ordenada_ordenaM:
"ordenada (ordenaM xs)"
by (induct xs rule: ordenaM.induct)
(auto simp add: ordenada_mezcla)

text {*
---------------------------------------------------------------------
Ejercicio 12. Demostrar que el número de apariciones de un elemento en
la concatenación de dos listas es la suma del número de apariciones en
cada una.
------------------------------------------------------------------ *}

lemma cuenta_conc:
"cuenta (xs @ ys) x = cuenta xs x + cuenta ys x"
by (induct xs) auto

text {*
---------------------------------------------------------------------
Ejercicio 13. Demostrar que las listas xs y (ordenaM xs) tienen los
mismos elementos.
------------------------------------------------------------------ *}

theorem cuenta_ordenaM:
"cuenta (ordenaM xs) x = cuenta xs x"
by (induct xs rule: ordenaM.induct)
(auto simp add: cuenta_mezcla
cuenta_conc [symmetric])

end
</source>

Tema 8b: Deducción natural proposicional con Isabelle/HOL

2018-07-16T11:11:24Z

WikiSysop: Texto reemplazado: «isar» por «isabelle»

<source lang="isabelle">
chapter {* Tema 8b: Deducción natural proposicional con Isabelle/HOL *}

theory T8b_Deduccion_natural_en_logica_proposicional_con_Isabelle
imports Main
begin

text {*
En este tema se presentan los ejemplos del tema de deducción natural
proposicional siguiendo la presentación de Huth y Ryan en su libro
"Logic in Computer Science" http://goo.gl/qsVpY y, más concretamente,
a la forma como se explica en la asignatura de "Lógica informática" (LI)
http://goo.gl/AwDiv

La página al lado de cada ejemplo indica la página de las transparencias
de LI donde se encuentra la demostración. *}

subsection {* Reglas de la conjunción *}

text {*
Ejemplo 1 (p. 4). Demostrar que
p ∧ q, r ⊢ q ∧ r.
*}

-- "La demostración detallada es"
lemma ejemplo_1_1:
assumes 1: "p ∧ q" and
2: "r"
shows "q ∧ r"
proof -
have 3: "q" using 1 by (rule conjunct2)
show 4: "q ∧ r" using 3 2 by (rule conjI)
qed

text {*
Notas sobre el lenguaje: En la demostración anterior se ha usado
· "assumes" para indicar las hipótesis,
· "and" para separar las hipótesis,
· "shows" para indicar la conclusión,
· "proof" para iniciar la prueba,
· "qed" para terminar la pruebas,
· "-" (después de "proof") para no usar el método por defecto,
· "have" para establecer un paso,
· "using" para usar hechos en un paso,
· "by (rule ..)" para indicar la regla con la que se peueba un hecho,
· "show" para establecer la conclusión.

Notas sobre la lógica: Las reglas de la conjunción son
· conjI: ⟦P; Q⟧ ⟹ P ∧ Q
· conjunct1: P ∧ Q ⟹ P
· conjunct2: P ∧ Q ⟹ Q
*}

text {* Se pueden dejar implícitas las reglas como sigue *}

lemma ejemplo_1_2:
assumes 1: "p ∧ q" and
2: "r"
shows "q ∧ r"
proof -
have 3: "q" using 1 ..
show 4: "q ∧ r" using 3 2 ..
qed

text {*
Nota sobre el lenguaje: En la demostración anterior se ha usado
· ".." para indicar que se prueba por la regla correspondiente. *}

text {* Se pueden eliminar las etiquetas como sigue *}

lemma ejemplo_1_3:
assumes "p ∧ q"
"r"
shows "q ∧ r"
proof -
have "q" using assms(1) ..
thus "q ∧ r" using assms(2) ..
qed

text {*
Nota sobre el lenguaje: En la demostración anterior se ha usado
· "assms(n)" para indicar la hipótesis n y
· "thus" para demostrar la conclusión usando el hecho anterior.
Además, no es necesario usar and entre las hipótesis. *}

text {* Se puede automatizar la demostración como sigue *}

lemma ejemplo_1_4:
assumes "p ∧ q"
"r"
shows "q ∧ r"
using assms
by auto

text {*
Nota sobre el lenguaje: En la demostración anterior se ha usado
· "assms" para indicar las hipótesis y
· "by auto" para demostrar la conclusión automáticamente. *}

text {* Se puede automatizar totalmente la demostración como sigue *}

lemma ejemplo_1_5:
"⟦p ∧ q; r⟧ ⟹ q ∧ r"
by auto

text {*
Nota sobre el lenguaje: En la demostración anterior se ha usado
· "⟦ ... ⟧" para representar las hipótesis,
· ";" para separar las hipótesis y
· "⟹" para separar las hipótesis de la conclusión. *}

text {* Se puede hacer la demostración por razonamiento hacia atrás,
como sigue *}

lemma ejemplo_1_6:
assumes "p ∧ q"
and "r"
shows "q ∧ r"
proof (rule conjI)
show "q" using assms(1) by (rule conjunct2)
next
show "r" using assms(2) by this
qed

text {*
Nota sobre el lenguaje: En la demostración anterior se ha usado
· "proof (rule r)" para indicar que se hará la demostración con la
regla r,
· "next" para indicar el comienzo de la prueba del siguiente
subobjetivo,
· "this" para indicar el hecho actual. *}

text {* Se pueden dejar implícitas las reglas como sigue *}

lemma ejemplo_1_7:
assumes "p ∧ q"
"r"
shows "q ∧ r"
proof
show "q" using assms(1) ..
next
show "r" using assms(2) .
qed

text {*
Nota sobre el lenguaje: En la demostración anterior se ha usado
· "." para indicar por el hecho actual. *}

subsection {* Reglas de la doble negación *}

text {*
La regla de eliminación de la doble negación es
· notnotD: ¬¬ P ⟹ P

Para ajustarnos al tema de LI vamos a introducir la siguiente regla de
introducción de la doble negación
· notnotI: P ⟹ ¬¬ P
aunque, de momento, no detallamos su demostración.
*}

lemma notnotI [intro!]: "P ⟹ ¬¬ P"
by auto

text {*
Ejemplo 2. (p. 5)
p, ¬¬(q ∧ r) ⊢ ¬¬p ∧ r
*}

-- "La demostración detallada es"
lemma ejemplo_2_1:
assumes 1: "p" and
2: "¬¬(q ∧ r)"
shows "¬¬p ∧ r"
proof -
have 3: "¬¬p" using 1 by (rule notnotI)
have 4: "q ∧ r" using 2 by (rule notnotD)
have 5: "r" using 4 by (rule conjunct2)
show 6: "¬¬p ∧ r" using 3 5 by (rule conjI)
qed

-- "La demostración estructurada es"
lemma ejemplo_2_2:
assumes "p"
"¬¬(q ∧ r)"
shows "¬¬p ∧ r"
proof -
have "¬¬p" using assms(1) ..
have "q ∧ r" using assms(2) by (rule notnotD)
hence "r" ..
with `¬¬p` show "¬¬p ∧ r" ..
qed

text {*
Nota sobre el lenguaje: En la demostración anterior se ha usado
· "hence" para indicar que se tiene por el hecho anterior,
· `...` para referenciar un hecho y
· "with P show Q" para indicar que con el hecho anterior junto con el
hecho P se demuestra Q. *}

-- "La demostración automática es"
lemma ejemplo_2_3:
assumes "p"
"¬¬(q ∧ r)"
shows "¬¬p ∧ r"
using assms
by auto

text {* Se puede demostrar hacia atrás *}

lemma ejemplo_2_4:
assumes "p"
"¬¬(q ∧ r)"
shows "¬¬p ∧ r"
proof (rule conjI)
show "¬¬p" using assms(1) by (rule notnotI)
next
have "q ∧ r" using assms(2) by (rule notnotD)
thus "r" by (rule conjunct2)
qed

text {* Se puede eliminar las reglas en la demostración anterior, como
sigue: *}

lemma ejemplo_2_5:
assumes "p"
"¬¬(q ∧ r)"
shows "¬¬p ∧ r"
proof
show "¬¬p" using assms(1) ..
next
have "q ∧ r" using assms(2) by (rule notnotD)
thus "r" ..
qed

subsection {* Regla de eliminación del condicional *}

text {*
La regla de eliminación del condicional es la regla del modus ponens
· mp: ⟦P ⟶ Q; P⟧ ⟹ Q
*}

text {*
Ejemplo 3. (p. 6) Demostrar que
¬p ∧ q, ¬p ∧ q ⟶ r ∨ ¬p ⊢ r ∨ ¬p
*}

-- "La demostración detallada es"
lemma ejemplo_3_1:
assumes 1: "¬p ∧ q" and
2: "¬p ∧ q ⟶ r ∨ ¬p"
shows "r ∨ ¬p"
proof -
show "r ∨ ¬p" using 2 1 by (rule mp)
qed

-- "La demostración estructurada es"
lemma ejemplo_3_2:
assumes "¬p ∧ q"
"¬p ∧ q ⟶ r ∨ ¬p"
shows "r ∨ ¬p"
proof -
show "r ∨ ¬p" using assms(2,1) ..
qed

-- "La demostración automática es"
lemma ejemplo_3_3:
assumes "¬p ∧ q"
"¬p ∧ q ⟶ r ∨ ¬p"
shows "r ∨ ¬p"
using assms
by auto

text {*
Ejemplo 4 (p. 6) Demostrar que
p, p ⟶ q, p ⟶ (q ⟶ r) ⊢ r
*}

-- "La demostración detallada es"
lemma ejemplo_4_1:
assumes 1: "p" and
2: "p ⟶ q" and
3: "p ⟶ (q ⟶ r)"
shows "r"
proof -
have 4: "q" using 2 1 by (rule mp)
have 5: "q ⟶ r" using 3 1 by (rule mp)
show 6: "r" using 5 4 by (rule mp)
qed

-- "La demostración estructurada es"
lemma ejemplo_4_2:
assumes "p"
"p ⟶ q"
"p ⟶ (q ⟶ r)"
shows "r"
proof -
have "q" using assms(2,1) ..
have "q ⟶ r" using assms(3,1) ..
thus "r" using `q` ..
qed

-- "La demostración automática es"
lemma ejemplo_4_3:
"⟦p; p ⟶ q; p ⟶ (q ⟶ r)⟧ ⟹ r"
by auto

subsection {* Regla derivada del modus tollens *}

text {*
Para ajustarnos al tema de LI vamos a introducir la regla del modus
tollens
· mt: ⟦F ⟶ G; ¬G⟧ ⟹ ¬F
aunque, de momento, sin detallar su demostración.
*}

lemma mt: "⟦F ⟶ G; ¬G⟧ ⟹ ¬F"
by auto

text {*
Ejemplo 5 (p. 7). Demostrar
p ⟶ (q ⟶ r), p, ¬r ⊢ ¬q
*}

-- "La demostración detallada es"
lemma ejemplo_5_1:
assumes 1: "p ⟶ (q ⟶ r)" and
2: "p" and
3: "¬r"
shows "¬q"
proof -
have 4: "q ⟶ r" using 1 2 by (rule mp)
show "¬q" using 4 3 by (rule mt)
qed

-- "La demostración estructurada es"
lemma ejemplo_5_2:
assumes "p ⟶ (q ⟶ r)"
"p"
"¬r"
shows "¬q"
proof -
have "q ⟶ r" using assms(1,2) ..
thus "¬q" using assms(3) by (rule mt)
qed

-- "La demostración automática es"
lemma ejemplo_5_3:
assumes "p ⟶ (q ⟶ r)"
"p"
"¬r"
shows "¬q"
using assms
by auto

text {*
Ejemplo 6. (p. 7) Demostrar
¬p ⟶ q, ¬q ⊢ p
*}

-- "La demostración detallada es"
lemma ejemplo_6_1:
assumes 1: "¬p ⟶ q" and
2: "¬q"
shows "p"
proof -
have 3: "¬¬p" using 1 2 by (rule mt)
show "p" using 3 by (rule notnotD)
qed

-- "La demostración estructurada es"
lemma ejemplo_6_2:
assumes "¬p ⟶ q"
"¬q"
shows "p"
proof -
have "¬¬p" using assms(1,2) by (rule mt)
thus "p" by (rule notnotD)
qed

-- "La demostración automática es"
lemma ejemplo_6_3:
"⟦¬p ⟶ q; ¬q⟧ ⟹ p"
by auto

text {*
Ejemplo 7. (p. 7) Demostrar
p ⟶ ¬q, q ⊢ ¬p
*}

-- "La demostración detallada es"
lemma ejemplo_7_1:
assumes 1: "p ⟶ ¬q" and
2: "q"
shows "¬p"
proof -
have 3: "¬¬q" using 2 by (rule notnotI)
show "¬p" using 1 3 by (rule mt)
qed

-- "La demostración detallada es"
lemma ejemplo_7_2:
assumes "p ⟶ ¬q"
"q"
shows "¬p"
proof -
have "¬¬q" using assms(2) by (rule notnotI)
with assms(1) show "¬p" by (rule mt)
qed

-- "La demostración estructurada es"
lemma ejemplo_7_3:
"⟦p ⟶ ¬q; q⟧ ⟹ ¬p"
by auto

subsection {* Regla de introducción del condicional *}

text {*
La regla de introducción del condicional es
· impI: (P ⟹ Q) ⟹ P ⟶ Q
*}

text {*
Ejemplo 8. (p. 8) Demostrar
p ⟶ q ⊢ ¬q ⟶ ¬p
*}

-- "La demostración detallada es"
lemma ejemplo_8_1:
assumes 1: "p ⟶ q"
shows "¬q ⟶ ¬p"
proof -
{ assume 2: "¬q"
have "¬p" using 1 2 by (rule mt) }
thus "¬q ⟶ ¬p" by (rule impI)
qed

text {*
Nota sobre el lenguaje: En la demostración anterior se ha usado
· "{ ... }" para representar una caja. *}

-- "La demostración estructurada es"
lemma ejemplo_8_2:
assumes "p ⟶ q"
shows "¬q ⟶ ¬p"
proof
assume "¬q"
with assms show "¬p" by (rule mt)
qed

-- "La demostración automática es"
lemma ejemplo_8_3:
assumes "p ⟶ q"
shows "¬q ⟶ ¬p"
using assms
by auto

text {*
Ejemplo 9. (p. 9) Demostrar
¬q ⟶ ¬p ⊢ p ⟶ ¬¬q
*}

-- "La demostración detallada es"
lemma ejemplo_9_1:
assumes 1: "¬q ⟶ ¬p"
shows "p ⟶ ¬¬q"
proof -
{ assume 2: "p"
have 3: "¬¬p" using 2 by (rule notnotI)
have "¬¬q" using 1 3 by (rule mt) }
thus "p ⟶ ¬¬q" by (rule impI)
qed

-- "La demostración estructurada es"
lemma ejemplo_9_2:
assumes "¬q ⟶ ¬p"
shows "p ⟶ ¬¬q"
proof
assume "p"
hence "¬¬p" by (rule notnotI)
with assms show "¬¬q" by (rule mt)
qed

-- "La demostración automática es"
lemma ejemplo_9_3:
assumes "¬q ⟶ ¬p"
shows "p ⟶ ¬¬q"
using assms
by auto

text {*
Ejemplo 10 (p. 9). Demostrar
⊢ p ⟶ p
*}

-- "La demostración detallada es"
lemma ejemplo_10_1:
"p ⟶ p"
proof -
{ assume 1: "p"
have "p" using 1 by this }
thus "p ⟶ p" by (rule impI)
qed

-- "La demostración estructurada es"
lemma ejemplo_10_2:
"p ⟶ p"
proof (rule impI)
qed

-- "La demostración automática es"
lemma ejemplo_10_3:
"p ⟶ p"
by auto

text {*
Ejemplo 11 (p. 10) Demostrar
⊢ (q ⟶ r) ⟶ ((¬q ⟶ ¬p) ⟶ (p ⟶ r))
*}

-- "La demostración detallada es"
lemma ejemplo_11_1:
"(q ⟶ r) ⟶ ((¬q ⟶ ¬p) ⟶ (p ⟶ r))"
proof -
{ assume 1: "q ⟶ r"
{ assume 2: "¬q ⟶ ¬p"
{ assume 3: "p"
have 4: "¬¬p" using 3 by (rule notnotI)
have 5: "¬¬q" using 2 4 by (rule mt)
have 6: "q" using 5 by (rule notnotD)
have "r" using 1 6 by (rule mp) }
hence "p ⟶ r" by (rule impI) }
hence "(¬q ⟶ ¬p) ⟶ p ⟶ r" by (rule impI) }
thus "(q ⟶ r) ⟶ ((¬q ⟶ ¬p) ⟶ p ⟶ r)" by (rule impI)
qed

-- "La demostración hacia atrás es"
lemma ejemplo_11_2:
"(q ⟶ r) ⟶ ((¬q ⟶ ¬p) ⟶ (p ⟶ r))"
proof (rule impI)
assume 1: "q ⟶ r"
show "(¬q ⟶ ¬p) ⟶ (p ⟶ r)"
proof (rule impI)
assume 2: "¬q ⟶ ¬p"
show "p ⟶ r"
proof (rule impI)
assume 3: "p"
have 4: "¬¬p" using 3 by (rule notnotI)
have 5: "¬¬q" using 2 4 by (rule mt)
have 6: "q" using 5 by (rule notnotD)
show "r" using 1 6 by (rule mp)
qed
qed
qed

-- "La demostración hacia atrás con reglas implícitas es"
lemma ejemplo_11_3:
"(q ⟶ r) ⟶ ((¬q ⟶ ¬p) ⟶ (p ⟶ r))"
proof
assume 1: "q ⟶ r"
show "(¬q ⟶ ¬p) ⟶ (p ⟶ r)"
proof
assume 2: "¬q ⟶ ¬p"
show "p ⟶ r"
proof
assume 3: "p"
have 4: "¬¬p" using 3 ..
have 5: "¬¬q" using 2 4 by (rule mt)
have 6: "q" using 5 by (rule notnotD)
show "r" using 1 6 ..
qed
qed
qed

-- "La demostración sin etiquetas es"
lemma ejemplo_11_4:
"(q ⟶ r) ⟶ ((¬q ⟶ ¬p) ⟶ (p ⟶ r))"
proof
assume "q ⟶ r"
show "(¬q ⟶ ¬p) ⟶ (p ⟶ r)"
proof
assume "¬q ⟶ ¬p"
show "p ⟶ r"
proof
assume "p"
hence "¬¬p" ..
with `¬q ⟶ ¬p` have "¬¬q" by (rule mt)
hence "q" by (rule notnotD)
with `q ⟶ r` show "r" ..
qed
qed
qed

-- "La demostración automática es"
lemma ejemplo_11_5:
"(q ⟶ r) ⟶ ((¬q ⟶ ¬p) ⟶ (p ⟶ r))"
by auto

subsection {* Reglas de la disyunción *}

text {*
Las reglas de la introducción de la disyunción son
· disjI1: P ⟹ P ∨ Q
· disjI2: Q ⟹ P ∨ Q
La regla de elimación de la disyunción es
· disjE: ⟦P ∨ Q; P ⟹ R; Q ⟹ R⟧ ⟹ R
*}

text {*
Ejemplo 12 (p. 11). Demostrar
p ∨ q ⊢ q ∨ p
*}

-- "La demostración detallada es"
lemma ejemplo_12_1:
assumes "p ∨ q"
shows "q ∨ p"
proof -
have "p ∨ q" using assms by this
moreover
{ assume 2: "p"
have "q ∨ p" using 2 by (rule disjI2) }
moreover
{ assume 3: "q"
have "q ∨ p" using 3 by (rule disjI1) }
ultimately show "q ∨ p" by (rule disjE)
qed

text {*
Nota sobre el lenguaje: En la demostración anterior se ha usado
· "moreover" para separar los bloques y
· "ultimately" para unir los resultados de los bloques. *}

-- "La demostración detallada con reglas implícitas es"
lemma ejemplo_12_2:
assumes "p ∨ q"
shows "q ∨ p"
proof -
note `p ∨ q`
moreover
{ assume "p"
hence "q ∨ p" .. }
moreover
{ assume "q"
hence "q ∨ p" .. }
ultimately show "q ∨ p" ..
qed

text {*
Nota sobre el lenguaje: En la demostración anterior se ha usado
· "note" para copiar un hecho. *}

-- "La demostración hacia atrás es"
lemma ejemplo_12_3:
assumes 1: "p ∨ q"
shows "q ∨ p"
using 1
proof (rule disjE)
{ assume 2: "p"
show "q ∨ p" using 2 by (rule disjI2) }
next
{ assume 3: "q"
show "q ∨ p" using 3 by (rule disjI1) }
qed

-- "La demostración hacia atrás con reglas implícitas es"
lemma ejemplo_12_4:
assumes "p ∨ q"
shows "q ∨ p"
using assms
proof
{ assume "p"
thus "q ∨ p" .. }
next
{ assume "q"
thus "q ∨ p" .. }
qed

-- "La demostración automática es"
lemma ejemplo_12_5:
assumes "p ∨ q"
shows "q ∨ p"
using assms
by auto

text {*
Ejemplo 13. (p. 12) Demostrar
q ⟶ r ⊢ p ∨ q ⟶ p ∨ r
*}

-- "La demostración detallada es"
lemma ejemplo_13_1:
assumes 1: "q ⟶ r"
shows "p ∨ q ⟶ p ∨ r"
proof (rule impI)
assume 2: "p ∨ q"
thus "p ∨ r"
proof (rule disjE)
{ assume 3: "p"
show "p ∨ r" using 3 by (rule disjI1) }
next
{ assume 4: "q"
have 5: "r" using 1 4 by (rule mp)
show "p ∨ r" using 5 by (rule disjI2) }
qed
qed

-- "La demostración estructurada es"
lemma ejemplo_13_2:
assumes "q ⟶ r"
shows "p ∨ q ⟶ p ∨ r"
proof
assume "p ∨ q"
thus "p ∨ r"
proof
{ assume "p"
thus "p ∨ r" .. }
next
{ assume "q"
have "r" using assms `q` ..
thus "p ∨ r" .. }
qed
qed

-- "La demostración automática es"
lemma ejemplo_13_3:
assumes "q ⟶ r"
shows "p ∨ q ⟶ p ∨ r"
using assms
by auto

subsection {* Regla de copia *}

text {*
Ejemplo 14 (p. 13). Demostrar
⊢ p ⟶ (q ⟶ p)
*}

-- "La demostración detallada es"
lemma ejemplo_14_1:
"p ⟶ (q ⟶ p)"
proof (rule impI)
assume 1: "p"
show "q ⟶ p"
proof (rule impI)
assume "q"
show "p" using 1 by this
qed
qed

-- "La demostración estructurada es"
lemma ejemplo_14_2:
"p ⟶ (q ⟶ p)"
proof
assume "p"
thus "q ⟶ p" ..
qed

-- "La demostración automática es"
lemma ejemplo_14_3:
"p ⟶ (q ⟶ p)"
by auto

subsection {* Reglas de la negación *}

text {*
La regla de eliminación de lo falso es
· FalseE: False ⟹ P
La regla de eliminación de la negación es
· notE: ⟦¬P; P⟧ ⟹ R
La regla de introducción de la negación es
· notI: (P ⟹ False) ⟹ ¬P
*}

text {*
Ejemplo 15 (p. 15). Demostrar
¬p ∨ q ⊢ p ⟶ q
*}

-- "La demostración detallada es"
lemma ejemplo_15_1:
assumes 1: "¬p ∨ q"
shows "p ⟶ q"
proof (rule impI)
assume 2: "p"
note 1
thus "q"
proof (rule disjE)
{ assume 3: "¬p"
show "q" using 3 2 by (rule notE) }
next
{ assume 4: "q"
show "q" using 4 by this}
qed
qed

-- "La demostración estructurada es"
lemma ejemplo_15_2:
assumes "¬p ∨ q"
shows "p ⟶ q"
proof
assume "p"
note `¬p ∨ q`
thus "q"
proof
assume "¬p"
thus "q" using `p` ..
next
assume "q"
thus "q" .
qed
qed

-- "La demostración automática es"
lemma ejemplo_15_3:
assumes "¬p ∨ q"
shows "p ⟶ q"
using assms
by auto

text {*
Ejemplo 16 (p. 16). Demostrar
p ⟶ q, p ⟶ ¬q ⊢ ¬p
*}

-- "La demostración detallada es"
lemma ejemplo_16_1:
assumes 1: "p ⟶ q" and
2: "p ⟶ ¬q"
shows "¬p"
proof (rule notI)
assume 3: "p"
have 4: "q" using 1 3 by (rule mp)
have 5: "¬q" using 2 3 by (rule mp)
show False using 5 4 by (rule notE)
qed

-- "La demostración estructurada es"
lemma ejemplo_16_2:
assumes "p ⟶ q"
"p ⟶ ¬q"
shows "¬p"
proof
assume "p"
have "q" using assms(1) `p` ..
have "¬q" using assms(2) `p` ..
thus False using `q` ..
qed

-- "La demostración automática es"
lemma ejemplo_16_3:
assumes "p ⟶ q"
"p ⟶ ¬q"
shows "¬p"
using assms
by auto

subsection {* Reglas del bicondicional *}

text {*
La regla de introducción del bicondicional es
· iffI: ⟦P ⟹ Q; Q ⟹ P⟧ ⟹ P ⟷ Q
Las reglas de eliminación del bicondicional son
· iffD1: ⟦Q ⟷ P; Q⟧ ⟹ P
· iffD2: ⟦P ⟷ Q; Q⟧ ⟹ P
*}

text {*
Ejemplo 17 (p. 17) Demostrar
(p ∧ q) ⟷ (q ∧ p)
*}

-- "La demostración detallada es"
lemma ejemplo_17_1:
"(p ∧ q) ⟷ (q ∧ p)"
proof (rule iffI)
{ assume 1: "p ∧ q"
have 2: "p" using 1 by (rule conjunct1)
have 3: "q" using 1 by (rule conjunct2)
show "q ∧ p" using 3 2 by (rule conjI) }
next
{ assume 4: "q ∧ p"
have 5: "q" using 4 by (rule conjunct1)
have 6: "p" using 4 by (rule conjunct2)
show "p ∧ q" using 6 5 by (rule conjI) }
qed

-- "La demostración estructurada es"
lemma ejemplo_17_2:
"(p ∧ q) ⟷ (q ∧ p)"
proof
{ assume 1: "p ∧ q"
have "p" using 1 ..
have "q" using 1 ..
show "q ∧ p" using `q` `p` .. }
next
{ assume 2: "q ∧ p"
have "q" using 2 ..
have "p" using 2 ..
show "p ∧ q" using `p` `q` .. }
qed

-- "La demostración automática es"
lemma ejemplo_17_3:
"(p ∧ q) ⟷ (q ∧ p)"
by auto

text {*
Ejemplo 18 (p. 18). Demostrar
p ⟷ q, p ∨ q ⊢ p ∧ q
*}

-- "La demostración detallada es"
lemma ejemplo_18_1:
assumes 1: "p ⟷ q" and
2: "p ∨ q"
shows "p ∧ q"
using 2
proof (rule disjE)
{ assume 3: "p"
have 4: "q" using 1 3 by (rule iffD1)
show "p ∧ q" using 3 4 by (rule conjI) }
next
{ assume 5: "q"
have 6: "p" using 1 5 by (rule iffD2)
show "p ∧ q" using 6 5 by (rule conjI) }
qed

-- "La demostración estructurada es"
lemma ejemplo_18_2:
assumes "p ⟷ q"
"p ∨ q"
shows "p ∧ q"
using assms(2)
proof
{ assume "p"
with assms(1) have "q" ..
with `p` show "p ∧ q" .. }
next
{ assume "q"
with assms(1) have "p" ..
thus "p ∧ q" using `q` .. }
qed

-- "La demostración automática es"
lemma ejemplo_18_3:
assumes "p ⟷ q"
"p ∨ q"
shows "p ∧ q"
using assms
by auto

subsection {* Reglas derivadas *}

subsubsection {* Regla del modus tollens *}

text {*
Ejemplo 19 (p. 20) Demostrar la regla del modus tollens a partir de
las reglas básicas.
*}

-- "La demostración detallada es"
lemma ejemplo_20_1:
assumes 1: "F ⟶ G" and
2: "¬G"
shows "¬F"
proof (rule notI)
assume 3: "F"
have 4: "G" using 1 3 by (rule mp)
show False using 2 4 by (rule notE)
qed

-- "La demostración estructurada es"
lemma ejemplo_20_2:
assumes "F ⟶ G"
"¬G"
shows "¬F"
proof
assume "F"
with assms(1) have "G" ..
with assms(2) show False ..
qed

-- "La demostración automática es"
lemma ejemplo_20_3:
assumes "F ⟶ G"
"¬G"
shows "¬F"
using assms
by auto

subsubsection {* Regla de la introducción de la doble negación *}

text {*
Ejemplo 21 (p. 21) Demostrar la regla de introducción de la doble
negación a partir de las reglas básicas.
*}

-- "La demostración detallada es"
lemma ejemplo_21_1:
assumes 1: "F"
shows "¬¬F"
proof (rule notI)
assume 2: "¬F"
show False using 2 1 by (rule notE)
qed

-- "La demostración estructurada es"
lemma ejemplo_21_2:
assumes "F"
shows "¬¬F"
proof
assume "¬F"
thus False using assms ..
qed

-- "La demostración automática es"
lemma ejemplo_21_3:
assumes "F"
shows "¬¬F"
using assms
by auto

subsubsection {* Regla de reducción al absurdo *}

text {*
La regla de reducción al absurdo en Isabelle se correponde con la
regla clásica de contradicción
· ccontr: (¬P ⟹ False) ⟹ P
*}

subsubsection {* Ley del tercio excluso *}

text {*
La ley del tercio excluso es
· excluded_middle: ¬P ∨ P
*}

text {*
Ejemplo 22 (p. 23). Demostrar la ley del tercio excluso a partir de
las reglas básicas.
*}

-- "La demostración detallada es"
lemma ejemplo_22_1:
"F ∨ ¬F"
proof (rule ccontr)
assume 1: "¬(F ∨ ¬F)"
thus False
proof (rule notE)
show "F ∨ ¬F"
proof (rule disjI2)
show "¬F"
proof (rule notI)
assume 2: "F"
hence 3: "F ∨ ¬F" by (rule disjI1)
show False using 1 3 by (rule notE)
qed
qed
qed
qed

-- "La demostración estructurada es"
lemma ejemplo_22_2:
"F ∨ ¬F"
proof (rule ccontr)
assume "¬(F ∨ ¬F)"
thus False
proof (rule notE)
show "F ∨ ¬F"
proof (rule disjI2)
show "¬F"
proof (rule notI)
assume "F"
hence "F ∨ ¬F" ..
with `¬(F ∨ ¬F)`show False ..
qed
qed
qed
qed

-- "La demostración automática es"
lemma ejemplo_22_3:
"F ∨ ¬F"
using assms
by auto

text {*
Ejemplo 23 (p. 24). Demostrar
p ⟶ q ⊢ ¬p ∨ q
*}

-- "La demostración detallada es"
lemma ejemplo_23_1:
assumes 1: "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 2: "p"
have "q" using 1 2 by (rule mp)
thus "¬p ∨ q" by (rule disjI2) }
qed
qed

-- "La demostración estructurada es"
lemma ejemplo_23_2:
assumes "p ⟶ q"
shows "¬p ∨ q"
proof -
have "¬p ∨ p" ..
thus "¬p ∨ q"
proof
{ assume "¬p"
thus "¬p ∨ q" .. }
next
{ assume "p"
with assms have "q" ..
thus "¬p ∨ q" .. }
qed
qed

-- "La demostración automática es"
lemma ejemplo_23_3:
assumes "p ⟶ q"
shows "¬p ∨ q"
using assms
by auto

subsection {* Demostraciones por contradicción *}

text {*
Ejemplo 24. Demostrar que
¬p, p ∨ q ⊢ q
*}

-- "La demostración detallada es"
lemma ejemplo_24_1:
assumes "¬p"
"p ∨ q"
shows "q"
using `p ∨ q`
proof (rule disjE)
assume "p"
with assms(1) show "q" by contradiction
next
assume "q"
thus "q" by assumption
qed

-- "La demostración estructurada es"
lemma ejemplo_24_2:
assumes "¬p"
"p ∨ q"
shows "q"
using `p ∨ q`
proof
assume "p"
with assms(1) show "q" ..
next
assume "q"
thus "q" .
qed

-- "La demostración automática es"
lemma ejemplo_24_3:
assumes "¬p"
"p ∨ q"
shows "q"
using assms
by auto

end
</source>

Tema 3: Razonamiento estructurado sobre programas en Isabelle/HOL

2018-07-16T11:11:23Z

WikiSysop: Texto reemplazado: «isar» por «isabelle»

<source lang="isabelle">
chapter {* Tema 3: Razonamiento estructurado sobre programas *}

theory T3_Razonamiento_sobre_programas
imports Main
begin

text {*
En este tema se demuestra con Isabelle las propiedades de los
programas funcionales como se expone en el tema 2a y se demostraron
automáticamente en el tema 2b. A diferencia del tema 2b, ahora
nos fijamos no sólo en el método de demostración sino en la estructura
de la prueba resaltando su semejanza con las del tema 2a. *}

section {* Razonamiento ecuacional *}

text {* ----------------------------------------------------------------
Ejemplo 1. Definir, por recursión, la función
longitud :: 'a list ⇒ nat
tal que (longitud xs) es la longitud de la listas xs. Por ejemplo,
longitud [a,c,d] = 3
------------------------------------------------------------------- *}

fun longitud :: "'a list ⇒ nat" where
"longitud [] = 0"
| "longitud (x#xs) = 1 + longitud xs"

value "longitud [a,c,d]" -- "= 3"

text {* ---------------------------------------------------------------
Ejemplo 2. Demostrar que
longitud [a,c,d] = 3
------------------------------------------------------------------- *}

lemma "longitud [a,c,d] = 3"
by simp

text {* ---------------------------------------------------------------
Ejemplo 3. Definir la función
fun intercambia :: 'a × 'b ⇒ 'b × 'a
tal que (intercambia p) es el par obtenido intercambiando las
componentes del par p. Por ejemplo,
intercambia (u,v) = (v,u)
------------------------------------------------------------------ *}

fun intercambia :: "'a × 'b ⇒ 'b × 'a" where
"intercambia (x,y) = (y,x)"

value "intercambia (u,v)" -- "= (v,u)"

text {*
La definición de la función intercambia genera una regla de
simplificación
· intercambia.simps: intercambia (x,y) = (y,x)

Se puede ver con
· thm intercambia.simps
*}

text {* ---------------------------------------------------------------
Ejemplo 4. (p.6) Demostrar que
intercambia (intercambia (x,y)) = (x,y)
------------------------------------------------------------------- *}

(* Demostración aplicativa *)
lemma "intercambia (intercambia (x,y)) = (x,y)"
apply (simp only: intercambia.simps)
done

(* Demostración declarativa *)
lemma "intercambia (intercambia (x,y)) = (x,y)"
proof -
have "intercambia (intercambia (x,y)) = intercambia (y,x)"
by (simp only: intercambia.simps)
also have "... = (x,y)"
by (simp only: intercambia.simps)
finally show "intercambia (intercambia (x,y)) = (x,y)" by simp
qed

text {*
Notas sobre el lenguaje: En la demostración anterior se ha usado
· "proof" para iniciar la prueba,
· "-" (después de "proof") para no usar el método por defecto,
· "have" para establecer un paso,
· "by (simp only: intercambia.simps)" para indicar que sólo se usa
como regla de escritura la correspondiente a la definición de
intercambia,
· "also" para encadenar pasos ecuacionales,
· "..." para representar la igualdad anterior en un razonamiento
ecuacional,
· "finally" para indicar el último pasa de un razonamiento ecuacional,
· "show" para establecer la conclusión.
· "by simp" para indicar el método de demostración por simplificación y
· "qed" para terminar la pruebas,
*}

(* Demostración declarativa simplificada *)
lemma "intercambia (intercambia (x,y)) = (x,y)"
proof -
have "intercambia (intercambia (x,y)) = intercambia (y,x)" by simp
also have "... = (x,y)" by simp
finally show "intercambia (intercambia (x,y)) = (x,y)" by simp
qed

text {*
Nota: La diferencia entre las dos demostraciones es que en los dos
primeros pasos no se explicita la regla de simplificación.
*}

-- "La demostración automática es"
lemma "intercambia (intercambia (x,y)) = (x,y)"
by simp

text {* ---------------------------------------------------------------
Ejemplo 5. Definir, por recursión, la función
inversa :: 'a list ⇒ 'a list
tal que (inversa xs) es la lista obtenida invirtiendo el orden de los
elementos de xs. Por ejemplo,
inversa [a,d,c] = [c,d,a]
------------------------------------------------------------------ *}

fun inversa :: "'a list ⇒ 'a list" where
"inversa [] = []"
| "inversa (x#xs) = inversa xs @ [x]"

value "inversa [a,d,c]" -- "= [c,d,a]"

text {* ---------------------------------------------------------------
Ejemplo 6. (p. 9) Demostrar que
inversa [x] = [x]
------------------------------------------------------------------- *}

-- "La demostración aplicativa es"
lemma "inversa [x] = [x]"
apply simp
done

text {*
En la demostración anterior se usaron las siguientes reglas:
· inversa.simps(1): inversa [] = []
· inversa.simps(2): inversa (x#xs) = inversa xs @ [x]
· append_Nil: [] @ ys = ys
Vamos a explicitar su aplicación.
*}

-- "La demostración aplicativa detallada es"
lemma "inversa [x] = [x]"
apply (simp only: inversa.simps(2))
apply (simp only: inversa.simps(1))
apply (simp only: append_Nil)
done

-- "La demostración declarativa es"
lemma "inversa [x] = [x]"
proof -
have "inversa [x] = inversa (x#[])" by simp
also have "... = (inversa []) @ [x]" by (simp only: inversa.simps(2))
also have "... = [] @ [x]" by (simp only: inversa.simps(1))
also have "... = [x]" by (simp only: append_Nil)
finally show "inversa [x] = [x]" by simp
qed

-- "La demostración declarativa simplificada es"
lemma "inversa [x] = [x]"
proof -
have "inversa [x] = inversa (x#[])" by simp
also have "... = (inversa []) @ [x]" by simp
also have "... = [] @ [x]" by simp
also have "... = [x]" by simp
finally show "inversa [x] = [x]" by simp
qed

-- "La demostración automática es"
lemma "inversa [x] = [x]"
by simp

section {* Razonamiento por inducción sobre los naturales *}

text {*
[Principio de inducción sobre los naturales] Para demostrar una
propiedad P para todos los números naturales basta probar que el 0
tiene la propiedad P y que si n tiene la propiedad P, entonces n+1
también la tiene.
⟦P 0; ⋀n. P n ⟹ P (Suc n)⟧ ⟹ P m

En Isabelle el principio de inducción sobre los naturales está
formalizado en el teorema nat.induct y puede verse con
thm nat.induct
*}

text {* ---------------------------------------------------------------
Ejemplo 7. Definir la función
repite :: nat ⇒ 'a ⇒ 'a list
tal que (repite n x) es la lista formada por n copias del elemento
x. Por ejemplo,
repite 3 a = [a,a,a]
------------------------------------------------------------------ *}

fun repite :: "nat ⇒ 'a ⇒ 'a list" where
"repite 0 x = []"
| "repite (Suc n) x = x # (repite n x)"

value "repite 3 a" -- "= [a,a,a]"

text {* ---------------------------------------------------------------
Ejemplo 8. (p. 18) Demostrar que
longitud (repite n x) = n
------------------------------------------------------------------- *}

-- "La demostración aplicativa es"
lemma "longitud (repite n x) = n"
apply (induct n)
apply auto
done

-- "La demostración estructurada es"
lemma "longitud (repite n x) = n"
proof (induct n)
show "longitud (repite 0 x) = 0" by simp
next
fix n
assume HI: "longitud (repite n x) = n"
have "longitud (repite (Suc n) x) = longitud (x # (repite n x))"
by simp
also have "... = 1 + longitud (repite n x)" by simp
also have "... = 1 + n" using HI by simp
finally show "longitud (repite (Suc n) x) = Suc n" by simp
qed

text {*
Comentarios sobre la demostración anterior:
· A la derecha de proof se indica el método de la demostración.
· (induct n) indica que la demostración se hará por inducción en n.
· Se generan dos subobjetivos correspondientes a la base y el paso de
inducción:
1. longitud (repite 0 x) = 0
2. ⋀n. longitud (repite n x) = n ⟹ longitud (repite (Suc n) x) = Suc n
donde ⋀n se lee "para todo n".
· "next" indica el siguiente subobjetivo.
· "fix n" indica "sea n un número natural cualquiera"
· assume HI: "longitud (repite n x) = n" indica «supongamos que
"longitud (repite n x) = n" y sea HI la etiqueta de este supuesto».
· "using HI" usando la propiedad etiquetada con HI.
*}

-- "La demostración automática es"
lemma "longitud (repite n x) = n"
by (induct n) auto

section {* Razonamiento por inducción sobre listas *}

text {*
Para demostrar una propiedad para todas las listas basta demostrar
que la lista vacía tiene la propiedad y que al añadir un elemento a una
lista que tiene la propiedad se obtiene otra lista que también tiene la
propiedad.

En Isabelle el principio de inducción sobre listas está formalizado
mediante el teorema list.induct
⟦P [];
⋀x xs. P xs ⟹ P (x#xs)⟧
⟹ P xs
*}

text {* ---------------------------------------------------------------
Ejemplo 9. Definir la función
conc :: 'a list ⇒ 'a list ⇒ 'a list
tal que (conc xs ys) es la concatención de las listas xs e ys. Por
ejemplo,
conc [a,d] [b,d,a,c] = [a,d,b,d,a,c]
------------------------------------------------------------------ *}

fun conc :: "'a list ⇒ 'a list ⇒ 'a list" where
"conc [] ys = ys"
| "conc (x#xs) ys = x # (conc xs ys)"

value "conc [a,d] [b,d,a,c]" -- "= [a,d,b,d,a,c]"

text {* ---------------------------------------------------------------
Ejemplo 10. (p. 24) Demostrar que
conc xs (conc ys zs) = (conc xs ys) zs
------------------------------------------------------------------- *}

-- "La demostración estructurada es"
lemma "conc xs (conc ys zs) = conc (conc xs ys) zs"
proof (induct xs)
show "conc [] (conc ys zs) = conc (conc [] ys) zs" by simp
next
fix x xs
assume HI: "conc xs (conc ys zs) = conc (conc xs ys) zs"
have "conc (x # xs) (conc ys zs) = x # (conc xs (conc ys zs))" by simp
also have "... = x # (conc (conc xs ys) zs)" using HI by simp
also have "... = conc (conc (x # xs) ys) zs" by simp
finally show "conc (x # xs) (conc ys zs) = conc (conc (x # xs) ys) zs" by simp
qed

text {*
Comentario sobre la demostración anterior
· (induct xs) genera dos subobjetivos:
1. conc [] (conc ys zs) = conc (conc [] ys) zs
2. ⋀a xs. conc xs (conc ys zs) = conc (conc xs ys) zs ⟹
conc (a#xs) (conc ys zs) = conc (conc (a#xs) ys) zs
*}

-- "La demostración automática es"
lemma "conc xs (conc ys zs) = conc (conc xs ys) zs"
by (induct xs) auto

text {* ---------------------------------------------------------------
Ejemplo 11. Refutar que
conc xs ys = conc ys xs
------------------------------------------------------------------- *}

lemma "conc xs ys = conc ys xs"
quickcheck
oops

text {* Encuentra el contraejemplo,
xs = [a2]
ys = [a1] *}

text {* ---------------------------------------------------------------
Ejemplo 12. (p. 28) Demostrar que
conc xs [] = xs
------------------------------------------------------------------- *}

-- "La demostración estructurada es"
lemma "conc xs [] = xs"
proof (induct xs)
show "conc [] [] = []" by simp
next
fix x xs
assume HI: "conc xs [] = xs"
have "conc (x # xs) [] = x # (conc xs [])" by simp
also have "... = x # xs" using HI by simp
finally show "conc (x # xs) [] = x # xs" by simp
qed

-- "La demostración automática es"
lemma "conc xs [] = xs"
by (induct xs) auto

text {* ---------------------------------------------------------------
Ejemplo 13. (p. 30) Demostrar que
longitud (conc xs ys) = longitud xs + longitud ys
------------------------------------------------------------------- *}

-- "La demostración automática es"
lemma "longitud (conc xs ys) = longitud xs + longitud ys"
proof (induct xs)
show "longitud (conc [] ys) = longitud [] + longitud ys" by simp
next
fix x xs
assume HI: "longitud (conc xs ys) = longitud xs + longitud ys"
have "longitud (conc (x # xs) ys) = longitud (x # (conc xs ys))"
by simp
also have "... = 1 + longitud (conc xs ys)" by simp
also have "... = 1 + longitud xs + longitud ys" using HI by simp
also have "... = longitud (x # xs) + longitud ys" by simp
finally show "longitud (conc (x # xs) ys) =
longitud (x # xs) + longitud ys" by simp
qed

-- "La demostración automática es"
lemma "longitud (conc xs ys) = longitud xs + longitud ys"
by (induct xs) auto

section {* Inducción correspondiente a la definición recursiva *}

text {* ---------------------------------------------------------------
Ejemplo 14. Definir la función
coge :: nat ⇒ 'a list ⇒ 'a list
tal que (coge n xs) es la lista de los n primeros elementos de xs. Por
ejemplo,
coge 2 [a,c,d,b,e] = [a,c]
------------------------------------------------------------------ *}

fun coge :: "nat ⇒ 'a list ⇒ 'a list" where
"coge n [] = []"
| "coge 0 xs = []"
| "coge (Suc n) (x#xs) = x # (coge n xs)"

value "coge 2 [a,c,d,b,e]" -- "= [a,c]"

text {* ---------------------------------------------------------------
Ejemplo 15. Definir la función
elimina :: nat ⇒ 'a list ⇒ 'a list
tal que (elimina n xs) es la lista obtenida eliminando los n primeros
elementos de xs. Por ejemplo,
elimina 2 [a,c,d,b,e] = [d,b,e]
------------------------------------------------------------------ *}

fun elimina :: "nat ⇒ 'a list ⇒ 'a list" where
"elimina n [] = []"
| "elimina 0 xs = xs"
| "elimina (Suc n) (x#xs) = elimina n xs"

value "elimina 2 [a,c,d,b,e]" -- "= [d,b,e]"

text {*
La definición coge genera el esquema de inducción coge.induct:
⟦⋀n. P n [];
⋀x xs. P 0 (x#xs);
⋀n x xs. P n xs ⟹ P (Suc n) (x#xs)⟧
⟹ P n x

Puede verse usando "thm coge.induct". *}

text {* ---------------------------------------------------------------
Ejemplo 16. (p. 35) Demostrar que
conc (coge n xs) (elimina n xs) = xs
------------------------------------------------------------------- *}

-- "La demostración estructurada es"
lemma "conc (coge n xs) (elimina n xs) = xs"
proof (induct rule: coge.induct)
fix n
show "conc (coge n []) (elimina n []) = []" by simp
next
fix x xs
show "conc (coge 0 (x#xs)) (elimina 0 (x#xs)) = x#xs" by simp
next
fix n x xs
assume HI: "conc (coge n xs) (elimina n xs) = xs"
have "conc (coge (Suc n) (x#xs)) (elimina (Suc n) (x#xs)) =
conc (x#(coge n xs)) (elimina n xs)" by simp
also have "... = x#(conc (coge n xs) (elimina n xs))" by simp
also have "... = x#xs" using HI by simp
finally show "conc (coge (Suc n) (x#xs)) (elimina (Suc n) (x#xs)) =
x#xs"
by simp
qed

text {*
Comentario sobre la demostración anterior:
· (induct rule: coge.induct) indica que el método de demostración es
por el esquema de inducción correspondiente a la definición de la
función coge.
· Se generan 3 subobjetivos:
· 1. ⋀n. conc (coge n []) (elimina n []) = []
· 2. ⋀x xs. conc (coge 0 (x#xs)) (elimina 0 (x#xs)) = x#xs
· 3. ⋀n x xs.
conc (coge n xs) (elimina n xs) = xs ⟹
conc (coge (Suc n) (x#xs)) (elimina (Suc n) (x#xs)) = x#xs
*}

-- "La demostración automática es"
lemma "conc (coge n xs) (elimina n xs) = xs"
by (induct rule: coge.induct) auto

section {* Razonamiento por casos *}

text {* ---------------------------------------------------------------
Ejemplo 17. Definir la función
esVacia :: 'a list ⇒ bool
tal que (esVacia xs) se verifica si xs es la lista vacía. Por ejemplo,
esVacia [] = True
esVacia [1] = False
------------------------------------------------------------------ *}

fun esVacia :: "'a list ⇒ bool" where
"esVacia [] = True"
| "esVacia (x#xs) = False"

value "esVacia []" -- "= True"
value "esVacia [1]" -- "= False"

text {* ---------------------------------------------------------------
Ejemplo 18 (p. 39) . Demostrar que
esVacia xs = esVacia (conc xs xs)
------------------------------------------------------------------- *}

-- "La demostración estructurada es"
lemma "esVacia xs = esVacia (conc xs xs)"
proof (cases xs)
assume "xs = []"
then show "esVacia xs = esVacia (conc xs xs)" by simp
next
fix y ys
assume "xs = y#ys"
then show "esVacia xs = esVacia (conc xs xs)" by simp
qed

text {*
Comentarios sobre la demostración anterior:
· "(cases xs)" es el método de demostración por casos según xs.
· Se generan dos subobjetivos correspondientes a los dos
constructores de listas:
· 1. xs = [] ⟹ esVacia xs = esVacia (conc xs xs)
· 2. ⋀y ys. xs = y#ys ⟹ esVacia xs = esVacia (conc xs xs)
· "then" indica "usando la propiedad anterior"
*}

-- "La demostración estructurada simplificada es"
lemma "esVacia xs = esVacia (conc xs xs)"
proof (cases xs)
case Nil
then show "esVacia xs = esVacia (conc xs xs)" by simp
next
case Cons
then show "esVacia xs = esVacia (conc xs xs)" by simp
qed

text {*
Comentarios sobre la demostración anterior:
· "case Nil" es una abreviatura de "assume xs = []"
· "case Cons" es una abreviatura de "fix y ys assume xs = y#ys"
· "thus" es una abreviatura de "then show".
*}

-- "La demostración automática es"
lemma "esVacia xs = esVacia (conc xs xs)"
by (cases xs) auto

section {* Heurística de generalización *}

text {*
Heurística de generalización: Cuando se use demostración estructural,
cuantificar universalmente las variables libres (o, equivalentemente,
considerar las variables libres como variables arbitrarias). *}

text {* ---------------------------------------------------------------
Ejemplo 19. Definir la función
inversaAc :: 'a list ⇒ 'a list
tal que (inversaAc xs) es a inversa de xs calculada usando
acumuladores. Por ejemplo,
inversaAc [a,c,b,e] = [e,b,c,a]
------------------------------------------------------------------ *}

fun inversaAcAux :: "'a list ⇒ 'a list ⇒ 'a list" where
"inversaAcAux [] ys = ys"
| "inversaAcAux (x#xs) ys = inversaAcAux xs (x#ys)"

fun inversaAc :: "'a list ⇒ 'a list" where
"inversaAc xs = inversaAcAux xs []"

value "inversaAc [a,c,b,e]" -- "= [e,b,c,a]"

text {* ---------------------------------------------------------------
Ejemplo 20. (p. 44) Demostrar que
inversaAcAux xs ys = (inversa xs) @ ys
------------------------------------------------------------------- *}

-- "La demostración estructurada es"
lemma inversaAcAux_es_inversa:
"inversaAcAux xs ys = (inversa xs) @ ys"
proof (induct xs arbitrary: ys)
show "⋀ys. inversaAcAux [] ys = inversa [] @ ys" by simp
next
fix a xs
assume HI: "⋀ys. inversaAcAux xs ys = inversa xs@ys"
show "⋀ys. inversaAcAux (a#xs) ys = inversa (a#xs)@ys"
proof -
fix ys
have "inversaAcAux (a#xs) ys = inversaAcAux xs (a#ys)" by simp
also have "… = inversa xs@(a#ys)" using HI by simp
also have "… = inversa (a#xs)@ys" by simp
finally show "inversaAcAux (a#xs) ys = inversa (a#xs)@ys" by simp
qed
qed

text {*
Comentarios sobre la demostración anterior:
· "(induct xs arbitrary: ys)" es el método de demostración por
inducción sobre xs usando ys como variable arbitraria.
· Se generan dos subobjetivos:
· 1. ⋀ys. inversaAcAux [] ys = inversa [] @ ys
· 2. ⋀a xs ys. (⋀ys. inversaAcAux xs ys = inversa xs @ ys) ⟹
inversaAcAux (a # xs) ys = inversa (a # xs) @ ys
· Dentro de una demostración se pueden incluir otras demostraciones.
· Para demostrar la propiedad universal "⋀ys. P(ys)" se elige una
lista arbitraria (con "fix ys") y se demuestra "P(ys)".
*}

-- "La demostración automática es"
lemma "inversaAcAux xs ys = (inversa xs)@ys"
by (induct xs arbitrary: ys) auto

text {* ---------------------------------------------------------------
Ejemplo 21. (p. 43) Demostrar que
inversaAc xs = inversa xs
------------------------------------------------------------------- *}

-- "La demostración automática es"
corollary "inversaAc xs = inversa xs"
by (simp add: inversaAcAux_es_inversa)

text {*
Comentario de la demostración anterior:
· "(simp add: inversaAcAux_es_inversa)" es el método de demostración
por simplificación usando como regla de simplificación la propiedad
inversaAcAux_es_inversa.
*}

section {* Demostración por inducción para funciones de orden superior *}

text {* ---------------------------------------------------------------
Ejemplo 22. Definir la función
sum :: nat list ⇒ nat
tal que (sum xs) es la suma de los elementos de xs. Por ejemplo,
sum [3,2,5] = 10
------------------------------------------------------------------ *}

fun sum :: "nat list ⇒ nat" where
"sum [] = 0"
| "sum (x#xs) = x + sum xs"

value "sum [3,2,5]" -- "= 10"

text {* ---------------------------------------------------------------
Ejemplo 23. Definir la función
map :: ('a ⇒ 'b) ⇒ 'a list ⇒ 'b list
tal que (map f xs) es la lista obtenida aplicando la función f a los
elementos de xs. Por ejemplo,
map (λx. 2*x) [3,2,5] = [6,4,10]
------------------------------------------------------------------ *}

fun map :: "('a ⇒ 'b) ⇒ 'a list ⇒ 'b list" where
"map f [] = []"
| "map f (x#xs) = (f x) # map f xs"

value "map (λx. 2*x) [3::nat,2,5]" -- "= [6,4,10]"

text {* ---------------------------------------------------------------
Ejemplo 24. (p. 45) Demostrar que
sum (map (λx. 2*x) xs) = 2 * (sum xs)
------------------------------------------------------------------- *}

-- "La demostración estructurada es"
lemma "sum (map (λx. 2*x) xs) = 2 * (sum xs)"
proof (induct xs)
show "sum (map (λx. 2*x) []) = 2 * (sum [])" by simp
next
fix a xs
assume HI: "sum (map (λx. 2*x) xs) = 2 * (sum xs)"
have "sum (map (λx. 2*x) (a#xs)) = sum ((2*a)#(map (λx. 2*x) xs))"
by simp
also have "... = 2*a + sum (map (λx. 2*x) xs)" by simp
also have "... = 2*a + 2*(sum xs)" using HI by simp
also have "... = 2*(a + sum xs)" by simp
also have "... = 2*(sum (a#xs))" by simp
finally show "sum (map (λx. 2*x) (a#xs)) = 2*(sum (a#xs))" by simp
qed

-- "La demostración automática es"
lemma "sum (map (λx. 2*x) xs) = 2 * (sum xs)"
by (induct xs) auto

text {* ---------------------------------------------------------------
Ejemplo 25. (p. 48) Demostrar que
longitud (map f xs) = longitud xs
------------------------------------------------------------------- *}

-- "La demostración estructurada es"
lemma "longitud (map f xs) = longitud xs"
proof (induct xs)
show "longitud (map f []) = longitud []" by simp
next
fix a xs
assume HI: "longitud (map f xs) = longitud xs"
have "longitud (map f (a#xs)) = longitud (f a # (map f xs))" by simp
also have "... = 1 + longitud (map f xs)" by simp
also have "... = 1 + longitud xs" using HI by simp
also have "... = longitud (a#xs)" by simp
finally show "longitud (map f (a#xs)) = longitud (a#xs)" by simp
qed

-- "La demostración automática es"
lemma "longitud (map f xs) = longitud xs"
by (induct xs) auto

section {* Referencias *}

text {*
· J.A. Alonso. "Razonamiento sobre programas" http://goo.gl/R06O3
· G. Hutton. "Programming in Haskell". Cap. 13 "Reasoning about
programms".
· S. Thompson. "Haskell: the Craft of Functional Programming, 3rd
Edition. Cap. 8 "Reasoning about programms".
· L. Paulson. "ML for the Working Programmer, 2nd Edition". Cap. 6.
"Reasoning about functional programs".
*}

end
</source>

Tema 4: Razonamiento por casos y por inducción

2018-07-16T11:11:23Z

WikiSysop: Texto reemplazado: «isar» por «isabelle»

<source lang="isabelle">
chapter {* Tema 4: Razonamiento por casos y por inducción *}

theory T4_Razonamiento_por_casos_y_por_induccion
imports Main Parity
begin

text {*
En este tema se amplían los métodos de demostración por casos y por
inducción iniciados en el tema anterior.
*}

section {* Razonamiento por distinción de casos *}

subsection {* Distinción de casos booleanos *}

text {*
Ejemplo de demostración por distinción de casos booleanos:
Demostrar "¬A ∨ A".
*}

-- "La demostración estructurada es"
lemma "¬A ∨ A"
proof cases
assume "A"
then show "¬A ∨ A" ..
next
assume "¬A"
then show "¬A ∨ A" ..
qed

text {*
Comentarios de la demostración anterior:
· "proof cases" indica que el método de demostración será por
distinción de casos.
· Se generan 2 casos:
1. ?P ⟹ ¬A ∨ A
2. ¬?P ⟹ ¬A ∨ A
donde ?P es una variable sobre las fórmulas.
· (assume "A") indica que se está usando "A" en lugar de la variable
?P.
· "then" indica usando la fórmula anterior.
· ".." indica usando la regla lógica necesaria (las reglas lógicas se
estudiarán en los siguientes temas).
· "next" indica el siguiente caso (se puede observar cómo ha
sustituido ¬?P por ¬A.
*}

-- "La demostración automática es"
lemma "¬A ∨ A"
by auto

text {*
Ejemplo de demostración por distinción de casos booleanos con nombres:
Demostrar "¬A ∨ A".
*}

-- "La demostración estructurada es"
lemma "¬A ∨ A"
proof (cases "A")
case True
then show "¬A ∨ A" ..
next
case False
thus "¬A ∨ A" ..
qed

text {*
Comentarios sobre la demostración anterior:
· (cases "A") indica que la demostración se hará por casos según los
distintos valores de "A".
· Como "A" es una fórmula, sus posibles valores son verdadero o falso.
· "case True" indica que se está suponiendo que A es verdadera. Es
equivalente a "assume A".
· "case False" indica que se está suponiendo que A es falsa. Es
equivalente a "assume ¬A".
· En general,
· el método (cases F) es una abreviatura de la aplicación de la regla
⟦F ⟹ Q; ¬F ⟹ Q⟧ ⟹ Q
· La expresión "case True" es una abreviatura de F.
· La expresión "case False" es una abreviatura de ¬F.
· Ventajas de "cases" con nombre:
· reduce la escritura de la fórmula y
· es independiente del orden de los casos.
*}

subsection {* Distinción de casos sobre otros tipos de datos *}

text {*
Ejemplo de distinción de casos sobre listas:
Demostrar que la longitud del resto de una lista es la longitud de la
lista menos 1.
*}

-- "La demostración detallada es"
lemma "length (tl xs) = length xs - 1"
proof (cases xs)
assume "xs = []"
then show "length (tl xs) = length xs - 1" by simp
next
fix y ys
assume "xs = y#ys"
then show "length(tl xs) = length xs - 1" by simp
qed

text {*
Comentarios sobre la demostración anterior:
· "(cases xs)" indica que la demostración se hará por casos sobre los
posibles valores de xs.
· Como xs es una lista, sus posibles valores son la lista vacía ([]) o
una lista no vacía (de la forma (y#ys)).
· Se generan 2 casos:
1. xs = [] ⟹ length (tl xs) = length xs - 1
2. ⋀a list. xs = a # list ⟹ length (tl xs) = length xs - 1
*}

-- "La demostración simplificada es"
lemma "length (tl xs) = length xs - 1"
proof (cases xs)
case Nil
then show ?thesis by simp
next
case Cons
then show ?thesis by simp
qed

text {*
Comentarios sobre la dmostración anterior:
· "case Nil" es una abreviatura de
"assume xs =[]".
· "case Cons" es una abreviatura de
"fix y ys assume xs = y#ys"
· ?thesis es una abreviatura de la conclusión del lema.
*}

-- "La demostración automática es"
lemma "length (tl xs) = length xs - 1"
by auto

text {*
Een el siguiente ejemplo vamos a demostrar una propiedad de la función
drop que está definida en la teoría List de forma que (drop n xs) la
lista obtenida eliminando en xs} los n primeros elementos. Su
definición es la siguiente
drop_Nil: "drop n [] = []"
drop_Cons: "drop n (x#xs) = (case n of
0 => x#xs |
Suc(m) => drop m xs)"
*}

text {*
Ejemplo de análisis de casos:
Demostrar que el resultado de eliminar los n+1 primeros elementos de
xs es el mismo que eliminar los n primeros elementos del resto de xs.
*}

-- "La demostración detallada es"
lemma "drop (n + 1) xs = drop n (tl xs)"
proof (cases xs)
case Nil
then show ?thesis by simp
next
case Cons
then show ?thesis by simp
qed

-- "La demostración automática es"
lemma "drop (n + 1) xs = drop n (tl xs)"
by (cases xs) auto

section {* Inducción matemática *}

text {*
[Principio de inducción matemática]
Para demostrar una propiedad P para todos los números naturales basta
probar que el 0 tiene la propiedad P y que si n tiene la propiedad P,
entonces n+1 también la tiene.
⟦P 0; ⋀n. P n ⟹ P (Suc n)⟧ ⟹ P m

En Isabelle el principio de inducción matemática está formalizado en
el teorema nat.induct y puede verse con
thm nat.induct
*}

text {*
Ejemplo de demostración por inducción: Usaremos el principio de
inducción matemática para demostrar que
1 + 3 + ... + (2n-1) = n^2

Definición. [Suma de los primeros impares]
(suma_impares n) la suma de los n números impares. Por ejemplo,
suma_impares 3 = 9
*}

fun suma_impares :: "nat ⇒ nat" where
"suma_impares 0 = 0"
| "suma_impares (Suc n) = (2*(Suc n) - 1) + suma_impares n"

value "suma_impares 3"

text {*
Ejemplo de demostración por inducción matemática:
Demostrar que la suma de los n primeros números impares es n^2.
*}

-- "Demostración del lema anterior por inducción y razonamiento ecuacional"
lemma "suma_impares n = n * n"
proof (induct n)
show "suma_impares 0 = 0 * 0" by simp
next
fix n assume HI: "suma_impares n = n * n"
have "suma_impares (Suc n) = (2 * (Suc n) - 1) + suma_impares n"
by simp
also have "… = (2 * (Suc n) - 1) + n * n" using HI by simp
also have "… = n * n + 2 * n + 1" by simp
finally show "suma_impares (Suc n) = (Suc n) * (Suc n)" by simp
qed

-- "Demostración del lema anterior con patrones y razonamiento ecuacional"
lemma "suma_impares n = n * n" (is "?P n")
proof (induct n)
show "?P 0" by simp
next
fix n
assume HI: "?P n"
have "suma_impares (Suc n) = (2 * (Suc n) - 1) + suma_impares n"
by simp
also have "… = (2 * (Suc n) - 1) + n * n" using HI by simp
also have "… = n * n + 2 * n + 1" by simp
finally show "?P (Suc n)" by simp
qed

text {*
Comentario sobre la demostración anterior:
· Con la expresión
"suma_impares n = n * n" (is "?P n")
se abrevia "suma_impares n = n * n" como "?P n". Por tanto,
"?P 0" es una abreviatura de "suma_impares 0 = 0 * 0"
"?P (Suc n)" es una abreviatura de "suma_impares (Suc n) = (Suc n) * (Suc n)"
· En general, cualquier fórmula seguida de (is patrón) equipara el
patrón con la fórmula.
*}

-- "La demostración usando patrones es"
lemma "suma_impares n = n * n" (is "?P n")
proof (induct n)
show "?P 0" by simp
next
fix n
assume "?P n"
then show "?P (Suc n)" by simp
qed

-- "La demostración automática es"
lemma "suma_impares n = n * n"
by (induct n) auto

text {*
Ejemplo de definición con existenciales.
Un número natural n es par si existe un natural m tal que n=m+m.
*}

definition par :: "nat ⇒ bool" where
"par n ≡ ∃m. n=m+m"

text {*
Ejemplo de inducción y existenciales:
Demostrar que para todo número natural n, se verifica que n*(n+1) par.
*}

-- "Demostración detallada por inducción"
lemma
fixes n :: "nat"
shows "par (n*(n+1))"
proof (induct n)
show "par (0*(0+1))" by (simp add: par_def)
next
fix n
assume "par (n*(n+1))"
then have "∃m. n*(n+1) = m+m" by (simp add:par_def)
then obtain m where m: "n*(n+1) = m+m" ..
then have "(Suc n)*((Suc n)+1) = (m+n+1)+(m+n+1)" by auto
then have "∃m. (Suc n)*((Suc n)+1) = m+m" ..
then show "par ((Suc n)*((Suc n)+1))" by (simp add:par_def)
qed

text {*
Comentarios sobre la demostración anterior:
· (fixes n :: "nat") es una abreviatura de "sea n un número natural".
*}

text {*
En Isabelle puede demostrarse de manera más simple un lema equivalente
usando en lugar de la función "par" la función "even" definida en la
teoría Parity por
even x ⟷ x mod 2 = 0"
*}

lemma
fixes n :: "nat"
shows "even (n*(n+1))"
by auto

text {*
Comentarios sobre la demostración anterior:
· Para poder usar la función "even" de la librería Parity es necesario
importar dicha librería. Por ello, antes del inicio de la teoría
aparece
imports Main Parity
*}

text {*
Para completar la demostración basta demostrar la equivalencia de las
funciones "par" y "even".
*}

lemma
fixes n :: "nat"
shows "par n = even n"
proof -
have "par n = (∃m. n = m+m)" by (simp add:par_def)
then show "par n = even n" by presburger
qed

text {*
Comentarios sobre la demostración anterior:
· "by presburger" indica que se use como método de demostración el
algoritmo de decisión de la aritmética de Presburger.
*}

section {* Inducción estructural *}

text {*
Inducción estructural:
· En Isabelle puede hacerse inducción estructural sobre cualquier tipo
recursivo.
· La inducción matemática es la inducción estructural sobre el tipo de
los naturales.
· El esquema de inducción estructural sobre listas es
· list.induct: ⟦P []; ⋀x ys. P ys ⟹ P (x # ys)⟧ ⟹ P zs
· Para demostrar una propiedad para todas las listas basta demostrar
que la lista vacía tiene la propiedad y que al añadir un elemento a una
lista que tiene la propiedad se obtiene una lista que también tiene la
propiedad.
· En Isabelle el principio de inducción sobre listas está formalizado
mediante el teorema list.induct que puede verse con
thm list.induct
*}

text {*
Concatenación de listas:
En la teoría List.thy está definida la concatenación de listas (que
se representa por @) como sigue
append_Nil: "[]@ys = ys"
append_Cons: "(x#xs)@ys = x#(xs@ys)"
*}

text {*
Lema. [Ejemplo de inducción sobre listas]
Demostrar que la concatenación de listas es asociativa.
*}

-- "La demostración estructurada es"
lemma conc_asociativa: "xs @ (ys @ zs) = (xs @ ys) @ zs"
proof (induct xs)
show "[] @ (ys @ zs) = ([] @ ys) @ zs"
proof -
have "[] @ (ys @ zs) = ys @ zs" by simp
also have "… = ([] @ ys) @ zs" by simp
finally show ?thesis .
qed
next
fix x xs
assume HI: "xs @ (ys @ zs) = (xs @ ys) @ zs"
show "(x#xs) @ (ys @ zs) = ((x#xs) @ ys) @ zs"
proof -
have "(x#xs) @ (ys @ zs) = x#(xs @ (ys @ zs))" by simp
also have "… = x#((xs @ ys) @ zs)" using HI by simp
also have "… = (x#(xs @ ys)) @ zs" by simp
also have "… = ((x#xs) @ ys) @ zs" by simp
finally show ?thesis .
qed
qed

-- "La demostración automática es"
lemma conc_asociativa_1: "xs @ (ys @ zs) = (xs @ ys) @ zs"
by (induct xs) auto

section {* Heurísticas para la inducción *}

text {*
Definición. [Definición recursiva de inversa]
(inversa xs) la inversa de la lista xs. Por ejemplo,
inversa [a,b,c] = [c,b,a]
*}

fun inversa :: "'a list ⇒ 'a list" where
"inversa [] = []"
| "inversa (x#xs) = (inversa xs) @ [x]"

value "inversa [a,b,c]"

text {*
Definición. [Definición de inversa con acumuladores]
(inversaAc xs) es la inversa de la lista xs calculada con
acumuladores. Por ejemplo,
inversaAc [a,b,c] = [c,b,a]
inversaAcAux [a,b,c] [] = [c,b,a]
*}

fun inversaAcAux :: "'a list ⇒ 'a list ⇒ 'a list" where
"inversaAcAux [] ys = ys"
| "inversaAcAux (x#xs) ys = inversaAcAux xs (x#ys)"

definition inversaAc :: "'a list ⇒ 'a list" where
"inversaAc xs ≡ inversaAcAux xs []"

value "inversaAcAux [a,b,c] []"
value "inversaAc [a,b,c]"

text {*
Lema. [Ejemplo de equivalencia entre las definiciones]
La inversa de [a,b,c] es lo mismo calculada con la primera definición
que con la segunda.
*}

lemma "inversaAc [a,b,c] = inversa [a,b,c]"
by (simp add: inversaAc_def)

text {*
Nota. [Ejemplo fallido de demostración por inducción]
El siguiente intento de demostrar que para cualquier lista xs, se
tiene que "inversaAc xs = inversa xs" falla.
*}

lemma "inversaAc xs = inversa xs"
proof (induct xs)
show "inversaAc [] = inversa []" by (simp add: inversaAc_def)
next
fix a xs assume HI: "inversaAc xs = inversa xs"
have "inversaAc (a#xs) = inversaAcAux (a#xs) []"
by (simp add: inversaAc_def)
also have "… = inversaAcAux xs [a]" by simp
also have "… = inversa (a#xs)"
-- "Problema: la hipótesis de inducción no es aplicable."
oops

text {*
Nota. [Heurística de generalización]
Cuando se use demostración estructural, cuantificar universalmente las
variables libres (o, equivalentemente, considerar las variables libres
como variables arbitrarias).

Lema. [Lema con generalización]
Para toda lista ys se tiene
inversaAcAux xs ys = (inversa xs) @ ys
*}

-- "La demostración estructurada es"
lemma inversaAcAux_es_inversa:
"inversaAcAux xs ys = (inversa xs)@ys"
proof (induct xs arbitrary: ys)
show "⋀ys. inversaAcAux [] ys = (inversa [])@ys" by simp
next
fix a xs
assume HI: "⋀ys. inversaAcAux xs ys = inversa xs@ys"
show "⋀ys. inversaAcAux (a#xs) ys = inversa (a#xs)@ys"
proof -
fix ys
have "inversaAcAux (a#xs) ys = inversaAcAux xs (a#ys)" by simp
also have "… = inversa xs@(a#ys)" using HI by simp
also have "… = inversa (a#xs)@ys" using [[simp_trace]] by simp
finally show "inversaAcAux (a#xs) ys = inversa (a#xs)@ys" by simp
qed
qed

-- "La demostración automática es"
lemma inversaAcAux_es_inversa_1:
"inversaAcAux xs ys = (inversa xs)@ys"
by (induct xs arbitrary: ys) auto

text {*
Corolario. Para cualquier lista xs, se tiene que
inversaAc xs = inversa xs
*}

corollary "inversaAc xs = inversa xs"
by (simp add: inversaAcAux_es_inversa inversaAc_def)

text {*
Nota. En el paso "inversa xs@(a#ys) = inversa (a#xs)@ys" se usan
lemas de la teoría List. Se puede observar, insertano
using [[simp_trace]]
entre la igualdad y by simp, que los lemas usados son
· List.append_simps_1: []@ys = ys
· List.append_simps_2: (x#xs)@ys = x#(xs@ys)
· List.append_assoc: (xs @ ys) @ zs = xs @ (ys @ zs)
Las dos primeras son las ecuaciones de la definición de append.

En la siguiente demostración se detallan los lemas utilizados.
*}

lemma "(inversa xs)@(a#ys) = (inversa (a#xs))@ys"
proof -
have "(inversa xs)@(a#ys) = (inversa xs)@(a#([]@ys))"
by (simp only: append.simps(1))
also have "… = (inversa xs)@([a]@ys)" by (simp only: append.simps(2))
also have "… = ((inversa xs)@[a])@ys" by (simp only: append_assoc)
also have "… = (inversa (a#xs))@ys" by (simp only: inversa.simps(2))
finally show ?thesis .
qed

section {* Recursión general. La función de Ackermann *}

text {*
El objetivo de esta sección es mostrar el uso de las definiciones
recursivas generales y sus esquemas de inducción. Como ejemplo se usa la
función de Ackermann (se puede consultar información sobre dicha función en
http://en.wikipedia.org/wiki/Ackermann_function).

Definición. La función de Ackermann se define por
A(m,n) = n+1, si m=0,
A(m-1,1), si m>0 y n=0,
A(m-1,A(m,n-1)), si m>0 y n>0
para todo los números naturales.

La función de Ackermann es recursiva, pero no es primitiva recursiva.
*}

fun ack :: "nat ⇒ nat ⇒ nat" where
"ack 0 n = n+1"
| "ack (Suc m) 0 = ack m 1"
| "ack (Suc m) (Suc n) = ack m (ack (Suc m) n)"

-- "Ejemplo de evaluación"
value "ack 2 3" (* devuelve 9 *)

text {*
Esquema de inducción correspondiente a una función:
· Al definir una función recursiva general se genera una regla de
inducción. En la definición anterior, la regla generada es
ack.induct:
⟦⋀n. P 0 n;
⋀m. P m 1 ⟹ P (Suc m) 0;
⋀m n. ⟦P (Suc m) n; P m (ack (Suc m) n)⟧ ⟹ P (Suc m) (Suc n)⟧
⟹ P a b
*}

text {*
Ejemplo de demostración por la inducción correspondiente a una función:
Demostrar que para todos m y n, A(m,n) > n.
*}

-- "La demostración detallada es"
lemma "ack m n > n"
proof (induct m n rule: ack.induct)
fix n
show "ack 0 n > n" by simp
next
fix m
assume "ack m 1 > 1"
then show "ack (Suc m) 0 > 0" by simp
next
fix m n
assume "n < ack (Suc m) n" and
"ack (Suc m) n < ack m (ack (Suc m) n)"
then show "Suc n < ack (Suc m) (Suc n)" by simp
qed

text {*
Comentarios sobre la demostración anterior:
· (induct m n rule: ack.induct) indica que el método de demostración
es el esquema de recursión correspondiente a la definición de
(ack m n).
· Se generan 3 casos:
1. ⋀n. n < ack 0 n
2. ⋀m. 1 < ack m 1 ⟹ 0 < ack (Suc m) 0
3. ⋀m n. ⟦n < ack (Suc m) n;
ack (Suc m) n < ack m (ack (Suc m) n)⟧
⟹ Suc n < ack (Suc m) (Suc n)
*}

-- "La demostración automática es"
lemma "ack m n > n"
by (induct m n rule: ack.induct) auto

end
</source>

Tema 5: Razonamiento sobre árboles y bosques

2018-07-16T11:11:23Z

WikiSysop: Texto reemplazado: «isar» por «isabelle»

<source lang="isabelle">
chapter {* Tema 5: Razonamiento sobre árboles *}

theory T5_Razonamiento_sobre_arboles
imports Main Parity
begin

text {*
En este tema se estudia razonamiento sobre otras estructuras
recursivas como árboles binarios, árboles generales y bosques.

También se muestra cómo definir tipos de datos por recursión cruzada y
la demostración de sus propiedades por inducción.
*}

section {* Razonamiento sobre árboles binarios *}

text {*
Ejemplo de definición de tipos recursivos:
Definir un tipo de dato para los árboles binarios.
*}

datatype 'a arbolB = Hoja "'a"
| Nodo "'a" "'a arbolB" "'a arbolB"

text {*
Ejemplo de definición sobre árboles binarios:
Definir la función "espejo" que aplicada a un árbol devuelve su imagen
especular.
*}

fun espejo :: "'a arbolB ⇒ 'a arbolB" where
"espejo (Hoja x) = (Hoja x)"
| "espejo (Nodo x i d) = (Nodo x (espejo d) (espejo i))"

value "espejo (Nodo a (Nodo b (Hoja c) (Hoja d)) (Hoja e))"
-- "= Nodo a (Hoja e) (Nodo b (Hoja d) (Hoja c))"

text {*
Ejemplo de demostración sobre árboles binarios:
Demostrar que la función "espejo" es involutiva; es decir, para
cualquier árbol a, se tiene que
espejo(espejo(a)) = a.
*}

-- "La demostración estructurada es"
lemma espejo_involutiva:
fixes a :: "'b arbolB"
shows "espejo (espejo a) = a" (is "?P a")
proof (induct a)
fix x
show "?P (Hoja x)" by simp
next
fix x
fix i assume h1: "?P i"
fix d assume h2: "?P d"
show "?P (Nodo x i d)"
proof -
have "espejo(espejo(Nodo x i d)) =
espejo(Nodo x (espejo d) (espejo i))" by simp
also have "… = Nodo x (espejo (espejo i)) (espejo (espejo d))"
by simp
also have "… = Nodo x i d" using h1 h2 by simp
finally show ?thesis .
qed
qed

text {*
Comentarios sobre la demostración anterior:
· (fixes a :: "'b arbolB") es una abreviatura de "sea a1 un árbol binario
cuyos elementos son de tipo b".
· (induct a) indica que el método de demostración es por inducción
en el árbol binario a.
· Se generan dos casos:
1. ⋀a. espejo (espejo (Hoja a)) = Hoja a
2. ⋀a1 a2 a3. ⟦espejo (espejo a2) = a2;
espejo (espejo a3) = a3⟧
⟹ espejo (espejo (Nodo a1 a2 a3)) = Nodo a1 a2 a3
*}

-- "La demostración automática es"
lemma espejo_involutiva_1:
"espejo (espejo a ) = a"
by (induct a) auto

text {*
Ejemplo. [Aplanamiento de árboles]
Definir la función "aplana" que aplane los árboles recorriéndolos en
orden infijo.
*}

fun aplana :: "'a arbolB ⇒ 'a list" where
"aplana (Hoja x) = [x]"
| "aplana (Nodo x i d) = (aplana i) @ [x] @ (aplana d)"

value "aplana (Nodo a (Nodo b (Hoja c) (Hoja d)) (Hoja e))"
-- "= [c, b, d, a, e]"

text {*
Ejemplo. [Aplanamiento de la imagen especular] Demostrar que
aplana (espejo a) = rev (aplana a)
*}

-- "La demostración estructurada es"
lemma
fixes a :: "'b arbolB"
shows "aplana (espejo a) = rev (aplana a)" (is "?P a")
proof (induct a)
fix x
show "?P (Hoja x)" by simp
next
fix x
fix i assume h1: "?P i"
fix d assume h2: "?P d"
show "?P (Nodo x i d)"
proof -
have "aplana (espejo (Nodo x i d)) =
aplana (Nodo x (espejo d) (espejo i))" by simp
also have "… = (aplana(espejo d))@[x]@(aplana(espejo i))" by simp
also have "… = (rev(aplana d))@[x]@(rev(aplana i))"
using h1 h2 by simp
also have "… = rev((aplana i)@[x]@(aplana d))" by simp
also have "… = rev(aplana (Nodo x i d))" by simp
finally show ?thesis .
qed
qed

-- "La demostración automática es"
lemma "aplana (espejo a) = rev (aplana a)"
by (induct a) auto

section {* Árboles y bosques. Recursión mutua e inducción *}

text {*
Nota. [Ejemplo de definición de tipos mediante recursión cruzada]
· Un árbol de tipo a es una hoja o un nodo de tipo a junto con un
bosque de tipo a.
· Un bosque de tipo a es el boque vacío o un bosque contruido añadiendo
un árbol de tipo a a un bosque de tipo a.
*}

datatype 'a arbol = Hoja | Nodo "'a" "'a bosque"
and 'a bosque = Vacio | ConsB "'a arbol" "'a bosque"

text {*
Regla de inducción correspondiente a la recursión cruzada:
La regla de inducción sobre árboles y bosques es arbol_bosque.induct:
⟦P1 Hoja;
⋀x b. P2 b ⟹ P1 (Nodo x b);
P2 Vacio;
⋀a b. ⟦P1 a; P2 b⟧ ⟹ P2 (ConsB a b)⟧
⟹ P1 a ∧ P2 b
*}

text {*
Ejemplos de definición por recursión cruzada:
· aplana_arbol a) es la lista obtenida aplanando el árbol a.
· (aplana_bosque b) es la lista obtenida aplanando el bosque b.
· (map_arbol a h) es el árbol obtenido aplicando la función h a
todos los nodos del árbol a.
· (map_bosque b h) es el bosque obtenido aplicando la función h a
todos los nodos del bosque b.
*}

fun aplana_arbol :: "'a arbol ⇒ 'a list" and
aplana_bosque :: "'a bosque ⇒ 'a list" where
"aplana_arbol Hoja = []"
| "aplana_arbol (Nodo x b) = x#(aplana_bosque b)"
| "aplana_bosque Vacio = []"
| "aplana_bosque (ConsB a b) = (aplana_arbol a) @ (aplana_bosque b)"

fun map_arbol :: "('a ⇒ 'b) ⇒ 'a arbol ⇒ 'b arbol" and
map_bosque :: "('a ⇒ 'b) ⇒ 'a bosque ⇒ 'b bosque" where
"map_arbol f Hoja = Hoja"
| "map_arbol f (Nodo x b) = Nodo (f x) (map_bosque f b)"
| "map_bosque f Vacio = Vacio"
| "map_bosque f (ConsB a b) = ConsB (map_arbol f a) (map_bosque f b)"

text {*
Ejemplo de demostración por inducción cruzada:
Demostrar que:
· aplana_arbol (map_arbol f a) = map f (aplana_arbol a)
· aplana_bosque (map_bosque f b) = map f (aplana_bosque b)
*}

-- "La demostración detallada es"
lemma "aplana_arbol (map_arbol f a) = map f (aplana_arbol a)
∧ aplana_bosque (map_bosque f b) = map f (aplana_bosque b)"
proof (induct_tac a and b)
show "aplana_arbol (map_arbol f Hoja ) = map f (aplana_arbol Hoja)"
by simp
next
fix x b
assume HI: "aplana_bosque (map_bosque f b) = map f (aplana_bosque b)"
have "aplana_arbol (map_arbol f (Nodo x b)) =
aplana_arbol (Nodo (f x) (map_bosque f b))" by simp
also have "… = (f x)#(aplana_bosque (map_bosque f b))" by simp
also have "… = (f x)#(map f (aplana_bosque b))" using HI by simp
also have "… = map f (aplana_arbol (Nodo x b))" by simp
finally show "aplana_arbol (map_arbol f (Nodo x b))
= map f (aplana_arbol (Nodo x b))" .
next
show "aplana_bosque (map_bosque f Vacio) = map f (aplana_bosque Vacio)"
by simp
next
fix a b
assume HI1: "aplana_arbol (map_arbol f a) = map f (aplana_arbol a)"
and HI2: "aplana_bosque (map_bosque f b) = map f (aplana_bosque b)"
have "aplana_bosque (map_bosque f (ConsB a b)) =
aplana_bosque (ConsB (map_arbol f a) (map_bosque f b))" by simp
also have "… = aplana_arbol(map_arbol f a)@aplana_bosque(map_bosque f b)"
by simp
also have "… = (map f (aplana_arbol a))@(map f (aplana_bosque b))"
using HI1 HI2 by simp
also have "… = map f (aplana_bosque (ConsB a b))" by simp
finally show "aplana_bosque (map_bosque f (ConsB a b))
= map f (aplana_bosque (ConsB a b))" by simp
qed

text {*
Comentarios sobre la demostración anterior:
· (induct_tac a and b) indica que el método de demostración es por
inducción cruzada sobre a y b.
· Se generan 4 casos:
1. aplana_arbol (map_arbol arbol.Hoja h) = map h (aplana_arbol arbol.Hoja)
2. ⋀a bosque.
aplana_bosque (map_bosque bosque h) = map h (aplana_bosque bosque) ⟹
aplana_arbol (map_arbol (arbol.Nodo a bosque) h) =
map h (aplana_arbol (arbol.Nodo a bosque))
3. aplana_bosque (map_bosque Vacio h) = map h (aplana_bosque Vacio)
4. ⋀arbol bosque.
⟦aplana_arbol (map_arbol arbol h) = map h (aplana_arbol arbol);
aplana_bosque (map_bosque bosque h) = map h (aplana_bosque bosque)⟧
⟹ aplana_bosque (map_bosque (ConsB arbol bosque) h) =
map h (aplana_bosque (ConsB arbol bosque))
*}

-- "La demostración automática es"
lemma "aplana_arbol (map_arbol f a) = map f (aplana_arbol a)
∧ aplana_bosque (map_bosque f b) = map f (aplana_bosque b)"
by (induct_tac a and b) auto

end
</source>

Tema 1: Programación funcional en Isabelle

2018-07-16T11:11:22Z

WikiSysop: Texto reemplazado: «isar» por «isabelle»

<source lang="isabelle">
chapter {* Tema 1: Programación funcional en Isabelle *}

theory T1_Programacion_funcional_en_Isabelle
imports Main
begin

section {* Introducción *}

text {* En este tema se presenta el lenguaje funcional que está
incluido en Isabelle. El lenguaje funcional es muy parecido a
Haskell. *}

section {* Números naturales, enteros y booleanos *}

text {* En Isabelle están definidos los número naturales con la sintaxis
de Peano usando dos constructores: 0 (cero) y Suc (el sucesor).

Los números como el 1 son abreviaturas de los correspondientes en la
notación de Peano, en este caso "Suc 0".

El tipo de los números naturales es nat.

Por ejemplo, el siguiente del 0 es el 1. *}

value "Suc 0 = 1"
(* ↝ "True" :: "bool"*)

text {* En Isabelle está definida la suma de los números naturales:
(x + y) es la suma de x e y.

Por ejemplo, la suma de los números naturales 1 y 2 es el número
natural 3. *}

value "(1::nat) + 2"
(* ↝ "Suc (Suc (Suc 0))" :: "nat" *)

value "(1::nat) + 2 = 3"
(* ↝ "True" :: "bool" *)

text {* La notación del par de dos puntos se usa para asignar un tipo a
un término (por ejemplo, (1::nat) significa que se considera que 1 es
un número natural).

En Isabelle está definida el producto de los números naturales:
(x * y) es el producto de x e y.

Por ejemplo, el producto de los números naturales 2 y 3 es el número
natural 6. *}

value "(2::nat) * 3"
(* ↝ "Suc (Suc (Suc (Suc (Suc (Suc 0)))))" :: "nat"*)

value "(2::nat) * 3 = 6"
(* ↝ "True" :: "bool" *)

text {* En Isabelle está definida la división de números naturales:
(n div m) es el cociente entero de x entre y.

Por ejemplo, la división natural de 7 entre 3 es 2. *}

value "(7::nat) div 3"
(* ↝ "Suc (Suc 0)" :: "nat" *)

value "(7::nat) div 3 = 2"
(* ↝ "True" :: "bool" *)

text {* En Isabelle está definida el resto de división de números
naturales: (n mod m) es el resto de dividir n entre m.

Por ejemplo, el resto de dividir 7 entre 3 es 1. *}

value "(7::nat) mod 3"
(* ↝ "Suc 0" :: "nat" *)

text {* En Isabelle también están definidos los números enteros. El tipo
de los enteros se representa por int.

Por ejemplo, la suma de 1 y -2 es el número entero -1. *}

value "(1::int) + -2" -- "= -1"
(* ↝ "- 1" :: "int"*)

text {* Los numerales están sobrecargados. Por ejemplo, el 1 puede ser
un natural o un entero, dependiendo del contexto.

Isabelle resuelve ambigüedades mediante inferencia de tipos.

A veces, es necesario usar declaraciones de tipo para resolver la
ambigüedad.

En Isabelle están definidos los valores booleanos (True y False), las
conectivas (¬, ∧, ∨, ⟶ y ↔) y los cuantificadores (∀ y ∃).

El tipo de los booleanos es bool. *}

text {* La conjunción de dos fórmulas verdaderas es verdadera. *}
value "True ∧ True"
(* ↝ "True" :: "bool" *)

text {* La conjunción de un fórmula verdadera y una falsa es falsa. *}
value "True ∧ False"
(* ↝ "False" :: "bool" *)

text {* La disyunción de una fórmula verdadera y una falsa es
verdadera. *}
value "True ∨ False"
(* ↝ "True" :: "bool" *)

text {* La disyunción de dos fórmulas falsas es falsa. *}
value "False ∨ False"
(* ↝ "False" :: "bool"*)

text {* La negación de una fórmula verdadera es falsa. *}
value "¬True"
(* ↝ "False" :: "bool"*)

text {* Una fórmula falsa implica una fórmula verdadera. *}
value "False ⟶ True"
(* ↝ "True" :: "bool"*)

text {* Un lema introduce una proposición seguida de una demostración.

Isabelle dispone de varios procedimientos automáticos para generar
demostraciones, uno de los cuales es el de simplificación (llamado
simp).

El procedimiento simp aplica un conjunto de reglas de reescritura, que
inicialmente contiene un gran número de reglas relativas a los objetos
definidos. *}

text {* Ej. de simp: Todo elemento es igual a sí mismo. *}
lemma "∀x. x = x"
by simp

text {* Ej. de simp: Existe un elemento igual a 1. *}
lemma "∃x. x = 1"
by simp

section {* Definiciones no recursivas *}

text {* La disyunción exclusiva de A y B se verifica si una es verdadera
y la otra no lo es. *}

definition xor :: "bool ⇒ bool ⇒ bool" where
"xor A B ≡ (A ∧ ¬B) ∨ (¬A ∧ B)"

text {* Prop.: La disyunción exclusiva de dos fórmulas verdaderas es
falsa.

Dem.: Por simplificación, usando la definición de la disyunción
exclusiva.
*}

lemma "xor True True = False"
by (simp add: xor_def)

text {* Se añade la definición de la disyunción exclusiva al conjunto de
reglas de simplificación automáticas. *}

declare xor_def [simp]

lemma "xor True False = True"
by simp

section {* Definiciones locales *}

text {* Se puede asignar valores a variables locales mediante 'let' y
usarlo en las expresiones dentro de 'in'.

Por ejemplo, si x es el número natural 3, entonces "x*x = 9". *}

value "let x = 3::int in x * x"
(* ↝ "9" :: "int" *)

section {* Pares *}

text {* Un par se representa escribiendo los elementos entre paréntesis
y separados por coma.

El tipo de los pares es el producto de los tipos.

La función fst devuelve el primer elemento de un par y la snd el
segundo.

Por ejemplo, si p es el par de números naturales (2,3), entonces la
suma del primer elemento de p y 1 es igual al segundo elemento de
p. *}

value "let p = (2,3)::nat × nat in fst p + 1 = snd p"
(* ↝ "True" :: "bool" *)

section {* Listas *}

text {* Una lista se representa escribiendo los elementos entre
corchetes y separados por comas.

La lista vacía se representa por [].

Todos los elementos de una lista tienen que ser del mismo tipo.

El tipo de las listas de elementos del tipo a es (a list).

El término (x#xs) representa la lista obtenida añadiendo el elemento x
al principio de la lista xs.

Por ejemplo, la lista obtenida añadiendo sucesivamente a la lista
vacía los elementos z, y y x a es [x,y,z]. *}

value "x#(y#(z#[]))"
(* ↝ "[x, y, z]" :: "'a list" *)

value "(1::int)#(2#(3#[]))"
(* ↝ "[1, 2, 3]" :: "int list" *)

text {* Funciones de descomposición de listas:
· (hd xs) es el primer elemento de la lista xs.
· (tl xs) es el resto de la lista xs.

Por ejemplo, si xs es la lista [a,b,c], entonces el primero de xs es a
y el resto de xs es [b,c]. *}

value "let xs = [a,b,c] in hd xs = a ∧ tl xs = [b,c]"
(* ↝ "True" :: "bool" *)

text {* (length xs) es la longitud de la lista xs. Por ejemplo, la
longitud de la lista [1,2,5] es 3. *}

value "length [1,2,5]"
(* ↝ "Suc (Suc (Suc 0))" :: "nat" *)

text {* En la página 8 de "What's in Main" http://bit.ly/2eyY0zI y
en la sesión 66 de "Isabelle/HOL — Higher-Order Logic"
http://bit.ly/2eyYoy5 se encuentran más definiciones y propiedades de
las listas. *}

section {* Funciones anónimas *}

text {* En Isabelle pueden definirse funciones anónimas.

Por ejemplo, el valor de la función que a un número le asigna su doble
aplicada a 1 es 2. *}

value "(λx. x + x) 1::int"
(* ↝ "2" :: "int" *)

section {* Condicionales *}

text {* El valor absoluto del entero x es x, si "x ≥ 0" y es -x en caso
contrario. *}

definition absoluto :: "int ⇒ int" where
"absoluto x ≡ (if x ≥ 0 then x else -x)"

text {* Ejemplo, el valor absoluto de -3 es 3. *}

value "absoluto(-3)"
(* ↝ "3" :: "int" *)

text {* Def.: Un número natural n es un sucesor si es de la forma
(Suc m). *}

definition es_sucesor :: "nat ⇒ bool" where
"es_sucesor n ≡ (case n of
0 ⇒ False
| Suc m ⇒ True)"

text {* Ejemplo, el número 3 es sucesor. *}

value "es_sucesor 3"
(* ↝ "True" :: "bool" *)

section {* Tipos de datos y definiciones recursivas *}

text {* Una lista de elementos de tipo a es la lista Vacia o se obtiene
añadiendo, con Cons, un elemento de tipo a a una lista de elementos de
tipo a. *}

datatype 'a Lista = Vacia | Cons 'a "'a Lista"

text {* (conc xs ys) es la concatenación de las lista xs e ys. Por
ejemplo,
conc (Cons a (Cons b Vacia)) (Cons c Vacia)
= Cons a (Cons b (Cons c Vacia))
*}

fun conc :: "'a Lista ⇒ 'a Lista ⇒ 'a Lista" where
"conc Vacia ys = ys"
| "conc (Cons x xs) ys = Cons x (conc xs ys)"

value "conc (Cons a (Cons b Vacia)) (Cons c Vacia)"
(* ↝ Lista.Cons a (Lista.Cons b (Lista.Cons c Vacia)) *)

text {* Se puede declarar que acorte los nombres. *}

declare [[names_short]]

value "conc (Cons a (Cons b Vacia)) (Cons c Vacia)"
(* ↝ Cons a (Cons b (Cons c Vacia) *)

text {* (suma n) es la suma de los primeros n números naturales. Por
ejemplo,
suma 3 = 6
*}

fun suma :: "nat ⇒ nat" where
"suma 0 = 0"
| "suma (Suc m) = (Suc m) + suma m"

value "suma 3"
(* ↝ Suc (Suc (Suc (Suc (Suc (Suc 0)))) *)

text {* (sumaImpares n) es la suma de los n primeros números impares.
Por ejemplo,
sumaImpares 3 = 9
*}

fun sumaImpares :: "nat ⇒ nat" where
"sumaImpares 0 = 0"
| "sumaImpares (Suc n) = (2 * (Suc n) - 1) + sumaImpares n"

value "sumaImpares 3"
(* ↝ Suc (Suc (Suc (Suc (Suc (Suc (Suc (Suc (Suc 0)))))))) *)

end
</source>

Tema 2b: Razonamiento automático sobre programas en Isabelle/HOL

2018-07-16T11:11:22Z

WikiSysop: Texto reemplazado: «isar» por «isabelle»

<source lang="isabelle">
chapter {* Tema 2: Razonamiento automático sobre programas en Isabelle/HOL *}

theory T2b_Razonamiento_automatico_sobre_programas_en_IsabelleHOL
imports Main
begin

text {*
En este tema se demuestra con Isabelle las propiedades de los
programas funcionales como se expone en el tema 8 del curso
"Informática" que puede leerse en http://goo.gl/Imvyt *}

text {*
Se usarán nombres cortos. *}

declare [[names_short]]

section {* Razonamiento ecuacional *}

text {* ----------------------------------------------------------------
Ejemplo 1. Definir, por recursión, la función
longitud :: 'a list ⇒ nat
tal que (longitud xs) es la longitud de la listas xs. Por ejemplo,
longitud [4,2,5] = 3
------------------------------------------------------------------- *}

fun longitud :: "'a list ⇒ nat" where
"longitud [] = 0"
| "longitud (x#xs) = 1 + longitud xs"

value "longitud [4,2,5] = 3"

text {* ---------------------------------------------------------------
Ejemplo 2. Demostrar que
longitud [4,2,5] = 3
------------------------------------------------------------------- *}

lemma "longitud [4,2,5] = 3"
by simp

text {* ---------------------------------------------------------------
Ejemplo 3. Definir la función
fun intercambia :: 'a × 'b ⇒ 'b × 'a
tal que (intercambia p) es el par obtenido intercambiando las
componentes del par p. Por ejemplo,
intercambia (u,v) = (v,u)
------------------------------------------------------------------ *}

fun intercambia :: "'a × 'b ⇒ 'b × 'a" where
"intercambia (x,y) = (y,x)"

value "intercambia (u,v) = (v,u)"

text {* ---------------------------------------------------------------
Ejemplo 4. (p.6) Demostrar que
intercambia (intercambia (x,y)) = (x,y)
------------------------------------------------------------------- *}

lemma "intercambia (intercambia (x,y)) = (x,y)"
by simp

text {* ---------------------------------------------------------------
Ejemplo 5. Definir, por recursión, la función
inversa :: 'a list ⇒ 'a list
tal que (inversa xs) es la lista obtenida invirtiendo el orden de los
elementos de xs. Por ejemplo,
inversa [a,d,c] = [c,d,a]
------------------------------------------------------------------ *}

fun inversa :: "'a list ⇒ 'a list" where
"inversa [] = []"
| "inversa (x#xs) = inversa xs @ [x]"

value "inversa [a,d,c] = [c,d,a]"

text {* ---------------------------------------------------------------
Ejemplo 6. (p. 9) Demostrar que
inversa [x] = [x]
------------------------------------------------------------------- *}

lemma "inversa [x] = [x]"
by simp

section {* Razonamiento por inducción sobre los naturales *}

text {*
[Principio de inducción sobre los naturales] Para demostrar una
propiedad P para todos los números naturales basta probar que el 0
tiene la propiedad P y que si n tiene la propiedad P, entonces n+1
también la tiene.
⟦P 0; ⋀n. P n ⟹ P (Suc n)⟧ ⟹ P m

En Isabelle el principio de inducción sobre los naturales está
formalizado en el teorema nat.induct y puede verse con
thm nat.induct
*}

thm nat.induct

text {* ---------------------------------------------------------------
Ejemplo 7. Definir la función
repite :: nat ⇒ 'a ⇒ 'a list
tal que (repite n x) es la lista formada por n copias del elemento
x. Por ejemplo,
repite 3 a = [a,a,a]
------------------------------------------------------------------ *}

fun repite :: "nat ⇒ 'a ⇒ 'a list" where
"repite 0 x = []"
| "repite (Suc n) x = x # (repite n x)"

value "repite 3 a = [a,a,a]"

text {* ---------------------------------------------------------------
Ejemplo 8. (p. 18) Demostrar que
longitud (repite n x) = n
------------------------------------------------------------------- *}

(* 1ª demostración (procedimental) *)
lemma "longitud (repite n x) = n"
apply (induct n)
apply auto
done

(* 2ª demostración (declarativa) *)
lemma "longitud (repite n x) = n"
by (induct n) auto

section {* Razonamiento por inducción sobre listas *}

text {*
Para demostrar una propiedad para todas las listas basta demostrar
que la lista vacía tiene la propiedad y que al añadir un elemento a
una lista que tiene la propiedad se obtiene otra lista que también
tiene la propiedad.

En Isabelle el principio de inducción sobre listas está formalizado
mediante el teorema list.induct que puede verse con
thm list.induct
*}

thm list.induct

text {* ---------------------------------------------------------------
Ejemplo 9. Definir la función
conc :: 'a list ⇒ 'a list ⇒ 'a list
tal que (conc xs ys) es la concatención de las listas xs e ys. Por
ejemplo,
conc [a,d] [b,d,a,c] = [a,d,b,d,a,c]
------------------------------------------------------------------ *}

fun conc :: "'a list ⇒ 'a list ⇒ 'a list" where
"conc [] ys = ys"
| "conc (x#xs) ys = x # (conc xs ys)"

value "conc [a,d] [b,d,a,c] = [a,d,b,d,a,c]"

text {* ---------------------------------------------------------------
Ejemplo 10. (p. 24) Demostrar que
conc xs (conc ys zs) = (conc xs ys) zs
------------------------------------------------------------------- *}

(* 1ª demostración *)
lemma "conc xs (conc ys zs) = conc (conc xs ys) zs"
apply (induct xs)
apply auto
done

(* 2ª demostración *)
lemma "conc xs (conc ys zs) = conc (conc xs ys) zs"
by (induct xs) auto

text {* ---------------------------------------------------------------
Ejemplo 11. Refutar que
conc xs ys = conc ys xs
------------------------------------------------------------------- *}

lemma "conc xs ys = conc ys xs"
quickcheck
oops

text {* Encuentra el contraejemplo,
xs = [a\<^isub>2]
ys = [a\<^isub>1] *}

text {* ---------------------------------------------------------------
Ejemplo 12. (p. 28) Demostrar que
conc xs [] = xs
------------------------------------------------------------------- *}

(* 1ª demostración *)
lemma "conc xs [] = xs"
apply (induct xs)
apply auto
done

(* 2ª demostración *)
lemma "conc xs [] = xs"
by (induct xs) auto

text {* ---------------------------------------------------------------
Ejemplo 13. (p. 30) Demostrar que
longitud (conc xs ys) = longitud xs + longitud ys
------------------------------------------------------------------- *}

(* 1ª demostración *)
lemma "longitud (conc xs ys) = longitud xs + longitud ys"
apply (induct xs)
apply auto
done

(* 2ª demostración *)
lemma "longitud (conc xs ys) = longitud xs + longitud ys"
by (induct xs) auto

section {* Inducción correspondiente a la definición recursiva *}

text {* ---------------------------------------------------------------
Ejemplo 14. Definir la función
coge :: nat ⇒ 'a list ⇒ 'a list
tal que (coge n xs) es la lista de los n primeros elementos de xs. Por
ejemplo,
coge 2 [a,c,d,b,e] = [a,c]
------------------------------------------------------------------ *}

fun coge :: "nat ⇒ 'a list ⇒ 'a list" where
"coge n [] = []"
| "coge 0 xs = []"
| "coge (Suc n) (x#xs) = x # (coge n xs)"

value "coge 2 [a,c,d,b,e] = [a,c]"

text {* ---------------------------------------------------------------
Ejemplo 15. Definir la función
elimina :: nat ⇒ 'a list ⇒ 'a list
tal que (elimina n xs) es la lista obtenida eliminando los n primeros
elementos de xs. Por ejemplo,
elimina 2 [a,c,d,b,e] = [d,b,e]
------------------------------------------------------------------ *}

fun elimina :: "nat ⇒ 'a list ⇒ 'a list" where
"elimina n [] = []"
| "elimina 0 xs = xs"
| "elimina (Suc n) (x#xs) = elimina n xs"

value "elimina 2 [a,c,d,b,e] = [d,b,e]"

text {*
La definición coge genera el esquema de inducción coge.induct:
⟦⋀n. P n [];
⋀x xs. P 0 (x#xs);
⋀n x xs. P n xs ⟹ P (Suc n) (x#xs)⟧
⟹ P n x

Puede verse usando "thm coge.induct". *}

thm coge.induct

text {* ---------------------------------------------------------------
Ejemplo 16. (p. 35) Demostrar que
conc (coge n xs) (elimina n xs) = xs
------------------------------------------------------------------- *}

(* 1ª demostración *)
lemma "conc (coge n xs) (elimina n xs) = xs"
apply (induct rule: coge.induct)
apply auto
done

(* 2ª demostración *)
lemma "conc (coge n xs) (elimina n xs) = xs"
by (induct rule: coge.induct) auto

section {* Razonamiento por casos *}

text {*
Distinción de casos sobre listas:
· El método de distinción de casos se activa con (cases xs) donde xs
es del tipo lista.
· "case Nil" es una abreviatura de
"assume Nil: xs =[]".
· "case Cons" es una abreviatura de
"fix ? ?? assume Cons: xs = ? # ??"
donde ? y ?? son variables anónimas. *}

text {* ---------------------------------------------------------------
Ejemplo 17. Definir la función
esVacia :: 'a list ⇒ bool
tal que (esVacia xs) se verifica si xs es la lista vacía. Por ejemplo,
esVacia [] = True
esVacia [1] = False
------------------------------------------------------------------ *}

fun esVacia :: "'a list ⇒ bool" where
"esVacia [] = True"
| "esVacia (x#xs) = False"

value "esVacia [] = True"
value "esVacia [1] = False"

text {* ---------------------------------------------------------------
Ejemplo 18 (p. 39) . Demostrar que
esVacia xs = esVacia (conc xs xs)
------------------------------------------------------------------- *}

(* 1ª demostración *)
lemma "esVacia xs = esVacia (conc xs xs)"
apply (cases xs)
apply auto
done

(* 2ª demostración *)
lemma "esVacia xs = esVacia (conc xs xs)"
by (cases xs) auto

section {* Heurística de generalización *}

text {*
Heurística de generalización: Cuando se use demostración estructural,
cuantificar universalmente las variables libres (o, equivalentemente,
considerar las variables libres como variables arbitrarias). *}

text {* ---------------------------------------------------------------
Ejemplo 19. Definir la función
inversaAc :: 'a list ⇒ 'a list
tal que (inversaAc xs) es a inversa de xs calculada usando
acumuladores. Por ejemplo,
inversaAc [a,c,b,e] = [e,b,c,a]
------------------------------------------------------------------ *}

fun inversaAcAux :: "'a list ⇒ 'a list ⇒ 'a list" where
"inversaAcAux [] ys = ys"
| "inversaAcAux (x#xs) ys = inversaAcAux xs (x#ys)"

fun inversaAc :: "'a list ⇒ 'a list" where
"inversaAc xs = inversaAcAux xs []"

value "inversaAc [a,c,b,e] = [e,b,c,a]"

text {* ---------------------------------------------------------------
Ejemplo 20. (p. 44) Demostrar que
inversaAcAux xs ys = (inversa xs) @ ys
------------------------------------------------------------------- *}

(* 1ª demostración *)
lemma inversaAcAux_es_inversa:
"inversaAcAux xs ys = (inversa xs)@ys"
apply (induct xs arbitrary: ys)
apply auto
done

(* 2ª demostración *)
lemma inversaAcAux_es_inversa:
"inversaAcAux xs ys = (inversa xs)@ys"
by (induct xs arbitrary: ys) auto

text {* ---------------------------------------------------------------
Ejemplo 21. (p. 43) Demostrar que
inversaAc xs = inversa xs
------------------------------------------------------------------- *}

corollary "inversaAc xs = inversa xs"
by (simp add: inversaAcAux_es_inversa)

section {* Demostración por inducción para funciones de orden superior *}

text {* ---------------------------------------------------------------
Ejemplo 22. Definir la función
sum :: nat list ⇒ nat
tal que (sum xs) es la suma de los elementos de xs. Por ejemplo,
sum [3,2,5] = 10
------------------------------------------------------------------ *}

fun sum :: "nat list ⇒ nat" where
"sum [] = 0"
| "sum (x#xs) = x + sum xs"

value "sum [3,2,5] = 10"

text {* ---------------------------------------------------------------
Ejemplo 23. Definir la función
map :: ('a ⇒ 'b) ⇒ 'a list ⇒ 'b list
tal que (map f xs) es la lista obtenida aplicando la función f a los
elementos de xs. Por ejemplo,
map (λx. 2*x) [3,2,5] = [6,4,10]
------------------------------------------------------------------ *}

fun map :: "('a ⇒ 'b) ⇒ 'a list ⇒ 'b list" where
"map f [] = []"
| "map f (x#xs) = (f x) # map f xs"

value "map (λx. 2*x) [3::nat,2,5] = [6,4,10]"

text {* ---------------------------------------------------------------
Ejemplo 24. (p. 45) Demostrar que
sum (map (λx. 2*x) xs) = 2 * (sum xs)
------------------------------------------------------------------- *}

(* 1ª demostración *)
lemma "sum (map (λx. 2*x) xs) = 2 * (sum xs)"
apply (induct xs)
apply auto
done

(* 2ª demostración *)
lemma "sum (map (λx. 2*x) xs) = 2 * (sum xs)"
by (induct xs) auto

text {* ---------------------------------------------------------------
Ejemplo 25. (p. 48) Demostrar que
longitud (map f xs) = longitud xs
------------------------------------------------------------------- *}

(* 1ª demostración *)
lemma "longitud (map f xs) = longitud xs"
apply (induct xs)
apply auto
done

(* 2ª demostración *)
lemma "longitud (map f xs) = longitud xs"
by (induct xs) auto

section {* Referencias *}

text {*
· J.A. Alonso. "Razonamiento sobre programas" http://goo.gl/R06O3
· G. Hutton. "Programming in Haskell". Cap. 13 "Reasoning about
programms". http://bit.ly/1gMqK0X
· S. Thompson. "Haskell: the Craft of Functional Programming, 3rd
Edition. Cap. 8 "Reasoning about programms".
· L. Paulson. "ML for the Working Programmer, 2nd Edition". Cap. 6.
"Reasoning about functional programs". http://bit.ly/1gMqFKI
*}

end
</source>

Tema 10: Conjuntos, funciones y relaciones

2018-07-16T11:11:22Z

WikiSysop: Texto reemplazado: «isar» por «isabelle»

<source lang="isabelle">
chapter {* Tema 10: Conjuntos, funciones y relaciones *}

theory T10_Conjuntos_funciones_y_relaciones
imports Main
begin

section {* Conjuntos *}

subsection {* Operaciones con conjuntos *}

text {*
Nota. La teoría elemental de conjuntos es HOL/Set.thy.

Nota. En un conjunto todos los elemento son del mismo tipo (por
ejemplo, del tipo τ) y el conjunto tiene tipo (en el ejemplo, "τ set").

Reglas de la intersección:
· IntI: ⟦c ∈ A; c ∈ B⟧ ⟹ c ∈ A ∩ B
· IntD1: c ∈ A ∩ B ⟹ c ∈ A
· IntD2: c ∈ A ∩ B ⟹ c ∈ B

Nota. Propiedades del complementario:
· Compl_iff: (c ∈ - A) = (c ∉ A)
· Compl_Un: - (A ∪ B) = - A ∩ - B

Nota. El conjunto vacío se representa por {} y el universal por UNIV.

Nota. Propiedades de la diferencia y del complementario:
· Diff_disjoint: A ∩ (B - A) = {}
· Compl_partition: A ∪ - A = UNIV

Nota. Reglas de la relación de subconjunto:
· subsetI: (⋀x. x ∈ A ⟹ x ∈ B) ⟹ A ⊆ B
· subsetD: ⟦A ⊆ B; c ∈ A⟧ ⟹ c ∈ B
*}

text {*
Ejemplo: A ∪ B ⊆ C syss A ⊆ C ∧ B ⊆ C.
*}

lemma "(A ∪ B ⊆ C) = (A ⊆ C ∧ B ⊆ C)"
by blast

text {*
Ejemplo: A ⊆ -B syss B ⊆ -A.
*}

lemma "(A ⊆ -B) = (B ⊆ -A)"
by blast

text {*
Principio de extensionalidad de conjuntos:
· set_eqI: (⋀x. (x ∈ A) = (x ∈ B)) ⟹ A = B

Reglas de la igualdad de conjuntos:
· equalityI: ⟦A ⊆ B; B ⊆ A⟧ ⟹ A = B
· equalityD1: A = B ⟹ A ⊆ B
· equalityD2: A = B ⟹ B ⊆ A
· equalityE: ⟦A = B; ⟦A ⊆ B; B ⊆ A⟧ ⟹ P⟧ ⟹ P
*}

text {*
Lema. [Analogía entre intersección y conjunción]
"x ∈ A ∩ B" syss "x ∈ A" y "x ∈ B".
*}

lemma "(x ∈ A ∩ B) = (x ∈ A ∧ x ∈ B)"
by simp

text {*
Lema. [Analogía entre unión y disyunción]
x ∈ A ∪ B syss x ∈ A ó x ∈ B.
*}

lemma "(x ∈ A ∪ B) = (x ∈ A ∨ x ∈ B)"
by simp

text {*
Lema. [Analogía entre subconjunto e implicación]
A ⊆ B syss para todo x, si x ∈ A entonces x ∈ B.
*}

lemma "(A ⊆ B) = (∀ x. x ∈ A ⟶ x ∈ B)"
by auto

text {*
Lema. [Analogía entre complementario y negación]
x pertenece al complementario de A syss x no pertenece a A.
*}

lemma "(x ∈ -A) = (x ∉ A)"
by simp

subsection {* Notación de conjuntos finitos *}

text {*
Nota. La teoría de conjuntos finitos es HOL/Finite_Set.thy.

Nota. Los conjuntos finitos se definen por inducción a partir de las
siguientes reglas inductivas:
· El conjunto vacío es un conjunto finito.
· emptyI: "finite {}"
· Si se le añade un elemento a un conjunto finito se obtiene otro
conjunto finito.
· insertI: "finite A ⟹ finite (insert a A)"

A continuación se muestran ejemplos de conjuntos finitos.
*}

lemma
"insert 2 {} = {2} ∧
insert 3 {2} = {2,3} ∧
insert 2 {2,3} = {2,3} ∧
{2,3} = {3,2,3,2,2}"
by auto

text {*
Nota. Los conjuntos finitos se representan con la notación conjuntista
habitual: los elementos entre llaves y separados por comas.
*}

text {*
Ejemplo: {a,b} ∪ {c,d} = {a,b,c,d}
*}

lemma "{a,b} ∪ {c,d} = {a,b,c,d}"
by blast

text {*
Ejemplo de conjetura falsa y su refutación.
*}

lemma "{a,b} ∩ {b,c} = {b}"
oops

text {*
Ejemplo con la conjetura corregida.
*}

lemma "{a,b} ∩ {b,c} = (if a = c then {a,b} else {b})"
by auto

text {*
Sumas de conjuntos finitos:
· ∑A es la suma de los elementos del conjunto finito A. Por ejemplo,
value "∑{1,2,3}::int" -- "= 6"
· (setsum f A) es la suma de la aplicación de f a los elementos del
conjunto finito A, Por ejemplo,
value "setsum (λx. x*x) {1,2,3}::int" -- "= 14"
*}

text {*
Ejemplos de definiciones recursivas sobre conjuntos finitos:
Sea A un conjunto finito de números naturales.
· sumaConj A es la suma de los elementos A.
· sumaCuadradosConj A es la suma de los cuadrados de los elementos A.
*}

definition sumaConj :: "nat set ⇒ nat" where
"sumaConj S ≡ ∑S"

value "sumaConj {2,5,3} = 10"

definition sumaCuadradosConj :: "nat set ⇒ nat" where
"sumaCuadradosConj S ≡ setsum (λx. x*x) S"

value "sumaCuadradosConj {2,5,3} = 38"

text {*
Nota. Para simplificar lo que sigue, declaramos las anteriores
definiciones como reglas de simplificación.
*}

declare sumaConj_def[simp]
declare sumaCuadradosConj_def[simp]

text {*
Ejemplos de evaluación de las anteriores definiciones recursivas.
*}

lemma
"sumaConj {1,2,3,4} = 10 ∧
sumaCuadradosConj {1,2,3,4} = 30"
by simp

text {*
Inducción sobre conjuntos finitos: Para demostrar que todos los
conjuntos finitos tienen una propiedad P basta probar que
· El conjunto vacío tiene la propiedad P.
· Si a un conjunto finito que tiene la propiedad P se le añade un
nuevo elemento, el conjunto obtenido sigue teniendo la propiedad P.
En forma de regla
· finite_induct: ⟦finite F;
P {};
⋀x F. ⟦finite F; x ∉ F; P F⟧ ⟹ P ({x} ∪ F)⟧
⟹ P F
*}

text {*
Ejemplo de inducción sobre conjuntos finitos: Sea S un conjunto finito
de números naturales. Entonces todos los elementos de S son menores o
iguales que la suma de los elementos de S.
*}

-- "La demostración automática es"
lemma "finite S ⟹ ∀x∈S. x ≤ sumaConj S"
by (induct rule: finite_induct) auto

-- "La demostración estructurada es"
lemma sumaConj_acota:
"finite S ⟹ ∀x∈S. x ≤ sumaConj S"
proof (induct rule: finite_induct)
show "∀x ∈ {}. x ≤ sumaConj {}" by simp
next
fix x and F
assume fF: "finite F"
and xF: "x ∉ F"
and HI: "∀ x∈F. x ≤ sumaConj F"
show "∀y ∈ insert x F. y ≤ sumaConj (insert x F)"
proof
fix y
assume "y ∈ insert x F"
show "y ≤ sumaConj (insert x F)"
proof (cases "y = x")
assume "y = x"
hence "y ≤ x + (sumaConj F)" by simp
also have "… = sumaConj (insert x F)" using fF xF by simp
finally show ?thesis .
next
assume "y ≠ x"
hence "y ∈ F" using `y ∈ insert x F` by simp
hence "y ≤ sumaConj F" using HI by blast
also have "… ≤ x + (sumaConj F)" by simp
also have "… = sumaConj (insert x F)" using fF xF by simp
finally show ?thesis .
qed
qed
qed

subsection {* Definiciones por comprensión *}

text {*
El conjunto de los elementos que cumple la propiedad P se representa
por {x. P}.

Reglas de comprensión (relación entre colección y pertenencia):
· mem_Collect_eq: (a ∈ {x. P x}) = P a
· Collect_mem_eq: {x. x ∈ A} = A
*}

text {*
Ejemplo de comprensión: {x. P x ∨ x ∈ A} = {x. P x} ∪ A
*}

lemma "{x. P x ∨ x ∈ A} = {x. P x} ∪ A"
by blast

text {*
Ejemplo de comprensión: {x. P x ⟶ Q x} = -{x. P x} ∪ {x. Q x}
*}

lemma "{x. P x ⟶ Q x} = -{x. P x} ∪ {x. Q x}"
by blast

text {*
Ejemplo con la sintaxis general de comprensión.
{p*q | p q. p ∈ prime ∧ q ∈ prime} =
{z. ∃p q. z = p*q ∧ p ∈ prime ∧ q ∈ prime}
*}

lemma
"{p*q | p q. p ∈ prime ∧ q ∈ prime} =
{z. ∃p q. z = p*q ∧ p ∈ prime ∧ q ∈ prime}"
by blast

text {*
En HOL, la notación conjuntista es azúcar sintáctica:
· x ∈ A es equivalente a A(x).
· {x. P} es equivalente a λx. P.
*}

text {*
Ejemplo de definición por comprensión: El conjunto de los pares es el
de los números n para los que existe un m tal que n = 2*m.
*}

definition Pares :: "nat set" where
"Pares ≡ {n. ∃m. n = 2*m }"

text {*
Ejemplo. Los números 2 y 34 son pares.
*}

lemma
"2 ∈ Pares ∧
34 ∈ Pares"
by (simp add: Pares_def)

text {*
Definición. El conjunto de los impares es el de los números n para los
que existe un m tal que n = 2*m + 1.
*}

definition Impares :: "nat set" where
"Impares ≡ {n. ∃m. n = 2*m + 1}"

text {*
Ejemplo con las reglas de intersección y comprensión: El conjunto de
los pares es disjunto con el de los impares.
*}

-- "La demostración detallada es"
lemma "x ∉ (Pares ∩ Impares)"
proof
fix x assume S: "x ∈ (Pares ∩ Impares)"
hence "x ∈ Pares" by (rule IntD1)
hence "∃m. x = 2 * m" by (simp only: Pares_def mem_Collect_eq)
then obtain p where p: "x = 2 * p" ..
from S have "x ∈ Impares" by (rule IntD2)
hence "∃ m. x = 2 * m + 1" by (simp only: Impares_def mem_Collect_eq)
then obtain q where q: "x = 2 * q + 1" ..
from p and q show "False" by arith
qed

-- "La demostración estructurada es"
lemma "x ∉ (Pares ∩ Impares)"
proof
fix x assume S: "x ∈ (Pares ∩ Impares)"
hence "x ∈ Pares" ..
hence "∃m. x = 2 * m" by (simp only: Pares_def mem_Collect_eq)
then obtain p where p: "x = 2 * p" ..
from S have "x ∈ Impares" ..
hence "∃ m. x = 2 * m + 1" by (simp only: Impares_def mem_Collect_eq)
then obtain q where q: "x = 2 * q + 1" ..
from p and q show "False" by arith
qed

-- "La demostración automática es"
lemma "x ∉ (Pares ∩ Impares)"
by (auto simp add: Pares_def Impares_def, arith)

subsection {* Cuantificadores acotados *}

text {*
Reglas de cuantificador universal acotado ("bounded"):
· ballI: (⋀x. x ∈ A ⟹ P x) ⟹ ∀x∈A. P x
· bspec: ⟦∀x∈A. P x; x ∈ A⟧ ⟹ P x

Reglas de cuantificador existencial acotado ("bounded"):
· bexI: ⟦P x; x ∈ A⟧ ⟹ ∃x∈A. P x
· bexE: ⟦∃x∈A. P x; ⋀x. ⟦x ∈ A; P x⟧ ⟹ Q⟧ ⟹ Q

Reglas de la unión indexada:
· UN_iff: (b ∈ (⋃x∈A. B x)) = (∃x∈A. b ∈ B x)
· UN_I: ⟦a ∈ A; b ∈ B a⟧ ⟹ b ∈ (⋃x∈A. B x)
· UN_E: ⟦b ∈ (⋃x∈A. B x); ⋀x. ⟦x ∈ A; b ∈ B x⟧ ⟹ R⟧ ⟹ R

Reglas de la unión de una familia:
· Union_def: ⋃S = (⋃x∈S. x)
· Union_iff: (A ∈ ⋃C) = (∃X∈C. A ∈ X)

Reglas de la intersección indexada:
· INT_iff: (b ∈ (⋂x∈A. B x)) = (∀x∈A. b ∈ B x)
· INT_I: (⋀x. x ∈ A ⟹ b ∈ B x) ⟹ b ∈ (⋂x∈A. B x)
· INT_E: ⟦b ∈ (⋂x∈A. B x); b ∈ B a ⟹ R; a ∉ A ⟹ R⟧ ⟹ R

Reglas de la intersección de una familia:
· Inter_def: ⋂S = (⋂x∈S. x)
· Inter_iff: (A ∈ ⋂C) = (∀X∈C. A ∈ X)

Abreviaturas:
· "Collect P" es lo mismo que "{x. P}".
· "All P" es lo mismo que "∀x. P x".
· "Ex P" es lo mismo que "∃x. P x".
· "Ball A P" es lo mismo que "∀x∈A. P x".
· "Bex A P" es lo mismo que "∃x∈A. P x".
*}

subsection {* Conjuntos finitos y cardinalidad *}

text {*
El número de elementos de un conjunto finito A es el cardinal de A y
se representa por "card A".
*}

text {*
Ejemplos de cardinales de conjuntos finitos.
*}

lemma
"card {} = 0 ∧
card {4} = 1 ∧
card {4,1} = 2 ∧
x ≠ y ⟹ card {x,y} = 2"
by simp

text {*
Propiedades de cardinales:
· Cardinal de la unión de conjuntos finitos:
card_Un_Int: ⟦finite A; finite B⟧
⟹ card A + card B = card (A ∪ B) + card (A ∩ B)"
· Cardinal del conjunto potencia:
card_Pow: finite A ⟹ card (Pow A) = 2 ^ card A
*}

section {* Funciones *}

text {*
La teoría de funciones es HOL/Fun.thy.
*}

subsection {* Nociones básicas de funciones *}

text {*
Principio de extensionalidad para funciones:
· ext: (⋀x. f x = g x) ⟹ f = g

Actualización de funciones
· fun_upd_apply: (f(x := y)) z = (if z = x then y else f z)
· fun_upd_upd: f(x := y, x := z) = f(x := z)

Función identidad
· id_def: id ≡ λx. x

Composición de funciones:
· o_def: f ∘ g = (λx. f (g x))

Asociatividad de la composición:
· o_assoc: f ∘ (g ∘ h) = (f ∘ g) ∘ h
*}

subsection {* Funciones inyectivas, suprayectivas y biyectivas *}

text {*
Función inyectiva sobre A:
· inj_on_def: inj_on f A ≡ ∀x∈A. ∀y∈A. f x = f y ⟶ x = y

Nota. "inj f" es una abreviatura de "inj_on f UNIV".

Función suprayectiva:
· surj_def: surj f ≡ ∀y. ∃x. y = f x

Función biyectiva:
· bij_def: bij f ≡ inj f ∧ surj f

Propiedades de las funciones inversas:
· inv_f_f: inj f ⟹ inv f (f x) = x
· surj_f_inv_f: surj f ⟹ f (inv f y) = y
· inv_inv_eq: bij f ⟹ inv (inv f) = f

Igualdad de funciones (por extensionalidad):
· fun_eq_iff: (f = g) = (∀x. f x = g x)
*}

text {*
Ejemplo de lema de demostración de propiedades de funciones: Una
función inyectiva puede cancelarse en el lado izquierdo de la
composición de funciones.
*}

-- "La demostración detallada es"
lemma
assumes "inj f"
shows "(f ∘ g = f ∘ h) = (g = h)"
proof
assume "f ∘ g = f ∘ h"
show "g = h"
proof
fix x
have "(f ∘ g)(x) = (f ∘ h)(x)" using `f ∘ g = f ∘ h` by simp
hence "f(g(x)) = f(h(x))" by simp
then show "g(x) = h(x)" using `inj f` by (simp add:inj_on_def)
qed
next
assume "g = h"
show "f ∘ g = f ∘ h"
proof
fix x
have "(f ∘ g) x = f(g(x))" by simp
also have "… = f(h(x))" using `g = h` by simp
also have "… = (f ∘ h) x" by simp
finally show "(f ∘ g) x = (f ∘ h) x" by simp
qed
qed

-- "La demostración estructurada es"
lemma
assumes "inj f"
shows "(f ∘ g = f ∘ h) = (g = h)"
proof
assume "f ∘ g = f ∘ h"
thus "g = h" using `inj f` by (simp add: inj_on_def fun_eq_iff)
next
assume "g = h"
thus "f ∘ g = f ∘ h" by auto
qed

-- "La demostración automática es"
lemma
assumes "inj f"
shows "(f ∘ g = f ∘ h) = (g = h)"
using assms
by (auto simp add: inj_on_def fun_eq_iff)

subsubsection {* Función imagen *}

text {*
Imagen de un conjunto mediante una función:
· image_def: f ` A = {y. (∃x∈A. y = f x)}

Propiedades de la imagen:
· image_compose: (f ∘ g)`r = f`g`r
· image_Un: f`(A ∪ B) = f`A ∪ f`B
· image_Int: inj f ⟹ f`(A ∩ B) = f`A ∩ f`B"
*}

text {*
Ejemplo de demostración de propiedades de la imagen:
f`A ∪ g`A = (⋃x∈A. {f x, g x})
*}

lemma "f`A ∪ g`A = (⋃x∈A. {f x, g x})"
by auto

text {*
Ejemplo de demostración de propiedades de la imagen:
f`{(x,y). P x y} = {f(x,y) | x y. P x y}
*}

lemma "f`{(x,y). P x y} = {f(x,y) | x y. P x y}"
by auto

text {*
El rango de una función ("range f") es la imagen del universo
("f`UNIV").

Imagen inversa de un conjunto:
· vimage_def: f -` B ≡ {x. f x ∈ B}

Propiedad de la imagen inversa de un conjunto:
· vimage_Compl: f -` (-A) = -(f -` A)
*}

section {* Relaciones *}

subsection {* Relaciones básicas *}

text {*
La teoría de relaciones es HOL/Relation.thy.

Las relaciones son conjuntos de pares.

Relación identidad:
· Id_def: Id ≡ {p. ∃x. p = (x,x)}

Composición de relaciones:
· rel_comp_def: r O s ≡ {(x,z). ∃y. (x, y) ∈ r ∧ (y, z) ∈ s}

Propiedades:
· R_O_Id: R O Id = R
· rel_comp_mono: ⟦r' ⊆ r; s' ⊆ s⟧ ⟹ (r' O s') ⊆ (r O s)

Imagen inversa de una relación:
· converse_iff: ((a,b) ∈ r^-1) = ((b,a) ∈ r)

Propiedad de la imagen inversa de una relación:
· converse_rel_comp: (r O s)^-1 = s^-1 O r^-1

Imagen de un conjunto mediante una relación:
· Image_iff: (b ∈ r``A) = (∃x:A. (x, b) ∈ r)

Dominio de una relación:
· Domain_iff: (a ∈ Domain r) = (∃y. (a, y) ∈ r)

Rango de una relación:
· Range_iff: (a ∈ Range r) = (∃y. (y,a) ∈ r)
*}

end
</source>

Relación 7

2018-07-16T11:11:21Z

WikiSysop: Texto reemplazado: «isar» por «isabelle»

<source lang="isabelle">
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 marcarmor13 josgarsan fracorjim1 jeamacpov *)
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 serrodcal crigomgom rubgonmar danrodcha ferrenseg
manmorjim1 juacabsou lucnovdos dancorgar bowma antsancab1 pabrodmac *)
(* 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 marcarmor13
dancorgar antsancab1 jeamacpov pabrodmac *)
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 serrodcal crigomgom rubgonmar
danrodcha ferrenseg josgarsan juacabsou lucnovdos bowma pabrodmac *)
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 profundidad3 :: "arbol => nat" where
"profundidad3 H = 0"
| "profundidad3 (N i d) = 1 + maximo (profundidad3 i, profundidad3 d)"

(* ivamenjim: llamando a la función anterior profundidad3 *)
lemma "profundidad a = profundidad3 a"
by (induct a) auto

(* ivamenjim manmorjim1 fracorjim1 *)
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
serrodcal crigomgom rubgonmar danrodcha ferrenseg josgarsan
manmorjim1 juacabsou fracorjim1 lucnovdos dancorgar bowma antsancab1 pabrodmac *)
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 marcarmor13 jeamacpov *)
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 serrodcal crigomgom rubgonmar
danrodcha ferrenseg juacabsou josgarsan fracorjim1 lucnovdos bowma *)
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 marcarmor13 manmorjim1
dancorgar antsancab1 jeamacpov pabrodmac *)
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 serrodcal
fracorjim1 josgarsan lucnovdos bowma antsancab1 jeamacpov *)
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 crigomgom ivamenjim paupeddeg juacabsou *)
lemma AUX7: "es_abc profundidad a ⟶ (hojas a = 2^(profundidad a))"
by (induct a) simp_all

(* Otra forma de escribir lo mismo *)
(* danrodcha *)
lemma aux1: "es_abc profundidad (a::arbol) ⟹ (hojas a = 2^ (profundidad a))"
by (induct a) simp_all

(* fraortmoy marpoldia1 migtermor anaprarod wilmorort serrodcal
crigomgom rubgonmar ivamenjim danrodcha marcarmor13 paupeddeg
juacabsou bowma antsancab1 *)
(* También funciona con AUX7 *)
lemma lej7: "es_abc profundidad a = es_abc hojas a"
by (induct a) (auto simp add: abc_prof_num_hojas)

(* danrodcha *)
lemma 7: "es_abc profundidad a = es_abc hojas a"
apply (induct a)
apply simp
apply (auto simp add: aux1)
done

(* ferrenseg *)
lemma [simp]: "es_abc profundidad a ⟶ hojas a = 2 ^ (profundidad a)"
by (induct a) auto

(* Comentario: Uso de la declaración simp *)

theorem es_abc_profundidad_hojas:
"es_abc profundidad a = es_abc hojas a"
by (induct a) auto

(* Comentario: Dependencia de la declaración simp. *)

(* dancorgar *)
lemma rel_hojas_prof: "es_abc hojas a ∧ es_abc profundidad a
⟹ hojas a = 2 ^ profundidad a"
by (induct a) (auto)

theorem "es_abc hojas a = es_abc profundidad a"
by (induct a) (auto simp add: rel_hojas_prof)

(* pabrodmac *)
lemma lemaaux1:"es_abc hojas a⟶(hojas a = 2^(profundidad a))"
by (induct a) auto

lemma 1:"es_abc profundidad a = es_abc hojas a "
by (induct a) (auto simp add: lemaaux1)

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 serrodcal
marcarmor13 josgarsan lucnovdos bowma antsancab1 jeamacpov *)
lemma abc_hojas_num_nodos:
assumes "es_abc hojas a"
shows "Suc (size a) = hojas a"
using assms
by (induct a) auto

(* Comentario sobre orientación de igualdades. *)

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

(* fraortmoy marpoldia1 migtermor anaprarod wilmorort pablucoto
serrodcal antsancab1 *)
lemma lej8: "es_abc hojas a = es_abc size a"
by (induct a) (auto simp add:abc_hojas_num_nodos [symmetric])

(* Comentario sobre orientación de igualdades. *)

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

(* ivamenjim *)
(* Teorema auxiliar *)
lemma auxEj8: "hojas a = size a + 1"
by (induct a) auto

(* ivamenjim *)
lemma lej8b: "es_abc hojas a = es_abc size a"
by (induct a) (auto simp add: auxEj8)

(* danrodcha *)
lemma aux3: "es_abc hojas a ⟹ (hojas a = 1 + size a)"
by (induct a) simp_all

(* danrodcha *)
lemma 8: "es_abc hojas a = es_abc size a"
apply (induct a)
apply simp
apply (auto simp add: aux3)
done

(* ferrenseg *)
lemma [simp]: "hojas a = size a + 1"
by (induct a) auto

theorem es_abc_hojas_size: "es_abc hojas a = es_abc size a"
by (induct a) auto

(* fracorjim1 *)
lemma aux10 : "hojas (a::arbol) = Suc(size a)"
apply (induct a)
apply auto
done

lemma es_abc_hojas_size_10 : "es_abc hojas (a::arbol) = es_abc size a"
apply (induct a)
apply (auto simp add: aux10)
done

(* dancorgar *)
lemma rel_hojas_size: "es_abc hojas a ∧ es_abc size a
⟹ hojas a = (size a) + 1"
by (induct a) (auto)

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

(* pabrodmac *)

lemma lemaaux2:"es_abc size a⟶(hojas a = Suc(size a)) "
by (induct a) auto

lemma 2:"es_abc hojas a = es_abc size a "
by (induct a) (auto simp add: lemaaux2)

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
serrodcal crigomgom rubgonmar danrodcha ivamenjim marcarmor13
paupeddeg juacabsou josgarsan bowma lucnovdos antsancab1 jeamacpov *)
lemma lej9: "es_abc profundidad a = es_abc size a"
by (simp add: lej7 lej8)

(* ferrenseg *)
corollary es_abc_size_profundidad: "es_abc size a = es_abc profundidad a"
by (simp add: es_abc_profundidad_hojas es_abc_hojas_size)

(* pabrodmac *)

lemma lemaaux3:"es_abc profundidad a⟶(Suc(size a)= 2^(profundidad a)) "
by (induct a) auto

lemma "es_abc profundidad a = es_abc size a "
by (simp add: 1 2 )

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

(* fraortmoy marpoldia1 migtermor wilmorort pablucoto serrodcal
crigomgom marcarmor13 paupeddeg juacabsou dancorgar lucnovdos jeamacpov pabrodmac *)
lemma lej10: "es_abc profundidad (abc n)"
by (induct n) auto

(* anaprarod rubgonmar danrodcha ferrenseg ivamenjim
paupeddeg juacabsou bowma antsancab1 *)
(* con un demostrador más débil *)
(* y en general para cualquier medida *)
lemma L10: "es_abc f (abc a)"
by (induct a) simp_all

(* ivamenjim *)
(* Igual que el anterior pero usando auto *)
lemma lej10b: "es_abc f (abc n)"
by (induct n) auto

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

(* fraortmoy marpoldia1 migtermor wilmorort pablucoto serrodcal
marcarmor13 antsancab1 jeamacpov *)
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 crigomgom rubgonmar ferrenseg ivamenjim paupeddeg juacabsou
dancorgar bowma lucnovdos pabrodmac *)
lemma "es_abc profundidad a ⟶ (a = (abc (profundidad a)))"
by (induct a) auto

(* danrodcha*)
lemma 11:"es_abc profundidad a ⟹ (a = (abc (profundidad a)))"
by (induct a) simp_all

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 serrodcal danrodcha marcarmor13
ferrenseg ivamenjim paupeddeg juacabsou rubgonmar bowma lucnovdos antsancab1 jeamacpov *)
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
(* el contraejemplo que encuentra es la medida constante a1 *)

(*crigomgom *)
(* Como en la primera de las soluciones he usado la función constante 0
pero he usado una expresión lambda *)
lemma "es_abc (λx. 0::nat) a = es_abc size a"
quickcheck
oops

(* dancorgar *)
fun medida :: "arbol => bool" where
"medida H = True"
| "medida (N i d) = ((profundidad i) = (size d))"

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

(* pabrodmac *)

fun todossoniguales :: "arbol => nat" where
"todossoniguales H = 1"
| "todossoniguales (N i d) = 1"

value "todossoniguales (H) = 1"
value "todossoniguales (N (N H H) (N H H)) = 1"
value "todossoniguales (N (N (N H H) (N H H)) (N (N H H) (N H H))) = 1"

lemma "es_abc todossoniguales a = es_abc size a"
oops

end
</source>

Relación 8

2018-07-16T11:11:21Z

WikiSysop: Texto reemplazado: «isar» por «isabelle»

<source lang="isabelle">
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 *)
--"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 jeamacpov migtermor *)
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 marpoldia1 manmorjim1 crigomgom juancabsou ferrenseg fraortmoy rubgonmar *)
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 paupeddeg pabrodmac fracorjim1 *)
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

(* josgarsan *)
lemma ejercicio_1_6:
assumes 1: "¬q ⟶ ¬p"
shows "p ⟶ q"
proof (cases)
assume "p"
then show "p⟶q" using 1 by simp
next
assume "¬p"
then show "p⟶q" by simp
qed

(* dancorgar *)
lemma ejercicio_1_7:
assumes "¬q ⟶ ¬p"
shows "p ⟶ q"
proof-
{assume "p"
assume "¬q"
have "¬p" using assms(1) `¬q` by (rule mp)}
thus "p ⟶ q"
apply simp
apply (rule ccontr)
apply simp
done
qed

(* antsancab1 *)
lemma ejercicio_1:
assumes 1: "¬q ⟶ ¬p"
shows "p ⟶ q"
proof -
{assume p
hence "¬¬p" by (rule notnotI)
with 1 have "¬¬q" by (rule mt)
hence "q" by (rule notnotD)
}
then show "p ⟶ q" by (rule impI)
qed

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

(* marcarmor13 *)
-- "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

(* Comentario: No se corresponde con el enunciado. *)

(* ivamenjim serrodcal marpoldia1 antsancab1 *)
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

(* Comentario: Usa auto. *)

(* pablucoto jeamacpov *)
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 bowma *)
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 crigomgom juancabsou ferrenseg fraortmoy fracorjim1 rubgonmar *)
(* Igual que el anterior pero con etiquetas *)
lemma ejercicio_2_4:
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

(* migtermor *)
lemma lem:
shows "p ∨ ¬p"
proof -
{assume 1: "¬(p∨¬p)"
{assume 2: "p"
then have 3: "p∨¬p" by (rule disjI1)
also have 4: "False" using 1 3 by (rule notE)}
then have 5: "¬p" by (rule notI)
then have 6: "p∨¬p" by (rule disjI2)
also have 7: "False" using 1 6 by (rule notE)}
thus "p∨¬p" by (rule ccontr)
qed

lemma ejercicio_2_5:
assumes 1: "¬(¬p∧¬q)"
shows "p∨q"
proof -
have 2: "p∨¬p" by (rule lem)
moreover
{assume 3: p
then have 4: "p∨q" by (rule disjI1)}
moreover
{assume 6: "¬p"
{assume 7: "¬q"
also have 8: "¬p∧¬q" using 6 7 by (rule conjI)
have 9: "False" using 1 8 by (rule notE)}
then have 10: "q" by (rule ccontr)
then have 11: "p∨q" by (rule disjI2)}
ultimately show "p∨q" by (rule disjE)
qed

(* paupeddeg pabrodmac *)
lemma aux1:
assumes "¬(p ∧ q)"
shows "¬p ∨ ¬q"
proof -
have "¬p ∨ p" by (rule excluded_middle)
thus "¬p ∨ ¬q"
proof
{assume "¬p"
thus "¬p ∨ ¬q" by (rule disjI1)}
next
{assume "p"
have "¬q"
proof
{assume "q"
have "(p ∧ q)" using `p` `q` by (rule conjI)
show "False" using `¬(p ∧ q)` `p ∧ q` by (rule notE)}
qed
thus "¬p ∨ ¬q" by (rule disjI2)}
qed
qed

lemma ejercicio_2_5b:
assumes "¬(¬p ∧ ¬q)"
shows "p ∨ q"
proof -
have "¬¬p ∨ ¬¬q" using assms by (rule aux1)
moreover
{assume "¬¬p"
hence "p" by (rule notnotD)
hence "p ∨ q" by (rule disjI1) }
moreover
{assume "¬¬q"
hence "q" by (rule notnotD)
hence "p ∨ q" by (rule disjI2) }
ultimately show "p ∨ q" by (rule disjE)
qed

(* pabrodmac *)
lemma ejercicio_2_6:
assumes "¬(¬p ∧ ¬q)"
shows "p ∨ q"
proof -
have "¬¬p ∨ ¬¬q" using assms by (rule Meson.not_conjD)
moreover
{ assume "¬¬p"
hence "p" by (rule notnotD)
hence "p ∨ q" by (rule disjI1) }
moreover
{ assume "¬¬q"
then have "q" by (rule notnotD)
hence "p ∨ q" by (rule disjI2) }
ultimately show "p ∨ q" by (rule disjE)
qed

(* dancorgar *)
lemma ejercicio_2_7:
assumes "¬(¬p ∧ ¬q)"
shows "p ∨ q"
proof-
{assume "(¬p ∧ ¬q)"
have "p" using assms `(¬p ∧ ¬q)` by (rule notE)
then have "p ∨ q" by (rule disjI1)}
then show "p ∨ q"
apply simp
apply (rule ccontr)
apply simp
done
qed

(* josgarsan *)
lemma ejercicio_2_8:
assumes 1: " ¬(¬p ∧ ¬q)"
shows "p ∨ q"
proof (cases)
assume "p"
then show "p ∨ q" using disjI1 by simp
next
assume "¬p"
then have "q" using 1 conjunct2 by simp
then show "p ∨ q" using disjI2 by simp
qed

(* Comentario: Uso de simp. *)

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

(* marcarmor13 serrodcal marpoldia1 antsancab1 *)
--"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

(* Comentario: Uso de la negación de la hipótesis. *)

(* pablucoto jeamacpov *)
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

(* Comentario: Uso de auxiliar con auto. *)

(* danrodcha migtermor bowma *)
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 crigomgom juancabsou ferrenseg fraortmoy fracorjim1 rubgonmar *)
(* Igual que el anterior pero con etiquetas *)
lemma ejercicio_3_4:
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

(* paupeddeg pabrodmac *)
lemma aux2:
assumes "¬(p ∨ q)"
shows "¬p ∧ ¬q"
proof -
have "¬p ∨ p" by (rule excluded_middle)
thus "¬p ∧ ¬q"
proof
{assume "¬p"
have "¬q ∨ q" by (rule excluded_middle)
thus "¬p ∧ ¬q"
proof
{assume "¬q"
show "¬p ∧ ¬q" using `¬p` `¬q` by (rule conjI)}
{assume "q"
hence "(p ∨ q)" by (rule disjI2)
have "False" using `¬(p ∨ q)` `p ∨ q` by (rule notE)
thus "¬p ∧ ¬q" by (rule FalseE) }
qed}
{assume "p"
hence "(p ∨ q)" by (rule disjI1)
have "False" using `¬(p ∨ q)` `p ∨ q` by (rule notE)
thus "¬p ∧ ¬q" by (rule FalseE) }
qed
qed

lemma ejercicio_3_5:
assumes "¬(¬p ∨ ¬q)"
shows "p ∧ q"
proof -
have "¬¬p ∧ ¬¬q" using assms by (rule aux2)
hence "¬¬p" by (rule conjunct1)
hence "p" by (rule notnotD)
have "¬¬q" using `¬¬p ∧ ¬¬q` by (rule conjunct2)
hence "q" by (rule notnotD)
show "p ∧ q" using `p` `q` by (rule conjI)
qed

(* pabrodmac *)
lemma ejercicio_3_6:
assumes "¬(¬p ∨ ¬q)"
shows "p ∧ q"
proof
have "¬¬p ∧ ¬¬q" using assms by (rule Meson.not_disjD)
hence "¬¬p" by (rule conjunct1)
show "p" using `¬¬p` by (rule notnotD)
next
have "¬¬p ∧ ¬¬q" using assms by (rule Meson.not_disjD)
hence "¬¬q" by (rule conjunct2)
show "q" using `¬¬q` by (rule notnotD)
qed

(* pabrodmac *)
lemma ejercicio_3_7:
assumes "¬(¬p ∨ ¬q)"
shows "p ∧ q"
proof
have "¬¬p ∧ ¬¬q" using assms by (rule Meson.not_disjD)
have "¬¬p" using `¬¬p ∧ ¬¬q` by (rule conjunct1)
have "¬¬q" using `¬¬p ∧ ¬¬q` by (rule conjunct2)
show "p" using `¬¬p` by (rule notnotD)
show "q" using `¬¬q` by (rule notnotD)
qed

(* dancorgar *)
lemma ejercicio_3_8:
assumes "¬(¬p ∨ ¬q)"
shows "p ∧ q"
proof -
{assume "(¬p ∨ ¬q)"
have "p" using assms `(¬p ∨ ¬q)` by (rule notE)
have "q" using assms `(¬p ∨ ¬q)` by (rule notE)
have "p ∧ q" using `p` `q` by (rule conjI)}
then show "p ∧ q"
apply simp
apply (rule ccontr)
apply simp
done
qed

(* josgarsan *)
lemma ejercicio_3_9:
assumes 1: "¬(¬p ∨ ¬q)"
shows "p ∧ q"
proof (rule ccontr)
assume "¬(p ∧ q)"
then show False using 1 by simp
qed

(* Comentario: Uso de simp. *)

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

(* marcarmor13 serrodcal *)
--"usando un supuesto p ∧ q"
lemma ejercicio_4_1:
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

(* Comentario: Uso de la negación de la hipótesis.*)

(* ivamenjim marpoldia1 ferrenseg fraortmoy rubgonmar antsancab1 *)
lemma ejercicio_4_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

(* Comentario: Uso de auto. *)

(* pablucoto jeamacpov crigomgom juancabsou *)
lemma ejercicio_4_3:
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 bowma *)
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_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

(* migtermor *)
lemma ejercicio_4_5:
assumes 1: "¬(p∧q)"
shows "¬p∨¬q"
proof -
have 2: "p∨¬p" by (rule lem)
moreover
{assume 3: "p"
{assume 4: "q"
also have 5: "p∧q" using 3 4 by (rule conjI)
have 6: "False" using assms 5 by (rule notE)}
then have 7: "¬q" by (rule notI)
then have 8: "¬p∨¬q" by (rule disjI2)}
moreover
{assume 9: "¬p"
then have "¬p∨¬q" by (rule disjI1)}
ultimately show "¬p∨¬q" by (rule disjE)
qed

(* paupeddeg pabrodmac*)
lemma ejercicio_4_6:
assumes "¬(p ∧ q)"
shows "¬p ∨ ¬q"
proof -
have "¬p ∨ p" by (rule excluded_middle)
thus "¬p ∨ ¬q"
proof
{ assume "¬p"
thus "¬p ∨ ¬q" by (rule disjI1)}
next
{ assume "p"
have "¬q"
proof
{ assume "q"
have "(p ∧ q)" using `p` `q` by (rule conjI)
show "False" using `¬(p ∧ q)` `p ∧ q` by (rule notE)}
qed
thus "¬p ∨ ¬q" by (rule disjI2)}
qed
qed

(* pabrodmac*)
lemma ejercicio_4_7:
assumes "¬(p ∧ q)"
shows "¬p ∨ ¬q"
proof -
show "¬p ∨ ¬q" using assms by (rule Meson.not_conjD)
qed

(* dancorgar *)
lemma ejercicio_4_8:
assumes "¬(p ∧ q)"
shows "¬p ∨ ¬q"
proof-
{assume "(p ∧ q)"
have "¬p" using assms `(p ∧ q)` by (rule notE)
then have "¬p ∨ ¬q" by (rule disjI1)}
then show "¬p ∨ ¬q"
apply simp
apply (rule ccontr)
apply simp
done
qed

(* josgarsan *)
lemma ejercicio_4:
assumes 1: "¬(p ∧ q)"
shows "¬p ∨ ¬q"
proof (rule ccontr)
assume "¬(¬ p ∨ ¬ q)"
then show False using 1 by simp
qed

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

(* marcarmor13 jeamacpov serrodcal juancabsou *)
--"usando un supuesto q"
lemma ejercicio_5_1:
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

(* Comentario: Hipótesis extra.*)

(* ivamenjim marpoldia1 ferrenseg fraortmoy rubgonmar antsancab1 *)
lemma ejercicio_5_2:
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

(* Comentario: Uso de auto. *)

(* 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

(* anaprarod *)
(* Muy parecida a la anterior pero con algunas etiquetas
y con algunas implicaciones más detalladas *)
lemma ejercicio_5_3:
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 1: "¬p ⟶ ¬q" by (rule impI)
{assume "q"
hence 2: "¬¬q" by (rule notnotI)
have "¬¬p" using 1 2 by (rule mt)
hence "p" by (rule notnotD)}
hence "q ⟶ p" by (rule impI)
thus "(p ⟶ q) ∨ (q ⟶ p)" by (rule disjI2)}
next
{assume 3: "q"
{assume "p"
have "q" using 3 by this}
hence "p ⟶ q" by (rule impI)
thus "(p ⟶ q) ∨ (q ⟶ p)" by (rule disjI1)}
qed}
next
{assume 4: "p"
{assume "q"
have "p" using 4 by this}
hence "q ⟶ p" by (rule impI)
thus "(p ⟶ q) ∨ (q ⟶ p)" by (rule disjI2)}
qed
qed

(* migtermor *)
lemma ejercicio_5_4:
shows "(p ⟶ q) ∨ (q ⟶ p)"
proof -
have 1: "q∨¬q" by (rule lem)
moreover
{assume 2: "q"
have 3: "p ⟶ q" using 2 by (rule impI)
then have 4: "(p ⟶ q) ∨ (q ⟶ p)" by (rule disjI1)}
moreover
{assume 5: "¬q"
have 6: "¬p⟶¬q" using 5 by (rule impI)
then have 7: "q⟶p" by (rule ejercicio_1_2)
then have "(p ⟶ q) ∨ (q ⟶ p)" by (rule disjI2)}
ultimately show "(p ⟶ q) ∨ (q ⟶ p)" by (rule disjE)
qed

(* paupeddeg pabrodmac crigomgom bowma *)
lemma aux3:
assumes "¬q ∨ p"
shows "q ⟶ p"
proof -
note `¬q ∨ p`
moreover
{assume "¬q"
have "q ⟶ p"
proof
assume "q"
show "p" using `¬q``q`by (rule notE)
qed}
moreover
{assume "p"
have "q ⟶ p"
proof
assume "q"
show "p" using `p` by this
qed}
ultimately show "q ⟶ p" by (rule disjE)
qed

lemma ejercicio_5_5:
shows "(p ⟶ q) ∨ (q ⟶ p)"
proof -
have "¬p ∨ p" by (rule excluded_middle)
thus "(p ⟶ q) ∨ (q ⟶ p)"
proof
{assume "¬p"
hence "¬p ∨ q" by (rule disjI1)
hence "p ⟶ q" by (rule aux3)
thus "(p ⟶ q) ∨ (q ⟶ p)" by (rule disjI1)}
{assume "p"
hence "¬q ∨ p" by (rule disjI2)
hence "q ⟶ p" by (rule aux3)
thus "(p ⟶ q) ∨ (q ⟶ p)" by (rule disjI2)}
qed
qed

(*pabrodmac*)
lemma ejercicio_5_6:
"(p ⟶ q) ∨ (q ⟶ p)"
proof -
have "¬(p ⟶ q) ∨ (p ⟶ q)" by (rule excluded_middle)
thus "(p ⟶ q) ∨ (q ⟶ p)"
proof
{ assume "¬(p ⟶ q)"
hence "p ∧ ¬ q" by (rule Meson.not_impD)
have "(q ⟶ p)"
proof
assume "q"
show "p" using `p ∧ ¬ q` by (rule conjunct1)
qed
thus "(p ⟶ q) ∨ (q ⟶ p)" by (rule disjI2) }
next
{ assume "p ⟶ q"
then show "(p ⟶ q) ∨ (q ⟶ p)" by (rule disjI1) }
qed
qed

(* josgarsan *)
lemma ejercicio_5_7:
shows "(p ⟶ q) ∨ (q ⟶ p)"
proof (cases)
assume "p"
then show "(p ⟶ q) ∨ (q ⟶ p)" by simp
next
assume "¬p"
then show "(p ⟶ q) ∨ (q ⟶ p)" by simp
qed

(* Comentario: Uso de simp. *)

(* dancorgar *)
lemma ejercicio_5_8:
shows "(p ⟶ q) ∨ (q ⟶ p)"
proof (cases)
assume "p"
have "(q ⟶ p)" using `p` by (rule impI)
then show "(p ⟶ q) ∨ (q ⟶ p)" by (rule disjI2)
next
assume "¬p"
have "(p ⟶ q)" using `¬p` by simp
then show "(p ⟶ q) ∨ (q ⟶ p)" by (rule disjI1)
qed

end
</source>

Relación 9

2018-07-16T11:11:21Z

WikiSysop: Texto reemplazado: «isar» por «isabelle»

<source lang="isabelle">
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 marcarmor13 jeamacpov*)
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 *)

(* antsancab1 *)
lemma ejercicio_1_4:
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)
obtain b where "Q b" using 2 by (rule exE)
then have "P a ⟶ Q b" by (rule impI)
then have "∃x. P a ⟶ Q x" by (rule exI)
}
then show "∃x. P a ⟶ Q x" by simp
qed

(* pabrodmac *)
lemma ejercicio_1_5:
fixes P Q :: "'b ⇒ bool"
assumes "P a ⟶ (∃x. Q x)"
shows "∃x. P a ⟶ Q x"
proof -
have "¬(P a) ∨ P a" by (rule excluded_middle)
thus "∃x. P a ⟶ Q x"
proof (rule disjE)
assume "¬(P a)"
have "P a ⟶ Q a"
proof (rule impI)
assume "P a"
with `¬(P a)` show "Q a" by (rule notE)
qed
thus "∃x. P a ⟶ Q x" by (rule exI)
next
assume "P a"
with assms have "∃x. Q x" by (rule mp)
then obtain b where "Q b" by (rule exE)
have "P a ⟶ Q b"
proof (rule impI)
assume "P a"
note `Q b`
thus "Q b" by this
qed
thus "∃x. P a ⟶ Q x" by (rule exI)
qed
qed

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 antsancab1 jeamacpov *)
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

(* pabrodmac *)

lemma ejercicio_2_2:
assumes "∀x y z. R x y ∧ R y z ⟶ R x z"
"∀x. ¬(R x x)"
shows "∀x y. R x y ⟶ ¬(R y x)"
proof
fix a
show "∀y. R a y ⟶ ¬(R y a)"
proof
fix b
show "R a b ⟶ ¬(R b a)"
proof
assume "R a b"
show "¬(R b a)"
proof
assume "R b a"
have "∀y z. R a y ∧ R y z ⟶ R a z" using assms(1) by (rule allE)
hence "∀z. R a b ∧ R b z ⟶ R a z" by (rule allE)
hence "R a b ∧ R b a ⟶ R a a" by (rule allE)
have "R a b ∧ R b a" using `R a b` `R b a` by (rule conjI)
have "R a a" using `R a b ∧ R b a ⟶ R a a` `R a b ∧ R b a` by (rule mp)
have "¬(R a a)" using assms(2) by (rule allE)
show False using `¬(R a a)` `R a a` by (rule notE)
qed
qed
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 marcarmor13 antsancab1 pabrodmac jeamacpov *)
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 marcarmor13 jeamacpov*)
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

(* antsancab1 *)
lemma ejercicio_4:
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 -
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)
also have "P (f a) (f a) (f (f a))" using 6 3 by (rule mp)
then show "∃z. P (f a) z (f (f a))" ..
qed

(* pabrodmac *)
lemma ejercicio_4_4:
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 "P a (f a) (f a)" using assms(1) by (rule allE)
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)
thus "P (f a) (f a) (f (f a))" using `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 marcarmor13 antsancab1 jeamacpov *)
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

(* pabrodmac *)
lemma ejercicio_5:
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 "Q a (s a)" using assms(1) by (rule allE)
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)
hence " Q (s a) (s (s a))" using `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 5

2018-07-16T11:11:20Z

WikiSysop: Texto reemplazado: «isar» por «isabelle»

<source lang="isabelle">
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 marcarmor13*)
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 antsancab1 *)
(* 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 marcarmor13 antsancab1 *)
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 marcarmor13 antsancab1 *)
fun borraDuplicados :: "'a list ⇒ 'a list" where
"borraDuplicados [] = []"
| "borraDuplicados (x#xs) = (if estaEn x xs
then borraDuplicados xs
else x#borraDuplicados 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]"

(* 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 [])"

value "borraDuplicados4 [1::nat,2,4,2,3] = [1,4,2,3]"

(* Comentario: Falla en el ejemplo anterior. Su valor es [3,4,2,1] *)

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 antsancab1 *)
lemma length_borraDuplicados2:
"length ( borraDuplicados xs ) ≤ length xs"
by (induct xs) auto

(* manmorjim1 marcarmor13 *)
(* 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 marcarmor13*)
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

(* antsancab1 *)
(* En este ejercicio probé a mantener el elemento 'a' dentro del métodos length y funciona
Para no tener que hacer 1 + length xs *)
(* Duda:
¿Por qué aparece este mensaje en Isabelle al asumir la hipótesis de inducción?
Introduced fixed type variable(s): 'b in "xsa__" *)

lemma length_borraDuplicados_2j: "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 # borraDuplicados (xs))" by simp
also have "... ≤ length (a # xs)" using HI 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 marcarmor13 antsancab1 *)
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

(* Comentario: Tiene pasos incompletos.*)

(* 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 marcarmor13*)
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
oops

(* Comentario: Demostración incompleta. *)

(* 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 rubgonmar marcarmor13 *)
lemma sinDuplicados_borraDuplicados:
"sinDuplicados (borraDuplicados xs)"
by (induct xs) (auto simp add: estaEn_borraDuplicados)

(* migtermor bowma *)
lemma sinDuplicados_borraDuplicados_2:
"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_3:
"sinDuplicados (borraDuplicados xs)"
apply (induct xs)
apply simp
apply (induct xs)
apply auto
apply (simp add: noEsta_tras_borrarDuplicados)
done

(* antsancab1 *)
lemma sinDuplicados_borraDuplicados_4:
"sinDuplicados (borraDuplicados xs)"
apply (induct xs)
apply simp
apply (simp add:estaEn_borraDuplicados)
done

text {*
---------------------------------------------------------------------
Ejercicio 6.2. Demostrar o refutar detalladamente
sinDuplicados (borraDuplicados xs)
---------------------------------------------------------------------
*}

(* wilmorort pablucoto marcarmor13*)
lemma sinDuplicados_borraDuplicados_2a:
"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 antsancab1 *)
lemma sinDuplicados_borraDuplicados_2b:
"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_2c:
"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_2d:
"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`
using 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 marcarmor13 antsancab1 *)
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 6

2018-07-16T11:11:20Z

WikiSysop: Texto reemplazado: «isar» por «isabelle»

<source lang="isabelle">
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 3
se representa por "N e (N c (H a) (H d)) (N g (H f) (H h))".
---------------------------------------------------------------------
*}

(* ivamenjim marpoldia1 manmorjim1 bowma migtermor wilmorort
juacabsou serrodcal pabrodmac ferrenseg rubgonmar paupeddeg
crigomgom danrodcha jeamacpov marcarmor13 josgarsan fraortmoy
dancorgar fracorjim1 anaprarod antsancab1 *)

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 fraortmoy migtermor
wilmorort lucnovdos serrodcal pabrodmac jeamacpov paupeddeg
marcarmor13 josgarsan dancorgar anaprarod antsancab1 *)

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 juacabsou ferrenseg
rubgonmar paupeddeg fracorjim1 *)
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 fraortmoy migtermor wilmorort lucnovdos
juacabsou serrodcal pabrodmac ferrenseg jeamacpov
rubgonmar paupeddeg marcarmor13 josgarsan dancorgar fracorjim1 anaprarod antsancab1 *)

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 fraortmoy
migtermor wilmorort lucnovdos juacabsou serrodcal pabrodmac
ferrenseg jeamacpov rubgonmar paupeddeg marcarmor13 josgarsan
dancorgar anaprarod antsancab1 *)
fun inOrden :: "'a arbol ⇒ 'a list" where
"inOrden (H t) = [t]"
| "inOrden (N t i d) = (inOrden i) @ [t] @ (inOrden d)"

(* danrodcha manmorjim1 fracorjim1 *)
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 fraortmoy migtermor wilmorort lucnovdos
juacabsou serrodcal pabrodmac ferrenseg jeamacpov rubgonmar
paupeddeg marcarmor13 josgarsan dancorgar fracorjim1 anaprarod antsancab1 *)
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 migtermor wilmorort juacabsou serrodcal dancorgar josgarsan
*)
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 paupeddeg anaprarod *)
lemma "preOrden (espejo a) = rev (postOrden a)"
by (induct a) simp_all

(* danrodcha crigomgom fracorjim1 *)
lemma "preOrden1 (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 "preOrden1 (espejo (N x i d)) =
preOrden1 (N x (espejo d) (espejo i))"
by (simp only: espejo.simps(2))
also have "… = x#preOrden1 (espejo d) @ preOrden1 (espejo i)"
by (simp only: preOrden1.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 fraortmoy anaprarod *)
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 marpoldia1 jeamacpov paupeddeg marcarmor13 anaprarod *)
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

(* Comentario sobre tipo inducido. *)

(* fraortmoy lucnovdos pabrodmac*)
lemma "preOrden (espejo a) = rev (postOrden a)"
by (induct a) auto

(*pabrodmac*)
lemma
fixes a ::"'b arbol"
shows "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"
show "?P (N x i d)"
proof -
have "preOrden (espejo (N x i d)) =
preOrden(N x (espejo d) (espejo i))" by simp
also have "… = [x]@rev(postOrden d)@rev(postOrden i)"
using h1 h2 by simp
also have "… = rev(postOrden (N x i d))" by simp
finally show ?thesis .
qed
qed

(* ferrenseg rubgonmar *)
lemma "preOrden (espejo a) = rev (postOrden a)" (is "?P a")
proof (induct a)
fix x
show "?P (H x)" by simp
next
fix x l r
assume H1: "?P l"
assume H2: "?P r"
show "?P (N x l r)"
proof -
have "preOrden (espejo (N x l r)) =
preOrden (N x (espejo r) (espejo l))" by simp
also have "… = x # (preOrden (espejo r) @ preOrden (espejo l))" by simp
also have "… = x # (rev (postOrden r) @ rev (postOrden l))"
using H1 H2 by simp
also have "… = x # rev (postOrden l @ postOrden r)" by simp
also have "… = rev ((postOrden l) @ (postOrden r) @ [x])" by simp
also have "… = rev (postOrden (N x l r))" by simp
finally show ?thesis .
qed
qed

(* antsancab1 *)
(* Le he puesto el nombre para utilizarlo en la siguiente demostración *)
lemma pre_es_rev_post: "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

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

(* ivamenjim crigomgom bowma migtermor wilmorort juacabsou serrodcal
dancorgar josgarsan antsancab1 *)
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 fraortmoy anaprarod *)
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 marpoldia1 jeamacpov paupeddeg rubgonmar anaprarod *)
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

(* fraortmoy lucnovdos pabrodmac paupeddeg marcarmor13 *)
lemma "postOrden (espejo a) = rev (preOrden a)"
by (induct a) auto

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

(*pabrodmac*)
lemma
fixes a ::"'b arbol"
shows "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"
show "?P (N x i d)"
proof -
have "postOrden (espejo (N x i d)) =
postOrden(N x (espejo d) (espejo i))" by simp
also have "… = rev (preOrden d) @ rev (preOrden i) @ [x]"
using h1 h2 by simp
also have "… = rev (preOrden (N x i d))" by simp
finally show ?thesis.
qed
qed

(* ferrenseg *)
lemma "postOrden (espejo a) = rev (preOrden a)" (is "?P a")
proof (induct a)
fix x
show "?P (H x)" by simp
next
fix x l r
assume H1: "?P l"
assume H2: "?P r"
show "?P (N x l r)"
proof -
have "postOrden (espejo (N x l r)) =
postOrden (N x (espejo r) (espejo l))" by simp
also have "… = postOrden (espejo r) @ postOrden (espejo l) @ [x]"
by simp
also have "… = rev (preOrden r) @ rev (preOrden l) @ [x]"
using H1 H2 by simp
also have "… = rev (preOrden l @ preOrden r) @ [x]" by simp
also have "… = rev ([x] @ preOrden l @ preOrden r)" by simp
also have "… = rev (preOrden (N x l r))" by simp
finally show ?thesis .
qed
qed

(* antsancab1 *)
(* Después de hacerlo como en la anterior demostración se me ocurrió cómo
relacionar ambas demostraciones.
Como hemos demostrado que
preOrden (espejo a) = rev (postOrden a)
a la inversa también queda demostrado *)
lemma "postOrden (espejo a) = rev (preOrden a)"
apply (induct a)
apply simp
apply (simp add:pre_es_rev_post)
done

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

(* ivamenjim crigomgom bowma migtermor wilmorort juacabsou serrodcal
dancorgar josgarsan antsancab1 *)
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 anaprarod *)
theorem "inOrden (espejo a) = rev (inOrden a)"
by (induct a) simp_all

(* danrodcha anaprarod *)
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 marpoldia1 jeamacpov paupeddeg marcarmor13 anaprarod *)
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

(* lucnovdos pabrodmac paupeddeg fraortmoy *)
theorem "inOrden (espejo a) = rev (inOrden a)"
by (induct a) auto

(* pabrodmac *)
lemma
fixes a ::"'b arbol"
shows "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"
show "?P (N x i d)"
proof -
have "inOrden (espejo (N x i d)) =
inOrden(N x (espejo d) (espejo i))" by simp
also have "… =rev(inOrden d)@[x]@rev(inOrden i)" using h1 h2 by simp
also have "… =rev(inOrden (N x i d))" by simp
finally show ?thesis .
qed
qed

(* ferrenseg rubgonmar *)
theorem "inOrden (espejo a) = rev (inOrden a)" (is "?P a")
proof (induct a)
fix x
show "?P (H x)" by simp
next
fix x l r
assume H1: "?P l"
assume H2: "?P r"
show "?P (N x l r)"
proof -
have "inOrden (espejo (N x l r)) =
inOrden (N x (espejo r) (espejo l))" by simp
also have "… = inOrden (espejo r) @ [x] @ inOrden (espejo l)"
by simp
also have "… = rev (inOrden r) @ [x] @ rev (inOrden l)"
using H1 H2 by simp
also have "… = rev (inOrden l @ [x] @ inOrden r)" by simp
also have "… = rev (inOrden (N x l r))" by simp
finally show ?thesis .
qed
qed

(* fraortmoy *)
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"
show "inOrden (espejo (N x i d)) = rev (inOrden (N x i d))"
using h1 h2 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
migtermor marpoldia1 wilmorort lucnovdos juacabsou serrodcal
ferrenseg paupeddeg rubgonmar jeamacpov marcarmor13 fraortmoy
fracorjim1 josgarsan dancorgar anaprarod antsancab1 *)
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
migtermor marpoldia1 wilmorort lucnovdos juacabsou serrodcal
pabrodmac ferrenseg jeamacpov paupeddeg rubgonmar marcarmor13
fraortmoy fracorjim1 josgarsan dancorgar anaprarod antsancab1 *)
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"

(* bowma *)
fun extremo_izquierda_1 :: "'a arbol ⇒ 'a" where
"extremo_izquierda_1 (H t) = t"
| "extremo_izquierda_1 (N t i d) = hd (inOrden (N t i d))"

(* ivamenjim *)
(* Metaejercicio de demostración.
Llamando teorema_13 al teorema del ejercicio 13 *)

lemma "extremo_izquierda a = extremo_izquierda_1 a"
by (induct a, simp_all add: aux_ej12_1 teorema_13)

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
migtermor marpoldia1 wilmorort lucnovdos juacabsou pabrodmac
serrodcal ferrenseg jeamacpov paupeddeg rubgonmar marcarmor13
fraortmoy fracorjim1 josgarsan dancorgar anaprarod antsancab1 *)
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))"

(* ivamenjim *)
(* Metaejercicio de demostración.
Llamando teorema_12 al teorema del ejercicio 12 *)

(*
lemma "extremo_derecha a = extremo_derecha_1 a"
by (induct a, simp_all add: aux_ej12_1 teorema_12)
*)

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

(* danrodcha anaprarod *)
lemma aux_ej12: "inOrden a ≠ []"
apply (induct a)
apply simp (* poniendo simp_all se agrupan estos dos *)
apply simp
done

(* danrodcha pablucoto crigomgom wilmorort juacabsou serrodcal
rubgonmar jeamacpov marcarmor13*)
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 marpoldia1 paupeddeg anaprarod antsancab1 *)
(* 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

(* lucnovdos*)
(* El mismo que el anterior,pero sin usar patrones *)
theorem "last (inOrden a) = extremo_derecha a"
proof (induct a)
fix x ::"'a"
show "last (inOrden (H x)) = extremo_derecha (H x)" by simp
next
fix x ::"'a"
fix i d ::"'a arbol"
assume HI: "last (inOrden d) = extremo_derecha 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)
also have "… = extremo_derecha d" using HI by simp
also have "… = extremo_derecha (N x i d)" by simp
finally show "last (inOrden (N x i d)) = extremo_derecha (N x i d)"
by simp
qed

(* migtermor *)
theorem "last (inOrden a) = extremo_derecha a" (is "?P a")
proof (induct a)
fix h
show "?P (H h)" by simp
next
fix n i
fix d assume HId: "?P d"
have AUX: "¬ (inOrden d = [])" (is "?Q d")
proof (induct d)
fix hd
show "?Q (H hd)" by simp
next
fix nd
fix id assume HIid: "?Q id"
fix dd assume HIdd: "?Q dd"
show "?Q (N nd id dd)" using HIid HIdd by simp
qed
have "last (inOrden (N n i d)) = last (inOrden i @[n]@inOrden d)"
by simp
also have "… = last (inOrden d)" using AUX by simp
also have "… = extremo_derecha d" using HId by simp
finally show "?P (N n i d)" by simp
qed

(* pabrodmac fraortmoy *)
lemma Aux_ej12: "inOrden a ≠ []"
by (induct a) auto

(* pabrodmac fraortmoy *)
theorem "last (inOrden a) = extremo_derecha a"
by (induct a)(auto simp add: Aux_ej12)

(* pabrodmac *)
lemma
fixes a ::"'b arbol"
shows "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"
show "?P (N x i d)"
proof -
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)
also have "… = extremo_derecha d" using h1 h2 by simp
also have "… = extremo_derecha (N x i d)" by simp
finally show ?thesis .
qed
qed

(* ferrenseg *)
theorem "last (inOrden a) = extremo_derecha a" (is "?P a")
proof (induct a)
fix x
show "?P (H x)" by simp
next
fix x l r
assume HI: "?P r"
show "?P (N x l r)"
proof -
have "last (inOrden (N x l r)) =
last (inOrden r @ [x] @ inOrden r)" by simp
also have "… = last (inOrden r)" by (simp add: inOrden)
also have "… = extremo_derecha r" using HI by simp
also have "… = extremo_derecha (N x l r)" by simp
finally show ?thesis .
qed
qed

(* fraortmoy *)
lemma
fixes a ::"'b arbol"
shows "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"
show "?P (N x i d)" using h1 h2 by (simp add: Aux_ej12)
qed

(* dancorgar *)
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 H2: "?P d"
have "last (inOrden (N t i d)) = last ((inOrden i)@[t]@(inOrden d))"
by simp
also have "… = last (inOrden d)"
proof (induct d)
fix x
show "last (inOrden i @ [t] @ inOrden (H x)) = last (inOrden (H x))"
by simp
next
fix x1a d1 d2
show "last (inOrden i @ [t] @ inOrden (N x1a d1 d2)) =
last (inOrden (N x1a d1 d2))" by simp
qed
finally show "last (inOrden (N t i d)) = extremo_derecha (N t i d)"
using H2 by simp
qed

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

(* danrodcha pablucoto crigomgom juacabsou serrodcal jeamacpov
marcarmor13 *)
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 lucnovdos dancorgar anaprarod *)
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

(* migtermor *)
theorem "hd (inOrden a) = extremo_izquierda a" (is "?P a")
proof (induct a)
fix h
show "?P (H h)" by simp
next
fix n d
fix i assume HId: "?P i"
have AUX: "¬ (inOrden i = [])" (is "?Q i")
proof (induct i)
fix hi
show "?Q (H hi)" by simp
next
fix ni
fix ii assume HIid: "?Q ii"
fix di assume HIdd: "?Q di"
show "?Q (N ni ii di)" using HIid HIdd by simp
qed
have "hd (inOrden (N n i d)) = hd (inOrden i @[n]@inOrden d)" by simp
also have "… = hd (inOrden i)" using AUX by simp
also have "… = extremo_izquierda i" using HId by simp
finally show "?P (N n i d)" by simp
qed

(* ivamenjim marpoldia1 wilmorort paupeddeg rubgonmar antsancab1 *)
theorem "hd (inOrden a) = extremo_izquierda 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 "hd (inOrden (N x i d)) = hd ((inOrden i) @ [x] @ (inOrden d))"
by simp
also have "... = hd (inOrden i)" by (simp add: aux_ej12_1)
also have "... = extremo_izquierda i" using h1 by simp
finally show "hd (inOrden (N x i d)) = extremo_izquierda (N x i d)"
by simp
qed

(* pabrodmac paupeddeg*)
theorem "hd (inOrden a) = extremo_izquierda a"
by (induct a)(auto simp add: Aux_ej12)

(* pabrodmac *)
lemma
fixes a ::"'b arbol"
shows "hd (inOrden a) = extremo_izquierda 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"
show "?P (N x i d)"
proof -
have "hd (inOrden (N x i d)) = hd((inOrden i)@ [x] @ (inOrden d))"
by simp
also have "… = hd (inOrden i)" by (simp add: Aux_ej12)
also have "… = extremo_izquierda i" using h1 h2 by simp
also have "… = extremo_izquierda (N x i d)" by simp
finally show ?thesis .
qed
qed

(* ferrenseg *)
theorem "hd (inOrden a) = extremo_izquierda a" (is "?P a")
proof (induct a)
fix x
show "?P (H x)" by simp
next
fix x l r
assume HI: "?P l"
show "?P (N x l r)"
proof -
have "hd (inOrden (N x l r)) =
hd (inOrden l @ [x] @ inOrden r)" by simp
also have "… = hd (inOrden l)" by (simp add: Aux_ej12)
also have "… = extremo_izquierda l" using HI by simp
also have "… = extremo_izquierda (N x l r)" by simp
finally show ?thesis .
qed
qed

(* fraortmoy *)
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 h1: "?P i"
assume h2: "?P d"
show "?P (N x i d)" using h1 h2 by (simp add: Aux_ej12)
qed

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

(* danrodcha pabrodmac dancorgar anaprarod *)
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 (preOrden1 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 (preOrden1 (N x i d)) = hd (x#preOrden1 i @ preOrden1 d)"
by (simp only: preOrden1.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 crigomgom bowma marpoldia1 wilmorort lucnovdos juacabsou
jeamacpov paupeddeg rubgonmar marcarmor13 anaprarod *)
(*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

(* migtermor *)
theorem "hd (preOrden a) = last (postOrden a)" (is "?P a")
proof (induct a)
fix h
show "?P (H h)" by simp
next
fix n i d
have "hd (preOrden (N n (i :: 'a arbol) (d :: 'a arbol))) =
hd ([n]@preOrden i@preOrden d)" by simp
(* Si no especifico que i y d son árboles, salta un error de tipo.
Supongo que será por no haber asumido hipótesis sobre ellos *)
also have "… = last (postOrden (N n i d))" by simp
show "?P (N n i d)" by simp
qed

(* ivamenjim serrodcal antsancab1 *)
theorem "hd (preOrden a) = last (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 "hd (preOrden (N x i d)) = hd ([x] @ (preOrden i) @ (preOrden d))"
by simp
also have "... = hd ([x])" by simp
finally show "hd (preOrden (N x i d)) = last (postOrden (N x i d))"
by simp
qed

(* pabrodmac paupeddeg fraortmoy *)
theorem "hd (preOrden a) = last (postOrden a)"
by (induct a) auto

(* pabrodmac *)
lemma
fixes a ::"'b arbol"
shows "hd (preOrden a) = last (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"
show "?P (N x i d)"
proof -
have "hd (preOrden (N x i d) )= x" by simp
also have "… = last (postOrden (N x i d))" by simp
finally show ?thesis .
qed
qed

(* Me he dado cuenta que no es necesario asumir ninguna hipótesis de
inducción puesto que no es necesario utilizarlas, así que no se si
está bien hecho puesto que no se aplicaría inducción *)

(* ferrenseg *)
theorem "hd (preOrden a) = last (postOrden a)" (is "?P a")
proof (cases a)
fix x
assume "a = H x"
then show "?P a" by simp
next
fix x l r
assume H: "a = N x l r"
show "?P a"
proof -
have "hd (preOrden a) = hd (preOrden (N x l r))" using H by simp
also have "… = hd (x # (preOrden l @ preOrden r))" by simp
also have "… = x" by simp
also have "… = last (postOrden l @ postOrden r @ [x])" by simp
also have "… = last (postOrden (N x l r))" by simp
also have "… = last (postOrden a)" using H by simp
finally show ?thesis .
qed
qed

(* fraortmoy *)
theorem "hd (preOrden a) = last (postOrden a)"
apply (induct a)
apply simp
apply simp
done

(* fraortmoy *)
(* como ya se ha comentado antes, no se usan hipótesis de inducción *)
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
show "?P (N x i d)" by simp
qed

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

(* danrodcha *)
theorem "hd (preOrden1 a) = raiz 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 (preOrden1 (N x i d)) = hd (x#preOrden1 i @ preOrden1 d)"
by (simp only: preOrden1.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 pabrodmac dancorgar anaprarod *)
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 crigomgom ivamenjim marpoldia1 wilmorort lucnovdos
juacabsou serrodcal jeamacpov paupeddeg rubgonmar marcarmor13 anaprarod *)
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

(* bowma *)
(* similar al anterior pero sin suponer "?p d" *)
theorem "hd (preOrden a) = last (postOrden 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 (preOrden (N t i d)) = hd ([t] @ preOrden i @ preOrden d)"
by simp
also have "… = t" by simp
also have "… = last (postOrden i @ postOrden d @ [t])" by simp
also have "… = last (postOrden (N t i d))" by simp
finally show "?p (N t i d)" by simp
qed

(* migtermor *)
theorem "hd (preOrden a) = raiz a" (is "?P a")
proof (induct a)
fix h
show "?P (H h)" by simp
next
fix n i d
have "hd (preOrden (N n (i :: 'a arbol) (d :: 'a arbol))) =
hd ([n]@preOrden i@preOrden d)" by simp
also have "… = raiz (N n i d)" by simp
finally show "?P (N n i d)" by simp
qed

(* ivamenjim: sin usar patrones *)
theorem "hd (preOrden a) = raiz a"
proof (induct a)
fix x ::"'a"
show "hd (preOrden (H x)) = raiz (H x)" by simp
next
fix x ::"'a"
fix i ::"'a arbol" assume h1: "hd (preOrden i) = raiz i"
fix d ::"'a arbol" assume h2: "hd (preOrden d) = raiz d"
have "hd (preOrden (N x i d)) = hd ([x] @ (preOrden i) @ (preOrden d))"
by simp
also have "... = hd ([x])" by simp
finally show "hd (preOrden (N x i d)) = raiz (N x i d)" by simp
qed

(* pabrodmac paupeddeg fraortmoy *)
theorem "hd (preOrden a) = raiz a"
by (induct a) auto

(* pabrodmac *)
lemma
fixes a ::"'b arbol"
shows "hd (preOrden a) = raiz 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"
show "?P (N x i d)"
proof -
have "hd (preOrden (N x i d) )= x" by simp
also have "… = raiz (N x i d)" by simp
finally show ?thesis .
qed
qed

(* Me he dado cuenta que no es necesario asumir ninguna hipótesis de
inducción puesto que no es necesario utilizarlas, así que no se si
está bien hecho puesto que no se aplicaría inducción *)

(* ferrenseg *)
theorem "hd (preOrden a) = raiz a" (is "?P a")
proof (cases a)
fix x
assume "a = H x"
then show "?P a" by simp
next
fix x l r
assume H: "a = N x l r"
show "?P a"
proof -
have "hd (preOrden a) = hd (preOrden (N x l r))" using H by simp
also have "… = hd (x#(preOrden l @ preOrden r))" by simp
also have "… = x" by simp
also have "… = raiz (N x l r)" by simp
also have "… = raiz a" using H by simp
finally show ?thesis .
qed
qed

(* fraortmoy *)
theorem "hd (preOrden a) = raiz a"
apply (induct a)
apply simp
apply simp
done

(* fraortmoy *)
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
show "?P (N x i d)" by simp
qed

(* antsancab1 *)
theorem "hd (preOrden a) = raiz 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 "hd (preOrden (N x i d)) = hd ([x] @ preOrden i @ preOrden d)" by simp
also have "... = hd [x]" by simp
also have "... = raiz (H x)" by simp
also have "... = raiz (N x i d)" by simp (* estos dos últimos no eran necesarios pero me queda más claro *)
finally show "hd (preOrden (N x i d)) = raiz (N x i d)" by simp
qed

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

(* crigomgom pablucoto bowma migtermor ivamenjim wilmorort lucnovdos
juacabsou pabrodmac serrodcal ferrenseg jeamacpov paupeddeg rubgonmar
marcarmor13 fraortmoy dancorgar anaprarod antsancab1 *)
theorem "hd (inOrden a) = raiz a"
quickcheck
oops

(* danrodcha anaprarod:
Auto Quickcheck found a counterexample:
a = N a1 (H a2) (H a1)
Evaluated terms:
hd (inOrden a) = a2
raiz a = a1 *)

(* ivamenjim marpoldia1 *)
theorem "hd (inOrden a) = raiz 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 "hd (inOrden (N x i d)) = hd ((inOrden i) @ [x] @ (inOrden d))" by simp
also have "... = hd (inOrden i)" by (simp add: aux_ej12_1)
(* Perdemos la x, luego se refuta el enunciado del teorema *)
oops

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

(* danrodcha pabrodmac dancorgar anaprarod *)
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 crigomgom ivamenjim marpoldia1 wilmorort lucnovdos
juacabsou serrodcal jeamacpov paupeddeg rubgonmar marcarmor13 anaprarod *)
(* 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

(* bowma *)
(* También sin usar el supuesto "?p d" *)
theorem "last (postOrden a) = raiz a" (is "?p a")
proof (induct a)
fix t
show "?p (H t)" by simp
next
fix t i d
assume "?p i"
(* si quito este supuesto, hay error pero no sé dónde se lo está
usando *)
have "last (postOrden (N t i d)) =
last (postOrden i @ postOrden d @ [t])" by simp
also have "... = t" by simp
also have "... = raiz (N t i d)" by simp
finally show "?p (N t i d)" by simp
qed
(*
Comentario de antsancab1:
No tengo muy claro por qué es, pero si en lugar de utilizar ?p al final lo pones completo:
finally show "last (postOrden (N t i d)) = raiz (N t i d)" by simp
No devuelve ningún error. No se si será que después de ?p espera un único elemento
*)

(* migtermor *)
theorem "last (postOrden a) = raiz a" (is "?P a")
proof (induct a)
fix h
show "?P (H h)" by simp
next
fix n i d
have "last (postOrden (N n (i :: 'a arbol) (d :: 'a arbol))) =
last (postOrden i@postOrden d@[n])" by simp
also have "… = raiz (N n i d)" by simp
finally show "?P (N n i d)" by simp
qed

(* ivamenjim: sin usar patrones *)
theorem "last (postOrden a) = raiz a"
proof (induct a)
fix x::"'a"
show "last (postOrden (H x)) = raiz (H x)" by simp
next
fix x::"'a"
fix i::"'a arbol" assume h1: "last (postOrden i) = raiz i"
fix d::"'a arbol" assume h2: "last (postOrden d) = raiz d"
have "last (postOrden (N x i d)) =
last ((postOrden i) @ (postOrden d) @ [x])" by simp
also have "... = last ([x])" by simp
finally show "last (postOrden (N x i d)) = raiz (N x i d)" by simp
qed

(* pabrodmac paupeddeg fraortmoy *)
theorem "last (postOrden a) = raiz a"
by (induct a) auto

(* pabrodmac *)
lemma
fixes a ::"'b arbol"
shows "last (postOrden a) = raiz 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"
show "?P (N x i d)"
proof -
have "last (postOrden (N x i d))= x" by simp
also have "… = raiz (N x i d)" by simp
finally show ?thesis .
qed
qed

(* Me he dado cuenta que no es necesario asumir ninguna hipótesis de
inducción puesto que no es necesario utilizarlas, así que no se si
está bien hecho puesto que no se aplicaría inducción *)

(* ferrenseg *)

theorem "last (postOrden a) = raiz a" (is "?P a")
proof (cases a)
fix x
assume "a = H x"
then show "?P a" by simp
next
fix x l r
assume H: "a = N x l r"
show "?P a"
proof -
have "last (postOrden a) = last (postOrden (N x l r))" using H by simp
also have "… = last (postOrden l @ preOrden r @ [x])" by simp
also have "… = x" by simp
also have "… = raiz a" using H by simp
finally show ?thesis .
qed
qed

(* fraortmoy *)
theorem "last (postOrden a) = raiz a"
apply (induct a)
apply simp
apply simp
done

(* fraortmoy *)
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
show "?P (N x i d)" by simp
qed

end

(* antsancab1 *)
theorem "last (postOrden a) = raiz 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 (postOrden (N x i d)) = last (postOrden i @ postOrden d @ [x])" by simp
also have "... = last [x]" by simp
also have "... = x" by simp
also have "... = raiz (H x)" by simp
also have "... = raiz (N x i d)" by simp
finally show "last (postOrden (N x i d)) = raiz (N x i d)" by simp
qed

</source>

Relación 10

2018-07-16T11:11:19Z

WikiSysop: Texto reemplazado: «isar» por «isabelle»

<source lang="isabelle">
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 bowma dancorgar antsancab1 *)
(* 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 marpoldia1 juacabsou jeamacpov rubgonmar*)

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 paupeddeg*)
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

(* fracorjim1 *)

lemma ejercicio_1_3:
assumes H1: "(A ∨ P) → (R ∧ F)" and
H2: "(F ∨ N) → O"
shows "A → O"
proof -
assume "A"
have "A ∨ P" by (rule disjI1)
hence "R ∧ F" using H1 by (rule mp)
hence "F" by (rule conjI2)
hence "F ∨ N" by (rule disjI1)
hence "O" using H2 by (rule mp)
thus "A → O"
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 fracorjim1 bowma paupeddeg marpoldia1
juacabsou jeamacpov dancorgar rubgonmar antsancab1 *)

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 marpoldia1 rubgonmar *)

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 fracorjim1 juacabsou bowma jeamacpov*)
(* 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 dancorgar *)
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 antsancab1 *)

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

(* paupeddeg *)
lemma ejercicio_3_3:
assumes "∀x y. P(x,y) ⟶ (∀z. (H(z,y) ⟶ P(x,z)))"
"H(j,l)"
"P(c,l)"
shows "P(c,j)"
proof-
have "∀ 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)
have "∀z. (H(z,l) ⟶ P(c,z))" using `P(c,l) ⟶ (∀z. (H(z,l) ⟶ P(c,z)))` `P(c,l)`
by (rule mp)
hence "H(j,l) ⟶ P(c,j)" by (rule allE)
show "P(c,j)" using `H(j,l) ⟶ P(c,j)` `H(j,l)` by (rule mp)
qed

(* fracorjim1 *)
(* Realizo una eliminación sucesiva del cuantificador como migtermor *)
lemma ejercicio_3_4:
assumes H1: "∀x y z. (P(x,y) ∧ H(y,z)) ⟶ P(x,z)"
assumes H2: "H(l,j)"
assumes H3: "P(c,l)"
shows "P(c,j)"
proof -
have H: "P(c,l) ∧ H(l,j)" using H3 H2 by (rule conjI)
have "∀y z. (P(c,y) ∧ H(y,z)) ⟶ P(c,z)" using H1 by (rule allE)
hence "∀z. (P(c,l) ∧ H(l,z)) ⟶ P(c,z)" by (rule allE)
hence "(P(c,l) ∧ H(l,j)) ⟶ P(c,j)" by (rule allE)
thus "P(c,j)" using H 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 jeamacpov*)
(* 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 bowma juacabsou marpoldia1 rubgonmar *)

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 josgarsan antsancab1 *)

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

(* paupeddeg *)
lemma ejercicio_4_2:
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
assume "∃x. E(x)∧N(x)"
then obtain b where "E(b) ∧ N(b)" by (rule exE)
hence "E(b)" by (rule conjunct1)
have "E(b) ⟶¬Ap(j,b)" using assms(2) by (rule allE)
have "¬Ap(j,b)" using `E(b) ⟶¬Ap(j,b)` `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)
have "¬ (Af(j)∧E(b))" using `Af(j)∧E(b) ⟶ Ap(j,b)` `¬ Ap(j,b)` by (rule mt)
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

(* dancorgar *)
lemma ej_4_3:
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-
{assume 3: "(∀x. E(x) ∧ N(x))"
have "E(x) ∧ N(x)" using 3 by (rule allE)
then have 5: "E(x)" by (rule conjunct1)
have "E(x) ⟶ ¬Ap(j, x)" using 2 by (rule allE)
then have 6: "¬Ap(j, x)" using 5 by (rule mp)
have "∀y. (Af(j) ∧ E(y)) ⟶ Ap(j, y)" using 1 by (rule allE)
then have 7: "(Af(j) ∧ E(x)) ⟶ Ap(j, x)" by (rule allE)
have 8: "¬(Af(j) ∧ E(x))" using 7 6 by (rule mt)
then have "¬Af(j)"
proof-
{assume "¬(¬Af(j))"
then have "Af(j)" by (rule notnotD)
then have 9: "Af(j) ∧ E(x)" using 5 by (rule conjI)
have False using 8 9 by (rule notE)}
then show "¬Af(j)" by (rule ccontr)
qed}
then show "(∀x. E(x) ∧ N(x)) ⟶ ¬Af(j)" by (rule impI)
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 paupeddeg juacabsou fracorjim1 jeamacpov
dancorgar rubgonmar josgarsan antsancab1 *)
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 bowma *)
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 2

2018-07-16T11:11:19Z

WikiSysop: Texto reemplazado: «isar» por «isabelle»

<source lang="isabelle">
chapter {* R2: Razonamiento sobre programas en Isabelle/HOL *}

theory R2_Razonamiento_automatico_sobre_programas
imports Main
begin

declare [[names_short]]

text {* ---------------------------------------------------------------
Ejercicio 1.1. Definir la función
sumaImpares :: nat ⇒ nat
tal que (sumaImpares n) es la suma de los n primeros números
impares. Por ejemplo,
sumaImpares 5 = 25
------------------------------------------------------------------ *}

(* fracorjim1 *)
fun sumaImpares0 :: "nat ⇒ nat" where
"sumaImpares0 0 = 0"
| "sumaImpares0 n = (n mod 2)*n + sumaImpares(n-1)"

value "sumaImpares0 5 = 25"

(* Comentario: Debe de definirse usando como patrones los dos
constructores de nat: 0 y Suc *)

(* ivamenjim wilmorort*)
fun sumaImpares :: "nat ⇒ nat" where
"sumaImpares 0 = 0"
| "sumaImpares n = 2*(n-1) + 1 + sumaImpares (n-1)"

value "sumaImpares 5 = 25"

(* wilmorort *)

(*Notar: Por la propiedad de Gauss se puede deducir que:
la suma de los n números impares es igual a la suma de los
n y n-1 números consecutivos*)
fun suma :: "nat ⇒ nat" where
"suma 0 = 0"
| "suma n = n + suma(n-1)"

fun sumaImpares1 :: "nat ⇒ nat" where
"sumaImpares1 0 = 0"
| "sumaImpares1 n = suma(n) + suma(n-1)"

value "sumaImpares1 5 = 25"

(* Comentario: La definición sumaImpares1 no es recursiva y, por tanto,
no es apropiada para demostraciones por inducción.

Comentario: La definición sumaImpares1 no es recursiva pero la
demostración, está realizada por inducción, por lo tanto no se puede
concluir que: "Si una def. no es recuriva no se puede demostrar por
inducción"

Comentario: Las definiciones no recursivas no generan esquemas de
inducción.
*)

(* fraortmoy antsancab1 *)
fun sumaImpares2 :: "nat ⇒ nat" where
"sumaImpares2 0 = 0"
| "sumaImpares2 (Suc(n)) = ((Suc(n)*2)-1) + sumaImpares2 n"

value "sumaImpares2 5 = 25"

(* anaprarod danrodcha crigomgom migtermor pabrodmac juacabsou *)
fun sumaImpares3 :: "nat ⇒ nat" where
"sumaImpares3 0 = 0"
|"sumaImpares3 (Suc n) = (2*n +1) + (sumaImpares3 n)"

value "sumaImpares3 5 = 25"

(* manmorjim1 *)
fun sumaImpares4 :: "nat ⇒ nat" where
"sumaImpares4 0 = 0"
| "sumaImpares4 (Suc n) = (Suc (2 * n)) + (sumaImpares4 n)"

value "sumaImpares4 5 = 25"

(* pablucoto serrodcal marcarmor13 rubgonmar marpoldia1 jeamacpov*)
fun sumaImpares5 :: "nat ⇒ nat" where
"sumaImpares5 0 = 0"
| "sumaImpares5 n = (2*n-1) + sumaImpares5 (n-1) "

value "sumaImpares5 5 = 25"

(* ferrenseg paupeddeg josgarsan dancorgar *)

fun sumaImpares6 :: "nat ⇒ nat" where
"sumaImpares6 0 = 0"
| "sumaImpares6 (Suc n) = (2 * (Suc n) - 1) + sumaImpares6 n"

value "sumaImpares6 5 = 25"

text {* ---------------------------------------------------------------
Ejercicio 1.2. Demostrar que
sumaImpares n = n*n
------------------------------------------------------------------- *}

(* ivamenjim wilmorort anaprarod crigomgom manmorjim1 pablucoto
serrodcal marcarmor13 rubgonmar pabrodmac marpoldia1 josgarsan
jeamacpov dancorgar antsancab1 fracorjim1 juacabsou *)
lemma "sumaImpares n = n*n"
apply (induct n)
apply auto
done

(* wilmorort *)
(* Demostración estructurada *)

lemma aux1 : "suma(Suc n) + suma(n) = (n+1) + suma(n) + n + suma(n-1)"
apply (induct n)
apply auto
done

lemma aux2: "(n+1) + suma(n) + n + suma(n-1) = (n+1) + n + sumaImpares1 n "
apply (induct n)
apply auto
done

lemma "sumaImpares1 n = n*n"
proof (induct n)
show "sumaImpares1 0 = 0*0" by simp
next
fix n
assume HI:"sumaImpares1 n = n*n"
have "sumaImpares1 (Suc n) = suma(Suc n) + suma(n)" by simp
have "suma(Suc n) + suma(n) = (n+1) + suma(n) + n + suma(n-1)" using
"aux1" by simp
also have "... = (n+1) + n + sumaImpares1 n " using "aux2" by simp
also have "... = 2*n +1 + n*n" using HI by simp
also have "... = (n+1)*(n+1)" by simp
finally show "sumaImpares1 (Suc n) = Suc n * Suc n" by simp
qed

(* fraortmoy *)
(* La demostración para la función sumaImpares2 es la misma que la de
sumaImpares*)

lemma "sumaImpares2 n = n*n"
apply (induct n)
apply auto
done

(* danrodcha migtermor paupeddeg *)
lemma "sumaImpares n = n*n"
by (induct n) auto

(* danrodcha *)
lemma "sumaImpares n = n*n" (is "?P n")
proof (induct n)
show "?P 0" by simp
next
fix n assume "?P n"
thus "?P (Suc n)" by simp
qed

(* ferrenseg *)
lemma "sumaImpares3 n = n*n"
by (induct n) simp_all

text {* ---------------------------------------------------------------
Ejercicio 2.1. Definir la función
sumaPotenciasDeDosMasUno :: nat ⇒ nat
tal que
(sumaPotenciasDeDosMasUno n) = 1 + 2^0 + 2^1 + 2^2 + ... + 2^n.
Por ejemplo,
sumaPotenciasDeDosMasUno 3 = 16
------------------------------------------------------------------ *}

(* wilmorort ivamenjim pablucoto serrodcal ferrenseg rubgonmar
pabrodmac *)
fun sumaPotenciasDeDosMasUno :: "nat ⇒ nat" where
"sumaPotenciasDeDosMasUno 0 = 2"
| "sumaPotenciasDeDosMasUno n = 2^n + sumaPotenciasDeDosMasUno (n-1)"

value "sumaPotenciasDeDosMasUno 3 = 16"

(* fraortmoy anaprarod danrodcha crigomgom migtermor manmorjim1
fracorjim1 paupeddeg josgarsan dancorgar antsancab1 juacabsou *)
fun sumaPotenciasDeDosMasUno2 :: "nat ⇒ nat" where
"sumaPotenciasDeDosMasUno2 0 = 2"
| "sumaPotenciasDeDosMasUno2 (Suc(n)) =
(2^(Suc n)) + sumaPotenciasDeDosMasUno2 n"

value "sumaPotenciasDeDosMasUno2 3 = 16"

(* marcarmor13 jeamacpov *)
fun sumaPotenciasDeDosMasUno3 :: "nat ⇒ nat" where
"sumaPotenciasDeDosMasUno3 0 = 2"
| "sumaPotenciasDeDosMasUno3 n = 2*sumaPotenciasDeDosMasUno3 (n-1)"

value "sumaPotenciasDeDosMasUno3 3 = 16"

text {* ---------------------------------------------------------------
Ejercicio 2.2. Demostrar que
sumaPotenciasDeDosMasUno n = 2^(n+1)
------------------------------------------------------------------- *}

(* wilmorort ivamenjim fraortmoy anaprarod crigomgom pablucoto serrodcal
manmorjim1 marcarmor13 rubgonmar pabrodmac josgarsan jeamacpov
dancorgar antsancab1 juacabsou *)
(* esta demostración funciona con sumaPotenciasDeDosMasUno2 *)

lemma "sumaPotenciasDeDosMasUno n = 2^(n+1)"
apply (induct n)
apply auto
done

(* danrodcha migtermor ferrenseg fracorjim1 paupeddeg *)
lemma "sumaPotenciasDeDosMasUno n = 2^(n+1)"
by (induct n) auto

(* danrodcha *)
lemma "sumaPotenciasDeDosMasUno n = 2^(n+1)" (is "?P n")
proof (induct n)
show "?P 0" by simp
next
fix n assume "?P n"
thus "?P (Suc n)"by simp
qed

text {* ---------------------------------------------------------------
Ejercicio 3.1. Definir la función
copia :: nat ⇒ 'a ⇒ 'a list
tal que (copia n x) es la lista formado por n copias del elemento
x. Por ejemplo,
copia 3 x = [x,x,x]
------------------------------------------------------------------ *}

(* ivamenjim wilmorort pablucoto rubgonmar marcarmor13 jeamacpov *)
fun copia :: "nat ⇒ 'a ⇒ 'a list" where
"copia 0 x = [] "
| "copia n x = x # (copia (n-1) x)"

(* wilmorort anaprarod *)
fun copia1 :: "nat ⇒ 'a ⇒ 'a list" where
"copia1 0 x = []"
| "copia1 (Suc n) x = [x] @ (copia1 n x) "

(* Comentario: La dedinición copia1 se puede simplificar eliminando @ *)

(* danrodcha crigomgom migtermor manmorjim1 fracorjim1 pabrodmac
paupeddeg josgarsan dancorgar antsancab1 juacabsou *)
fun copia2 :: "nat ⇒ 'a ⇒ 'a list" where
"copia2 0 x = []"
| "copia2 (Suc n) x = x # copia2 n x"

(* serrodcal ferrenseg *)
fun copia3 :: "nat ⇒ 'a ⇒ 'a list" where
"copia3 0 x = []"
| "copia3 n x = [x] @ copia3 (n-1) x"

value "copia 3 x = [x,x,x]"

(* Comentario: La definición copia3 se puede simplificar eliminando @ *)

text {* ---------------------------------------------------------------
Ejercicio 3.2. Definir la función
todos :: ('a ⇒ bool) ⇒ 'a list ⇒ bool
tal que (todos p xs) se verifica si todos los elementos de xs cumplen
la propiedad p. Por ejemplo,
todos (λx. x>(1::nat)) [2,6,4] = True
todos (λx. x>(2::nat)) [2,6,4] = False
Nota: La conjunción se representa por ∧
----------------------------------------------------------------- *}

(* wilmorort *)
fun todos :: "('a ⇒ bool) ⇒ 'a list ⇒ bool" where
"todos p [] = True"
| "todos p (x#xs) = (p x = True ∧ todos p xs) "

(* Comentario: La definición todos se puede simplificar eliminando True
*)

value "todos (λx. x>(1::nat)) [2,6,4] = True"
value "todos (λx. x>(2::nat)) [2,6,4] = False"

(* fraortmoy anaprarod danrodcha crigomgom migtermor ivamenjim serrodcal
pablucoto ferrenseg rubgonmar pabrodmac paupeddeg marcarmor13
jeamacpov dancorgar josgarsan antsancab1 fracorjim1 juacabsou *)
fun todos1 :: "('a ⇒ bool) ⇒ 'a list ⇒ bool" where
"todos1 p [] = True"
| "todos1 p (x#xs) = (p x ∧ todos1 p xs )"

value "todos1 (λx. x>(1::nat)) [2,6,4] = True"
value "todos1 (λx. x>(2::nat)) [2,6,4] = False"

(* manmorjim1 *)
fun todos2 :: "('a ⇒ bool) ⇒ 'a list ⇒ bool" where
"todos2 p xs = ((filter p xs) = xs)"

value "todos2 (λx. x>(1::nat)) [2,6,4] = True"
value "todos2 (λx. x>(2::nat)) [2,6,4] = False"

text {* ---------------------------------------------------------------
Ejercicio 3.3. Demostrar que todos los elementos de (copia n x) son
iguales a x.
------------------------------------------------------------------- *}

(* wilmorort fraortmoy anaprarod crigomgom ivamenjim serrodcal
manmorjim1 pablucoto rubgonmar pabrodmac marcarmor13 dancorgar
josgarsan antsancab1 juacabsou *)
lemma "todos (λy. y=x) (copia n x)"
apply (induct n)
apply auto
done

(* danrodcha ferrenseg paupeddeg jeamacpov *)
lemma "todos (λy. y=x) (copia n x)"
by (induct n) auto

(* danrodcha *)
lemma "todos (λy. y=x) (copia n x)" (is "?P n")
proof (induct n)
show "?P 0" by simp
next
fix n assume "?P n"
thus "?P (Suc n)" by simp
qed

(* fracorjim1 *)
lemma "todos (λy. y = x)(copia n x) = True"
apply (induct n)
apply auto
done

text {* ---------------------------------------------------------------
Ejercicio 4.1. Definir, recursivamente y sin usar (@), la función
amplia :: 'a list ⇒ 'a ⇒ 'a list
tal que (amplia xs y) es la lista obtenida añadiendo el elemento y al
final de la lista xs. Por ejemplo,
amplia [d,a] t = [d,a,t]
------------------------------------------------------------------ *}

(* wilmorort fraortmoy anaprarod crigomgom migtermor ivamenjim serrodcal
manmorjim1 pablucoto ferrenseg rubgonmar pabrodmac paupeddeg
marcarmor13 jeamacpov dancorgar josgarsan antsancab1 fracorjim1 juacabsou *)
fun amplia :: "'a list ⇒ 'a ⇒ 'a list" where
"amplia [] y = [y] "
| "amplia (x#xs) y = x # amplia xs y"

value "amplia [d,a] t = [d,a,t]"

text {* ---------------------------------------------------------------
Ejercicio 4.2. Demostrar que
amplia xs y = xs @ [y]
------------------------------------------------------------------- *}

(* wilmorort fraortmoy anaprarod crigomgom ivamenjim serrodcal
manmorjim1 pablucoto rubgonmar pabrodmac marcarmor13 jeamacpov
dancorgar josgarsan antsancab1 fracorjim1 juacabsou *)
lemma "amplia xs y = xs @ [y]"
apply (induct xs)
apply auto
done

(* danrodcha migtermor ferrenseg paupeddeg *)
lemma "amplia xs y = xs @ [y]"
by (induct xs) auto

(* danrodcha *)
lemma "amplia xs y = xs @ [y]" (is "?P xs")
proof (induct xs)
show "?P []" by simp
next
fix a xs assume "?P xs"
thus "?P (a#xs)" by simp
qed

end
</source>

Relación 3

2018-07-16T11:11:19Z

WikiSysop: Texto reemplazado: «isar» por «isabelle»

<source lang="isabelle">
chapter {* R3: Razonamiento sobre programas *}

theory R3_Razonamiento_sobre_programas
imports Main
begin

text {* ---------------------------------------------------------------
Ejercicio 1.1. Definir la función
sumaImpares :: nat ⇒ nat
tal que (sumaImpares n) es la suma de los n primeros números
impares. Por ejemplo,
sumaImpares 5 = 25
------------------------------------------------------------------ *}

fun sumaImpares :: "nat ⇒ nat" where
"sumaImpares 0 = 0"
| "sumaImpares (Suc n) = sumaImpares n + (2*n+1)"

text {* ---------------------------------------------------------------
Ejercicio 1.2. Escribir la demostración detallada de
sumaImpares n = n*n
------------------------------------------------------------------- *}

-- "La demostración detallada es"
(* crigomgom fraortmoy marpoldia1 ivamenjim serrodcal rubgonmar
ferrenseg juacabsou wilmorort josgarsan lucnovdos paupeddeg bowma
pabrodmac dancorgar antsancab1 *)
lemma "sumaImpares n = n*n"
proof (induct n)
show "sumaImpares 0 = 0 * 0" by simp
next
fix n
assume HI: "sumaImpares n = n * n"
have "sumaImpares (Suc n) = sumaImpares n + 2*n + 1" by simp
also have "... = n*n + 2*n + 1" using HI by simp
also have "... = Suc n * Suc n" by simp
finally show "sumaImpares (Suc n) = Suc n * Suc n" by simp
qed

(* danrodcha: Es la misma demostración que la anterior pero uso ?P para
sustituir la propiedad.*)
lemma "sumaImpares n = n*n" (is "?P n")
proof (induct n)
show "?P 0" by simp
next
fix n
assume HI: "?P n"
have "sumaImpares (Suc n) = sumaImpares n + 2*n + 1" by simp
also have "... = n*n + 2*n + 1" using HI by simp
also have "... = Suc n * Suc n" by simp
finally show "?P (Suc n)" by simp
qed

(* migtermor *)
lemma "sumaImpares n = n*n"
proof (induct n)
show "sumaImpares 0 = 0*0" by simp
next
fix n
assume HI: "sumaImpares n = n*n"
have "sumaImpares (Suc n) = sumaImpares n + 2*n+1"
by (simp only: sumaImpares.simps(2))
also have "sumaImpares (Suc n) = n*n + 2*n+1" using HI by simp
finally show "sumaImpares (Suc n) = (Suc n)*(Suc n)" by simp
qed

(* fracorjim1: Especifico la regla de simplificación y el paso del
desarrollo al cuadrado para hacerlo más legible. *)
lemma "sumaImpares n = n*n"
proof (induct n)
show "sumaImpares 0 = 0 * 0" by simp
next
fix n
assume HI : "sumaImpares n = n * n"
have "sumaImpares (Suc n) = sumaImpares n + (2 * n + 1)"
by (simp only: sumaImpares.simps(2))
also have "... = n * n + 2 * n + 1" using HI by simp
also have "... = Suc n * Suc n" by simp
finally show "sumaImpares (Suc n) = Suc n * Suc n" by simp
qed

(* palucoto anaprarod marcarmor13 jeamacpov *)
lemma "sumaImpares n = n*n"
proof (induct n)
show "sumaImpares 0 = 0 * 0" by simp
next
fix n
assume HI : "sumaImpares n = n * n"
have " sumaImpares (Suc n) = sumaImpares n + 2*n+1" by simp
also have "... = n*n + 2*n+1" using HI by simp
finally show "sumaImpares (Suc n) = Suc n * Suc n" by simp
qed

text {* ---------------------------------------------------------------
Ejercicio 2.1. Definir la función
sumaPotenciasDeDosMasUno :: nat ⇒ nat
tal que
(sumaPotenciasDeDosMasUno n) = 1 + 2^0 + 2^1 + 2^2 + ... + 2^n.
Por ejemplo,
sumaPotenciasDeDosMasUno 3 = 16
------------------------------------------------------------------ *}

fun sumaPotenciasDeDosMasUno :: "nat ⇒ nat" where
"sumaPotenciasDeDosMasUno 0 = 2"
| "sumaPotenciasDeDosMasUno (Suc n) =
sumaPotenciasDeDosMasUno n + 2^(n+1)"

text {* ---------------------------------------------------------------
Ejercicio 2.2. Escribir la demostración detallada de
sumaPotenciasDeDosMasUno n = 2^(n+1)
------------------------------------------------------------------- *}

(*crigomgom ivamenjim danrodcha serrodcal rubgonmar ferrenseg juacabsou
wilmorort anaprarod marpoldia1 josgarsan paupeddeg pabrodmac bowma
dancorgar antsancab1 *)
lemma "sumaPotenciasDeDosMasUno n = 2^(n+1)"
proof (induct n)
show "sumaPotenciasDeDosMasUno 0 = 2 ^ (0 + 1)" by simp
next
fix n
assume HI: "sumaPotenciasDeDosMasUno n = 2 ^ (n + 1)"
have "sumaPotenciasDeDosMasUno (Suc n) =
sumaPotenciasDeDosMasUno n + 2^(n+1)" by simp
also have "... = 2 ^ (n + 1) + 2 ^ (n + 1)" using HI by simp
also have "... = 2 ^ ((Suc n) + 1)" by simp
finally show "sumaPotenciasDeDosMasUno (Suc n) = 2 ^ (Suc n + 1)"
by simp
qed

(* fraortmoy : Es la misma demostración, pero quise probar a delimitar
lo que se usa en el "by simp" *)
lemma "sumaPotenciasDeDosMasUno n = 2^(n+1)"
proof (induct n)
show "sumaPotenciasDeDosMasUno 0 = 2 ^ (0 + 1)" by simp
next
fix n
assume H1:" sumaPotenciasDeDosMasUno n = 2 ^ (n + 1)"
have "sumaPotenciasDeDosMasUno (Suc n) =
sumaPotenciasDeDosMasUno n + 2^(n + 1)"
by (simp only : sumaPotenciasDeDosMasUno.simps(2))
also have "... = 2 ^ (n + 1) + 2 ^ (n + 1)" using H1 by simp
also have "... = 2 ^ (Suc n + 1)" by simp
finally show "sumaPotenciasDeDosMasUno (Suc n) = 2 ^ (Suc n + 1)"
by simp
qed

(* migtermor *)
lemma "sumaPotenciasDeDosMasUno n = 2^(n+1)"
proof (induct n)
show "sumaPotenciasDeDosMasUno 0 = 2^(0+1)" by simp
next
fix n
assume HI: "sumaPotenciasDeDosMasUno n = 2^(n+1)"
have "sumaPotenciasDeDosMasUno (Suc n) =
sumaPotenciasDeDosMasUno n + 2^(n+1)"
by (simp only: sumaPotenciasDeDosMasUno.simps(2))
also have "sumaPotenciasDeDosMasUno (Suc n) = 2^(n+1)+2^(n+1)"
using HI by simp
finally show "sumaPotenciasDeDosMasUno (Suc n) = 2^((Suc n)+1)"
by simp
qed

(* pablucoto marcarmor13 lucnovdos jeamacpov *)
lemma "sumaPotenciasDeDosMasUno n = 2^(n+1)"
proof (induct n)
show "sumaPotenciasDeDosMasUno 0 = 2 ^ (0 + 1)" by simp
next
fix n
assume HI : "sumaPotenciasDeDosMasUno n = 2 ^ (n + 1)"
have "sumaPotenciasDeDosMasUno (Suc n) =
sumaPotenciasDeDosMasUno n + 2^(n+1) " by simp
also have "... = 2^(n + 1) + 2^( n + 1) " using HI by simp
finally show " sumaPotenciasDeDosMasUno (Suc n) = 2 ^ (Suc n + 1)"
by simp
qed

(* fracorjim1 - Hago explícita toda la manipulación algebraica. Quizás
excesivo.*)
lemma "sumaPotenciasDeDosMasUno n = 2^(n+1)"
proof (induction n)
show "sumaPotenciasDeDosMasUno 0 = 2 ^ (0 + 1)" by simp
next
fix n
assume HI : "sumaPotenciasDeDosMasUno n = 2 ^ (n + 1)"
have "sumaPotenciasDeDosMasUno (Suc n) =
sumaPotenciasDeDosMasUno n + 2 ^ (n + 1)"
by (simp only:sumaPotenciasDeDosMasUno.simps(2))
also have "... = 2 ^ (n + 1) + 2 ^ (n + 1)" using HI by simp
also have "... = 2 ^ (n + 1) * 2" by simp
also have "... = 2 ^ ((n + 1) + 1)" by simp
also have "... = 2 ^ (Suc n + 1)" by simp
finally show "sumaPotenciasDeDosMasUno (Suc n) = 2 ^ (Suc n + 1)"
by simp
qed

text {* ---------------------------------------------------------------
Ejercicio 3.1. Definir la función
copia :: nat ⇒ 'a ⇒ 'a list
tal que (copia n x) es la lista formado por n copias del elemento
x. Por ejemplo,
copia 3 x = [x,x,x]
------------------------------------------------------------------ *}

fun copia :: "nat ⇒ 'a ⇒ 'a list" where
"copia 0 x = []"
| "copia (Suc n) x = x # copia n x"

text {* ---------------------------------------------------------------
Ejercicio 3.2. Definir la función
todos :: ('a ⇒ bool) ⇒ 'a list ⇒ bool
tal que (todos p xs) se verifica si todos los elementos de xs cumplen
la propiedad p. Por ejemplo,
todos (λx. x>(1::nat)) [2,6,4] = True
todos (λx. x>(2::nat)) [2,6,4] = False
------------------------------------------------------------------ *}

fun todos :: "('a ⇒ bool) ⇒ 'a list ⇒ bool" where
"todos p [] = True"
| "todos p (x#xs) = (p x ∧ todos p xs)"

text {* ---------------------------------------------------------------
Ejercicio 3.2. Demostrar detalladamente que todos los elementos de
(copia n x) son iguales a x.
------------------------------------------------------------------- *}

(* crigomgom ivamenjim serrodcal ferrenseg wilmorort juacabsou josgarsan
lucnovdos *)
lemma "todos (λy. y=x) (copia n x)"
proof (induct n)
show "todos (λy. y = x) (copia 0 x)" by simp
next
fix n
assume HI: "todos (λy. y = x) (copia n x)"
have "todos (λy. y = x) (copia (Suc n) x) =
todos (λy. y = x) (x # (copia n x))" by simp
also have "... = ((x = x) ∧ (todos (λy. y = x) (copia n x)))" by simp
also have "... = True" using HI by simp
finally show "todos (λy. y = x) (copia (Suc n) x)" by simp
qed

(* fraortmoy *)
lemma "todos (λy. y=x) (copia n x)"
proof (induct n)
have "todos (λy. y = x) (copia 0 x) =
todos (λy. y = x) []" by (simp only: copia.simps(1))
also have "... = True" by (simp only: todos.simps(1))
show "todos (λy. y = x) (copia 0 x)" by simp
next
fix n
assume H1 : " todos (λy. y = x) (copia n x) "
have "todos (λy. y = x) (copia (Suc n) x) =
((todos (λy. y = x) (x#[])) ∧ (todos (λy. y = x) (copia n x)))"
by simp
also have " ... = (todos (λy. y = x) (x#[]))" using H1 by simp
also have " ... = ((λy. y = x) x ∧ todos (λy. y = x) [])" by simp
also have " ... = True" by simp
finally show " todos (λy. y = x) (copia (Suc n) x)" by simp
qed

(* danrodcha rubgonmar *)
lemma "todos (λy. y=x) (copia n x)" (is "?P n")
proof (induct n)
show "?P 0" by simp
next
fix n
assume HI: "?P n"
have "todos (λy. y=x) (copia (Suc n) x) =
todos (λy. y=x) (x # copia n x)" by simp
also have "... = ((λy. y=x) x ∧ todos (λy. y=x) (copia n x))" by simp
also have "... = todos (λy. y=x) (copia n x)" by simp
finally show "?P (Suc n)" using HI by simp
qed

(* migtermor *)
lemma "todos (λy. y=x) (copia n x)"
proof (induct n)
have "todos (λy. y=x) (copia 0 x) = todos (λy. y=x) []"
by (simp only: copia.simps(1))
also have "todos (λy. y=x) []" by (simp only: todos.simps(1))
show "todos (λy. y=x) (copia 0 x)" by simp
next
fix n
assume HI: "todos (λy. y=x) (copia n x)"
have "todos (λy. y=x) (copia (Suc n) x) =
todos (λy. y=x) (x # (copia n x))"
by (simp only: copia.simps(2))
also have "todos (λy. y=x) (x#(copia n x)) =
((λy. y=x) x ∧ todos (λy. y=x) (copia n x))"
by (simp only: todos.simps(2))
also have "todos (λy. y=x) (copia (Suc n) x) = ((λy. y=x) x)"
using HI by simp
also have "((λy. y=x) x) = True" by simp
finally show "(todos (λy. y=x) (copia (Suc n) x))" by simp
qed

(*pablucoto pabrodmac marcarmor13 jeamacpov *)
lemma "todos (λy. y=x) (copia n x)"
proof (induct n)
show "todos (λy. y = x) (copia 0 x)" by simp
next
fix n
assume HI: "todos (λy. y=x) (copia n x)"
have "todos (λy. y = x) (copia (Suc n) x) =
todos (λy. y=x) (x # copia n x)" by simp
also have "... = ( (λy. y=x) x ∧ todos (λy. y=x) (copia (Suc n) x))"
by simp
also have "... = True" using HI by simp
finally show " todos (λy. y = x) (copia (Suc n) x) " by simp
qed

(* anaprarod marpoldia1 *)
lemma "todos (λy. y=x) (copia n x)"
proof (induct n)
show "todos (λy. y = x) (copia 0 x)" by simp
next
fix n
assume HI: "todos (λy. y = x) (copia n x)"
have "todos (λy. y = x) (copia (Suc n) x) =
todos (λy. y = x) (x # copia n x)" by simp
also have "... = ((λy. y = x) x ∧ todos (λy. y = x) ( copia n x))"
by simp
finally show "todos (λy. y = x) (copia (Suc n) x)" using HI by simp
qed

(* paupeddeg antsancab1 *)
lemma "todos (λy. y=x) (copia n x)"
proof (induct n)
show "todos (λy. y = x) (copia 0 x)" by simp
next
fix n
assume HI: "todos (λy. y = x) (copia n x)"
have "todos (λy. y = x) (copia (Suc n) x) =
(todos (λy. y = x) (x # copia n x))" by simp
also have "... = ((todos (λy. y = x) [x]) ∧
(todos (λy. y = x) (copia n x)))" by simp
also have "... = True" using HI by simp
finally show "todos(λy. y = x) (copia (Suc n) x)" by simp
qed

(* dancorgar *)
lemma "todos (λy. y=x) (copia n x)"
proof (induct n)
show "todos (λy. y=x) (copia 0 x)" by simp
next
fix n
assume HI: "todos (λy. y=x) (copia n x)"
have "todos (λy. y=x) (x#(copia n x))" using HI by simp
also have "... = (copia (Suc n) x)" by simp
finally show "todos (λy. y=x) (copia (Suc n) x)" by simp
qed

(* fracorjim1 *)
lemma "todos (λy. y=x) (copia n x)"
proof (induct n)
have "todos (λy. y=x) (copia 0 x) = todos (λy. y=x) []"
by (simp only: copia.simps(1))
also have "... = True" by (simp only: todos.simps(1))
finally show "todos (λy. y=x) (copia 0 x)" by simp
next
fix n
assume HI : "todos (λy. y=x) (copia n x)"
have "todos (λy. y=x) (copia (Suc n) x) =
todos (λy. y=x)(x # copia n x)" by (simp only: copia.simps(2))
also have "... = ((λy. y=x) x ∧ todos (λy. y=x) (copia n x))"
by (simp only: todos.simps(2))
also have "... = True ∧ True" using HI by simp
finally show "todos (λy. y=x) (copia (Suc n) x)" by simp
qed

text {* ---------------------------------------------------------------
Ejercicio 4.1. Definir la función
factR :: nat ⇒ nat
tal que (factR n) es el factorial de n. Por ejemplo,
factR 4 = 24
------------------------------------------------------------------ *}

fun factR :: "nat ⇒ nat" where
"factR 0 = 1"
| "factR (Suc n) = Suc n * factR n"

text {* ---------------------------------------------------------------
Ejercicio 4.2. Se considera la siguiente definición iterativa de la
función factorial
factI :: "nat ⇒ nat" where
factI n = factI' n 1

factI' :: nat ⇒ nat ⇒ nat" where
factI' 0 x = x
factI' (Suc n) x = factI' n (Suc n)*x
Demostrar que, para todo n y todo x, se tiene
factI' n x = x * factR n
Indicación: La propiedad mult_Suc es
(Suc m) * n = n + m * n
Puede que se necesite desactivarla en un paso con
(simp del: mult_Suc)
------------------------------------------------------------------- *}

fun factI' :: "nat ⇒ nat ⇒ nat" where
"factI' 0 x = x"
| "factI' (Suc n) x = factI' n (x * Suc n)"

fun factI :: "nat ⇒ nat" where
"factI n = factI' n 1"

(* fraortmoy danrodcha serrodcal pablucoto wilmorort anaprarod
marpoldia1 juacabsou josgarsan rubgonmar paupeddeg pabrodmac bowma
fracorjim1 dancorgar antsancab1 *)
lemma fact: "factI' n x = x * factR n"
proof (induct n)
show "factI' 0 x = x * factR 0" by simp
next
fix n
assume H1 : "factI' n x = x * factR n"
have "factI' (Suc n) x = factI' n (x * Suc n)"
by (simp only:factI'.simps(2))
also have "... = (x * Suc n) * factR n" using H1 by simp
(* no entiendo por qué no hace esto bien y luego todo funciona *)
(* Creo que es porque no le estás dando la hipótesis de inducción
sino otra que se le parece *)
also have "... = x * factR (Suc n)" by (simp del: mult_Suc)
finally show "factI' (Suc n) x = x * factR (Suc n)" by simp
qed

(* crigomgom ivamenjim ferrenseg marcarmor13 jeamacpov anaprarod *)
lemma fact: "factI' n x = x * factR n"
proof (induct n arbitrary: x)
show "⋀x. factI' 0 x = x * factR 0" by simp
next
fix n
assume HI: "⋀x. factI' n x = x * factR n"
show "⋀x. factI' (Suc n) x = x * factR (Suc n)"
proof -
fix x
have "factI' (Suc n) x = factI' n (x * Suc n)" by simp
also have "... = (x * Suc n) * factR n" using HI by simp
also have "... = x * (Suc n * factR n)" by (simp del: mult_Suc)
also have "... = x * factR (Suc n)" by simp
finally show "factI' (Suc n) x = x * factR (Suc n)" by simp
qed
qed

(* migtermor *)
lemma fact: "factI' n x = x * factR n"
proof (induct n)
have "x * factR 0 = x" by (simp only: factR.simps(1))
also have "factI' 0 x = x" by (simp only: factI'.simps(1))
show "factI' 0 x = x * factR 0" by simp
next
fix n
assume HI: "∀x. factI' n x = x * factR n"
fix x
have "factI' (Suc n) x = factI' n (x * Suc n)"
by (simp only: factI'.simps(2))
also have "factI' (Suc n) x = factI' n (x * Suc n)" by simp
have "factI' (Suc n) x = x * Suc n * factR n" using HI by simp
also have "... = x * factR (Suc n)" by (simp del: mult_Suc)
finally show "factI' (Suc n) x = x * factR (Suc n)" by simp
(* No entiendo por qué no acepta esto como demostrado *)
oops

(* ivamenjim *)

lemma fact: "factI' n x = x * factR n"
proof (induct n arbitrary: x)
show "⋀x. factI' 0 x = x * factR 0" by simp
next
fix n x
assume HI: "⋀x. factI' n x = x * factR n"
have "factI' (Suc n) x = factI' n (Suc n)*x" by simp
(* Primero (Suc n) y luego multiplicado por x, como dice el enunciado del ejercicio *)
also have "... = x * factI' n (Suc n)" by simp
also have "... = x * ((Suc n) * factR n)" using HI by simp
also have "... = x * factR (Suc n)" by simp
finally show "factI' (Suc n) x = x * factR (Suc n)" by simp
qed

text {* ---------------------------------------------------------------
Ejercicio 4.3. Escribir la demostración detallada de
factI n = factR n
------------------------------------------------------------------- *}

(* fraortmoy danrodcha crigomgom ivamenjim serrodcal ferrenseg pablucoto
wilmorort rubgonmar marpoldia1 juacabsou josgarsan paupeddeg bowma anaprarod marcarmor13 antsancab1 *)
corollary "factI n = factR n"
proof -
have "factI n = factI' n 1" by simp
also have "... = 1 * factR n" by (simp add: fact)
finally show "factI n = factR n" by simp
qed

(* migtermor *)
corollary "factI n = factR n"
proof -
have "factI n = factI' n 1" by simp
also have "... = 1 * factI' n 1" by simp
also have "... = 1 * factR n" using fact by simp
finally show "factI n = factR n" by simp
qed

(* pabrodmac *)
corollary "factI n = factR n"
proof -
have "factI n = factI' n 1" by simp
also have "... = 1 * factR n" using fact by simp
finally show "factI n = factR n" by simp
qed

(* dancorgar *)
corollary "factI n = factR n"
proof (induct n)
show "factI 0 = factR 0" by simp
next
fix n
assume "factI n = factR n"
have "factI (Suc n) = factI' (Suc n) 1" by simp
also have "... = 1 * factR (Suc n)" by (simp add: fact)
also have "... = factR (Suc n)" by simp
finally show "factI (Suc n) = factR (Suc n)" by simp
qed

text {* ---------------------------------------------------------------
Ejercicio 5.1. Definir, recursivamente y sin usar (@), la función
amplia :: 'a list ⇒ 'a ⇒ 'a list
tal que (amplia xs y) es la lista obtenida añadiendo el elemento y al
final de la lista xs. Por ejemplo,
amplia [d,a] t = [d,a,t]
------------------------------------------------------------------ *}

fun amplia :: "'a list ⇒ 'a ⇒ 'a list" where
"amplia [] y = [y]"
| "amplia (x#xs) y = x # amplia xs y"

text {* ---------------------------------------------------------------
Ejercicio 5.2. Escribir la demostración detallada de
amplia xs y = xs @ [y]
------------------------------------------------------------------- *}

(* crigomgom fraortmoy rubgonmar ivamenjim serrodcal pablucoto wilmorort
anaprarod marpoldia1 juacabsou paupeddeg pabrodmac lucnovdos
dancorgar bowma marcarmor13 antsancab1 *)
lemma "amplia xs y = xs @ [y]"
proof (induct xs)
show "amplia [] y = [] @ [y]" by simp
next
fix x xs
assume HI: "amplia xs y = xs @ [y]"
have "amplia (x # xs) y = x # (amplia xs y)" by simp
also have "... = x # (xs @ [y])" using HI by simp
also have "... = (x # xs) @ [y]" by simp
finally show "amplia (x # xs) y = (x # xs) @ [y]" by simp
qed

(* danrodcha ferrenseg *)
lemma "amplia xs y = xs @ [y]" (is "?P xs")
proof (induct xs)
show "?P []" by simp
next
fix x xs
assume HI: "?P xs"
have "amplia (x # xs) y = x # amplia xs y" by simp
also have "... = (x # xs) @ [y]" using HI by simp
finally show "?P (x#xs)" by simp
qed

(* migtermor *)
lemma "amplia xs y = xs @ [y]"
proof (induct xs)
have "amplia [] y = [y]" by (simp only: amplia.simps(1))
have "[] @ [y] = [y]" by simp
show "amplia [] y = [] @ [y]" by simp
next
fix x xs
assume HI: "amplia xs y = xs @ [y]"
have "amplia (x # xs) y = x # amplia xs y"
by (simp only: amplia.simps(2))
also have "... = x # (xs @ [y])" using HI by simp
also have "... = (x # xs) @ [y]" by simp
finally show "amplia (x # xs) y = (x # xs) @ [y]" by simp
qed

end
</source>

R8

2018-07-16T11:11:18Z

WikiSysop: Texto reemplazado: «isar» por «isabelle»

R9

2018-07-16T11:11:18Z

WikiSysop: Texto reemplazado: «isar» por «isabelle»

Relación 1

2018-07-16T11:11:18Z

WikiSysop: Texto reemplazado: «isar» por «isabelle»

<source lang="isabelle">
chapter {* R1: Programación funcional en Isabelle *}

theory R1
imports Main
begin

text {* ----------------------------------------------------------------
Ejercicio 0. Definir, por recursión, la función
factorial :: nat ⇒ nat
tal que (factorial n) es el factorial de n. Por ejemplo,
factorial 4 = 24
------------------------------------------------------------------- *}

(* danrodcha anaprarod ivamenjim serrodcal manmorjim1 fraortmoy
dancorgar ferrenseg paupeddeg pabrodmac antsancab1 fracorjim1 juacabsou *)
fun factorial :: "nat ⇒ nat" where
"factorial 0 = 1"
| "factorial (Suc n) = (Suc n) * factorial n"

value "factorial 4 = 24"

(* wilmorort pablucoto marcarmor13 crigomgom rubgonmar jeamacpov
marpoldia1*)
fun factorial1 :: "nat ⇒ nat" where
"factorial1 0 = 1 "
| "factorial1 n = n * factorial1 (n-1)"

value "factorial1 4 = 24"

text {* ----------------------------------------------------------------
Ejercicio 1. Definir, por recursión, la función
longitud :: 'a list ⇒ nat
tal que (longitud xs) es la longitud de la listas xs. Por ejemplo,
longitud [4,2,5] = 3
------------------------------------------------------------------- *}

(* wilmorort *)
(* Para usar las lista en forma de [a,b,c] *)
translations
"[x, xs]" == "x#[xs]"
"[x]" == "x#[]"

(* wilmorort serrodcal crigomgom rubgonmar anaprarod dancorgar ferrenseg
paupeddeg pabrodmac antsancab1 juacabsou *)
fun longitud :: "'a list ⇒ nat" where
"longitud [] = 0" |
"longitud (x # xs) = 1 + longitud xs "

value "longitud [4,2,5] = 3"

(* pablucoto *)
fun longitud2 :: "'a list ⇒ nat " where
"longitud2 [] = 0"
| "longitud2 xs = 1 + longitud2 (butlast xs) "

value "longitud2 [4,2,5] = 3"

(* pablucoto *)
fun longitud3 :: "'a list ⇒ nat " where
"longitud3 xs = (if xs = []
then 0
else 1 + longitud3 (butlast xs))"

value "longitud3 [4,2,5] = 3"

(* marcarmor13 manmorjim1 jeamacpov marpoldia1 fraortmoy *)
fun longitud4 :: "'a list ⇒ nat" where
"longitud4 [] = 0 "
| "longitud4 xs = (1 + longitud4 (tl xs))"

value "longitud4 [4,2,5] = 3"

(* danrodcha fracorjim1 *)
fun longitud5 :: "'a list ⇒ nat" where
"longitud5 [] = 0"
| "longitud5 (x#xs) = Suc (longitud5 xs)"

value "longitud5 [4,2,5] = 3"

(* serrodcal *)
fun longitud6 :: "'a list ⇒ nat" where
"longitud xs = length xs"

value "longitud6 [4,2,5] " -- "= 3"

(* Comentario: El objetivo es explicitar cómo está definida la función
length *)

(* ivamenjim *)
fun longitud7 :: "'a list ⇒ nat" where
"longitud7 [] = 0"
| "longitud7 xs = longitud7 (drop 1 xs) + 1"

value "longitud7 [4,2,5] = 3"

text {* ---------------------------------------------------------------
Ejercicio 2. Definir la función
fun intercambia :: 'a × 'b ⇒ 'b × 'a
tal que (intercambia p) es el par obtenido intercambiando las
componentes del par p. Por ejemplo,
intercambia (u,v) = (v,u)
------------------------------------------------------------------ *}

(* wilmorort marcarmor13 danrodcha crigomgom pablucoto rubgonmar
manmorjim1 serrodcal fraortmoy anaprarod dancorgar ferrenseg
paupeddeg pabrodmac antsancab1 fracorjim1 juacabsou*)
fun intercambia :: "'a × 'b ⇒ 'b × 'a" where
"intercambia (x,y) = (y,x)"

value "intercambia (u,v) = (v,u)"

(* jeamacpov ivamenjim marpoldia1*)
fun intercambia2 :: "'a × 'b ⇒ 'b × 'a" where
"intercambia2 (x,y) = (snd (x,y), fst (x,y))"

value "intercambia2 (u,v) = (v,u)"

text {* ---------------------------------------------------------------
Ejercicio 3. Definir, por recursión, la función
inversa :: 'a list ⇒ 'a list
tal que (inversa xs) es la lista obtenida invirtiendo el orden de los
elementos de xs. Por ejemplo,
inversa [a,d,c] = [c,d,a]
------------------------------------------------------------------ *}

(* wilmorort fraortmoy *)
(* @ :: "'a list => 'a list => 'a list", función agregación definida
en Theory Main, concatena dos listas: [a,b] @ [c,d] = [a,b,c,d] *)
fun inversa :: "'a list ⇒ 'a list" where
"inversa [] = []" |
"inversa (x # xs) = (inversa xs) @ (x#[]) "

value "inversa [a,d,c] = [c,d,a]"

(*marcarmor13 manmorjim1*)
fun inversa1 :: "'a list ⇒ 'a list" where
"inversa1 [] = []"
| "inversa1 xs = inversa1 (tl xs)@ ((hd xs)#[])"

value "inversa1 [a,d,c] = [c,d,a]"

(* danrodcha pablucoto rubgonmar anaprarod dancorgar ferrenseg paupeddeg
pabrodmac fracorjim1 *)
(* es igual que inversa sustituyendo x#[] por [x] *)
fun inversa2 :: "'a list ⇒ 'a list" where
"inversa2 [] = []"
| "inversa2 (x#xs) = (inversa2 xs) @ [x]"

value "inversa2 [a,d,c] = [c,d,a]"

(* danrodcha ivamenjim antsancab1 *)
fun inversa3 :: "'a list ⇒ 'a list" where
"inversa3 [] = []"
| "inversa3 (x#xs) = concat [(inversa3 xs),[x]] "

value "inversa3 [a,d,c] = [c,d,a]"

(* crigomgom serrodcal marpoldia1 juacabsou *)
fun inversa4 :: "'a list ⇒ 'a list" where
"inversa4 [] = []" |
"inversa4 xs = (last xs) # (inversa4 (butlast xs)) "

value "inversa4 [a,d,c] = [c,d,a]"

(* jeamacpov *)
fun inversa5 :: "'a list ⇒ 'a list" where
"inversa5 [] = []"
| "inversa5 xs =
(drop (length(xs)-1) xs) @ inversa5 (take (length(xs)-1) xs)"

value "inversa5 [a,d,c] = [c,d,a]"

text {* ---------------------------------------------------------------
Ejercicio 4. Definir la función
repite :: nat ⇒ 'a ⇒ 'a list
tal que (repite n x) es la lista formada por n copias del elemento
x. Por ejemplo,
repite 3 a = [a,a,a]
------------------------------------------------------------------ *}

(* wilmorort marcarmor13 crigomgom pablucoto rubgonmar manmorjim1
jeamacpov ivamenjim marpoldia1 juacabsou *)
fun repite :: "nat ⇒ 'a ⇒ 'a list" where
"repite 0 x = [] " |
"repite n x = x # (repite (n-1) x)"

value "repite 3 a = [a,a,a]"

(* danrodcha fraortmoy dancorgar ferrenseg paupeddeg pabrodmac antsancab1 fracorjim1 *)
fun repite1 :: "nat ⇒ 'a ⇒ 'a list" where
"repite1 0 x = []"
| "repite1 (Suc n) x = x # (repite1 n x)"

value "repite1 3 a = [a,a,a]"

(*serrodcal anaprarod *)
fun repite2 :: "nat ⇒ 'a ⇒ 'a list" where
"repite2 0 x = []"
| "repite2 (Suc n) x = [x] @ (repite2 n x)"

value "repite2 3 a = [a,a,a]"

(* ivamenjim *)
fun repite3 :: "nat ⇒ 'a ⇒ 'a list" where
"repite3 n x = replicate n x"

value "repite3 3 a = [a,a,a]"

(* Comentario: uso de replicate *)

text {* ---------------------------------------------------------------
Ejercicio 5. Definir la función
conc :: 'a list ⇒ 'a list ⇒ 'a list
tal que (conc xs ys) es la concatención de las listas xs e ys. Por
ejemplo,
conc [a,d] [b,d,a,c] = [a,d,b,d,a,c]
------------------------------------------------------------------ *}

(* marcarmor13 serrodcal *)
fun conc :: "'a list ⇒ 'a list ⇒ 'a list" where
"conc xs ys = xs@ys"

value "conc [a,d] [b,d,a,c] = [a,d,b,d,a,c]"

(* Comentario: El objetivo del ejercicio es explicitar la definición de
@ *)

(* danrodcha crigomgom rubgonmar paupeddeg *)
fun conc1 :: "'a list ⇒ 'a list ⇒ 'a list" where
"conc1 [] ys = ys"
| "conc1 xs [] = xs" (* esta no hace falta *)
| "conc1 (x#xs) ys = x# (conc1 xs ys)"

value "conc1 [a,d] [b,d,a,c] = [a,d,b,d,a,c]"

(* pablucoto jeamacpov fraortmoy anaprarod ferrenseg pabrodmac antsancab1 fracorjim1 juacabsou *)
fun conc2 :: "'a list ⇒ 'a list ⇒ 'a list" where
"conc2 [] ys = ys" |
"conc2 (x#xs) ys = x # (conc2 xs ys)"

value "conc2 [a,d] [b,d,a,c] = [a,d,b,d,a,c]"

(* manmorjim1 *)
fun conc3 :: "'a list ⇒ 'a list ⇒ 'a list" where
"conc3 [] ys = ys"
| "conc3 xs ys = (hd xs)#[] @ (conc3 (tl xs) ys)"

value "conc3 [a,d] [b,d,a,c] = [a,d,b,d,a,c]"

(* ivamenjim *)
fun conc4 :: "'a list ⇒ 'a list ⇒ 'a list" where
"conc4 xs ys = concat [xs,ys]"

value "conc4 [a,d] [b,d,a,c] = [a,d,b,d,a,c]"

(* Comentario: uso de concat *)

(* dancorgar *)
fun conc5 :: "'a list ⇒ 'a list ⇒ 'a list" where
"conc5 [] [] = []"
| "conc5 [] (y#ys) = y#(conc5 [] ys)"
| "conc5 (x#xs) ys = x#(conc5 xs ys)"

value "conc5 [a,d] [b,d,a,c] = [a,d,b,d,a,c]"

(* Comentario: La definición conc5 se puede simplificar. *)

(* marpoldia1 *)
fun conc6 :: "'a list ⇒ 'a list ⇒ 'a list" where
"conc6 [] [] = []"
| "conc6 [] ys = ys"
| "conc6 xs ys = (hd xs)#(conc6 (tl xs) ys)"

value "conc6 [a,d] [b,d,a,c] = [a,d,b,d,a,c]"

(* Comentario: La 2ª ecuación de conc6 contiene a la primera. Además, se
puede eliminar el uso de hd y tl. *)

text {* ---------------------------------------------------------------
Ejercicio 6. Definir la función
coge :: nat ⇒ 'a list ⇒ 'a list
tal que (coge n xs) es la lista de los n primeros elementos de xs. Por
ejemplo,
coge 2 [a,c,d,b,e] = [a,c]
------------------------------------------------------------------ *}

(* marcarmor13 manmorjim1*)
fun coge :: "nat ⇒ 'a list ⇒ 'a list" where
"coge 0 xs = []"
| "coge n xs = (hd xs)#(coge (n-1) (tl xs)) "

value "coge 2 [a,c,d,b,e] = [a,c]"
value "coge 2 [a,c,d,b,e] = [a,c]"

(* danrodcha fraortmoy anaprarod ferrenseg paupeddeg pabrodmac juacabsou *)
fun coge1 :: "nat ⇒ 'a list ⇒ 'a list" where
"coge1 0 _ = []"
| "coge1 _ [] = []"
| "coge1 (Suc n) (x#xs) = x#(coge1 n xs)"

value "coge1 2 [a,c,d,b,e] = [a,c]"
value "coge1 2 [a,c,d,b,e] = [a,c]"

(* crimgomgom jeamacpov rubgonmar ivamenjim *)
fun coge2 :: "nat ⇒ 'a list ⇒ 'a list" where
"coge2 0 _ = []"|
"coge2 _ [] = []" |
"coge2 n (x#xs) = x#(coge2 (n-1) xs) "

value "coge2 2 [a,c,d,b,e] = [a,c]"
value "coge2 2 [a,c,d,b,e] = [a,c]"

(* pablucoto *)
fun coge3 :: "nat ⇒ 'a list ⇒ 'a list" where
"coge3 0 xs = []" |
"coge3 n (x#xs) = (if n > length (x#xs)
then (x#xs)
else x # (coge3 (n-1) xs))"

value "coge3 2 [a,c,d,b,e] = [a,c]"
value "coge3 3 [a,c] = [a,c]"

(* serrodcal *)
fun coge4 :: "nat ⇒ 'a list ⇒ 'a list" where
"coge4 0 xs = []"
| "coge4 n xs = [(hd xs)]@(coge4 (n-1) (tl xs))"

value "coge4 2 [a,c,d,b,e] = [a,c]"

(* ivamenjim *)
fun coge5 :: "nat ⇒ 'a list ⇒ 'a list" where
"coge5 n xs = take n xs"

value "coge5 2 [a,c,d,b,e] = [a,c]"

(* Comentario: uso de take *)

(* dancorgar antsancab1 *)
fun coge6 :: "nat ⇒ 'a list ⇒ 'a list" where
"coge6 0 xs = []"
| "coge6 (Suc m) [] = []"
| "coge6 (Suc m) (x#xs) = (x#(coge6 m xs))"

value "coge6 2 [a,c,d,b,e] = [a,c]"

(* marpoldia1 *)
fun coge7 :: "nat ⇒ 'a list ⇒ 'a list" where
"coge7 0 xs = []"|
"coge7 n [] = []"|
"coge7 n xs = (hd xs) # (coge7 (n-1) (tl xs))"

value "coge7 0 [a,b,c] = []"
value "coge7 2 [] = []"
value "coge7 2 [a,c,d,b,e] = [a,c]"

(* fracorjim1 - Versión compacta, pero genera un esquema de inducción poco claro al no utilizar patrones *)
fun coge8 :: "nat ⇒ 'a list ⇒ 'a list" where
"coge8 0 xs = []"
| "coge8 (Suc n) xs = (hd xs)#(coge8 n (tl xs))"

text {* ---------------------------------------------------------------
Ejercicio 7. Definir la función
elimina :: nat ⇒ 'a list ⇒ 'a list
tal que (elimina n xs) es la lista obtenida eliminando los n primeros
elementos de xs. Por ejemplo,
elimina 2 [a,c,d,b,e] = [d,b,e]
------------------------------------------------------------------ *}

(* marcarmor13 manmorjim1 serrodcal marpoldia1*)
fun elimina :: "nat ⇒ 'a list ⇒ 'a list" where
"elimina 0 xs = xs"
| "elimina n xs = (elimina (n-1) (tl xs ))"

(* danrodcha crigomgom fraortmoy anaprarod paupeddeg pabrodmac juacabsou *)
fun elimina1 :: "nat ⇒ 'a list ⇒ 'a list" where
"elimina1 0 xs = xs"
| "elimina1 _ [] = []"
| "elimina1 (Suc n) (x#xs) = elimina1 n xs"

(* pablucoto *)
fun elimina2 :: "nat ⇒ 'a list ⇒ 'a list" where
"elimina2 0 xs = xs" |
"elimina2 n (x#xs) = (if n > length (x#xs)
then []
else (elimina2 (n-1) xs))"

(* jeamacpov ivamenjim *)
fun elimina3 :: "nat ⇒ 'a list ⇒ 'a list" where
"elimina3 0 xs = xs"
| "elimina3 n xs = (elimina3 (n-1) (drop 1 xs)) "

(* rubgonmar *)
fun elimina4 :: "nat ⇒ 'a list ⇒ 'a list" where
"elimina4 0 xs = xs" |
"elimina4 n xs = (if n > length(xs)
then []
else (elimina4 (n-1) (tl xs)))"

value "elimina4 2 [a,c,d,b,e] = [d,b,e]"

(* dancorgar antsancab1 *)
fun elimina5 :: "nat ⇒ 'a list ⇒ 'a list" where
"elimina5 0 xs = xs"
| "elimina5 (Suc m) [] = []"
| "elimina5 (Suc m) (x#xs) = elimina5 m xs"

value "elimina5 2 [a,c,d,b,e]= [d,b,e]" -- "= [d,b,e]"

(* ferrenseg *)

fun elimina6 :: "nat ⇒ 'a list ⇒ 'a list" where
"elimina6 0 xs = xs"
|"elimina6 n [] = []"
|"elimina6 (Suc n) (x#xs) = elimina6 n xs"

value "elimina6 2 [a,c,d,b,e]" -- "= [d,b,e]"

(* fracorjim1 - Versión compacta, pero genera un esquema de inducción poco claro al no utilizar patrones *)
fun elimina7 :: "nat ⇒ 'a list ⇒ 'a list" where
"elimina7 0 xs = xs"
| "elimina7 (Suc n) xs = elimina7 n (tl xs)"

text {* ---------------------------------------------------------------
Ejercicio 8. Definir la función
esVacia :: 'a list ⇒ bool
tal que (esVacia xs) se verifica si xs es la lista vacía. Por ejemplo,
esVacia [] = True
esVacia [1] = False
------------------------------------------------------------------ *}

(* marcarmor13 rubgonmar danrodcha crigomgom fraortmoy dancorgar
paupeddeg *)
fun esVacia :: "'a list ⇒ bool" where
"esVacia [] = True "
| "esVacia xs = False"

(* danrodcha anaprarod*)
fun esVacia1 :: "'a list ⇒ bool" where
"esVacia1 xs = (xs = [])"

(* pablucoto jeamacpov antsancab1 *)
fun esVacia2 :: "'a list ⇒ bool" where
"esVacia2 xs = (if xs = []
then True
else False)"

(*manmorjim1, serrodcal*)
fun esVacia3 :: "'a list ⇒ bool" where
"esVacia3 xs = (length xs = 0)"

value "esVacia3 [] = True"
value "esVacia3 [1] = False"

(* ivamenjim marpoldia1*)
fun esVacia4 :: "'a list ⇒ bool" where
"esVacia4 xs = (if length (xs) > 0
then False
else True)"

value "esVacia4 [] = True"
value "esVacia4 [1] = False"

(* ferrenseg fracorjim1 juacabsou *)
fun esVacia5 :: "'a list ⇒ bool" where
"esVacia5 [] = True"
|"esVacia5 (x#xs) = False"

value "esVacia5 []" -- "= True"
value "esVacia5 [1]" -- "= False"

(* pabrodmac *)
fun esVacia6 :: "'a list ⇒ bool" where
"esVacia6 [] = True"
| "esVacia6 _ = False"

value "esVacia6 []= True"
value "esVacia6 [1]= False"

text {* ---------------------------------------------------------------
Ejercicio 9. Definir la función
inversaAc :: 'a list ⇒ 'a list
tal que (inversaAc xs) es a inversa de xs calculada usando
acumuladores. Por ejemplo,
inversaAc [a,c,b,e] = [e,b,c,a]
------------------------------------------------------------------ *}

(* rubgonmar marcarmor13 mamnorjim1 serrodcal ivamenjim marpoldia1*)
fun inversaAcAux :: "'a list ⇒ 'a list ⇒ 'a list" where
"inversaAcAux [] ys = ys"
| "inversaAcAux xs ys = inversaAcAux (tl xs) (hd xs#ys) "

(* rubgonmar danrodcha crigomgom manmorjim1 serrodcal ivamenjim
anaprarod marpoldia1*)
fun inversaAc :: "'a list ⇒ 'a list" where
"inversaAc xs = inversaAcAux xs []"

value "inversaAc [a,c,b,e]= [e,b,c,a]"

(* danrodcha crigomgom anaprarod paupeddeg juacabsou *)
fun inversaAcAux1 :: "'a list ⇒ 'a list ⇒ 'a list" where
"inversaAcAux1 [] ys = ys"
| "inversaAcAux1 (x#xs) ys = inversaAcAux1 xs (x#ys)"

(* pablucoto *)
fun inversaAcAux2 :: "'a list ⇒ 'a list ⇒ 'a list" where
"inversaAcAux2 [] [] = []"|
"inversaAcAux2 xs (y#ys) = (inversaAcAux2 [] ys) @ [y]"

fun inversaAc2 :: "'a list ⇒ 'a list" where
"inversaAc2 xs = inversaAcAux2 [] xs "

value "inversaAc2 [a,c,b,e]= [e,b,c,a]"

(* jeamacpov *)
fun inversaAcAux3 :: "'a list ⇒ 'a list ⇒ 'a list" where
"inversaAcAux3 ys [] = ys"
| "inversaAcAux3 ys (x#xs) = inversaAcAux3 (x#ys) xs"

fun inversaAc3 :: "'a list ⇒ 'a list" where
"inversaAc3 xs = inversaAcAux3 [] xs"

value "inversaAc3 [a,c,b,e]= [e,b,c,a]"

(* fraortmoy *)
fun inversaAcAux4 :: "'a list ⇒ 'a list ⇒ 'a list" where
"inversaAcAux4 xs [] = xs"
| "inversaAcAux4 xs (x#ys) = inversaAcAux4 (x#xs) ys"

fun inversaAc4 :: "'a list ⇒ 'a list" where
"inversaAc4 [] = []"
| "inversaAc4 (x#xs)= (inversaAcAux4 [x] xs)"

value "inversaAc4 [a,c,b,e]" -- "= [e,b,c,a]"

(* dancorgar *)
fun inversaAcAux5 :: "'a list ⇒ 'a list ⇒ 'a list" where
"inversaAcAux5 [] [] = []"
| "inversaAcAux5 [] (y#ys) = y#(inversaAcAux5 [] ys)"
| "inversaAcAux5 (x#xs) ys = x#(inversaAcAux5 xs ys)"

(* dancorgar *)
fun inversaAc5 :: "'a list ⇒ 'a list" where
"inversaAc5 [] = []"
| "inversaAc5 (x#xs) = inversaAcAux5 (inversaAc5 xs) [x]"

value "inversaAc5 [a,c,b,e] = [e,b,c,a]" -- "= [e,b,c,a]"

(* ferrenseg *)
fun inversaAcAux6 :: "'a list ⇒ 'a list ⇒ 'a list" where
"inversaAcAux6 [] ys = ys"
|"inversaAcAux6 (x#xs) ys = inversaAcAux6 xs (x#ys)"

fun inversaAc6 :: "'a list ⇒ 'a list" where
"inversaAc6 xs = inversaAcAux6 xs []"

value "inversaAc6 [a,c,b,e]" -- "= [e,b,c,a]"

(* pabrodmac antsancab1 *)
fun inversaAcAux7 :: "'a list ⇒ 'a list ⇒ 'a list" where
"inversaAcAux7 [] ys = ys "
| "inversaAcAux7 (x#xs) ys =inversaAcAux7 xs (x#ys) "

fun inversaAc7 :: "'a list ⇒ 'a list" where
"inversaAc7 [] =[]"
| "inversaAc7 (x#xs) = inversaAcAux7 xs [x]"

value "inversaAc7 [a,c,b,e]= [e,b,c,a]"

text {* ---------------------------------------------------------------
Ejercicio 10. Definir la función
sum :: nat list ⇒ nat
tal que (sum xs) es la suma de los elementos de xs. Por ejemplo,
sum [3,2,5] = 10
------------------------------------------------------------------ *}

(* rubgonmar marcarmor13 manmorjim1 serrodcal ivamenjim marpoldia1*)
fun sum :: "nat list ⇒ nat" where
"sum [] = 0"
| "sum xs = hd xs + sum (tl xs)"

value "sum [3,2,5] = 10"

(* danrodcha crigomgom pablucoto fraortmoy anaprarod ferrenseg paupeddeg
pabrodmac antsancab1 fracorjim1 juacabsou *)
fun sum1 :: "nat list ⇒ nat" where
"sum1 [] = 0"
| "sum1 (x#xs) = x + sum1 xs"

value "sum1 [3,2,5] = 10"

(* danrodcha *)
fun sum2 :: "nat list ⇒ nat" where
"sum2 xs = fold (op +) xs 0"

value "sum2 [3,2,5] = 10"

(* jeamacpov *)
fun sum3 :: "nat list ⇒ nat" where
"sum3 [] = 0"
| "sum3 xs = (hd xs) + (sum3(drop 1 xs))"

value "sum3 [3,2,5] = 10"

(* ivamenjim *)
fun sum4 :: "nat list ⇒ nat" where
"sum4 xs = listsum xs"

value "sum4 [3,2,5] = 10"

(* Comentario: uso de listsum *)

(* dancorgar *)
fun sum5 :: "nat list ⇒ nat" where
"sum5 [] = 0"
| "sum5 (x#[]) = x"
| "sum5 (x#(y#xs)) = sum5 ((x + y)#xs)"

text {* ---------------------------------------------------------------
Ejercicio 11. Definir la función
map :: ('a ⇒ 'b) ⇒ 'a list ⇒ 'b list
tal que (map f xs) es la lista obtenida aplicando la función f a los
elementos de xs. Por ejemplo,
map (λx. 2*x) [3,2,5] = [6,4,10]
------------------------------------------------------------------ *}

(* rubgonmar marcarmor13 manmorjim1 serrodcal ivamenjim marpoldia1*)
fun map :: "('a ⇒ 'b) ⇒ 'a list ⇒ 'b list" where
"map f [] = []"
| "map f xs = f (hd xs) # map f (tl xs)"

value "map (λx. 2*x) [3::nat,2,5] = [6,4,10]"

(* wilmorort danrodcha crigomgom pablucoto fraortmoy anaprarod dancorgar
ferrenseg paupeddeg pabrodmac antsancab1 fracorjim1 juacabsou *)
fun map2 :: "('a ⇒ 'b) ⇒ 'a list ⇒ 'b list" where
"map2 f [] = []"
| "map2 f (x # xs) = f x # map2 f xs" (* yo pondría paréntesis, pero sin
ellos lo entiende*)

value "map2 (λx. 2*x) [3::nat,2,5] = [6,4,10]"

(* jeamacpov *)
fun map3 :: "('a ⇒ 'b) ⇒ 'a list ⇒ 'b list" where
"map3 f [] = []"
| "map3 f xs = f (hd xs) # (map3 f (drop 1 xs)) "

value "map3 (λx. 2*x) [3::nat,2,5] = [6,4,10]"

end
</source>

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Documentación

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En esta página se recogen en enlaces que sirven de documentación al curso de "Razonamiento automático"

== Visiones generales de la DAO ==

# J.A. Alonso. [http://goo.gl/NWk7b Razonamiento formalizado: Del sueño a la realidad de las pruebas]. ''Vestigium'', 26 de diciembre de 2012.
# J. Avigad. [http://www.andrew.cmu.edu/user/avigad/Talks/icerm.pdf Interactive theorem proving, automated reasoning, and mathematical computation]. ICERM, 14 de diciembre de 2012.
# M. Davis. [http://www.cs.nyu.edu/cs/faculty/davism/early.ps The early history of automated deduction].
# J.P. Delahaye [http://interstices.info/jcms/int_63417/du-reve-a-la-realite-des-preuves Du rêve à la réalité des preuves]. ''Interstices'', 8 de julio de 2012.
# J. Germoni [http://images.math.cnrs.fr/Coq-et-caracteres.html Coq et caractères: Preuve formelle du théorème de Feit et Thompson]. ''Images des Mathématiques'', CNRS, 23 de noviembre de 2012.
# H. Geuvers [http://www.ias.ac.in/article/fulltext/sadh/034/01/0003-0025 Proof assistants: History, ideas and future]. ''Sadhana'', Vol. 34-1, pp. 3-25, février 2009.
# G. Gonthier [http://www.ams.org/notices/200811/tx081101382p.pdf The four-color theorem]. ''Notices of the AMS'', Vol. 55, n° 11, pp. 1382-1393, 2008.
# T. Hales. [http://www.ams.org/notices/200811/tx081101370p.pdf Formal proof]. ''Notices of AMS'', Vol. 55, N. 11 (2008) pp. 1370-1380.
# J. Harrison. [http://www.cl.cam.ac.uk/~jrh13/papers/ab.html A short survey of automated reasoning]. ''Lecture Notes in Computer Science'', Vol. 4545, pp. 334-349, 2007.
# J. Harrison. [http://www.ams.org/notices/200811/tx081101395p.pdf Formal proof: Theory and practice]. ''Notices of the AMS'', Vol. 55, N. 11 (2008) p.1395-1406.
# G. Kolata. [http://www.nytimes.com/library/cyber/week/1210math.html Computer math proof shows reasoning power]. ''The New York Times'', 10 de diciembre de 1996.
# D. MacKenzie [http://www.bcs.org/server.php?show=ConWebDoc.4364 Computers and the sociology of mathematical proof].
# G. Sutcliffe. [http://www.cs.miami.edu/~tptp/OverviewOfATP.html What is automated theorem proving?].
# F. Wiedijk [http://www.cs.ru.nl/~freek/100/ Formalizing the «top 100» of mathematical theorems].
# F. Wiedijk [http://www.ams.org/notices/200811/tx081101408p.pdf Formal proof - Getting started]. ''Notices of the AMS'', Vol. 55, n° 11, pp. 1408-1414, 2008.
# F. Wiedijk, [http://mizar.org/trybulec65/8.pdf The QED manifesto revisited]. ''Studies in Logic, Grammar and Rhetoric'', Vol. 10(23), pp. 121-133, 2007.

== Referencias sobre Isabelle/HOL ==
# B. Grechuk [http://dream.inf.ed.ac.uk/projects/isabelle/Isabelle_Primer.pdf Isabelle primer for mathematicians].
# T. Nipkow [https://isabelle.in.tum.de/doc/prog-prove.pdf Programming and proving in Isabelle/HOL].
# T. Nipkow, M. Wenzel y L.C. Paulson [http://isabelle.in.tum.de/doc/tutorial.pdf A proof assistant for higher-order logic].
# [http://www.cl.cam.ac.uk/research/hvg/Isabelle/dist/library/HOL/HOL/document.pdf Isabelle/HOL — Higher-Order Logic].
# M. Wenzel [http://isabelle.in.tum.de/doc/isabelle-ref.pdf The Isabelle/Isar Reference Manual].
# M. Wenzel [http://typessummerschool07.cs.unibo.it/courses/wenzel-isabelle-quickref.pdf The Isabelle/Isar quick reference].
# J. Siek [http://ecee.colorado.edu/~siek/ecen3703/spring10/quick-ref.pdf Quick Reference for Isabelle/Isar Propositional Logic].
# J. Siek [http://ecee.colorado.edu/~siek/ecen3703/spring10/quick-ref2.pdf Quick Reference for Isabelle/Isar More Proof Techniques].
# J. Siek [http://ecee.colorado.edu/~siek/ecen3703/spring10/quick-ref3.pdf Quick Reference for Isabelle/Isar First-Order Logic].
# [http://www.cl.cam.ac.uk/research/hvg/Isabelle/documentation.html Tutorials and manuals for Isabelle].

== Lecturas complementarias ==
=== Programación funcional ===
# J.A. Alonso [https://www.cs.us.es/~jalonso/cursos/i1m-16/temas/2016-17-IM-temas-PF.pdf Temas de "Programación funcional"]. Publicaciones del Grupo de Lógica Computacional. Universidad de Sevilla, 2016.
# J.A. Alonso y M.J. Hidalgo [http://www.cs.us.es/~jalonso/publicaciones/Piensa_en_Haskell.pdf Piensa en Haskell (Ejercicios de programación funcional con Haskell)]. Publicaciones del Grupo de Lógica Computacional. Universidad de Sevilla, 2012.
# G. Hutton [http://goo.gl/pKqG Programming in Haskell]. Cambridge University Press, 2007.
# M. Lipovača [http://aprendehaskell.es ¡Aprende Haskell por el bien de todos!].

=== Lógica computacional ===
# J.A. Alonso [https://www.cs.us.es/~jalonso/cursos/li-15/temas/temas-LI-2015-16.pdf Temas de "Lógica informática" (2015-16)]. Publicaciones del Grupo de Lógica Computacional. Universidad de Sevilla, 2015.
# K. Broda, S. Eisenbach, H. Khoshnevisan y S. Vickers [http://pubs.doc.ic.ac.uk/reasoned-programming/reasoned-programming.pdf Reasoned programming]. Imperial College, 1994.
# K. Doets y J. van Eijck [http://www.ldc.usb.ve/~astorga/Haskell.Road.pdf The Haskell Road to Logic, Maths and Programming].
# M. Huth y M. Ryan [http://bit.ly/2e8dFEm Logic in computer science: Modelling and reasoning about systems]. Cambridge University Press, 2004. (Incluye el [http://www.cs.bham.ac.uk/research/lics/tutor/index.html tutor en la Red]).

== Cursos relacionados ==
=== Cursos con Isabelle/HOL ===
# Jeremy Avigad. [http://www.phil.cmu.edu/~avigad/formal/ Logic and Formal Verification]. (Carnegie Mellon, 2009).
# Clemens Ballarin. [http://cl-informatik.uibk.ac.at/teaching/ss08/atp/introduction.php Automatic Deduction]. (Univ de Innsbruck, 2008).
# Clemens Ballarin. [http://www4.in.tum.de/~ballarin/belgrade08-tut/ Introduction to the Isabelle Proof Assistant]. (Belgrado, 2008).
# Clemens Ballarin y Gerwin Klein [http://isabelle.in.tum.de/coursematerial/IJCAR04 Introduction to the Isabelle Proof Assistant]. (en el IJCAR-2004).
# Clemens Ballarin y Tjark Weber. [http://cl-informatik.uibk.ac.at/teaching/ws06/atp/introduction.php Automated Theorem Proving in Isabelle/HOL]. (Univ. de Innsbruck, 2006-07).
# Jasmin Blanchette, Mathias Fleury y Daniel Wand [http://people.mpi-inf.mpg.de/~jblanche/cswi/ss2015/ Concrete semantics with Isabelle/HOL]. (Univ. del Sarre, 2015-16).
# A.D. Brucker, D. Basin, J.G. Smaus y B. Wolff. [http://archiv.infsec.ethz.ch/education/permanent/csmr.html Computer-supported Modeling and Reasoning]. (ETH Zurich, 2011).
# Mads Dam. [http://www.csc.kth.se/utbildning/kth/kurser/DD2453/aform07/ Advanced formal methods]. (KTH Royal Institute of Technology, 2007).
# Jacques Fleuriot. [http://www.inf.ed.ac.uk/teaching/courses/ar Automated reasoning]. (Univ. de Edimburgo, 2016-17).
# Thomas Genet [http://www.irisa.fr/celtique/genet/ACF Software formal analysis and design]. (Univ. de Rennes)
# Gerwin Klein. [http://www.cse.unsw.edu.au/~kleing/teaching/thprv-04 Theorem Proving - Principles, Techniques, Applications]. (NICTA, 2004).
# Gerwin Klein. [http://www.cse.unsw.edu.au/~cs4161/index.html Advanced Topics in Software Verification]. (NICTA, 2012).
# Joao Marcos. [http://www.dimap.ufrn.br/~jmarcos/courses/LC/Ementa.htm Lógica computacional: Demonstração assistida e semi-automática de teoremas].(UFRN, 2000).
# Tobias Nipkow. [http://www21.in.tum.de/teaching/semantik/WS1617/ Semantics of programming languages]. (Univ. de Munich, 2012-13).
# Tobias Nipkow. [http://isabelle.in.tum.de/coursematerial/PSV2009-1 Theorem Proving with Isabelle/HOL An Intensive Course].
# Larry Paulson. [http://www.cl.cam.ac.uk/teaching/0910/L21/ Interactive Formal Verification]. (Univ. de Cambridge, 2009-10).
# Arnd Poetzsch-Heffter. [https://softech.informatik.uni-kl.de/homepage/de/teaching/SVHOL14/ Specification and Verification with Higher-Order Logic].
# Jeremy G. Siek. [http://www.cs.colorado.edu/~siek/7000/spring07/ Practical Theorem Proving with Isabelle/Isar]. (Univ. de Colorado, 2007).
# Jeremy G. Siek. [http://ecee.colorado.edu/~siek/ecen5013/spring11/ Theorem proving in Isabelle]. (Univ. de Colorado, 2011).
# Jan-Georg Smaus. [http://www.informatik.uni-freiburg.de/~ki/teaching/ws0910/csmr/lecture.html Computer-supported modeling and reasoning]. (Univ. de Feiburgo, 2009).
# Christian Sternagel [http://cl-informatik.uibk.ac.at/teaching/ss11/eve/content.php Experiments in Verification – Introduction to Isabelle/HOL]. (Univ. de Innsbruck, 2011-12).
# Tjark Weber. [http://www.cl.cam.ac.uk/teaching/1011/L21/ Interactive Formal Verification]. (Univ. de Cambridge, 2010-11).

=== Otros cursos ===
# José A. Alonso [http://www.cs.us.es/~jalonso/cursos/li/ Lógica informática] (Univ. de Sevilla, 2012-13).
# Thorsten Altenkirch y Peter Morris [http://www.cs.nott.ac.uk/~txa/g52ifr Introduction to formal reasoning] (Univ. de Nottingham, 2010-11).
# Yves Bertot, Pierre Casteran, Benjamin Gregoire, Pierre Letouzey y Assia Mahboubi [http://www.di.ens.fr/~zappa/teaching/coq/ecole11 Modelling and verifying algorithms in Coq: an introduction]. (INRIA Paris-Rocquencourt, 14-18 noviembre 2011).
# Pierre Castéran [http://www.labri.fr/perso/casteran/FM/Logique/index.html Logic (Master In Verification)] (Univ. de Burdeos, 2011-12).
# Adam Chlipala [http://stellar.mit.edu/S/course/6/fa11/6.892/ Interactive computer theorem proving]. (MIT, 2012-13).
# Adam Chlipala y Armando Solar Lezama [https://stellar.mit.edu/S/course/6/fa13/6.820/index.html Foundations of program analysis]. (MIT, 2013-14).
# Robby Findler [http://www.eecs.northwestern.edu/~robby/courses/395-495-2013-fall Certified programming with dependent types]. (Northwestern, 2013-14).
# Carlos Luna y Gustavo Betarte. [https://eva.fing.edu.uy/course/view.php?id=363 Construcción formal de programas en teoría de tipos]. (Univ. de la República, Uruguay, 2013-14).
# Ian Hodkinson [http://www.doc.ic.ac.uk/~imh/teaching/140_logic/logic.html Logic] (Imperial College, Londres, 2010-11).
# Peter Lucas [http://www.cs.ru.nl/~peterl/teaching/KeR/ Knowledge Representation and Reasoning] (Radboud University # egen, 2011-12).
# Larry Paulson [http://www.cl.cam.ac.uk/Teaching/current/LogicProof/ Logic and Proof] (Univ. de Cambridge, 2011-12).
# David Pichardie [http://www.irisa.fr/celtique/pichardie/teaching/M2/MDV/ Méthode de vérification] (Universidad de Rennes, 2006-07).

== Bibliotecas de ejemplos de verificación ==
# [http://afp.sourceforge.net Archive of Formal Proofs].
# [http://www.cs.ru.nl/~freek/100 Formalizing 100 Theorems].
# [http://toccata.lri.fr/gallery Gallery of verified programs].
# [http://automatedreasoning.net/ Larry Wos' Notebooks].
# [http://www.cs.miami.edu/~tptp/ The TPTP Problem Library for Automated Theorem Proving].
# [http://www.macs.hw.ac.uk/vstte10/Competition.html The 1st Verified Software Competition].
# [https://sites.google.com/site/vstte2012/compet The 2nd Verified Software Competition].
# [http://verifythis.cost-ic0701.org VerifyThis (A collection of verification benchmarks].

== Artículos recientes ==
Están en orden cronológico inverso a la fecha de su reseña en [http://www.glc.us.es/~jalonso/vestigium/tag/resena Vestigium]:
# [http://bit.ly/1iZjgqN Proof Pearl: A probabilistic proof for the Girth-Chromatic number theorem]. L. Noschinski
# [http://bit.ly/1iJ8uVz A graph library for Isabelle]. ~ L. Noschinski
# [http://bit.ly/I0CU80 Gödel’s incompleteness theorems]. ~ L.C. Paulson
# [http://bit.ly/I0CPRN The hereditarily finite sets]. ~ L.C. Paulson
# [http://bit.ly/HBiIJI Applications of real number theorem proving in PVS]. ~ H. Gottliebsen, R. Hardy, O. Lightfoot y U. Martin
# [http://bit.ly/1awnMLB A machine-assisted proof of Gödel’s incompleteness theorems for the theory of hereditarily finite sets]. ~ L.C. Paulson
# [http://bit.ly/19lWeYy Verified AIG algorithms in ACL2]. ~ J. Davis y S. Swords
# [http://bit.ly/GAhC00 A formal model and correctness proof for an access control policy framework]. ~ C. Wu, X. Zhang y C. Urban
# [http://bit.ly/16SMvSS The ontological argument in PVS]. ~ J. Rushby
# [http://bit.ly/1dRt9n0 Formalizing Moessner’s theorem and generalizations in Nuprl]. ~ M. Bickford, D. Kozen y A. Silva
# [http://bit.ly/1bSeDNB Formalization in PVS of balancing properties necessary for the security of the Dolev-Yao cascade protocol model]. ~ M. Ayala y Y. Santos
# [http://bit.ly/1feFqWE Proof assistant based on didactic considerations]. ~ J. Pais y A Tasistro
# [http://bit.ly/18tHNBi Theory exploration for interactive theorem proving]. ~ M. Johansson
# [http://bit.ly/1b0242s From Tarski to Hilbert]. ~ G. Braun y J. Narboux
# [http://bit.ly/18HaXaR Formal verification of language-based concurrent noninterference]. ~ A. Popescu, J. Hölzl y T. Nipkow
# [http://bit.ly/1aRTQsU A Traffic Alert and Collision Avoidance System(TCAS-II) Resolution Advisory Algorithm]. ~ C. Muñoz, A. Narkawicz y J. Chamberlain
# [http://bit.ly/1dNwhDI Formal verification of cryptographic security proofs]. ~ M. Berg
# [http://bit.ly/17muAUv Polygonal numbers in Mizar]. ~ A. Grabowski
# [http://bit.ly/1hk5z6L A mechanised proof of Gödel’s incompleteness theorems using Nominal Isabelle]. ~ L.C. Paulson
# [http://bit.ly/1cSL0wE Steps towards verified implementations of HOL Light]. ~ M.O. Myreen, S. Owens y R. Kumar
# [http://bit.ly/16Kbgm0 Generic datatypes à la carte]. ~ S. Keuchel y T. Schrijvers
# [http://bit.ly/1bqJGx4 Proof pearl: A verified bignum implementation in x86-64 machine code]. ~ M.O. Myreen y G. Curello
# [http://bit.ly/142ow2Q Mechanized metatheory for a λ λ-calculus with trust types]. ~ R. Ribeiro, C. Camarão y L. Figueiredo
# [http://bit.ly/15WZBDy Proving soundness of combinatorial Vickrey auctions and generating verified executable code]. ~ M.B. Caminati, M. Kerber, C. Lange y C. Rowat
# [http://bit.ly/198g4n9 A computer-assisted proof of correctness of a marching cubes algorithm]. ~ A.N. Chernikov y J. Xu
# [http://bit.ly/11QA5g7 Verifying the bridge between simplicial topology and algebra: the Eilenberg-Zilber algorithm]. ~ L. Lambán, J. Rubio, F.J. Martín y J.L. Ruiz
# [http://bit.ly/1cJAXYk The Königsberg bridge problem and the friendship theorem]. ~ W. Li
# [http://bit.ly/13DBK9R Formal verification of a proof procedure for the description logic ALC]. ~ M. Chaabani, M. Mezghiche y M. Strecker
# [http://bit.ly/1ep2ex9 Pratt’s primality certificates]. ~ S. Wimmer y L. Noschinski
# [http://bit.ly/13C95Ci Reasoning about higher-order relational specifications]. ~ Y. Wang, K. Chaudhuri, A. Gacek y G. Nadathur
# [http://bit.ly/18QQLcL Proofs you can believe in – Proving equivalences between Prolog semantics in Coq]. ~ J. Kriener, A. King y S. Blazy
# [http://bit.ly/19uc82J Certified, efficient and sharp univariate Taylor models in Coq]. ~ E. Martin-Dorel, L. Rideau, L. Théry, M. Mayero y I. Paşca
# [http://bit.ly/1c4Rzel Ordinals in HOL: Transfinite arithmetic up to (and beyond) ω₁]. ~ M. Norrish y B. Huffman
# [http://bit.ly/14b4Akz Program verification based on Kleene algebra in Isabelle/HOL] ~ A. Armstrong, G. Struth y T. Weber
# [http://bit.ly/1aOgRKx Reading an algebra textbook (by translating it to a formal document in the Isabelle/Isar language)]. ~ C. Ballarin
# [http://bit.ly/11HKixj Computational verification of network programs in Coq]. ~ G. Stewart
# [http://www.glc.us.es/~jalonso/vestigium/resena-certifying-homological-algorithms-to-study-biomedical-images Certifying homological algorithms to study biomedical images]. ~ M. Poza
# [http://bit.ly/16Nks9m Formalizing cut elimination of coalgebraic logics in Coq]. ~ H. Tews
# [http://www.glc.us.es/~jalonso/vestigium/resena-the-formalization-of-syntax-based-mathematical-algorithms-using-quotation-and-evaluation/ The formalization of syntax-based mathematical algorithms using quotation and evaluation]. ~ W.M. Farmer
# [http://www.glc.us.es/~jalonso/vestigium/resena-certified-symbolic-manipulation-bivariate-simplicial-polynomials/ Certified symbolic manipulation: Bivariate simplicial polynomials]. ~ L. Lambán, F.J. Martín, J. Rubio y J.L. Ruiz
# [http://www.glc.us.es/~jalonso/vestigium/resena-solveurs-cpfd-verifies-formellement/ Solveurs CP(FD) vérifiés formellement]. ~ C Dubois y A. Gotlieb
# [http://www.glc.us.es/~jalonso/vestigium/resena-formalizing-bounded-increase/ Formalizing bounded increase]. ~ R. Thiemann
# [http://www.glc.us.es/~jalonso/vestigium/resena-formal-mathematics-on-display-a-wiki-for-flyspeck/ Formal mathematics on display: A wiki for Flyspeck]. ~ C. Tankink, C. Kaliszyk, J. Urban y H. Geuvers
# [http://www.glc.us.es/~jalonso/vestigium/resena-theorem-of-three-circles-in-coq Theorem of three circles in Coq]. ~ J. Zsidó
# [http://bit.ly/19fiWAY Certified HLints with Isabelle/HOLCF-Prelude]. ~ J. Breitner, B. Huffman, N. Mitchell y C. Sternagel
# [http://bit.ly/ZfSQrQ Automatic data refinement]. ~ P. Lammich
# [http://bit.ly/18vyjm7 The rooster and the butterflies (a machine-checked proof of the Jordan-Hölder theorem for finite groups)]. ~ A. Mahboubi
# [http://bit.ly/114oyZV Mechanical verification of SAT refutations with extended resolution]. ~ N. Wetzler, M.J.H. Heule y W.A. Hunt Jr.
# [http://bit.ly/13H0REu Type classes and filters for mathematical analysis in Isabelle/HOL] ~ J. Hölzl, F. Immler y B. Huffman
# [http://bit.ly/10fv8wO Verifying refutations with extended resolution]. ~ M. J. H. Heule, W. A. Hunt, Jr. y N. Wetzler
# [http://bit.ly/10EcFWj A Web interface for Isabelle: The next generation]. ~ C. Lüth y M. Ring
# [http://bit.ly/18P9CSv On the formalization of continuous-time Markov chains in HOL]. ~ L. Liu, O. Hasan y S. Tahar
# [http://bit.ly/17H2mqy Formalizing Turing machines]. ~ A. Asperti y W. Ricciotti
# [http://bit.ly/YwuCeL Light-weight containers for Isabelle: efficient, extensible, nestable]. ~ A. Lochbihler
# [http://bit.ly/10XLrRA Graph theory]. ~ L. Noschinski
# [http://bit.ly/19kPEP4 A machine-checked proof of the odd order theorem]. ~ G. Gonthier et als.
# [http://goo.gl/LdihL A constructive theory of regular languages in Coq]. ~ C. Doczkal, J.O. Kaiser y G. Smolka
# [http://goo.gl/gwcwL A formal proof of Kruskal’s tree theorem in Isabelle/HOL]. ~ C. Sternagel
# [http://goo.gl/CUF3e Formalizing Knuth-Bendix orders and Knuth-Bendix completion]. ~ C. Sternagel y R. Thiemann
# [http://goo.gl/9JAfX Developing an auction theory toolbox]. ~ C. Lange, C. Rowat, W. Windsteiger y M. Kerber
# [http://goo.gl/6OfmQ Formalization of incremental simplex algorithm by stepwise refinement]. ~ M. Spasić y F. Marić
# [http://goo.gl/Guxky Coinductive pearl: Modular first-order logic completeness]. ~ J.C. Blanchette, A. Popescu y D. Traytel
# [http://goo.gl/HUOl8 A fully verified executable LTL model checker]. ~ J. Esparza et als.
# [http://goo.gl/RV54S ForMaRE - formal mathematical reasoning in economics]. ~ M. Kerber, C. Lange y C. Rowat.
# [http://goo.gl/Y7sIq AI over large formal knowledge bases: The first decade]. ~ J. Urban.
# [http://goo.gl/vvqNg Formalization of real analysis: A survey of proof assistants and libraries]. ~ S. Boldo, C. Lelay y G. Melquiond.
# [http://goo.gl/bEFYa Data refinement in Isabelle/HOL]. ~ F. Haftmann, A. Krauss, O. Kunčar y T. Nipkow
# [http://goo.gl/xTyvE Formalizing the confluence of orthogonal rewriting systems]. ~ A.C. Rocha y M. Ayala.
# [http://goo.gl/zCYHj Formalization of the complex number theory in HOL4]. ~ Z. Shi et als.
# [http://goo.gl/kM0dI Programming and reasonning with PowerLists in Coq]. ~ F. Loulergue y V. Niculescu
# [http://goo.gl/KkU6s A hierarchy of mathematical structures in ACL2]. ~ J. Heras, F.J. Martín y V. Pascual.
# [http://www.inf.kcl.ac.uk/staff/urbanc/Publications/tm.pdf Mechanising Turing Machines and Computability Theory in Isabelle/HOL] ~ J. Xu, X. Zhang y C. Urban

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<source lang="isabelle">
chapter {* Tema 7: Caso de estudio: Compilación de expresiones *}

theory T7_Caso_de_estudio_Compilacion_de_expresiones

imports Main
begin

declare [[names_short]]

text {*
El objetivo de este tema es contruir un compilador de expresiones
genéricas (construidas con variables, constantes y operaciones
binarias) a una máquina de pila y demostrar su corrección.
*}

section {* Las expresiones y el intérprete *}

text {*
Definición. Las expresiones son las constantes, las variables
(representadas por números naturales) y las aplicaciones de operadores
binarios a dos expresiones.
*}

type_synonym 'v binop = "'v ⇒ 'v ⇒ 'v"

datatype 'v expr =
Const 'v
| Var nat
| App "'v binop" "'v expr" "'v expr"

text {*
Definición. [Intérprete]
La función "valor" toma como argumentos una expresión y un entorno
(i.e. una aplicación de las variables en elementos del lenguaje) y
devuelve el valor de la expresión en el entorno.
*}

fun valor :: "'v expr ⇒ (nat ⇒ 'v) ⇒ 'v" where
"valor (Const b) ent = b"
| "valor (Var x) ent = ent x"
| "valor (App f e1 e2) ent = (f (valor e1 ent) (valor e2 ent))"

text {*
Ejemplo. A continuación mostramos algunos ejemplos de evaluación con
el intérprete.
*}

lemma
"valor (Const 3) id = 3 ∧
valor (Var 2) id = 2 ∧
valor (Var 2) (λx. x+1) = 3 ∧
valor (App (op +) (Const 3) (Var 2)) (λx. x+1) = 6 ∧
valor (App (op +) (Const 3) (Var 2)) (λx. x+4) = 9"
by simp

section {* La máquina de pila *}

text {*
Nota. La máquina de pila tiene tres clases de intrucciones:
· cargar en la pila una constante,
· cargar en la pila el contenido de una dirección y
· aplicar un operador binario a los dos elementos superiores de la pila.
*}

datatype 'v instr =
IConst 'v
| ILoad nat
| IApp "'v binop"

text {*
Definición. [Ejecución]
La ejecución de la máquina de pila se modeliza mediante la función
"ejec" que toma una lista de intrucciones, una memoria (representada
como una función de las direcciones a los valores, análogamente a los
entornos) y una pila (representada como una lista) y devuelve la pila
al final de la ejecución.
*}

fun ejec :: "'v instr list ⇒ (nat ⇒ 'v) ⇒ 'v list ⇒ 'v list" where
"ejec [] ent vs = vs"
| "ejec (i#is) ent vs =
(case i of
IConst v ⇒ ejec is ent (v#vs)
| ILoad x ⇒ ejec is ent ((ent x)#vs)
| IApp f ⇒ ejec is ent ((f (hd vs) (hd (tl vs)))#(tl(tl vs))))"

text {*
A continuación se muestran ejemplos de ejecución.
*}

lemma
"ejec [IConst 3] id [7] = [3,7] ∧
ejec [ILoad 2, IConst 3] id [7] = [3,2,7] ∧
ejec [ILoad 2, IConst 3] (λx. x+4) [7] = [3,6,7] ∧
ejec [ILoad 2, IConst 3, IApp (op +)] (λx. x+4) [7] = [9,7]"
by simp

section {* El compilador *}

text {*
Definición. El compilador "comp" traduce una expresión en una lista de
instrucciones.
*}

fun comp :: "'v expr ⇒ 'v instr list" where
"comp (Const v) = [IConst v]"
| "comp (Var x) = [ILoad x]"
| "comp (App f e1 e2) = (comp e2) @ (comp e1) @ [IApp f]"

text {*
A continuación se muestran ejemplos de compilación.
*}

lemma
"comp (Const 3) = [IConst 3] ∧
comp (Var 2) = [ILoad 2] ∧
comp (App (op +) (Const 3) (Var 2)) = [ILoad 2, IConst 3, IApp (op +)]"
by simp

section {* Corrección del compilador *}

text {*
Para demostrar que el compilador es correcto, probamos que el
resultado de compilar una expresión y a continuación ejecutarla es lo
mismo que interpretarla; es decir,
*}

theorem "ejec (comp e) ent [] = [valor e ent]"
apply (induct e)
apply auto
oops

text {*
El teorema anterior no puede demostrarse por inducción en e. Para
demostrarlo, lo generalizamos a
*}

theorem "∀vs. ejec (comp e) ent vs = (valor e ent)#vs"
oops

text {*
En la demostración del teorema anterior usaremos el siguiente lema.
*}

lemma ejec_append:
"∀ vs. ejec (xs@ys) ent vs = ejec ys ent (ejec xs ent vs)" (is "?P xs")
proof (induct xs)
show "?P []" by simp
next
fix a xs
assume "?P xs"
thus "?P (a#xs)" by (cases "a", auto)
qed

-- "La demostración detallada es"
lemma ejec_append_1:
"∀ vs. ejec (xs@ys) ent vs = ejec ys ent (ejec xs ent vs)" (is "?P xs")
proof (induct xs)
show "?P []" by simp
next
fix a xs
assume HI: "?P xs"
thus "?P (a#xs)"
proof (cases "a")
case IConst thus ?thesis using HI by simp
next
case ILoad thus ?thesis using HI by simp
next
case IApp thus ?thesis using HI by simp
qed
qed

text {*
Una demostración más detallada del lema es la siguiente:
*}

lemma ejec_append_2:
"∀vs. ejec (xs@ys) ent vs = ejec ys ent (ejec xs ent vs)" (is "?P xs")
proof (induct xs)
show "?P []" by simp
next
fix a xs
assume HI: "?P xs"
thus "?P (a#xs)"
proof (cases "a")
fix v assume C1: "a=IConst v"
show " ∀vs. ejec ((a#xs)@ys) ent vs = ejec ys ent (ejec (a#xs) ent vs)"
proof
fix vs
have "ejec ((a#xs)@ys) ent vs = ejec (((IConst v)#xs)@ys) ent vs"
using C1 by simp
also have "… = ejec (xs@ys) ent (v#vs)" by simp
also have "… = ejec ys ent (ejec xs ent (v#vs))" using HI by simp
also have "… = ejec ys ent (ejec ((IConst v)#xs) ent vs)" by simp
also have "… = ejec ys ent (ejec (a#xs) ent vs)" using C1 by simp
finally show "ejec ((a#xs)@ys) ent vs =
ejec ys ent (ejec (a#xs) ent vs)" .
qed
next
fix n assume C2: "a=ILoad n"
show " ∀vs. ejec ((a#xs)@ys) ent vs = ejec ys ent (ejec (a#xs) ent vs)"
proof
fix vs
have "ejec ((a#xs)@ys) ent vs = ejec (((ILoad n)#xs)@ys) ent vs"
using C2 by simp
also have "… = ejec (xs@ys) ent ((ent n)#vs)" by simp
also have "… = ejec ys ent (ejec xs ent ((ent n)#vs))" using HI by simp
also have "… = ejec ys ent (ejec ((ILoad n)#xs) ent vs)" by simp
also have "… = ejec ys ent (ejec (a#xs) ent vs)" using C2 by simp
finally show "ejec ((a#xs)@ys) ent vs =
ejec ys ent (ejec (a#xs) ent vs)" .
qed
next
fix f assume C3: "a=IApp f"
show "∀vs. ejec ((a#xs)@ys) ent vs = ejec ys ent (ejec (a#xs) ent vs)"
proof
fix vs
have "ejec ((a#xs)@ys) ent vs = ejec (((IApp f)#xs)@ys) ent vs"
using C3 by simp
also have "… = ejec (xs@ys) ent ((f (hd vs) (hd (tl vs)))#(tl(tl vs)))"
by simp
also have "… = ejec ys
ent
(ejec xs ent ((f (hd vs) (hd (tl vs)))#(tl(tl vs))))"
using HI by simp
also have "… = ejec ys ent (ejec ((IApp f)#xs) ent vs)" by simp
also have "… = ejec ys ent (ejec (a#xs) ent vs)" using C3 by simp
finally show "ejec ((a#xs)@ys) ent vs =
ejec ys ent (ejec (a#xs) ent vs)" .
qed
qed
qed

text {*
La demostración automática del teorema es
*}

theorem "∀vs. ejec (comp e) ent vs = (valor e ent)#vs"
by (induct e) (auto simp add: ejec_append)

text {*
La demostración estructurada del teorema es
*}

theorem "∀vs. ejec (comp e) ent vs = (valor e ent)#vs"
proof (induct e)
fix v
show "∀vs. ejec (comp (Const v)) ent vs = (valor (Const v) ent)#vs"
by simp
next
fix x
show "∀vs. ejec (comp (Var x)) ent vs = (valor (Var x) ent) # vs"
by simp
next
fix f e1 e2
assume HI1: "∀vs. ejec (comp e1) ent vs = (valor e1 ent) # vs"
and HI2: "∀vs. ejec (comp e2) ent vs = (valor e2 ent) # vs"
show "∀vs. ejec (comp (App f e1 e2)) ent vs =
(valor (App f e1 e2) ent) # vs"
proof
fix vs
have "ejec (comp (App f e1 e2)) ent vs
= ejec ((comp e2) @ (comp e1) @ [IApp f]) ent vs" by simp
also have "… = ejec ((comp e1) @ [IApp f]) ent (ejec (comp e2) ent vs)"
using ejec_append by blast
also have "… = ejec [IApp f]
ent
(ejec (comp e1) ent (ejec (comp e2) ent vs))"
using ejec_append by blast
also have "… = ejec [IApp f] ent (ejec (comp e1) ent ((valor e2 ent)#vs))"
using HI2 by simp
also have "… = ejec [IApp f] ent ((valor e1 ent)#((valor e2 ent)#vs))"
using HI1 by simp
also have "… = (f (valor e1 ent) (valor e2 ent))#vs" by simp
also have "… = (valor (App f e1 e2) ent) # vs" by simp
finally
show "ejec (comp (App f e1 e2)) ent vs = (valor (App f e1 e2) ent) # vs"
by blast
qed
qed

end
</source>

R1

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Relación 4

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<source lang="isabelle">
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 ivamenjim migtermor dancorgar wilmorort marpoldia1
ferrenseg paupeddeg pablucoto crigomgom anaprarod serrodcal juacabsou
rubgonmar josgarsan fraortmoy lucnovdos pabrodmac fracorjim1
marcarmor13 jeamacpov antsancab1 *)
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 pabrodmac marcarmor13 jeamacpov antsancab1*)
fun algunos :: "('a ⇒ bool) ⇒ 'a list ⇒ bool" where
"algunos p [] = False"
| "algunos p (x#xs) = (p x ∨ algunos p xs)"

(* fracorjim1. En esta versión el procesamiento se detiene al encontrar
una coincidencia *)
fun algunos2 :: "('a ⇒ bool) ⇒ 'a list ⇒ bool" where
"algunos2 p [] = False"
| "algunos2 p (x#xs) = (if p x then True else algunos2 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 pabrodmac marcarmor13 jeamacpov antsancab1 *)
lemma "todos (λx. P x ∧ Q x) xs = (todos P xs ∧ todos Q xs)"
by (induct xs) auto

(* fracorjim1 *)
lemma "todos (λx. P x ∧ Q x) xs = (todos P xs ∧ todos Q xs)"
apply (induct xs)
apply auto
done

text {*
---------------------------------------------------------------------
Ejercicio 3.2. Demostrar o refutar detalladamente
todos (λx. P x ∧ Q x) xs = (todos P xs ∧ todos Q xs)
---------------------------------------------------------------------
*}

(* danrodcha fracorjim1 *)
lemma "todos (λx. P x ∧ Q x) xs = (todos P xs ∧ todos Q xs)" (is "?R 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

(* Comentario: Uso de blast. *)

(* 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

(* Comentario: Uso de auto. *)

(* 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

(* Comentarios: Ruptura de la cadena de igualdades y uso de arith. *)

(* 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
oops

(* Comentario: Demostración incompleta. *)

(* 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
marcarmor13 jeamacpov *)
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

(* pabrodmac *)
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
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
finally show "todos (λx. P x ∧ Q x) (x#xs) =
(todos P (x#xs) ∧ todos Q (x#xs))" by simp
qed

(* antsancab1 *)
lemma "todos (λx. P x ∧ Q x) xs = (todos P xs ∧ todos Q xs)" (is "?R 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
finally show "todos (λx. P x ∧ Q x) (a # xs) = (todos P (a#xs) ∧ todos Q (a#xs))" by auto
qe

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 pabrodmac marcarmor13 jeamacpov antsancab1 *)
lemma "todos P (x @ y) = (todos P x ∧ todos P y)"
by (induct x) auto

(* anaprarod dancorgar fracorjim1 *)
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
rubgonmar *)
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 antsancab1 *)
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_append2:
"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 marcarmor13 jeamacpov *)
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

(* Comentario: Uso de auto. *)

(* danrodcha fracorjim1 *)
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

(* pabrodmac *)
lemma todos_append7:
"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

text {*
---------------------------------------------------------------------
Ejercicio 5.1. Demostrar o refutar automáticamente
todos P (rev xs) = todos P xs
---------------------------------------------------------------------
*}

(* migtermor ivamenjim marpoldia1 serrodcal anaprarod paupeddeg
dancorgar antsancab1 fracorjim1 *)
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
pabrodmac *)
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)

(* Comentario: Pasos dentro de by *)

(* juacabsou *)
lemma "todos P (rev xs) = todos P xs"
apply (induct xs, simp, simp add: todos_append, auto)
done

(* Comentario: Pasos dentro de apply *)

(* pablucoto marcarmor13 jeamacpov *)
lemma "todos P (rev xs) = todos P xs"
by (induct xs) (auto, simp_all add: todos_append)

(* Comentario: Uso de simp_all *)

lemma "todos P (rev xs) = todos P xs"
by (induct xs) (simp_all add: todos_append, auto)

(* Comentario: Uso de add en simp_all *)

text {*
---------------------------------------------------------------------
Ejercicio 5.2. Demostrar o refutar detalladamente
todos P (rev xs) = todos P xs
---------------------------------------------------------------------
*}

(* 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

(* Comentarios:
+ Ruptura de cadena de igualdades.
+ Uso de hechos auxiliares
+ Uso de arith
*)

(* marpoldia1 ferrenseg crigomgom serrodcal juacabsou rubgonmar
josgarsan pablucoto pabrodmac lucnovdos marcarmor13 jeamacpov *)
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 serrodcal *)
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

(* Comentario: Uso de HOL.conj_commute *)

(* anaprarod fracorjim1 *)
(* 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

(* antsancab1 *)
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)
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 rubgonmar anaprarod marpoldia1 fraortmoy josgarsan
danrodcha pablucoto lucnovdos marcarmor13 jeamacpov antsancab1 fracorjim1 pabrodmac *)
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 lucnovdos
marcarmor13 jeamacpov antsancab1 pabrodmac *)
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 serrodcal fracorjim1 *)
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
---------------------------------------------------------------------
*}

(* 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

(* Comentario: Se puede simplificar. *)

(* 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 by simp
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 josgarsan lucnovdos pabrodmac *)
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 marcarmor13 jeamacpov *)
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 serrodcal *)
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 fracorjim1 *)
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

(* antsancab1 *)
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 "... = (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 "... = 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 danrodcha pablucoto dancorgar josgarsan anaprarod juacabsou
lucnovdos marcarmor13 jeamacpov antsancab1 pabrodmac *)
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

(* anaprarod fracorjim1 *)
lemma "algunos P (xs @ ys) = (algunos P xs ∨ algunos P ys)"
apply (induct xs)
apply auto
done

text {*
---------------------------------------------------------------------
Ejercicio 8.2. Demostrar o refutar detalladamente
algunos P (xs @ ys) = (algunos P xs ∨ algunos P ys)
---------------------------------------------------------------------
*}

(* ivamenjim ferrenseg marpoldia1 wilmorort dancorgar josgarsan
juacabsou serrodcal lucnovdos rubgonmar *)
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 anaprarod marcarmor13 jeamacpov pabrodmac *)
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

(* antsancab1 *)
lemma algunos_append6: "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
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 dancorgar
anaprarod serrodcal antsancab1 fracorjim1 *)
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 josgarsan pabrodmac lucnovdos
marcarmor13 jeamacpov *)
lemma "algunos P (rev xs) = algunos P xs"
by (induct xs) (auto simp add: algunos_append)

(* wilmorort juacabsou *)
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
---------------------------------------------------------------------
*}

(* 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 by simp
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 serrodcal lucnovdos*)
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 anaprarod rubgonmar juacabsou marcarmor13 jeamacpov pabrodmac *)
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

(* antsancab1 *)
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 auto
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

(* CComentario: Ruptura de la cadena de igualdades. *)

(* 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 by simp
qed
qed

(* Comentario: Uso de auto *)

(* 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

(* Comentario: Uso de auto *)

(* wilmorort danrodcha anaprarod juacabsou rubgonmar*)
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 crigomgom serrodcal marcarmor13 jeamacpov pabrodmac *)
-- "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

(* anaprarod juacabsou rubgonmar *)
lemma "algunos (λx. P x ∨ Q x) xs = (algunos P xs ∨ algunos Q xs)"
(is "?P xs")
proof (induct xs)
show "?P []" by simp
next
fix x xs
assume HI: "?P xs"
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 arith
finally show "?P (x#xs)" by simp
qed

(* antsancab1 *)
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)"
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 "... = (algunos P (a # xs) ∨ algunos Q (a # xs))" by auto
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 josgarsan anaprarod juacabsou
lucnovdos serrodcal marcarmor13 jeamacpov antsancab1 pabrodmac *)
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

(* anaprarod *)
lemma "algunos P xs = (¬ todos (λx. (¬ P x)) xs)"
apply (induct xs)
apply auto
done

text {*
---------------------------------------------------------------------
Ejercicio 11.2. Demostrar o refutar datalladamente
algunos P xs = (¬ todos (λx. (¬ P x)) xs)
---------------------------------------------------------------------
*}

(* ivamenjim migtermor marpoldia1 crigomgom paupeddeg josgarsan
juacabsou serrodcal wilmorort *)
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 by simp
qed
qed

(* danrodcha pablucoto anaprarod rubgonmar marcarmor13 jeamacpov pabrodmac *)
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

(* lucnovdos*)
lemma "algunos P xs = (¬ todos (λx. (¬ P x)) xs)"
proof (induct xs)
show "algunos P [] = (¬ todos (λx. (¬ P x)) [])" by simp
next
fix n xs
assume HI: "algunos P xs = (¬ todos (λx. (¬ P x)) xs)"
have "algunos P (n#xs) = ((P n) ∨ (algunos P xs))" by simp
also have "… = ((P n) ∨ (¬ todos (λx. (¬ P x)) xs))" using HI by simp
also have "… = (¬((¬P n) ∧ todos (λx. (¬ P x)) xs))" by simp
also have "… = (¬ todos (λx. (¬ P x)) (n#xs))" by simp
finally show "algunos P (n#xs) = (¬ todos (λx. (¬ P x)) (n#xs))"
by simp
qed

(* antsancab1 *)
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
also have "... = (P a ∨ (¬ todos (λx. ¬ P x) (a#xs)))" by auto
finally show "algunos P (a # xs) = (¬ todos (λx. ¬ P x) (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 dancorgar pablucoto anaprarod
juacabsou lucnovdos marcarmor13 wilmorort jeamacpov fracorjim1 pabrodmac *)
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"

(* antsancab1 *)
fun estaEn1 :: "'a ⇒ 'a list ⇒ bool" where
"estaEn1 x [] = False"
| "estaEn1 x (a#xs) = (if x = a then True else estaEn1 x xs )" (* Se detiene al encontrar la primera coincidencia *)

value "estaEn1 (2::nat) [3,2,4] = True"
value "estaEn1 (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

(* Comentario: Ruptura de la cadena de igualdades. Se puede simplificar. *)

(* 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 by simp
qed
qed

(* Comentario: Uso de auto *)

(* crigomgom paupeddeg dancorgar juacabsou serrodcal lucnovdos wilmorort *)
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

(* Comentarios:
+ Uso de (op = x)
+ Uso de auto
*)

(* 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 marcarmor13 jeamacpov pabrodmac *)
-- "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

(* Comentario: Uso de auto *)

(* anaprarod *)
(* Automática*)
lemma "estaEn a xs = algunos (λx. a=x) xs" (is "?P xs")
by (induct xs) auto

(* Detallada (igual que las anteriores pero obviando el paso previo al
finally) *)
lemma "estaEn a xs = algunos (λx. a=x) xs" (is "?P xs")
proof (induct xs)
show "?P []" by simp
next
fix x xs
assume HI : "?P xs"
have "estaEn a (x#xs) = (x=a ∨ estaEn a xs)" by simp
also have "… = (x=a ∨ algunos (λx. a=x) xs)" using HI by simp
also have "… = (a=x ∨ algunos (λx. a=x) xs)" using HI by auto
finally show "?P (x#xs)" by simp
qed

(* Comentario: Uso de auto *)

(* antsancab1 *)
(* Demostrar que
algunos (λa. a=x) xs = estaEn x xs
es lo mismo que demostrar
lemma "estaEn a xs = algunos (λx. a=x) xs" *)

lemma "algunos (λa. a=x) xs = estaEn x xs"
by (induct xs) auto

lemma "algunos (λa. a=x) xs = estaEn x xs"
proof (induct xs)
show "algunos (λa. a = x) [] = estaEn x []" by simp
next
fix a xs
assume HI: "algunos (λa. a = x) xs = estaEn x xs"
have "algunos (λa. a = x) (a # xs) = ((λa. a = x) a ∨ algunos (λa. a = x) xs)" by simp
also have "... = ((λa. a = x) a ∨ estaEn x xs)" using HI by simp
also have "... = (estaEn x (a#xs))" by simp (* no tengo claro cómo simplifica el "(λa. a = x) a" *)
finally show "algunos (λa. a = x) (a # xs) = estaEn x (a # xs)" by simp
qed

end

(* fracorjim1 *)
lemma estaEnYAlgunos :
"estaEn a xs = algunos (λx. x = a) xs" (is "?P a xs")
proof (induct xs)
show "?P a []" by simp
next
fix a x xs
assume HI : "?P a xs"
have "estaEn a (x#xs) = ((a = x) ∨ estaEn a xs)"
by (simp only:estaEn.simps(2))
also have "… = (a = x ∨ algunos (λx. x = a) xs)" using HI by simp
also have "… = ((λx. x = a) x ∨ algunos (λx. x = a) xs)" by auto
also have "… = algunos (λx. x = a) (x # xs)" by simp
finally show "?P a (x#xs)" by simp
qed

</source>

Tema 9b: Deducción natural en lógica de primer orden con Isabelle/HOL

2018-07-16T11:05:14Z

WikiSysop: Texto reemplazado: «"isar"» por «"isabelle"»

<source lang="isabelle">
chapter {* Tema 9: Deducción natural en lógica de primer orden *}

theory T9b_Deduccion_natural_en_logica_de_primer_orden
imports Main
begin

text {*
El objetivo de este tema es presentar la deducción natural en
lógica de primer orden con Isabelle/HOL. La presentación se
basa en los ejemplos de tema 8 del curso LMF que se encuentra
en http://goo.gl/uJj8d (que a su vez se basa en el libro de
Huth y Ryan "Logic in Computer Science" http://goo.gl/qsVpY ).

La página al lado de cada ejemplo indica la página de las
transparencias de LMF donde se encuentra la demostración. *}

section {* Reglas del cuantificador universal *}

text {*
Las reglas del cuantificador universal son
· allE: ⟦∀x. P x; P a ⟹ R⟧ ⟹ R
· allI: (⋀x. P x) ⟹ ∀x. P x
*}

text {*
Ejemplo 1 (p. 10). Demostrar que
P(c), ∀x. (P(x) ⟶ ¬Q(x)) ⊢ ¬Q(c)
*}

-- "La demostración detallada es"
lemma ejemplo_1a:
assumes 1: "P(c)" and
2: "∀x. (P(x) ⟶ ¬Q(x))"
shows "¬Q(c)"
proof -
have 3: "P(c) ⟶ ¬Q(c)" using 2 by (rule allE)
show 4: "¬Q(c)" using 3 1 by (rule mp)
qed

-- "La demostración estructurada es"
lemma ejemplo_1b:
assumes "P(c)"
"∀x. (P(x) ⟶ ¬Q(x))"
shows "¬Q(c)"
proof -
have "P(c) ⟶ ¬Q(c)" using assms(2) ..
thus "¬Q(c)" using assms(1) ..
qed

-- "La demostración automática es"
lemma ejemplo_1c:
assumes "P(c)"
"∀x. (P(x) ⟶ ¬Q(x))"
shows "¬Q(c)"
using assms
by auto

text {*
Ejemplo 2 (p. 11). Demostrar que
∀x. (P x ⟶ ¬(Q x)), ∀x. P x ⊢ ∀x. ¬(Q x)
*}

-- "La demostración detallada es"
lemma ejemplo_2a:
assumes 1: "∀x. (P x ⟶ ¬(Q x))" and
2: "∀x. P x"
shows "∀x. ¬(Q x)"
proof -
{ fix a
have 3: "P a ⟶ ¬(Q a)" using 1 by (rule allE)
have 4: "P a" using 2 by (rule allE)
have 5: "¬(Q a)" using 3 4 by (rule mp) }
thus "∀x. ¬(Q x)" by (rule allI)
qed

-- "La demostración detallada hacia atrás es"
lemma ejemplo_2b:
assumes 1: "∀x. (P x ⟶ ¬(Q x))" and
2: "∀x. P x"
shows "∀x. ¬(Q x)"
proof (rule allI)
fix a
have 3: "P a ⟶ ¬(Q a)" using 1 by (rule allE)
have 4: "P a" using 2 by (rule allE)
show 5: "¬(Q a)" using 3 4 by (rule mp)
qed

-- "La demostración estructurada es"
lemma ejemplo_2c:
assumes "∀x. (P x ⟶ ¬(Q x))"
"∀x. P x"
shows "∀x. ¬(Q x)"
proof
fix a
have "P a" using assms(2) ..
have "P a ⟶ ¬(Q a)" using assms(1) ..
thus "¬(Q a)" using `P a` ..
qed

-- "La demostración automática es"
lemma ejemplo_2d:
assumes "∀x. (P x ⟶ ¬(Q x))"
"∀x. P x"
shows "∀x. ¬(Q x)"
using assms
by auto

section {* Reglas del cuantificador existencial *}

text {*
Las reglas del cuantificador existencial son
· exI: P a ⟹ ∃x. P x
· exE: ⟦∃x. P x; ⋀x. P x ⟹ Q⟧ ⟹ Q

En la regla exE la nueva variable se introduce mediante la declaración
"obtain ... where ... by (rule exE)"
*}

text {*
Ejemplo (p. 12). Demostrar que
∀x. P x ⊢ ∃x. P x
*}

-- "La demostración detallada es"
lemma ejemplo_3a:
assumes "∀x. P x"
shows "∃x. P x"
proof -
fix a
have "P a" using assms by (rule allE)
thus "∃x. P x" by (rule exI)
qed

-- "La demostración estructurada es"
lemma ejemplo_3b:
assumes "∀x. P x"
shows "∃x. P x"
proof -
fix a
have "P a" using assms ..
thus "∃x. P x" ..
qed

-- "La demostración estructurada se puede simplificar"
lemma ejemplo_3c:
assumes "∀x. P x"
shows "∃x. P x"
proof (rule exI)
fix a
show "P a" using assms ..
qed

-- "La demostración estructurada se puede simplificar aún más"
lemma ejemplo_3d:
assumes "∀x. P x"
shows "∃x. P x"
proof
fix a
show "P a" using assms ..
qed

-- "La demostración automática es"
lemma ejemplo_3e:
assumes "∀x. P x"
shows "∃x. P x"
using assms
by auto

text {*
Ejemplo 4 (p. 13). Demostrar
∀x. (P x ⟶ Q x), ∃x. P x ⊢ ∃x. Q x
*}

-- "La demostración detallada es"
lemma ejemplo_4a:
assumes 1: "∀x. (P x ⟶ Q x)" and
2: "∃x. P x"
shows "∃x. Q x"
proof -
obtain a where 3: "P a" using 2 by (rule exE)
have 4: "P a ⟶ Q a" using 1 by (rule allE)
have 5: "Q a" using 4 3 by (rule mp)
thus 6: "∃x. Q x" by (rule exI)
qed

-- "La demostración estructurada es"
lemma ejemplo_4b:
assumes "∀x. (P x ⟶ Q x)"
"∃x. P x"
shows "∃x. Q x"
proof -
obtain a where "P a" using assms(2) ..
have "P a ⟶ Q a" using assms(1) ..
hence "Q a" using `P a` ..
thus "∃x. Q x" ..
qed

-- "La demostración automática es"
lemma ejemplo_4c:
assumes "∀x. (P x ⟶ Q x)"
"∃x. P x"
shows "∃x. Q x"
using assms
by auto

section {* Demostración de equivalencias *}

text {*
Ejemplo 5.1 (p. 15). Demostrar
¬∀x. P x ⊢ ∃x. ¬(P x) *}

-- "La demostración detallada es"
lemma ejemplo_5_1a:
assumes "¬(∀x. P(x))"
shows "∃x. ¬P(x)"
proof (rule ccontr)
assume "¬(∃x. ¬P(x))"
have "∀x. P(x)"
proof (rule allI)
fix a
show "P(a)"
proof (rule ccontr)
assume "¬P(a)"
hence "∃x. ¬P(x)" by (rule exI)
with `¬(∃x. ¬P(x))` show False by (rule notE)
qed
qed
with assms show False by (rule notE)
qed

-- "La demostración estructurada es"
lemma ejemplo_5_1b:
assumes "¬(∀x. P(x))"
shows "∃x. ¬P(x)"
proof (rule ccontr)
assume "¬(∃x. ¬P(x))"
have "∀x. P(x)"
proof
fix a
show "P(a)"
proof (rule ccontr)
assume "¬P(a)"
hence "∃x. ¬P(x)" ..
with `¬(∃x. ¬P(x))` show False ..
qed
qed
with assms show False ..
qed

-- "La demostración automática es"
lemma ejemplo_5_1c:
assumes "¬(∀x. P(x))"
shows "∃x. ¬P(x)"
using assms
by auto

text {*
Ejemplo 5.2 (p. 16). Demostrar
∃x. ¬(P x) ⊢ ¬∀x. P x *}

-- "La demostración detallada es"
lemma ejemplo_5_2a:
assumes "∃x. ¬P(x)"
shows "¬(∀x. P(x))"
proof (rule notI)
assume "∀x. P(x)"
obtain a where "¬P(a)" using assms by (rule exE)
have "P(a)" using `∀x. P(x)` by (rule allE)
with `¬P(a)` show False by (rule notE)
qed

-- "La demostración estructurada es"
lemma ejemplo_5_2b:
assumes "∃x. ¬P(x)"
shows "¬(∀x. P(x))"
proof
assume "∀x. P(x)"
obtain a where "¬P(a)" using assms ..
have "P(a)" using `∀x. P(x)` ..
with `¬P(a)` show False ..
qed

-- "La demostración automática es"
lemma ejemplo_5_2c:
assumes "∃x. ¬P(x)"
shows "¬(∀x. P(x))"
using assms
by auto

text {*
Ejemplo 5.3 (p. 17). Demostrar
⊢ ¬∀x. P x ⟷ ∃x. ¬(P x) *}

-- "La demostración detallada es"
lemma ejemplo_5_3a:
"(¬(∀x. P(x))) ⟷ (∃x. ¬P(x))"
proof (rule iffI)
assume "¬(∀x. P(x))"
thus "∃x. ¬P(x)" by (rule ejemplo_5_1a)
next
assume "∃x. ¬P(x)"
thus "¬(∀x. P(x))" by (rule ejemplo_5_2a)
qed

-- "La demostración automática es"
lemma ejemplo_5_3b:
"(¬(∀x. P(x))) ⟷ (∃x. ¬P(x))"
by auto

text {*
Ejemplo 6.1 (p. 18). Demostrar
∀x. P(x) ∧ Q(x) ⊢ (∀x. P(x)) ∧ (∀x. Q(x)) *}

-- "La demostración detallada es"
lemma ejemplo_6_1a:
assumes "∀x. P(x) ∧ Q(x)"
shows "(∀x. P(x)) ∧ (∀x. Q(x))"
proof (rule conjI)
show "∀x. P(x)"
proof (rule allI)
fix a
have "P(a) ∧ Q(a)" using assms by (rule allE)
thus "P(a)" by (rule conjunct1)
qed
next
show "∀x. Q(x)"
proof (rule allI)
fix a
have "P(a) ∧ Q(a)" using assms by (rule allE)
thus "Q(a)" by (rule conjunct2)
qed
qed

-- "La demostración estructurada es"
lemma ejemplo_6_1b:
assumes "∀x. P(x) ∧ Q(x)"
shows "(∀x. P(x)) ∧ (∀x. Q(x))"
proof
show "∀x. P(x)"
proof
fix a
have "P(a) ∧ Q(a)" using assms ..
thus "P(a)" ..
qed
next
show "∀x. Q(x)"
proof
fix a
have "P(a) ∧ Q(a)" using assms ..
thus "Q(a)" ..
qed
qed

-- "La demostración automática es"
lemma ejemplo_6_1c:
assumes "∀x. P(x) ∧ Q(x)"
shows "(∀x. P(x)) ∧ (∀x. Q(x))"
using assms
by auto

text {*
Ejemplo 6.2 (p. 19). Demostrar
(∀x. P(x)) ∧ (∀x. Q(x)) ⊢ ∀x. P(x) ∧ Q(x) *}

-- "La demostración detallada es"
lemma ejemplo_6_2a:
assumes "(∀x. P(x)) ∧ (∀x. Q(x))"
shows "∀x. P(x) ∧ Q(x)"
proof (rule allI)
fix a
have "∀x. P(x)" using assms by (rule conjunct1)
hence "P(a)" by (rule allE)
have "∀x. Q(x)" using assms by (rule conjunct2)
hence "Q(a)" by (rule allE)
with `P(a)` show "P(a) ∧ Q(a)" by (rule conjI)
qed

-- "La demostración estructurada es"
lemma ejemplo_6_2b:
assumes "(∀x. P(x)) ∧ (∀x. Q(x))"
shows "∀x. P(x) ∧ Q(x)"
proof
fix a
have "∀x. P(x)" using assms ..
hence "P(a)" by (rule allE)
have "∀x. Q(x)" using assms ..
hence "Q(a)" ..
with `P(a)` show "P(a) ∧ Q(a)" ..
qed

-- "La demostración automática es"
lemma ejemplo_6_2c:
assumes "(∀x. P(x)) ∧ (∀x. Q(x))"
shows "∀x. P(x) ∧ Q(x)"
using assms
by auto

text {*
Ejemplo 6.3 (p. 20). Demostrar
⊢ ∀x. P(x) ∧ Q(x) ⟷ (∀x. P(x)) ∧ (∀x. Q(x)) *}

-- "La demostración detallada es"
lemma ejemplo_6_3a:
"(∀x. P(x) ∧ Q(x)) ⟷ ((∀x. P(x)) ∧ (∀x. Q(x)))"
proof (rule iffI)
assume "∀x. P(x) ∧ Q(x)"
thus "(∀x. P(x)) ∧ (∀x. Q(x))" by (rule ejemplo_6_1a)
next
assume "(∀x. P(x)) ∧ (∀x. Q(x))"
thus "∀x. P(x) ∧ Q(x)" by (rule ejemplo_6_2a)
qed

text {*
Ejemplo 7.1 (p. 21). Demostrar
(∃x. P(x)) ∨ (∃x. Q(x)) ⊢ ∃x. P(x) ∨ Q(x) *}

-- "La demostración detallada es"
lemma ejemplo_7_1a:
assumes "(∃x. P(x)) ∨ (∃x. Q(x))"
shows "∃x. P(x) ∨ Q(x)"
using assms
proof (rule disjE)
assume "∃x. P(x)"
then obtain a where "P(a)" by (rule exE)
hence "P(a) ∨ Q(a)" by (rule disjI1)
thus "∃x. P(x) ∨ Q(x)" by (rule exI)
next
assume "∃x. Q(x)"
then obtain a where "Q(a)" by (rule exE)
hence "P(a) ∨ Q(a)" by (rule disjI2)
thus "∃x. P(x) ∨ Q(x)" by (rule exI)
qed

-- "La demostración estructurada es"
lemma ejemplo_7_1b:
assumes "(∃x. P(x)) ∨ (∃x. Q(x))"
shows "∃x. P(x) ∨ Q(x)"
using assms
proof
assume "∃x. P(x)"
then obtain a where "P(a)" ..
hence "P(a) ∨ Q(a)" ..
thus "∃x. P(x) ∨ Q(x)" ..
next
assume "∃x. Q(x)"
then obtain a where "Q(a)" ..
hence "P(a) ∨ Q(a)" ..
thus "∃x. P(x) ∨ Q(x)" ..
qed

-- "La demostración automática es"
lemma ejemplo_7_1c:
assumes "(∃x. P(x)) ∨ (∃x. Q(x))"
shows "∃x. P(x) ∨ Q(x)"
using assms
by auto

text {*
Ejemplo 7.2 (p. 22). Demostrar
∃x. P(x) ∨ Q(x) ⊢ (∃x. P(x)) ∨ (∃x. Q(x)) *}

-- "La demostración detallada es"
lemma ejemplo_7_2a:
assumes "∃x. P(x) ∨ Q(x)"
shows "(∃x. P(x)) ∨ (∃x. Q(x))"
proof -
obtain a where "P(a) ∨ Q(a)" using assms by (rule exE)
thus "(∃x. P(x)) ∨ (∃x. Q(x))"
proof (rule disjE)
assume "P(a)"
hence "∃x. P(x)" by (rule exI)
thus "(∃x. P(x)) ∨ (∃x. Q(x))" by (rule disjI1)
next
assume "Q(a)"
hence "∃x. Q(x)" by (rule exI)
thus "(∃x. P(x)) ∨ (∃x. Q(x))" by (rule disjI2)
qed
qed

-- "La demostración estructurada es"
lemma ejercicio_7_2b:
assumes "∃x. P(x) ∨ Q(x)"
shows "(∃x. P(x)) ∨ (∃x. Q(x))"
proof -
obtain a where "P(a) ∨ Q(a)" using assms ..
thus "(∃x. P(x)) ∨ (∃x. Q(x))"
proof
assume "P(a)"
hence "∃x. P(x)" ..
thus "(∃x. P(x)) ∨ (∃x. Q(x))" ..
next
assume "Q(a)"
hence "∃x. Q(x)" ..
thus "(∃x. P(x)) ∨ (∃x. Q(x))" ..
qed
qed

-- "La demostración automática es"
lemma ejercicio_7_2c:
assumes "∃x. P(x) ∨ Q(x)"
shows "(∃x. P(x)) ∨ (∃x. Q(x))"
using assms
by auto

text {*
Ejemplo 7.3 (p. 23). Demostrar
⊢ ((∃x. P(x)) ∨ (∃x. Q(x))) ⟷ (∃x. P(x) ∨ Q(x)) *}

-- "La demostración detallada es"
lemma ejemplo_7_3a:
"((∃x. P(x)) ∨ (∃x. Q(x))) ⟷ (∃x. P(x) ∨ Q(x))"
proof (rule iffI)
assume "(∃x. P(x)) ∨ (∃x. Q(x))"
thus "∃x. P(x) ∨ Q(x)" by (rule ejemplo_7_1a)
next
assume "∃x. P(x) ∨ Q(x)"
thus "(∃x. P(x)) ∨ (∃x. Q(x))" by (rule ejemplo_7_2a)
qed

-- "La demostración automática es"
lemma ejemplo_7_3b:
"((∃x. P(x)) ∨ (∃x. Q(x))) ⟷ (∃x. P(x) ∨ Q(x))"
using assms
by auto

text {*
Ejemplo 8.1 (p. 24). Demostrar
∃x y. P(x,y) ⊢ ∃y x. P(x,y) *}

-- "La demostración detallada es"
lemma ejemplo_8_1a:
assumes "∃x y. P(x,y)"
shows "∃y x. P(x,y)"
proof -
obtain a where "∃y. P(a,y)" using assms by (rule exE)
then obtain b where "P(a,b)" by (rule exE)
hence "∃x. P(x,b)" by (rule exI)
thus "∃y x. P(x,y)" by (rule exI)
qed

-- "La demostración estructurada es"
lemma ejemplo_8_1b:
assumes "∃x y. P(x,y)"
shows "∃y x. P(x,y)"
proof -
obtain a where "∃y. P(a,y)" using assms ..
then obtain b where "P(a,b)" ..
hence "∃x. P(x,b)" ..
thus "∃y x. P(x,y)" ..
qed

-- "La demostración automática es"
lemma ejemplo_8_1c:
assumes "∃x y. P(x,y)"
shows "∃y x. P(x,y)"
using assms
by auto

text {*
Ejemplo 8.2. Demostrar
∃y x. P(x,y) ⊢ ∃x y. P(x,y) *}

-- "La demostración detallada es"
lemma ejemplo_8_2a:
assumes "∃y x. P(x,y)"
shows "∃x y. P(x,y)"
proof -
obtain b where "∃x. P(x,b)" using assms by (rule exE)
then obtain a where "P(a,b)" by (rule exE)
hence "∃y. P(a,y)" by (rule exI)
thus "∃x y. P(x,y)" by (rule exI)
qed

-- "La demostración estructurada es"
lemma ejemplo_8_2b:
assumes "∃y x. P(x,y)"
shows "∃x y. P(x,y)"
proof -
obtain b where "∃x. P(x,b)" using assms ..
then obtain a where "P(a,b)" ..
hence "∃y. P(a,y)" ..
thus "∃x y. P(x,y)" ..
qed

-- "La demostración estructurada es"
lemma ejemplo_8_2c:
assumes "∃y x. P(x,y)"
shows "∃x y. P(x,y)"
using assms
by auto

text {*
Ejemplo 8.3 (p. 25). Demostrar
⊢ (∃x y. P(x,y)) ⟷ (∃y x. P(x,y)) *}

-- "La demostración detallada es"
lemma ejemplo_8_3a:
"(∃x y. P(x,y)) ⟷ (∃y x. P(x,y))"
proof (rule iffI)
assume "∃x y. P(x,y)"
thus "∃y x. P(x,y)" by (rule ejemplo_8_1a)
next
assume "∃y x. P(x,y)"
thus "∃x y. P(x,y)" by (rule ejemplo_8_2a)
qed

-- "La demostración automática es"
lemma ejemplo_8_3b:
"(∃x y. P(x,y)) ⟷ (∃y x. P(x,y))"
by auto

section {* Reglas de la igualdad *}

text {*
Las reglas básicas de la igualdad son:
· refl: t = t
· subst: ⟦s = t; P s⟧ ⟹ P t
*}

text {*
Ejemplo 9 (p. 27). Demostrar
x+1 = 1+x, x+1 > 1 ⟶ x+1 > 0 ⊢ 1+x > 1 ⟶ 1+x > 0
*}

-- "La demostración detallada es"
lemma ejemplo_9a:
assumes "x+1 = 1+x"
"x+1 > 1 ⟶ x+1 > 0"
shows "1+x > 1 ⟶ 1+x > 0"
proof -
show "1+x > 1 ⟶ 1+x > 0" using assms by (rule subst)
qed

-- "La demostración estructurada es"
lemma ejemplo_9b:
assumes "x+1 = 1+x"
"x+1 > 1 ⟶ x+1 > 0"
shows "1+x > 1 ⟶ 1+x > 0"
using assms
by (rule subst)

-- "La demostración automática es"
lemma ejemplo_9c:
assumes "x+1 = 1+x"
"x+1 > 1 ⟶ x+1 > 0"
shows "1+x > 1 ⟶ 1+x > 0"
using assms
by auto

text {*
Ejemplo 10 (p. 27). Demostrar
x = y, y = z ⊢ x = z
*}

-- "La demostración detallada es"
lemma ejemplo_10a:
assumes "x = y"
"y = z"
shows "x = z"
proof -
show "x = z" using assms(2,1) by (rule subst)
qed

-- "La demostración estructurada es"
lemma ejemplo_10b:
assumes "x = y"
"y = z"
shows "x = z"
using assms(2,1)
by (rule subst)

-- "La demostración automática es"
lemma ejemplo_10c:
assumes "x = y"
"y = z"
shows "x = z"
using assms
by auto

text {*
Ejemplo 11 (p. 28). Demostrar
s = t ⊢ t = s
*}

-- "La demostración detallada es"
lemma ejemplo_11a:
assumes "s = t"
shows "t = s"
proof -
have "s = s" by (rule refl)
with assms show "t = s" by (rule subst)
qed

-- "La demostración automática es"
lemma ejemplo_11b:
assumes "s = t"
shows "t = s"
using assms
by auto

end
</source>

Razonamiento automático (2016-17)

2018-07-16T11:03:44Z

WikiSysop: /* Razonamiento automático (2016-17) */

Este sitio contiene materiales del curso ''Razonamiento automático'' del [http://master.cs.us.es/Máster_Universitario_en_Lógica,_Computación_e_Inteligencia_Artificial Máster Universitario en Lógica, Computación e Inteligencia Artificial] de la [http://www.us.es Universidad de Sevilla].

== Material para el curso ==
* [[Temas]]: Teorías de los temas.
* [[Ejercicios]]: Relaciones de ejercicios.
* [[Documentación]]: Lecturas recomendadas.
* [[Sistemas]]: Sistemas utilizados.
* [http://www.glc.us.es/~jalonso/vestigium/tag/ra2016 Diario]: Descripción diaria de las clases.
* [https://github.com/jaalonso/RA20116 GitHub]: Curso en GitHub.

MediaWiki:Common.css

2018-07-16T11:03:27Z

WikiSysop: Página creada con «/* Los estilos CSS colocados aquí se aplicarán a todas las apariencias */ @import url("/~jalonso/font-awesome-4.7.0/css/font-awesome.min.css");»

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MediaWiki:Sidebar

2016-10-15T13:48:52Z

WikiSysop:

Documentación

2016-10-15T13:48:00Z

WikiSysop: Página creada con 'En esta página se recogen en enlaces que sirven de documentación al curso de demostración asistida por ordenador (DAO). == Visiones generales de la DAO == # J.A. Alonso. [h...'

En esta página se recogen en enlaces que sirven de documentación al curso de demostración asistida por ordenador (DAO).

== Visiones generales de la DAO ==

# J.A. Alonso. [http://goo.gl/NWk7b Razonamiento formalizado: Del sueño a la realidad de las pruebas]. ''Vestigium'', 26 de diciembre de 2012.
# J. Avigad. [http://www.andrew.cmu.edu/user/avigad/Talks/icerm.pdf Interactive theorem proving, automated reasoning, and mathematical computation]. ICERM, 14 de diciembre de 2012.
# M. Davis. [http://www.cs.nyu.edu/cs/faculty/davism/early.ps The early history of automated deduction].
# J.P. Delahaye [http://interstices.info/jcms/int_63417/du-reve-a-la-realite-des-preuves Du rêve à la réalité des preuves]. ''Interstices'', 8 de julio de 2012.
# J. Germoni [http://images.math.cnrs.fr/Coq-et-caracteres.html Coq et caractères: Preuve formelle du théorème de Feit et Thompson]. ''Images des Mathématiques'', CNRS, 23 de noviembre de 2012.
# H. Geuvers [http://www.ias.ac.in/sadhana/Pdf2009Feb/3.pdf Proof assistants: History, ideas and future]. ''Sadhana'', Vol. 34-1, pp. 3-25, février 2009.
# G. Gonthier [http://www.ams.org/notices/200811/tx081101382p.pdf The four-color theorem]. ''Notices of the AMS'', Vol. 55, n° 11, pp. 1382-1393, 2008.
# T. Hales. [http://www.ams.org/notices/200811/tx081101370p.pdf Formal proof]. ''Notices of AMS'', Vol. 55, N. 11 (2008) pp. 1370-1380.
# J. Harrison. [http://www.cl.cam.ac.uk/~jrh13/papers/ab.html A short survey of automated reasoning]. ''Lecture Notes in Computer Science'', Vol. 4545, pp. 334-349, 2007.
# J. Harrison. [http://www.ams.org/notices/200811/tx081101395p.pdf Formal proof: Theory and practice]. ''Notices of the AMS'', Vol. 55, N. 11 (2008) p.1395-1406.
# G. Kolata. [http://www.nytimes.com/library/cyber/week/1210math.html Computer math proof shows reasoning power]. ''The New York Times'', 10 de diciembre de 1996.
# D. MacKenzie [http://www.bcs.org/server.php?show=ConWebDoc.4364 Computers and the sociology of mathematical proof].
# G. Sutcliffe. [http://www.cs.miami.edu/~tptp/OverviewOfATP.html What is automated theorem proving?].
# F. Wiedijk [http://www.cs.ru.nl/~freek/100/ Formalizing the «top 100» of mathematical theorems].
# F. Wiedijk [http://www.ams.org/notices/200811/tx081101408p.pdf Formal proof - Getting started]. ''Notices of the AMS'', Vol. 55, n° 11, pp. 1408-1414, 2008.
# F. Wiedijk, [http://www.cs.ru.nl/~freek/pubs/qed2.ps.gz The QED manifesto revisited]. ''Studies in Logic, Grammar and Rhetoric'', Vol. 10(23), pp. 121-133, 2007.

== Referencias sobre Isabelle/HOL ==
# B. Grechuk [http://dream.inf.ed.ac.uk/projects/isabelle/Isabelle_Primer.pdf Isabelle primer for mathematicians].
# T. Nipkow [http://www.cl.cam.ac.uk/research/hvg/Isabelle/dist/Isabelle2013-2/doc/prog-prove.pdf Programming and proving in Isabelle/HOL]. 5 de diciembre de 2013.
# T. Nipkow, M. Wenzel y L.C. Paulson [http://www.cl.cam.ac.uk/research/hvg/Isabelle/dist/Isabelle2013-2/doc/tutorial.pdf A proof assistant for higher-order logic]. Springer-Verlag. 5 de diciembre de 2013.
# [http://www.cl.cam.ac.uk/research/hvg/Isabelle/dist/library/HOL/HOL/document.pdf Isabelle/HOL — Higher-Order Logic]. 5 de diciembre de 2013.
# M. Wenzel [http://www.cl.cam.ac.uk/research/hvg/Isabelle/dist/Isabelle2013-2/doc/isar-ref.pdf The Isabelle/Isar Reference Manual]. 5 de diciembre de 2013.
# M. Wenzel [https://www.lri.fr/~wenzel/Isabelle2011-Paris/quickref.pdf The Isabelle/Isar quick reference].
# J. Siek [http://ecee.colorado.edu/~siek/ecen3703/spring10/quick-ref.pdf Quick Reference for Isabelle/Isar Propositional Logic].
# J. Siek [http://ecee.colorado.edu/~siek/ecen3703/spring10/quick-ref2.pdf Quick Reference for Isabelle/Isar More Proof Techniques].
# J. Siek [http://ecee.colorado.edu/~siek/ecen3703/spring10/quick-ref3.pdf Quick Reference for Isabelle/Isar First-Order Logic].
# [http://www.cl.cam.ac.uk/research/hvg/Isabelle/documentation.html Tutorials and manuals for Isabelle2013].

== Lecturas complementarias ==
=== Programación funcional ===
# J.A. Alonso [http://www.cs.us.es/~jalonso/cursos/i1m/temas/2012-13-IM-temas-PF.pdf Temas de "Programación funcional"]. Publicaciones del Grupo de Lógica Computacional. Universidad de Sevilla, 2012.
# J.A. Alonso y M.J. Hidalgo [http://www.cs.us.es/~jalonso/publicaciones/Piensa_en_Haskell.pdf Piensa en Haskell (Ejercicios de programación funcional con Haskell)]. Publicaciones del Grupo de Lógica Computacional. Universidad de Sevilla, 2012.
# G. Hutton [http://goo.gl/pKqG Programming in Haskell]. Cambridge University Press, 2007.
# M. Lipovača [http://aprendehaskell.es ¡Aprende Haskell por el bien de todos!].

=== Lógica computacional ===
# J.A. Alonso [http://www.cs.us.es/~jalonso/cursos/li/temas/temas-LI-2012-13.pdf Temas de "Lógica informática" (2012-13)]. Publicaciones del Grupo de Lógica Computacional. Universidad de Sevilla, 2012.
# R. Bornat [http://bit.ly/oithic Proof and disproof in formal logic: an introduction for programmers]. Oxford University Press, 2005.
# K. Broda, S. Eisenbach, H. Khoshnevisan y S. Vickers [http://pubs.doc.ic.ac.uk/reasoned-programming/reasoned-programming.pdf Reasoned programming]. Imperial College, 1994.
# K. Doets y J. van Eijck [http://www.ldc.usb.ve/~astorga/Haskell.Road.pdf The Haskell Road to Logic, Maths and Programming].
# M. Huth y M. Ryan [http://goo.gl/TMqOo Logic in computer science: Modelling and reasoning about systems]. Cambridge University Press, 2004. (Incluye el [http://www.cs.bham.ac.uk/research/lics/tutor/index.html tutor en la Red]).

== Cursos relacionados ==
=== Cursos con Isabelle/HOL ===
# Jeremy Avigad. [http://www.phil.cmu.edu/~avigad/formal/ Logic and Formal Verification]. (Carnegie Mellon, 2009).
# Clemens Ballarin. [http://cl-informatik.uibk.ac.at/teaching/ss08/atp/introduction.php Automatic Deduction]. (Univ de Innsbruck, 2008).
# Clemens Ballarin. [http://www4.in.tum.de/~ballarin/belgrade08-tut/ Introduction to the Isabelle Proof Assistant]. (Belgrado, 2008).
# Clemens Ballarin y Gerwin Klein [http://isabelle.in.tum.de/coursematerial/IJCAR04 Introduction to the Isabelle Proof Assistant]. (en el IJCAR-2004).
# Clemens Ballarin y Tjark Weber. [http://cl-informatik.uibk.ac.at/teaching/ws06/atp/introduction.php Automated Theorem Proving in Isabelle/HOL]. (Univ. de Innsbruck, 2006-07).
# A.D. Brucker, D. Basin, J.G. Smaus y B. Wolff. [http://archiv.infsec.ethz.ch/education/permanent/csmr.html Computer-supported Modeling and Reasoning]. (ETH Zurich, 2011).
# Mads Dam. [http://www.csc.kth.se/utbildning/kth/kurser/DD2453/aform07/ Advanced formal methods]. (KTH Royal Institute of Technology, 2007).
# Jacques Fleuriot y Paul Jackson. [http://www.inf.ed.ac.uk/teaching/courses/ar/slides/ Automated reasoning]. (Univ. de Edimburgo, 2012-13).
# Thomas Genet [http://www.irisa.fr/celtique/genet/ACF Software formal analysis and design]. (Univ. de Rennes)
# Gerwin Klein. [http://www.cse.unsw.edu.au/~kleing/teaching/thprv-04 Theorem Proving - Principles, Techniques, Applications]. (NICTA, 2004).
# Gerwin Klein. [http://www.cse.unsw.edu.au/~cs4161/index.html Advanced Topics in Software Verification]. (NICTA, 2012).
# Joao Marcos. [http://www.dimap.ufrn.br/~jmarcos/courses/LC/Ementa.htm Lógica computacional: Demonstração assistida e semi-automática de teoremas].(UFRN, 2000).
# Tobias Nipkow. [http://www4.informatik.tu-muenchen.de/~nipkow/semantics/ Semantics of programming languages]. (Univ. de Munich, 2012-13).
# Tobias Nipkow [http://isabelle.in.tum.de/coursematerial/PSV2009-1 Theorem Proving with Isabelle/HOL An Intensive Course].
# Larry Paulson. [http://www.cl.cam.ac.uk/teaching/0910/L21/ Interactive Formal Verification]. (Univ. de Cambridge, 2009-10).
# Arnd Poetzsch-Heffter. [https://softech.informatik.uni-kl.de/Homepage/SVHOL10 Specification and Verification with Higher-Order Logic].
# Viorel Preoteasa, Ralph-Johan Back y Charmi Panchal. [http://users.abo.fi/vpreotea/isabelle-2012 Introduction to mechanized reasoning with Isabelle/HOL]. (Åbo Akademi University, 2012).
# Jeremy G. Siek. [http://www.cs.colorado.edu/~siek/7000/spring07/ Practical Theorem Proving with Isabelle/Isar]. (Univ. de Colorado, 2007).
# Jeremy G. Siek. [http://ecee.colorado.edu/~siek/ecen5013/spring11/ Theorem proving in Isabelle]. (Univ. de Colorado, 2011).
# Jan-Georg Smaus. [http://www.informatik.uni-freiburg.de/~ki/teaching/ws0910/csmr/lecture.html Computer-supported modeling and reasoning]. (Univ. de Feiburgo, 2009).
# Christian Sternagel [http://cl-informatik.uibk.ac.at/teaching/ss11/eve/content.php Experiments in Verification – Introduction to Isabelle/HOL]. (Univ. de Innsbruck, 2011-12).
# Tjark Weber. [http://www.cl.cam.ac.uk/teaching/1011/L21/ Interactive Formal Verification]. (Univ. de Cambridge, 2010-11).

=== Otros cursos ===
# José A. Alonso [http://www.cs.us.es/~jalonso/cursos/li/ Lógica informática] (Univ. de Sevilla, 2012-13).
# Thorsten Altenkirch y Peter Morris [http://www.cs.nott.ac.uk/~txa/g52ifr Introduction to formal reasoning] (Univ. de Nottingham, 2010-11).
# Yves Bertot, Pierre Casteran, Benjamin Gregoire, Pierre Letouzey y Assia Mahboubi [http://www.di.ens.fr/~zappa/teaching/coq/ecole11 Modelling and verifying algorithms in Coq: an introduction]. (INRIA Paris-Rocquencourt, 14-18 noviembre 2011).
# Pierre Castéran [http://www.labri.fr/perso/casteran/FM/Logique/index.html Logic (Master In Verification)] (Univ. de Burdeos, 2011-12).
# Adam Chlipala [http://stellar.mit.edu/S/course/6/fa11/6.892/ Interactive computer theorem proving]. (MIT, 2012-13).
# Adam Chlipala y Armando Solar Lezama [https://stellar.mit.edu/S/course/6/fa13/6.820/index.html Foundations of program analysis]. (MIT, 2013-14).
# Robby Findler [http://www.eecs.northwestern.edu/~robby/courses/395-495-2013-fall Certified programming with dependent types]. (Northwestern, 2013-14).
# Nuno Gaspar [http://www-sop.inria.fr/members/Nuno.Gaspar/teaching/coq2012.php Verification with the Coq Proof Assistant] (INRIA Sophia Antipolis, 2012-13).
# Carlos Luna y Gustavo Betarte. [https://eva.fing.edu.uy/course/view.php?id=363 Construcción formal de programas en teoría de tipos]. (Univ. de la República, Uruguay, 2013-14).
# Michael Genesereth [http://logic.stanford.edu/classes/cs157/2011/cs157.html Computational Logic] (Univ. de Stanford, 2011-12).
# Ian Hodkinson [http://www.doc.ic.ac.uk/~imh/teaching/140_logic/logic.html Logic] (Imperial College, Londres, 2010-11).
# Peter Lucas [http://www.cs.ru.nl/~peterl/teaching/KeR/ Knowledge Representation and Reasoning] (Radboud University # egen, 2011-12).
# Larry Paulson [http://www.cl.cam.ac.uk/Teaching/current/LogicProof/ Logic and Proof] (Univ. de Cambridge, 2011-12).
# David Pichardie [http://www.irisa.fr/celtique/pichardie/teaching/M2/MDV/ Méthode de vérification] (Universidad de Rennes, 2006-07).
# Michael Winter [Logic in Computer Science] (Brock University, Ontario, Canada, 2010-11).

== Bibliotecas de ejemplos de verificación ==
# [http://afp.sourceforge.net Archive of Formal Proofs].
# [http://www.cs.ru.nl/~freek/100 Formalizing 100 Theorems].
# [http://toccata.lri.fr/gallery Gallery of verified programs].
# [http://www.cs.nott.ac.uk/~lad/research/challenges/ Induction Challenge Problems].
# [http://automatedreasoning.net/ Larry Wos' Notebooks].
# [http://www.cs.miami.edu/~tptp/ The TPTP Problem Library for Automated Theorem Proving].
# [http://www.macs.hw.ac.uk/vstte10/Competition.html The 1st Verified Software Competition].
# [https://sites.google.com/site/vstte2012/compet The 2nd Verified Software Competition].
# [http://verifythis.cost-ic0701.org VerifyThis (A collection of verification benchmarks].

== Artículos recientes ==
Están en orden cronológico inverso a la fecha de su reseña en [http://www.glc.us.es/~jalonso/vestigium/tag/resena Vestigium]:
# [http://bit.ly/1iZjgqN Proof Pearl: A probabilistic proof for the Girth-Chromatic number theorem]. L. Noschinski
# [http://bit.ly/1iJ8uVz A graph library for Isabelle]. ~ L. Noschinski
# [http://bit.ly/I0CU80 Gödel’s incompleteness theorems]. ~ L.C. Paulson
# [http://bit.ly/I0CPRN The hereditarily finite sets]. ~ L.C. Paulson
# [http://bit.ly/HBiIJI Applications of real number theorem proving in PVS]. ~ H. Gottliebsen, R. Hardy, O. Lightfoot y U. Martin
# [http://bit.ly/1awnMLB A machine-assisted proof of Gödel’s incompleteness theorems for the theory of hereditarily finite sets]. ~ L.C. Paulson
# [http://bit.ly/19lWeYy Verified AIG algorithms in ACL2]. ~ J. Davis y S. Swords
# [http://bit.ly/GAhC00 A formal model and correctness proof for an access control policy framework]. ~ C. Wu, X. Zhang y C. Urban
# [http://bit.ly/16SMvSS The ontological argument in PVS]. ~ J. Rushby
# [http://bit.ly/1dRt9n0 Formalizing Moessner’s theorem and generalizations in Nuprl]. ~ M. Bickford, D. Kozen y A. Silva
# [http://bit.ly/1bSeDNB Formalization in PVS of balancing properties necessary for the security of the Dolev-Yao cascade protocol model]. ~ M. Ayala y Y. Santos
# [http://bit.ly/1feFqWE Proof assistant based on didactic considerations]. ~ J. Pais y A Tasistro
# [http://bit.ly/18tHNBi Theory exploration for interactive theorem proving]. ~ M. Johansson
# [http://bit.ly/1b0242s From Tarski to Hilbert]. ~ G. Braun y J. Narboux
# [http://bit.ly/18HaXaR Formal verification of language-based concurrent noninterference]. ~ A. Popescu, J. Hölzl y T. Nipkow
# [http://bit.ly/1aRTQsU A Traffic Alert and Collision Avoidance System(TCAS-II) Resolution Advisory Algorithm]. ~ C. Muñoz, A. Narkawicz y J. Chamberlain
# [http://bit.ly/1dNwhDI Formal verification of cryptographic security proofs]. ~ M. Berg
# [http://bit.ly/17muAUv Polygonal numbers in Mizar]. ~ A. Grabowski
# [http://bit.ly/1hk5z6L A mechanised proof of Gödel’s incompleteness theorems using Nominal Isabelle]. ~ L.C. Paulson
# [http://bit.ly/1cSL0wE Steps towards verified implementations of HOL Light]. ~ M.O. Myreen, S. Owens y R. Kumar
# [http://bit.ly/16Kbgm0 Generic datatypes à la carte]. ~ S. Keuchel y T. Schrijvers
# [http://bit.ly/1bqJGx4 Proof pearl: A verified bignum implementation in x86-64 machine code]. ~ M.O. Myreen y G. Curello
# [http://bit.ly/142ow2Q Mechanized metatheory for a λ λ-calculus with trust types]. ~ R. Ribeiro, C. Camarão y L. Figueiredo
# [http://bit.ly/15WZBDy Proving soundness of combinatorial Vickrey auctions and generating verified executable code]. ~ M.B. Caminati, M. Kerber, C. Lange y C. Rowat
# [http://bit.ly/198g4n9 A computer-assisted proof of correctness of a marching cubes algorithm]. ~ A.N. Chernikov y J. Xu
# [http://bit.ly/11QA5g7 Verifying the bridge between simplicial topology and algebra: the Eilenberg-Zilber algorithm]. ~ L. Lambán, J. Rubio, F.J. Martín y J.L. Ruiz
# [http://bit.ly/1cJAXYk The Königsberg bridge problem and the friendship theorem]. ~ W. Li
# [http://bit.ly/13DBK9R Formal verification of a proof procedure for the description logic ALC]. ~ M. Chaabani, M. Mezghiche y M. Strecker
# [http://bit.ly/1ep2ex9 Pratt’s primality certificates]. ~ S. Wimmer y L. Noschinski
# [http://bit.ly/13C95Ci Reasoning about higher-order relational specifications]. ~ Y. Wang, K. Chaudhuri, A. Gacek y G. Nadathur
# [http://bit.ly/18QQLcL Proofs you can believe in – Proving equivalences between Prolog semantics in Coq]. ~ J. Kriener, A. King y S. Blazy
# [http://bit.ly/19uc82J Certified, efficient and sharp univariate Taylor models in Coq]. ~ E. Martin-Dorel, L. Rideau, L. Théry, M. Mayero y I. Paşca
# [http://bit.ly/1c4Rzel Ordinals in HOL: Transfinite arithmetic up to (and beyond) ω₁]. ~ M. Norrish y B. Huffman
# [http://bit.ly/14b4Akz Program verification based on Kleene algebra in Isabelle/HOL] ~ A. Armstrong, G. Struth y T. Weber
# [http://bit.ly/1aOgRKx Reading an algebra textbook (by translating it to a formal document in the Isabelle/Isar language)]. ~ C. Ballarin
# [http://bit.ly/11HKixj Computational verification of network programs in Coq]. ~ G. Stewart
# [http://www.glc.us.es/~jalonso/vestigium/resena-certifying-homological-algorithms-to-study-biomedical-images Certifying homological algorithms to study biomedical images]. ~ M. Poza
# [http://bit.ly/16Nks9m Formalizing cut elimination of coalgebraic logics in Coq]. ~ H. Tews
# [http://www.glc.us.es/~jalonso/vestigium/resena-the-formalization-of-syntax-based-mathematical-algorithms-using-quotation-and-evaluation/ The formalization of syntax-based mathematical algorithms using quotation and evaluation]. ~ W.M. Farmer
# [http://www.glc.us.es/~jalonso/vestigium/resena-certified-symbolic-manipulation-bivariate-simplicial-polynomials/ Certified symbolic manipulation: Bivariate simplicial polynomials]. ~ L. Lambán, F.J. Martín, J. Rubio y J.L. Ruiz
# [http://www.glc.us.es/~jalonso/vestigium/resena-solveurs-cpfd-verifies-formellement/ Solveurs CP(FD) vérifiés formellement]. ~ C Dubois y A. Gotlieb
# [http://www.glc.us.es/~jalonso/vestigium/resena-formalizing-bounded-increase/ Formalizing bounded increase]. ~ R. Thiemann
# [http://www.glc.us.es/~jalonso/vestigium/resena-formal-mathematics-on-display-a-wiki-for-flyspeck/ Formal mathematics on display: A wiki for Flyspeck]. ~ C. Tankink, C. Kaliszyk, J. Urban y H. Geuvers
# [http://www.glc.us.es/~jalonso/vestigium/resena-theorem-of-three-circles-in-coq Theorem of three circles in Coq]. ~ J. Zsidó
# [http://bit.ly/19fiWAY Certified HLints with Isabelle/HOLCF-Prelude]. ~ J. Breitner, B. Huffman, N. Mitchell y C. Sternagel
# [http://bit.ly/ZfSQrQ Automatic data refinement]. ~ P. Lammich
# [http://bit.ly/18vyjm7 The rooster and the butterflies (a machine-checked proof of the Jordan-Hölder theorem for finite groups)]. ~ A. Mahboubi
# [http://bit.ly/114oyZV Mechanical verification of SAT refutations with extended resolution]. ~ N. Wetzler, M.J.H. Heule y W.A. Hunt Jr.
# [http://bit.ly/13H0REu Type classes and filters for mathematical analysis in Isabelle/HOL] ~ J. Hölzl, F. Immler y B. Huffman
# [http://bit.ly/10fv8wO Verifying refutations with extended resolution]. ~ M. J. H. Heule, W. A. Hunt, Jr. y N. Wetzler
# [http://bit.ly/10EcFWj A Web interface for Isabelle: The next generation]. ~ C. Lüth y M. Ring
# [http://bit.ly/18P9CSv On the formalization of continuous-time Markov chains in HOL]. ~ L. Liu, O. Hasan y S. Tahar
# [http://bit.ly/17H2mqy Formalizing Turing machines]. ~ A. Asperti y W. Ricciotti
# [http://bit.ly/YwuCeL Light-weight containers for Isabelle: efficient, extensible, nestable]. ~ A. Lochbihler
# [http://bit.ly/10XLrRA Graph theory]. ~ L. Noschinski
# [http://bit.ly/19kPEP4 A machine-checked proof of the odd order theorem]. ~ G. Gonthier et als.
# [http://goo.gl/LdihL A constructive theory of regular languages in Coq]. ~ C. Doczkal, J.O. Kaiser y G. Smolka
# [http://goo.gl/gwcwL A formal proof of Kruskal’s tree theorem in Isabelle/HOL]. ~ C. Sternagel
# [http://goo.gl/CUF3e Formalizing Knuth-Bendix orders and Knuth-Bendix completion]. ~ C. Sternagel y R. Thiemann
# [http://goo.gl/9JAfX Developing an auction theory toolbox]. ~ C. Lange, C. Rowat, W. Windsteiger y M. Kerber
# [http://goo.gl/6OfmQ Formalization of incremental simplex algorithm by stepwise refinement]. ~ M. Spasić y F. Marić
# [http://goo.gl/Guxky Coinductive pearl: Modular first-order logic completeness]. ~ J.C. Blanchette, A. Popescu y D. Traytel
# [http://goo.gl/HUOl8 A fully verified executable LTL model checker]. ~ J. Esparza et als.
# [http://goo.gl/RV54S ForMaRE - formal mathematical reasoning in economics]. ~ M. Kerber, C. Lange y C. Rowat.
# [http://goo.gl/Y7sIq AI over large formal knowledge bases: The first decade]. ~ J. Urban.
# [http://goo.gl/vvqNg Formalization of real analysis: A survey of proof assistants and libraries]. ~ S. Boldo, C. Lelay y G. Melquiond.
# [http://goo.gl/bEFYa Data refinement in Isabelle/HOL]. ~ F. Haftmann, A. Krauss, O. Kunčar y T. Nipkow
# [http://goo.gl/xTyvE Formalizing the confluence of orthogonal rewriting systems]. ~ A.C. Rocha y M. Ayala.
# [http://goo.gl/zCYHj Formalization of the complex number theory in HOL4]. ~ Z. Shi et als.
# [http://goo.gl/kM0dI Programming and reasonning with PowerLists in Coq]. ~ F. Loulergue y V. Niculescu
# [http://goo.gl/KkU6s A hierarchy of mathematical structures in ACL2]. ~ J. Heras, F.J. Martín y V. Pascual.
# [http://www.inf.kcl.ac.uk/staff/urbanc/Publications/tm.pdf Mechanising Turing Machines and Computability Theory in Isabelle/HOL] ~ J. Xu, X. Zhang y C. Urban

Ejercicios

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== Relaciones de ejercicios ==

=== Relaciones de ejercicios propuestos ===

En esta sección se publicarán las relaciones de ejercicios. Las soluciones se escriben de forma colaborativa por los alumnos del curso y no deben tomarse como definitivas.

Temas

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== Temas de ''Razonamiento automático (2014-15)'' ==

MediaWiki:Mainpage

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Razonamiento automático (2016-17)

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