RA2012: Verificación de propiedades de la sustitución en Isabelle/HOL
En la clase de hoy del curso de Razonamiento automático se ha resuelto de manera colaborativa ejercicios sobre la demostración de propiedades de la función de sustitución con Isabelle/HOL. Su objetivo es ilustrar el uso del razonamiento por inducción y por casos en Isabelle. Para ca propiedad se presentan distintas demostraciones desde las automáticas a las detalladas.
La teoría con los ejemplos presentados en la clase es la siguiente:
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theory R11 imports Main begin text {* --------------------------------------------------------------------- Ejercicio 1. Definir la función sust :: "'a ⇒ 'a ⇒ 'a list ⇒ 'a list" tal que (sust x y zs) es la lista obtenida sustituyendo cada occurrencia de x por y en la lista zs. Por ejemplo, sust (1::nat) 2 [1,2,3,4,1,2,3,4] = [2,2,3,4,2,2,3,4] --------------------------------------------------------------------- *} fun sust :: "'a ⇒ 'a ⇒ 'a list ⇒ 'a list" where "sust x y [] = []" | "sust x y (z#zs) = (if z=x then y else z)#(sust x y zs)" value "sust (1::nat) 2 [1,2,3,4,1,2,3,4]" -- "= [2,2,3,4,2,2,3,4]" text {* --------------------------------------------------------------------- Ejercicio 2. Demostrar o refutar: sust x y (xs@ys) = (sust x y xs)@(sust x y ys)" --------------------------------------------------------------------- *} -- "La demostración automática es" lemma sust_append: "sust x y (xs@ys) = (sust x y xs)@(sust x y ys)" by (induct xs) auto -- "La demostración estructurada es" lemma sust_append_2: "sust x y (xs @ ys) = (sust x y xs)@(sust x y ys)" proof (induct xs) show "sust x y ([]@ys) = (sust x y [])@(sust x y ys)" by simp next fix a xs assume HI: "sust x y (xs@ys) = (sust x y xs)@(sust x y ys)" show "sust x y ((a#xs)@ys) = (sust x y (a#xs))@(sust x y ys)" proof (cases) assume "x=a" thus "sust x y ((a#xs)@ys) = (sust x y (a#xs))@(sust x y ys)" using HI by auto next assume "x≠a" thus "sust x y ((a#xs)@ys) = (sust x y (a#xs))@(sust x y ys)" using HI by auto qed qed -- "La demostración detallada es" lemma sust_append_3: "sust x y (xs @ ys) = (sust x y xs)@(sust x y ys)" proof (induct xs) show "sust x y ([]@ys) = (sust x y [])@(sust x y ys)" by simp next fix a xs assume HI: "sust x y (xs@ys) = (sust x y xs)@(sust x y ys)" show "sust x y ((a#xs)@ys) = (sust x y (a#xs))@(sust x y ys)" proof (cases) assume "x=a" hence "sust x y ((a#xs)@ys) = sust x y (a#(xs@ys))" by simp also have "… = y#(sust x y (xs@ys))" using `x=a` by simp also have "… = y#((sust x y xs)@(sust x y ys))" using HI by simp also have "… = (y#(sust x y xs))@(sust x y ys)" by simp also have "… = (sust x y (a#xs))@(sust x y ys)" using `x=a` by simp finally show "sust x y ((a#xs)@ys) = (sust x y (a#xs))@(sust x y ys)" . next assume "x≠a" hence "sust x y ((a#xs)@ys) = sust x y (a#(xs@ys))" by simp also have "… = a#(sust x y (xs@ys))" using `x≠a` by simp also have "… = a#((sust x y xs)@(sust x y ys))" using HI by simp also have "… = (a#(sust x y xs))@(sust x y ys)" by simp also have "… = (sust x y (a#xs))@(sust x y ys)" using `x≠a` by simp finally show "sust x y ((a#xs)@ys) = (sust x y (a#xs))@(sust x y ys)" . qed qed text {* --------------------------------------------------------------------- Ejercicio 3. Demostrar o refutar: rev (sust x y zs) = sust x y (rev zs) --------------------------------------------------------------------- *} -- "La demostración automática es" lemma rev_sust: "rev(sust x y zs) = sust x y (rev zs)" by (induct zs) (simp_all add: sust_append) -- "La demostración estructurada es" lemma rev_sust_2: "rev (sust x y zs) = sust x y (rev zs)" proof (induct zs) show "rev (sust x y []) = sust x y (rev [])" by simp next fix a zs assume HI: "rev (sust x y zs) = sust x y (rev zs)" show "rev (sust x y (a#zs)) = sust x y (rev (a#zs))" using HI by (auto simp add: sust_append) qed -- "La demostración detallada es" lemma rev_sust_3: "rev (sust x y zs) = sust x y (rev zs)" proof (induct zs) show "rev (sust x y []) = sust x y (rev [])" by simp next fix a zs assume HI: "rev (sust x y zs) = sust x y (rev zs)" show "rev (sust x y (a#zs)) = sust x y (rev (a#zs))" proof - have "rev (sust x y (a#zs)) = rev ((if x=a then y else a)#(sust x y zs))" by simp also have "… = (rev (sust x y zs))@[if x=a then y else a]" by simp also have "… = (sust x y (rev zs))@[if x=a then y else a]" using HI by simp also have "… = (sust x y (rev zs))@(sust x y [a])" by simp also have "… = sust x y ((rev zs)@[a])" by (simp add: sust_append) also have "… = sust x y (rev (a#zs))" by simp finally show "rev (sust x y (a#zs)) = sust x y (rev (a#zs))" . qed qed -- "La demostración detallada es con casos es" lemma rev_sust_4: "rev (sust x y zs) = sust x y (rev zs)" proof (induct zs) show "rev (sust x y []) = sust x y (rev [])" by simp next fix a zs assume HI: "rev (sust x y zs) = sust x y (rev zs)" show "rev (sust x y (a#zs)) = sust x y (rev (a#zs))" proof (cases "x=a") assume "x=a" hence "rev (sust x y (a#zs)) = rev (y # (sust x y zs))" by simp also have "... = (rev (sust x y zs)) @ [y]" by simp also have "... = (sust x y (rev zs)) @ [y]" using HI by simp also have "... = (sust x y (rev zs)) @ (sust x y [a])" using `x=a` by simp also have "... = sust x y ((rev zs) @ [a])" by (simp add: sust_append) also have "... = sust x y (rev (a#zs))" by simp finally show "rev (sust x y (a#zs)) = sust x y (rev (a#zs))" by simp next assume "x≠a" hence "rev (sust x y (a#zs)) = rev (a # (sust x y zs))" by simp also have "... = (rev (sust x y zs)) @ [a]" by simp also have "... = (sust x y (rev zs)) @ [a]" using HI by simp also have "... = (sust x y (rev zs)) @ (sust x y [a])" using `x≠a` by simp also have "... = sust x y ((rev zs) @ [a])" by (simp add: sust_append) also have "... = sust x y (rev (a#zs))" by simp finally show "rev (sust x y (a#zs)) = sust x y (rev (a#zs))" by simp qed qed text {* --------------------------------------------------------------------- Ejercicio 4. Demostrar o refutar: sust x y (sust u v zs) = sust u v (sust x y zs) --------------------------------------------------------------------- *} lemma "sust x y (sust u v zs) = sust u v (sust x y zs)" quickcheck oops text {* El contraejemplo encontrado es: u = -1 v = 0 x = -1 y = 1 zs = [-1] Efectivamente, sust (-1) 1 (sust (-1) 0 [(-1)]) = [0] sust (-1) 0 (sust (-1) 1 [(-1)]) = [1] *} text {* --------------------------------------------------------------------- Ejercicio 5. Demostrar o refutar: sust y z (sust x y zs) = sust x z zs --------------------------------------------------------------------- *} lemma "sust y z (sust x y zs) = sust x z zs" quickcheck oops text {* El contraejemplo encontrado es: x = 0 y = 1 z = 0 zs = [1] En efecto, sust 1 0 (sust 0 1 [1]) = [0] sust 0 0 [1] = [1] *} |