RA2019: Ejercicios de razonamiento estructurado sobre programas en Isabelle/HOL
En la segunda parte de la clase de hoy del curso de Razonamiento automático se han comentado las soluciones de la 3ª relación de ejercicios de razonamiento estructurado sobre programas. 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 ‹R3: Razonamiento estructurado sobre programas› theory R3_Razonamiento_estructurado_sobre_programas_sol imports Main begin text ‹Nota: De cada propiedad se debe de escribir primero la demostración automática, a continuación la estructurada y finalmente la detallada (en la que se indica todos los detalles usando sólo "by simp only: ..." y "by this").› 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 automática es› lemma "sumaImpares n = n*n" by (induct n) simp_all ― ‹La demostración estructurada es› 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 ― ‹La demostración detallada es› lemma "sumaImpares n = n*n" proof (induct n) have "sumaImpares 0 = 0" by (simp only: sumaImpares.simps(1)) also have "... = 0*0" by (simp only: mult_0) finally show "sumaImpares 0 = 0*0" by this 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)" by (simp only: HI) also have "... = n*(n+1)+1*(n+1)" by (simp only: add_mult_distrib2) also have "... = (n+1)*(n+1)" by (simp only: add_mult_distrib) also have "... = (Suc n)*(Suc n)" by (simp only: Suc_eq_plus1) finally show "sumaImpares (Suc n) = (Suc n)*(Suc n)" by this 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) -------------------------------------------------------------------› ― ‹La demostración automática es› lemma "sumaPotenciasDeDosMasUno n = 2^(n+1)" by (induct n) simp_all ― ‹La demostración estructurada es› 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 ― ‹La demostración detallada es› lemma "sumaPotenciasDeDosMasUno n = 2^(n+1)" proof (induct n) have "sumaPotenciasDeDosMasUno 0 = 2" by (simp only: sumaPotenciasDeDosMasUno.simps(1)) also have "... = 2^1" by (simp only: monoid_mult_class.power_one_right) also have "... = 2^(0+1)" by (simp only: add_0) finally show "sumaPotenciasDeDosMasUno 0 = 2^(0+1)" by this 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)" by (simp only: HI) also have "... = 2^(n+1)*2" by (simp only: mult_2_right) also have "... = 2^(Suc(n+1))" by (simp only: power_Suc2) also have "... = 2^((Suc n)+1)" by (simp only: Suc_eq_plus1) finally show "sumaPotenciasDeDosMasUno (Suc n) = 2^((Suc n)+1)" by this 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)" value "todos (λx. x>(1::nat)) [2,6,4] = True" value "todos (λx. x>(2::nat)) [2,6,4] = False" text ‹--------------------------------------------------------------- Ejercicio 3.2. Demostrar detalladamente que todos los elementos de (copia n x) son iguales a x. -------------------------------------------------------------------› ― ‹La demostración automática es› lemma "todos (λy. y=x) (copia n x)" by (induct n) simp_all ― ‹La demostración estructurada es› 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 "todos (λy. y = x) (copia n x)" then show "todos (λy. y = x) (copia (Suc n) x)" by simp qed ― ‹La demostración detallada es› lemma "todos (λy. y=x) (copia n x)" proof (induct n) have "todos (λy. y=x) []" by (simp only: todos.simps(1)) then show "todos (λy. y=x) (copia 0 x)" by (simp only: copia.simps(1)) next fix n assume HI: "todos (λy. y = x) (copia n x)" then have "todos (λy. y = x) (x # copia n x)" by (simp only: todos.simps(2)) then show "todos (λy. y = x) (copia (Suc n) x)" by (simp only: copia.simps(2)) 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" ― ‹La demostración automática es› lemma "factI' n x = x * factR n" by (induct n arbitrary: x) (auto simp del: mult_Suc) ― ‹La demostración estructurada es› lemma "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 ― ‹La demostración detallada es› lemma fact: "factI' n x = x * factR n" proof (induct n arbitrary: x) fix x have "factI' 0 x = x" by (simp only: factI'.simps(1)) also have "... = x * 1" by (simp only: mult_1_right) also have "... = x * factR 0" by (simp only: factR.simps(1)) finally show "factI' 0 x= x* factR 0" by this 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 only: factI'.simps(2)) also have "... = (x*Suc n)*factR n" by (simp only: HI) also have "... = x*(Suc n*factR n)" by (simp only: mult.assoc) also have "... = x*factR (Suc n)" by (simp only: factR.simps(2)) finally show "factI' (Suc n) x= x*factR (Suc n)" by this qed qed text ‹--------------------------------------------------------------- Ejercicio 4.3. Escribir la demostración detallada de factI n = factR n -------------------------------------------------------------------› ― ‹La demostración automática es› corollary "factI n = factR n" by (simp add: fact) ― ‹La demostración estructurada es› corollary "factI n = factR n" proof - have "factI n = factI' n 1" by simp also have "… = 1 * factR n" by (simp add: fact) also have "… = factR n" by simp finally show "factI n = factR n" by this qed ― ‹La demostración detallada es› corollary "factI n = factR n" proof - have "factI n = factI' n 1" by (simp only: factI.simps) also have "… = 1 * factR n" by (simp only: fact) also have "… = factR n" by (simp only: nat_mult_1) finally show "factI n = factR n" by this 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] -------------------------------------------------------------------› ― ‹La demostración automática es› lemma "amplia xs y = xs @ [y]" by (induct xs) simp_all ― ‹La demostración estructurada es› 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 ― ‹La demostración detallada es› lemma "amplia xs y = xs @ [y]" proof (induct xs) have "amplia [] y = [y]" by (simp only: amplia.simps(1)) also have "... = [] @ [y]" by (simp only: append_Nil) finally show "amplia [] y = [] @ [y]" by this 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])" by (simp only: HI) also have "... = (x # xs) @ [y]" by (simp only: append_Cons) finally show "amplia (x#xs) y = (x # xs) @ [y]" by this qed end |