RA2013: Ejercicios de cuantificadores sobre listas con Isabelle/HOL
En la primera parte de la clase de hoy del curso de Razonamiento automático se ha comentado las soluciones de la 5ª relación de ejercicios cuyo objetivo es automáticamente con Isabelle/HOL propiedades de programas de cuantificadores sobre listas.
Los ejercicios y sus soluciones se muestran a continuación
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header {* R5: Cuantificadores sobre listas *} theory R5_Cuantificadores_sobre_listas_sol 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. --------------------------------------------------------------------- *} fun todos :: "('a ⇒ bool) ⇒ 'a list ⇒ bool" where "todos p [] = True" | "todos p (y#ys) = ((p y) ∧ (todos p ys))" value "todos (λx. 1 < length x) [[2,1,4],[1,3]]" -- "= True" value "todos (λx. 1 < length x) [[2,1,4],[3]]" -- "= False" 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. --------------------------------------------------------------------- *} fun algunos :: "('a ⇒ bool) ⇒ 'a list ⇒ bool" where "algunos p [] = False" | "algunos p (x#xs) = ((p x) ∨ (algunos p xs))" value "algunos (λx. 1 < length x) [[2,1,4],[3]]" -- "= True" value "algunos (λx. 1 < length x) [[],[3]]" -- "= False" text {* --------------------------------------------------------------------- Ejercicio 3.1. Demostrar o refutar automáticamente todos (λx. P x ∧ Q x) xs = (todos P xs ∧ todos Q xs) --------------------------------------------------------------------- *} -- "La demostración automática es" lemma "todos (λx. P x ∧ Q x) xs = (todos P xs ∧ todos Q xs)" by (induct xs) auto text {* --------------------------------------------------------------------- Ejercicio 3.2. Demostrar o refutar detalladamente todos (λx. P x ∧ Q x) xs = (todos P xs ∧ todos Q xs) --------------------------------------------------------------------- *} -- "La demostración estructurada es" 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 "todos (λx. P x ∧ Q x) xs = (todos P xs ∧ todos Q xs)" thus "todos (λx. P x ∧ Q x) (a#xs) = (todos P (a#xs) ∧ todos Q (a#xs))" by auto qed -- "La demostración detallada es" 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)" show "todos (λx. P x ∧ Q x) (a#xs) = (todos P (a#xs) ∧ todos Q (a#xs))" proof - 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 auto also have "… = (todos P (a#xs) ∧ todos Q (a#xs))" by simp finally show ?thesis . qed qed text {* --------------------------------------------------------------------- Ejercicio 4.1. Demostrar o refutar automáticamente todos P (x @ y) = (todos P x ∧ todos P y) --------------------------------------------------------------------- *} -- "La demostración automática es" lemma todos_append: "todos P (x @ y) = (todos P x ∧ todos P y)" by (induct x) simp_all text {* --------------------------------------------------------------------- Ejercicio 4.2. Demostrar o refutar detalladamente todos P (x @ y) = (todos P x ∧ todos P y) --------------------------------------------------------------------- *} -- "La demostración estructurada es" lemma todos_append_2: "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 "todos P (x @ y) = (todos P x ∧ todos P y)" thus "todos P ((a#x) @ y) = (todos P (a#x) ∧ todos P y)" by auto qed -- "La demostración detallada es" lemma todos_append_3: "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)" show "todos P ((a#x) @ y) = (todos P (a#x) ∧ todos P y)" proof - 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 ?thesis . qed qed text {* --------------------------------------------------------------------- Ejercicio 5.1. Demostrar o refutar automáticamente todos P (rev xs) = todos P xs --------------------------------------------------------------------- *} -- "La demostración automática es" lemma "todos P (rev xs) = todos P xs" by (induct xs) (auto simp add: todos_append) text {* --------------------------------------------------------------------- Ejercicio 5.2. Demostrar o refutar detalladamente todos P (rev xs) = todos P xs --------------------------------------------------------------------- *} -- "La demostración estructurada es" lemma "todos P (rev xs) = todos P xs" proof (induct xs) show "todos P (rev []) = todos P []" by simp next fix a xs assume "todos P (rev xs) = todos P xs" thus "todos P (rev (a#xs)) = todos P (a#xs)" by (auto simp add: todos_append) qed -- "La demostración detallada es" 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" show "todos P (rev (a#xs)) = todos P (a#xs)" proof - 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 auto also have "… = (P a ∧ todos P xs)" by simp also have "… = todos P (a#xs)" by simp finally show ?thesis . qed qed text {* --------------------------------------------------------------------- Ejercicio 6. Demostrar o refutar: algunos (λx. P x ∧ Q x) xs = (algunos P xs ∧ algunos Q xs) --------------------------------------------------------------------- *} -- "Se busca un contraejemplo con nitpick" lemma "algunos (λx. P x ∧ Q x) xs = (algunos P xs ∧ algunos Q xs)" nitpick oops text {* El contraejemplo encontrado es Nitpick found a counterexample for card 'a = 3: Free variables: P = (λx. _)(a⇘1⇙ := True, a⇘2⇙ := True, a⇘3⇙ := False) Q = (λx. _)(a⇘1⇙ := False, a⇘2⇙ := False, a⇘3⇙ := True) xs = [a⇘3⇙, a⇘2⇙] *} text {* --------------------------------------------------------------------- Ejercicio 7.1. Demostrar o refutar automáticamente algunos P (map f xs) = algunos (P ∘ f) xs --------------------------------------------------------------------- *} -- "La demostración automática es" lemma "algunos P (map f xs) = algunos (P o f) xs" by (induct xs) simp_all -- "La demostración estructurada es" 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 a xs assume "algunos P (map f xs) = algunos (P ∘ f) xs" thus "algunos P (map f (a#xs)) = algunos (P ∘ f) (a#xs)" by auto qed text {* --------------------------------------------------------------------- Ejercicio 7.2. Demostrar o refutar detalladamente algunos P (map f xs) = algunos (P ∘ f) xs --------------------------------------------------------------------- *} -- "La demostración detallada es" 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 a xs assume HI: "algunos P (map f xs) = algunos (P ∘ f) xs" show "algunos P (map f (a#xs)) = algunos (P ∘ f) (a#xs)" proof - 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 ?thesis . qed qed text {* --------------------------------------------------------------------- Ejercicio 8.1. Demostrar o refutar automáticamente algunos P (xs @ ys) = (algunos P xs ∨ algunos P ys) --------------------------------------------------------------------- *} -- "La demostración automática es" lemma algunos_append: "algunos P (xs @ ys) = (algunos P xs ∨ algunos P ys)" by (induct xs) simp_all text {* --------------------------------------------------------------------- Ejercicio 8.2. Demostrar o refutar detalladamente algunos P (xs @ ys) = (algunos P xs ∨ algunos P ys) --------------------------------------------------------------------- *} -- "La demostración estructurada es" lemma algunos_append_2: "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 "algunos P (xs @ ys) = (algunos P xs ∨ algunos P ys)" thus "algunos P ((a#xs) @ ys) = (algunos P (a#xs) ∨ algunos P ys)" by auto qed -- "La demostración detallada es" lemma algunos_append_3: "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)" show "algunos P ((a#xs) @ ys) = (algunos P (a#xs) ∨ algunos P ys)" proof - have "algunos P ((a#xs) @ ys) = algunos P (a#(xs @ ys))" by simp also have "… = ((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 ?thesis . qed qed text {* --------------------------------------------------------------------- Ejercicio 9.1. Demostrar o refutar automáticamente algunos P (rev xs) = algunos P xs --------------------------------------------------------------------- *} -- "La demostración automática es" lemma "algunos P (rev xs) = algunos P xs" by (induct xs) (auto simp add: algunos_append) text {* --------------------------------------------------------------------- Ejercicio 9.2. Demostrar o refutar detalladamente algunos P (rev xs) = algunos P xs --------------------------------------------------------------------- *} -- "La demostración estructurada es" lemma "algunos P (rev xs) = algunos P xs" proof (induct xs) show "algunos P (rev []) = algunos P []" by simp next fix a xs assume "algunos P (rev xs) = algunos P xs" thus "algunos P (rev (a#xs)) = algunos P (a#xs)" by (auto simp add: algunos_append) qed -- "La demostración detallada es" 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" show "algunos P (rev (a#xs)) = algunos P (a#xs)" proof - 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 xs ∨ P a)" by simp also have "… = (P a ∨ algunos P xs)" by auto also have "… = algunos P (a#xs)" by simp finally show ?thesis . qed 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. --------------------------------------------------------------------- *} text {* Solución: La ecuación se verifica eligiendo como Z el término algunos P xs ∨ algunos Q xs En efecto, *} lemma "algunos (λx. P x ∨ Q x) xs = (algunos P xs ∨ algunos Q xs)" by (induct xs) auto -- "De forma estructurada" 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 "algunos (λx. (P x ∨ Q x)) xs = (algunos P xs ∨ algunos Q xs)" thus "algunos (λx. P x ∨ Q x) (a#xs) = (algunos P (a#xs) ∨ algunos Q (a#xs))" by auto qed -- "De forma 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)" show "algunos (λx. P x ∨ Q x) (a#xs) = (algunos P (a#xs) ∨ algunos Q (a#xs))" proof - 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 auto also have "… = (algunos P (a#xs) ∨ algunos Q (a#xs))" by simp finally show ?thesis . qed qed text {* --------------------------------------------------------------------- Ejercicio 11.1. Demostrar o refutar automáticamente algunos P xs = (¬ todos (λx. (¬ P x)) xs) --------------------------------------------------------------------- *} -- "La demostración automática es" lemma "algunos P xs = (¬ todos (λx. (¬ P x)) xs)" by (induct xs) simp_all text {* --------------------------------------------------------------------- Ejercicio 11.2. Demostrar o refutar detalladamente algunos P xs = (¬ todos (λx. (¬ P x)) xs) --------------------------------------------------------------------- *} -- "La demostración estructurada es" 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 "algunos P xs = (¬ todos (λx. (¬ P x)) xs)" thus "algunos P (a#xs) = (¬ todos (λx. (¬ P x)) (a#xs))" by auto qed -- "La demostración detallada es" 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)" show "algunos P (a#xs) = (¬ todos (λx. (¬ P x)) (a#xs))" proof - 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 ?thesis . qed qed text {* --------------------------------------------------------------------- Ejercicio 12. Definir la funcion primitiva recursiva estaEn :: 'a ⇒ 'a list ⇒ bool tal que (estaEn x xs) se verifica si el elemento x está en la lista xs. Por ejemplo, estaEn (2::nat) [3,2,4] = True estaEn (1::nat) [3,2,4] = False --------------------------------------------------------------------- *} fun estaEn :: "'a ⇒ 'a list ⇒ bool" where "estaEn x [] = False" | "estaEn x (a#xs) = (x=a ∨ estaEn x xs)" value "estaEn (2::nat) [3,2,4]" -- "= True" value "estaEn (1::nat) [3,2,4]" -- "= False" text {* --------------------------------------------------------------------- Ejercicio 13. Expresar la relación existente entre estaEn y algunos. Demostrar dicha relación de forma automática y detallada. --------------------------------------------------------------------- *} text {* Solución: La relación es estaEn y xs = algunos (λx. x=y) xs En efecto, *} lemma estaEn_algunos: "estaEn y xs = algunos (λx. x=y) xs" by (induct xs) auto -- "La demostración estructurada es" 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 a xs assume "estaEn y xs = algunos (λx. x=y) xs" thus "estaEn y (a#xs) = algunos (λx. x=y) (a#xs)" by auto qed -- "La demostración detallada es" lemma estaEn_algunos_3: "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" show "estaEn y (a#xs) = algunos (λx. x=y) (a#xs)" proof - 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 ?thesis . qed qed text {* --------------------------------------------------------------------- Ejercicio 14. 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 --------------------------------------------------------------------- *} fun uplicados :: "'a list ⇒ bool" where "uplicados [] = True" | "uplicados (a#xs) = ((¬ estaEn a xs) ∧ uplicados xs)" value "sinDuplicados [1::nat,4,2]" -- "= True" value "sinDuplicados [1::nat,4,2,4]" -- "= False" text {* --------------------------------------------------------------------- Ejercicio 15. 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. --------------------------------------------------------------------- *} fun borraDuplicados :: "'a list ⇒ 'a list" where "borraDuplicados [] = []" | "borraDuplicados (a#xs) = (if estaEn a xs then borraDuplicados xs else (a#borraDuplicados xs))" value "borraDuplicados [1::nat,2,4,2,3]" -- "= [1,4,2,3]" text {* --------------------------------------------------------------------- Ejercicio 16.1. Demostrar o refutar automáticamente length (borraDuplicados xs) ≤ length xs --------------------------------------------------------------------- *} -- "La demostración automática es" lemma length_borraDuplicados: "length (borraDuplicados xs) ≤ length xs" by (induct xs) simp_all text {* --------------------------------------------------------------------- Ejercicio 16.2. Demostrar o refutar detalladamente length (borraDuplicados xs) ≤ length xs --------------------------------------------------------------------- *} -- "La demostración estructurada es" lemma length_borraDuplicados_2: "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" thus "length (borraDuplicados (a#xs)) ≤ length (a#xs)" proof (cases) assume "estaEn a xs" thus "length (borraDuplicados (a#xs)) ≤ length (a#xs)" using HI by auto next assume "(¬ estaEn a xs)" thus "length (borraDuplicados (a#xs)) ≤ length (a#xs)" using HI by auto qed qed -- "La demostración detallada es" lemma length_borraDuplicados_3: "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" show "length (borraDuplicados (a#xs)) ≤ length (a#xs)" proof (cases) assume "estaEn a xs" hence "length (borraDuplicados (a#xs)) = length (borraDuplicados xs)" by simp also have "… ≤ length xs" using HI by simp also have "… ≤ length (a#xs)" by simp finally show ?thesis . next assume "(¬ estaEn a xs)" hence "length (borraDuplicados (a#xs)) = length (a#(borraDuplicados xs))" by simp also have "… = 1 + length (borraDuplicados xs)" by simp also have "… ≤ 1 + length xs" using HI by simp also have "… = length (a#xs)" by simp finally show ?thesis . qed qed text {* --------------------------------------------------------------------- Ejercicio 17.1. Demostrar o refutar automáticamente estaEn a (borraDuplicados xs) = estaEn a xs --------------------------------------------------------------------- *} -- "La demostración automática es" lemma estaEn_borraDuplicados: "estaEn a (borraDuplicados xs) = estaEn a xs" by (induct xs) auto text {* --------------------------------------------------------------------- Ejercicio 17.2. Demostrar o refutar detalladamente estaEn a (borraDuplicados xs) = estaEn a xs --------------------------------------------------------------------- *} -- "La demostración estructurada es" 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 c1: "estaEn a (borraDuplicados (b#xs))" show "estaEn a (b#xs)" proof (cases) assume "estaEn b xs" thus "estaEn a (b#xs)" using c1 HI by auto next assume "¬ estaEn b xs" thus "estaEn a (b#xs)" using c1 HI by auto qed next assume c2: "estaEn a (b#xs)" show "estaEn a (borraDuplicados (b#xs))" proof (cases) assume "a=b" thus "estaEn a (borraDuplicados (b#xs))" using HI by auto next assume "a≠b" thus "estaEn a (borraDuplicados (b#xs))" using `a≠b` c2 HI by auto qed qed qed -- "La demostración detallada es" lemma estaEn_borraDuplicados_3: "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 c1: "estaEn a (borraDuplicados (b#xs))" show "estaEn a (b#xs)" proof (cases) assume "estaEn b xs" hence "estaEn a (borraDuplicados xs)" using c1 by simp hence "estaEn a xs" using HI by simp thus "estaEn a (b#xs)" by simp next assume "¬ estaEn b xs" hence "estaEn a (b#(borraDuplicados xs))" using c1 by simp hence "a=b ∨ (estaEn a (borraDuplicados xs))" by simp hence "a=b ∨ (estaEn a xs)" using HI by simp thus "estaEn a (b#xs)" by simp qed next assume c2: "estaEn a (b#xs)" show "estaEn a (borraDuplicados (b#xs))" proof (cases) assume "a=b" thus "estaEn a (borraDuplicados (b#xs))" using HI by auto next assume "a≠b" hence "estaEn a xs" using c2 by simp hence "estaEn a (borraDuplicados xs)" using HI by simp thus "estaEn a (borraDuplicados (b#xs))" using `a≠b` by simp qed qed qed text {* --------------------------------------------------------------------- Ejercicio 18.1. Demostrar o refutar automáticamente sinDuplicados (borraDuplicados xs) --------------------------------------------------------------------- *} -- "La demostración automática" lemma sinDuplicados_borraDuplicados: "sinDuplicados (borraDuplicados xs)" by (induct xs) (auto simp add: estaEn_borraDuplicados) text {* --------------------------------------------------------------------- Ejercicio 18.2. Demostrar o refutar detalladamente sinDuplicados (borraDuplicados xs) --------------------------------------------------------------------- *} -- "La demostración estructurada es" lemma sinDuplicados_borraDuplicados_2: "sinDuplicados (borraDuplicados xs)" proof (induct xs) show "sinDuplicados (borraDuplicados [])" by simp next fix a xs assume HI: "sinDuplicados (borraDuplicados xs)" show "sinDuplicados (borraDuplicados (a#xs))" proof (cases) assume "estaEn a xs" thus "sinDuplicados (borraDuplicados (a#xs))" using HI by simp next assume "¬ estaEn a xs" thus "sinDuplicados (borraDuplicados (a#xs))" using `¬ estaEn a xs` HI by (auto simp add: estaEn_borraDuplicados) qed qed -- "La demostración detallada es" lemma sinDuplicados_borraDuplicados_3: "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" thus "sinDuplicados (borraDuplicados (a#xs))" using HI by simp next assume "¬ estaEn a xs" hence "¬ (estaEn a xs) ∧ sinDuplicados (borraDuplicados xs)" using HI by simp hence "¬ estaEn a (borraDuplicados xs) ∧ sinDuplicados (borraDuplicados xs)" by (simp add: estaEn_borraDuplicados) hence "sinDuplicados (a#borraDuplicados xs)" by simp thus "sinDuplicados (borraDuplicados (a#xs))" using `¬ estaEn a xs` by simp qed qed text {* --------------------------------------------------------------------- Ejercicio 19. Demostrar o refutar: borraDuplicados (rev xs) = rev (borraDuplicados xs) --------------------------------------------------------------------- *} -- "Se busca un contraejemplo con" lemma "borraDuplicados (rev xs) = rev (borraDuplicados xs)" quickcheck oops text {* El contraejemplo encontrado es xs = [3, 2, 3] En efecto, borraDuplicados (rev xs) = borraDuplicados (rev [3,2,3]) = [2,3] rev (borraDuplicados xs) = rev (borraDuplicados [3,2,3]) = [3,2] *} end |