RA2012: Razonamiento sobre programas con Isabelle/HOL (2)
En la primera parte de la clase de hoy del curso de Razonamiento automático se ha continuado la presentación (iniciada en la clase anterior) de cómo se puede demostrar propiedades de programas funcionales con Isabelle/HOL.
En la presentación se han usado los ejemplos del tema 8 del curso de Informática (de 1º del Grado en Matemáticas).
La teoría con los ejemplos presentados en la clase es la siguiente:
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header {* Tema 5: Razonamiento sobre programas *} theory T5 imports Main begin section {* Inducción correspondiente a una 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 -- "La demostración automática es" 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) ------------------------------------------------------------------- *} -- "La demostración estructurada es" lemma "esVacia xs = esVacia (conc xs xs)" proof (cases xs) assume "xs = []" thus "esVacia xs = esVacia (conc xs xs)" by simp next fix y ys assume "xs = y#ys" thus "esVacia xs = esVacia (conc xs xs)" by simp qed -- "La demostración estructurada simplificad es" lemma "esVacia xs = esVacia (conc xs xs)" proof (cases xs) case Nil thus "esVacia xs = esVacia (conc xs xs)" by simp next case Cons thus "esVacia xs = esVacia (conc xs xs)" by simp qed -- "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 -- "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) 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 |