RA2019: Ejercicios de cuantificadores sobre listas en Isabelle/HOL
En la segunda parte de la clase de hoy del curso de Razonamiento automático se han comentado las soluciones de la 4ª relación de ejercicios de cuantificadores sobre listas. Para cada propiedad se dan tres demostraciones en Isabelle/HOL: la primera automática, la segunda estructurada y la tercera totalmente detallada mostrando todos los lemas de HOL que se utilizan en cada paso.
La teoría con las soluciones de los ejercicios es la siguiente
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chapter ‹R4: Cuantificadores sobre listas› theory R4_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))" lemma shows "(λx. 1 < length x) [[2,1,4],[1,3]]" and "¬todos (λx. 1 < length x) [[2,1,4],[3]]" by simp+ 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))" lemma shows "algunos (λx. 1 < length x) [[2,1,4],[3]]" and "¬algunos (λx. 1 < length x) [[],[3]]" by simp+ 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)" then show "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) have "todos (λx. P x ∧ Q x) [] = True" by (simp only: todos.simps(1)) also have "... = (True ∧ True)" by (simp only: conj_absorb) also have "... = (todos P [] ∧ todos Q [])" by (simp only: todos.simps(1)) finally show "todos (λx. P x ∧ Q x) [] = (todos P [] ∧ todos Q [])" by this 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) = ((λx. P x ∧ Q x) x ∧ todos (λx. P x ∧ Q x) xs)" by (simp only: todos.simps(2)) also have "... = ((λx. P x ∧ Q x) x ∧ (todos P xs ∧ todos Q xs))" by (simp only: HI) also have "... = (P x ∧ Q x ∧ todos P xs ∧ todos Q xs)" by (simp only: conj_assoc) also have "... = (P x ∧ todos P xs ∧ Q x ∧ todos Q xs)" by (simp only: conj_left_commute) also have "... = ((P x ∧ todos P xs) ∧ Q x ∧ todos Q xs)" by (simp only: conj_assoc) also have "... = (todos P (x#xs) ∧ Q x ∧ todos Q xs)" by (simp only: todos.simps(2)) also have "... = (todos P (x#xs) ∧ todos Q (x#xs))" by (simp only: todos.simps(2)) finally show "todos (λx. P x ∧ Q x) (x#xs) = (todos P (x#xs) ∧ todos Q (x#xs))" by this 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)" then show "todos P ((a#x) @ y) = (todos P (a#x) ∧ todos P y)" by simp qed (* La demostración detallada es *) lemma todos_append_3: "todos P (x @ y) = (todos P x ∧ todos P y)" proof (induct x arbitrary:y) fix y have "todos P ([] @ y) = todos P y" by (simp only: append.simps(1)) also have "... = (True ∧ todos P y)" by (simp only: simp_thms(22)) also have "... = (todos P [] ∧ todos P y)" by (simp only: todos.simps(1)) finally show "todos P ([] @ y) = (todos P [] ∧ todos P y)" by this next fix a x assume HI:"⋀y. todos P (x @ y) = (todos P x ∧ todos P y)" fix y have "todos P ((a#x) @ y) = (todos P (a#(x @ y)))" by (simp only: append.simps(2)) also have "... = (P a ∧ todos P (x @ y))" by (simp only: todos.simps(2)) also have "... = (P a ∧ todos P x ∧ todos P y)" by (simp only: HI) also have "... = ((P a ∧ todos P x) ∧ todos P y)" by (simp only: conj_assoc) also have "... = (todos P (a#x) ∧ todos P y)" by (simp only: todos.simps(2)) finally show "todos P ((a#x) @ y) = (todos P (a#x) ∧ todos P y)" by this 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" then show "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 only: rev.simps(1)) next fix x xs assume HI: "todos P (rev xs) = todos P xs" have "todos P (rev (x#xs)) = todos P (rev xs@[x])" by (simp only: rev.simps(2)) also have "... = (todos P (rev xs) ∧ todos P [x])" by (simp only: todos_append) also have "... = (todos P xs ∧ todos P [x])" by (simp only: HI) also have "... = (todos P [x] ∧ todos P xs)" by (simp only: conj_commute) also have "... = (todos P ([x]@xs))" by (simp only: todos_append) also have "... = (todos P (x#xs))" by (simp only: append.simps) finally show "todos P (rev (x#xs)) = todos P (x#xs)" by this 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)" quickcheck oops text ‹ Quickcheck found a counterexample: P = {a⇩1} Q = {a⇩2} xs = [a⇩1, a⇩2] Evaluated terms: algunos (λx. P x ∧ Q x) xs = False algunos P xs ∧ algunos Q xs = True › text ‹------------------------------------------------------------------ Ejercicio 7.1. Demostrar o refutar automáticamente algunos P (map f xs) = algunos (P ∘ f) xs ---------------------------------------------------------------------› (* La demostración automática es *) lemma "algunos P (map f xs) = algunos (P o f) xs" by (induct xs) simp_all text ‹------------------------------------------------------------------ Ejercicio 7.2. Demostrar o refutar detalladamente algunos P (map f xs) = algunos (P ∘ f) xs ---------------------------------------------------------------------› (* 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" then show "algunos P (map f (a#xs)) = algunos (P ∘ f) (a#xs)" by auto qed (* La demostración detallada es *) lemma "algunos P (map f xs) = algunos (P o f) xs" proof (induct xs) have "algunos P (map f []) = algunos P []" by (simp only: list.map(1)) also have "... = algunos (P o f) []" by (simp only: algunos.simps(1)) finally show "algunos P (map f []) = algunos (P o f) []" by this next fix x xs assume HI: "algunos P (map f xs) = algunos (P o f) xs" have "algunos P (map f (x#xs)) = algunos P ((f x) # (map f xs))" by (simp only: list.map(2)) 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)" by (simp only: HI) also have "... = ((P o f) x ∨ algunos (P o f) xs)" by (simp only: o_apply) also have "... = algunos (P o f) (x#xs)" by (simp only: algunos.simps(2)) finally show "algunos P (map f (x#xs)) = algunos (P o f) (x#xs)" by this 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)" then show "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) have "algunos P ([] @ ys) = (algunos P ys)" by (simp only: append_Nil) also have "... = (False ∨ algunos P ys)" by (simp only: simp_thms(32)) also have "... = (algunos P [] ∨ algunos P ys)" by (simp only: algunos.simps(1)) finally show "algunos P ([] @ ys) = (algunos P [] ∨ algunos P ys)" by this next fix x xs assume HI: "algunos P (xs @ ys) = (algunos P xs ∨ algunos P ys)" have "algunos P ((x#xs) @ ys) = algunos P (x#(xs @ ys))" by (simp only: append_Cons) also have "... = (P x ∨ algunos P (xs @ ys))" by (simp only: algunos.simps(2)) also have "... = (P x ∨ algunos P xs ∨ algunos P ys)" by (simp only: HI) also have "... = ((P x ∨ algunos P xs) ∨ algunos P ys)" by (simp only: disj_assoc) also have "... = (algunos P (x#xs) ∨ algunos P ys)" by (simp only: algunos.simps(2)) finally show "algunos P ((x#xs) @ ys) = (algunos P (x#xs) ∨ algunos P ys)" by this 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" then show "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 only: rev.simps(1)) 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 only: rev.simps(2)) also have "... = (algunos P (rev xs) ∨ algunos P [x])" by (simp only: algunos_append) also have "... = (algunos P xs ∨ algunos P [x])" by (simp only: HI) also have "... = (algunos P xs ∨ P x ∨ algunos P [])" by (simp only: algunos.simps(2)) also have "... = (algunos P xs ∨ P x ∨ False)" by (simp only: algunos.simps(1)) also have "... = (algunos P xs ∨ P x)" by (simp only: simp_thms(31)) also have "... = (P x ∨ algunos P xs)" by (simp only: disj_commute) also have "... = algunos P (x#xs)" by (simp only: algunos.simps(2)) finally show "algunos P (rev (x#xs)) = algunos P (x#xs)" by this 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)" then show "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) have "algunos (λx. P x ∨ Q x) [] = False" by (simp only: algunos.simps(1)) also have "... = (False ∨ False)" by (simp only: simp_thms(31)) also have "... = (algunos P [] ∨ False)" by (simp only: algunos.simps(1)) also have "... = (algunos P [] ∨ algunos Q [])" by (simp only: algunos.simps(1)) finally show "algunos (λx. P x ∨ Q x) [] = (algunos P [] ∨ algunos Q [])" by this next fix x xs assume HI:"algunos (λx. P x ∨ Q x) xs = (algunos P xs ∨ algunos Q xs)" have "algunos (λx. P x ∨ Q x) (x#xs) =((P x ∨ Q x) ∨ algunos (λx. P x ∨ Q x) xs)" by (simp only: algunos.simps(2)) also have "... = ((P x ∨ Q x) ∨ algunos P xs ∨ algunos Q xs)" by (simp only: HI) also have "... = (P x ∨ Q x ∨ algunos P xs ∨ algunos Q xs)" by (simp only: disj_assoc) also have "... = (P x ∨ algunos P xs ∨ Q x ∨ algunos Q xs)" by (simp only: disj_left_commute) also have "... = ((P x ∨ algunos P xs) ∨ (Q x ∨ algunos Q xs))" by (simp only: disj_assoc) also have "... = (algunos P (x#xs) ∨ algunos Q (x#xs))" by (simp only: algunos.simps(2)) finally show "algunos (λx. P x ∨ Q x) (x#xs) = (algunos P (x#xs) ∨ algunos Q (x#xs))" by this 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)" then show "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) = (a=x ∨ estaEn x xs)" value "estaEn (2::nat) [3,2,4] = True" value "estaEn (1::nat) [3,2,4] = False" text ‹------------------------------------------------------------------ Ejercicio 13. Expresar la relación existente entre estaEn y algunos. Demostrar dicha relación de forma automática y detallada. ---------------------------------------------------------------------› text ‹Solución: La relación es estaEn y xs = algunos (λx. x=y) xs En efecto,› lemma "estaEn x xs = algunos (λy. y=x) xs" by (induct xs) auto (* La demostración estructurada es *) lemma "estaEn x xs = algunos (λy. y=x) xs" proof (induct xs) show "estaEn x [] = algunos (λy. y=x) []" by simp next fix a xs assume HI: "estaEn x xs = algunos (λy. y=x) xs" have "estaEn x (a#xs) = algunos (λy. y=x) (a#xs)" using HI by simp then show "estaEn x (a#xs) = algunos (λy. y=x) (a#xs)" by simp qed (* La demostración detallada es *) lemma "estaEn x xs = algunos (λy. y=x) xs" proof (induct xs) have "estaEn x [] = False" by (simp only: estaEn.simps(1)) also have "... = algunos (λy. y=x) []" by (simp only: algunos.simps(1)) finally show "estaEn x [] = algunos (λy. y=x) []" by this next fix a xs assume HI: "estaEn x xs = algunos (λy. y=x) xs" have "estaEn x (a#xs) = ((a=x) ∨ estaEn x xs)" by (simp only: estaEn.simps(2)) also have "... = ((a=x) ∨ algunos (λy. y=x) xs) " using HI by (simp only :) also have "... = algunos (λy. y=x) (a#xs)" by (simp only: algunos.simps(2)) finally show "estaEn x (a#xs) = algunos (λy. y=x) (a#xs)" by this qed end |