LMF2014: Razonamiento sobre programas con Isabelle/HOL
En la clase de hoy del curso Lógica matemática y fundamentos se ha estudiado cóm escribir programas funcionales y cómo demostrar sus propiedades 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 correspondiente es
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header {* Tema 13: Razonamiento sobre programas en Isabelle *} theory T13 imports Main begin text {* En este tema se demuestra con Isabelle las propiedades de los programas funcionales como se expone en el tema 8 del curso "Informática" que puede leerse en http://www.cs.us.es/~jalonso/cursos/i1m/temas/tema-8t.pdf *} section {* Razonamiento ecuacional *} text {* ---------------------------------------------------------------- Ejercicio 1. Definir, por recursión, la función longitud :: "'a list ⇒ nat" where tal que (longitud xs) es la longitud de la listas xs. Por ejemplo, longitud [4,2,5] = 3 ------------------------------------------------------------------- *} fun longitud :: "'a list ⇒ nat" where "longitud [] = 0" | "longitud (x#xs) = 1 + longitud xs" value "longitud [4,2,5]" -- "= 3" text {* --------------------------------------------------------------- Ejercicio 2. Demostrar que longitud [4,2,5] = 3 ------------------------------------------------------------------- *} lemma "longitud [4,2,5] = 3" by simp text {* --------------------------------------------------------------- Ejercicio 3. Definir la función fun intercambia :: "'a × 'b ⇒ 'b × 'a" tal que (intercambia p) es el par obtenido intercambiando las componentes del par p. Por ejemplo, "intercambia (2,3) = (3,2) ------------------------------------------------------------------ *} fun intercambia :: "'a × 'b ⇒ 'b × 'a" where "intercambia (x,y) = (y,x)" value "intercambia (2,3)" -- "= (3,2)" text {* --------------------------------------------------------------- Ejercicio 4. Demostrar que intercambia (intercambia (x,y)) = (x,y) ------------------------------------------------------------------- *} lemma "intercambia (intercambia (x,y)) = (x,y)" by simp text {* --------------------------------------------------------------- Ejercicio 5. Definir, por recursión, la función inversa :: "'a list ⇒ 'a list" tal que (inversa xs) es la lista obtenida invirtiendo el orden de los elementos de xs. Por ejemplo, inversa [3,2,5] = [5,2,3] ------------------------------------------------------------------ *} fun inversa :: "'a list ⇒ 'a list" where "inversa [] = []" | "inversa (x#xs) = inversa xs @ [x]" value "inversa [3,2,5]" -- "= [5,2,3]" text {* --------------------------------------------------------------- Ejercicio 6. Demostrar que inversa [x] = [x] ------------------------------------------------------------------- *} lemma "inversa [x] = [x]" by simp text {* --------------------------------------------------------------- Ejercicio 7. Definir la función repite :: "nat ⇒ 'a ⇒ 'a list" where tal que (repite n x) es la lista obtenida repitiendo n veces el elemento x. Por ejemplo, repite 3 5 = [5,5,5] ------------------------------------------------------------------ *} fun repite :: "nat ⇒ 'a ⇒ 'a list" where "repite 0 x = []" | "repite (Suc n) x = x # (repite n x)" value "repite 3 5" -- "= [5,5,5]" text {* --------------------------------------------------------------- Ejercicio 8. Demostrar que longitud (repite n x) = n ------------------------------------------------------------------- *} lemma "longitud (repite n x) = n" by (induct n) auto lemma longitud_repite: "longitud (repite n x) = n" proof (induct n) show "longitud (repite 0 x) = 0" by simp next fix n assume hi: "longitud (repite n x) = n" show "longitud (repite (Suc n) x) = Suc n" proof - have "longitud (repite (Suc n) x) = longitud (x # (repite n x))" by simp also have "... = 1 + longitud (repite n x)" by simp also have "... = 1 + n" using hi by simp also have "... = Suc n" by simp finally show ?thesis . qed qed text {* --------------------------------------------------------------- Ejercicio 9. Definir la función fun conc :: "'a list ⇒ 'a list ⇒ 'a list" tal que (conc xs ys) es la concatenación de las listas xs e ys. Por ejemplo, conc [2,3] [4,3,5] = [2,3,4,3,5] ------------------------------------------------------------------ *} fun conc :: "'a list ⇒ 'a list ⇒ 'a list" where "conc [] ys = ys" | "conc (x#xs) ys = x # (conc xs ys)" value "conc [2,3] [4,3,5]" -- "= [2,3,4,3,5]" text {* --------------------------------------------------------------- Ejercicio 10. Demostrar que conc xs (conc ys zs) = (conc xs ys) zs ------------------------------------------------------------------- *} lemma "conc xs (conc ys zs) = conc (conc xs ys) zs" by (induct xs) auto lemma conc_asociativa: "conc xs (conc ys zs) = conc (conc xs ys) zs" proof (induct xs) show "conc [] (conc ys zs) = conc (conc [] ys) zs" by simp next fix x xs assume hi: "conc xs (conc ys zs) = conc (conc xs ys) zs" show "conc (x # xs) (conc ys zs) = conc (conc (x # xs) ys) zs" proof - have "conc (x # xs) (conc ys zs) = x # (conc xs (conc ys zs))" by simp also have "... = x # (conc (conc xs ys) zs)" using hi by simp also have "... = conc (x#(conc xs ys)) zs" by simp also have "... = conc (conc (x # xs) ys) zs" by simp finally show ?thesis . qed qed text {* --------------------------------------------------------------- Ejercicio 11. Refutar que conc xs ys = conc ys xs ------------------------------------------------------------------- *} lemma "conc xs ys = conc ys xs" quickcheck oops text {* Encuentra el contraejemplo, xs = [a2] ys = [a1] *} text {* --------------------------------------------------------------- Ejercicio 12. Demostrar que conc xs [] = xs ------------------------------------------------------------------- *} lemma "conc xs [] = xs" by (induct xs) auto text {* --------------------------------------------------------------- Ejercicio 13. Demostrar que longitud (conc xs ys) = longitud xs + longitud ys ------------------------------------------------------------------- *} lemma "longitud (conc xs ys) = longitud xs + longitud ys" by (induct xs) auto lemma long_conc: "longitud (conc xs ys) = longitud xs + longitud ys" proof (induct xs) show "longitud (conc [] ys) = longitud [] + longitud ys" by simp next fix x xs assume hi: "longitud (conc xs ys) = longitud xs + longitud ys" show "longitud (conc (x # xs) ys) = longitud (x # xs) + longitud ys" proof - have "longitud (conc (x # xs) ys) = longitud (x # (conc xs ys))" by simp also have "... = 1 + longitud (conc xs ys)" by simp also have "... = 1 + (longitud xs + longitud ys)" using hi by simp also have "... = (1+ longitud xs) + longitud ys" by simp also have "... = longitud (x # xs) + longitud ys" by simp finally show ?thesis . qed qed text {* --------------------------------------------------------------- Ejercicio 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 [3,7,5,4] = [3,7] ------------------------------------------------------------------ *} 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 [3,7,5,4]" -- "= [3,7]" text {* --------------------------------------------------------------- Ejercicio 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 [3,7,5,4] = [5,4] ------------------------------------------------------------------ *} 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 [3,7,5,4]" -- "= [5,4]" text {* --------------------------------------------------------------- Ejercicio 16. Demostrar que conc (coge n xs) (elimina n xs) = xs ------------------------------------------------------------------- *} lemma "conc (coge n xs) (elimina n xs) = xs" by (induct rule: coge.induct) auto text {* coge.induct es el esquema de inducción asociado a la definición de la función coge. Puede verse como sigue: *} thm coge.induct lemma "conc (coge n xs) (elimina n xs) = xs" by (induct rule: elimina.induct) auto lemma conc_coge_elimina: "conc (coge n xs) (elimina n xs) = xs" proof (induct rule: coge.induct) show "⋀n. conc (coge n []) (elimina n []) = []" by simp next show "⋀v va. conc (coge 0 (v # va)) (elimina 0 (v # va)) = v # va" by simp next fix x xs n assume hi: "conc (coge n xs) (elimina n xs) = xs" show "conc (coge (Suc n) (x # xs)) (elimina (Suc n) (x # xs)) = x # xs" proof - have "conc (coge (Suc n) (x # xs)) (elimina (Suc n) (x # xs)) = conc (coge (Suc n) (x # xs)) (elimina n xs)" by simp also have "... = 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 ?thesis . qed qed text {* --------------------------------------------------------------- Ejercicio 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 {* --------------------------------------------------------------- Ejercicio 18. Demostrar que esVacia xs = esVacia (conc xs xs) ------------------------------------------------------------------- *} lemma "esVacia xs = esVacia (conc xs xs)" by (induct xs) auto lemma vacia_conc: "esVacia xs = esVacia (conc xs xs)" proof (induct xs) show "esVacia [] = esVacia (conc [] [])" by simp next fix x xs assume hi: "esVacia xs = esVacia (conc xs xs)" show "esVacia (x # xs) = esVacia (conc (x # xs) (x # xs))" proof - have "esVacia (conc (x # xs) (x # xs)) = esVacia (x# (conc xs (x#xs)))" by simp also have "... = esVacia (x # xs)" by simp finally show ?thesis by simp qed qed lemma vacia_conc': "esVacia xs = esVacia (conc xs xs)" proof (cases xs) case Nil thus ?thesis by simp next case Cons thus ?thesis by simp qed lemma vacia_conc'': "esVacia xs = esVacia (conc xs xs)" by (cases xs) simp_all text {* --------------------------------------------------------------- Ejercicio 19. Definir la función inversaAc :: "'a list ⇒ 'a list" tal que (inversaAc xs) es a inversa de xs calculada usando acumuladores. Por ejemplo, inversaAc [3,2,5] = [5,2,3] ------------------------------------------------------------------ *} 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 [3,2,5]" -- "= [5,2,3]" text {* --------------------------------------------------------------- Ejercicio 20. Demostrar que inversaAcAux xs ys = (inversa xs)@ys ------------------------------------------------------------------- *} lemma inversaAcAux_es_inversa: "inversaAcAux xs ys = (inversa xs)@ys" by (induct xs arbitrary: ys) auto lemma inversaAcAux_es_inversa_b: "inversaAcAux xs ys = (inversa xs)@ys" proof (induct xs arbitrary: ys) show "⋀ys. inversaAcAux [] ys = inversa [] @ ys" by simp next fix x xs zs assume hi: "⋀ys. inversaAcAux xs ys = inversa xs @ ys" show "inversaAcAux (x#xs) zs = inversa (x#xs) @ zs" proof - have "inversaAcAux (x#xs) zs = inversaAcAux xs (x#zs)" by simp also have "... = inversa xs @ (x#zs)" using hi by simp also have "... = inversa xs @ [x] @ zs" by simp also have "... = inversa (x#xs) @ zs" by simp finally show ?thesis . qed qed text {* --------------------------------------------------------------- Ejercicio 21. Demostrar que inversaAc xs = inversa xs ------------------------------------------------------------------- *} corollary "inversaAc xs = inversa xs" by (simp add: inversaAcAux_es_inversa) text {* --------------------------------------------------------------- Ejercicio 22. Definir la función sum :: "int list ⇒ int" tal que (sum xs) es la suma de los elementos de xs. Por ejemplo, sum [3,2,5] = 10 ------------------------------------------------------------------ *} fun sum :: "int list ⇒ int" where "sum [] = 0" | "sum (x#xs) = x + sum xs" value "sum [3,2,5]" -- "= 10" text {* --------------------------------------------------------------- Ejercicio 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::int,2,5]" -- "= [6,4,10]" value "map (λx. 6*x) [3::int,2,5]" -- "= [18,12,30]" text {* --------------------------------------------------------------- Ejercicio 24. Demostrar que sum (map (λx. 2*x) xs) = 2 * (sum xs) ------------------------------------------------------------------- *} lemma "sum (map (λx. 2*x) xs) = 2 * (sum xs)" by (induct xs) auto lemma sum_map: "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)" show "sum (map (λx. 2*x) (a#xs)) = 2 * (sum (a#xs))" proof - 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 ?thesis . qed qed text {* --------------------------------------------------------------- Ejercicio 25. Demostrar que longitud (map f xs) = longitud xs ------------------------------------------------------------------- *} lemma "longitud (map f xs) = longitud xs" by (induct xs) auto lemma long_map: "longitud (map f xs) = longitud xs" proof (induct xs) show "longitud (map f []) = longitud []" by simp next fix x xs assume hi: "longitud (map f xs) = longitud xs" show "longitud (map f (x#xs)) = longitud (x#xs)" proof - have "longitud (map f (x#xs)) = longitud ((f x)#(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 (x#xs)" by simp finally show ?thesis . qed qed end |