{"id":5856,"date":"2017-11-30T18:59:56","date_gmt":"2017-11-30T17:59:56","guid":{"rendered":"https:\/\/www.glc.us.es\/~jalonso\/vestigium\/?p=5856"},"modified":"2017-12-02T09:00:58","modified_gmt":"2017-12-02T08:00:58","slug":"ra2017-ejercicios-de-razonamiento-detallado-sobre-programas-en-isabellehol","status":"publish","type":"post","link":"https:\/\/www.glc.us.es\/~jalonso\/vestigium\/ra2017-ejercicios-de-razonamiento-detallado-sobre-programas-en-isabellehol\/","title":{"rendered":"RA2017: Ejercicios de razonamiento detallado sobre programas en Isabelle\/HOL"},"content":{"rendered":"<p>En la primera parte de la clase de hoy del curso de<br \/>\n<a href=\"http:\/\/www.cs.us.es\/~jalonso\/cursos\/m-ra-17\">Razonamiento autom\u00e1tico<\/a> se han comentado las soluciones de la 3\u00aa relaci\u00f3n de ejercicios sobre razonamiento detallado sobre programas en Isabelle\/HOL<\/p>\n<p>La teor\u00eda con las soluciones de los ejercicios es la siguiente<br \/>\n<!--more--><\/p>\n<pre lang=\"isar\">\nchapter {* R3: Razonamiento sobre programas *}\n\ntheory R3\nimports Main \nbegin\n\ntext {* --------------------------------------------------------------- \n  Ejercicio 1.1. Definir la funci\u00f3n\n     sumaImpares :: nat \u21d2 nat\n  tal que (sumaImpares n) es la suma de los n primeros n\u00fameros\n  impares. Por ejemplo,\n     sumaImpares 5  =  25\n  ------------------------------------------------------------------ *}\n\nfun sumaImpares :: \"nat \u21d2 nat\" where\n  \"sumaImpares 0 = 0\"\n| \"sumaImpares (Suc n) = sumaImpares n + (2*n+1)\"\n\ntext {* --------------------------------------------------------------- \n  Ejercicio 1.2. Escribir la demostraci\u00f3n detallada de \n     sumaImpares n = n*n\n  ------------------------------------------------------------------- *}\n\n-- \"La demostraci\u00f3n detallada es\"\nlemma \"sumaImpares n = n*n\"\nproof (induct n)\n  show \"sumaImpares 0 = 0 * 0\" by simp\nnext\n  fix n\n  assume HI: \"sumaImpares n = n * n\"\n  have \"sumaImpares (Suc n) = sumaImpares n + (2*n+1)\" by simp\n  also have \"... = n*n + (2*n+1)\" using HI by simp\n  also have \"... = Suc n * Suc n\" by simp\n  finally show \"sumaImpares (Suc n) = Suc n * Suc n\" by simp\nqed\n\n-- \"La demostraci\u00f3n autom\u00e1tica es\"\nlemma \"sumaImpares n = n*n\"\nby (induct n) auto\n\ntext {* --------------------------------------------------------------- \n  Ejercicio 2.1. Definir la funci\u00f3n\n     sumaPotenciasDeDosMasUno :: nat \u21d2 nat\n  tal que \n     (sumaPotenciasDeDosMasUno n) = 1 + 2^0 + 2^1 + 2^2 + ... + 2^n. \n  Por ejemplo, \n     sumaPotenciasDeDosMasUno 3  =  16\n  ------------------------------------------------------------------ *}\n\nfun sumaPotenciasDeDosMasUno :: \"nat \u21d2 nat\" where\n  \"sumaPotenciasDeDosMasUno 0 = 2\"\n| \"sumaPotenciasDeDosMasUno (Suc n) = \n      sumaPotenciasDeDosMasUno n + 2^(n+1)\"\n\ntext {* --------------------------------------------------------------- \n  Ejercicio 2.2. Escribir la demostraci\u00f3n detallada de \n     sumaPotenciasDeDosMasUno n = 2^(n+1)\n  ------------------------------------------------------------------- *}\n\n-- \"La demostraci\u00f3n detallada es\"\nlemma \"sumaPotenciasDeDosMasUno n = 2^(n+1)\"\nproof (induct n) \n  show \"sumaPotenciasDeDosMasUno 0 = 2^(0+1)\" by simp\nnext\n  fix n\n  assume HI: \"sumaPotenciasDeDosMasUno n = 2^(n+1)\"\n  have \"sumaPotenciasDeDosMasUno (Suc n) = \n        sumaPotenciasDeDosMasUno n + 2^(n+1)\" by simp\n  also have \"... = 2^(n+1) + 2^(n+1)\" using HI by simp\n  also have \"... = 2 ^ (Suc n + 1)\" by simp\n  finally show \"sumaPotenciasDeDosMasUno (Suc n) = 2 ^ (Suc n + 1)\"\n    by simp\nqed\n\n-- \"La demostraci\u00f3n autom\u00e1tica es\"\nlemma \"sumaPotenciasDeDosMasUno n = 2^(n+1)\"\nby (induct n) auto\n\ntext {* --------------------------------------------------------------- \n  Ejercicio 3.1. Definir la funci\u00f3n\n     copia :: nat \u21d2 'a \u21d2 'a list\n  tal que (copia n x) es la lista formado por n copias del elemento\n  x. Por ejemplo, \n     copia 3 x = [x,x,x]\n  ------------------------------------------------------------------ *}\n\nfun copia :: \"nat \u21d2 'a \u21d2 'a list\" where\n  \"copia 0 x       = []\"\n| \"copia (Suc n) x = x # copia n x\"\n\ntext {* --------------------------------------------------------------- \n  Ejercicio 3.2. Definir la funci\u00f3n\n     todos :: ('a \u21d2 bool) \u21d2 'a list \u21d2 bool\n  tal que (todos p xs) se verifica si todos los elementos de xs cumplen\n  la propiedad p. Por ejemplo,\n     todos (\u03bbx. x>(1::nat)) [2,6,4] = True\n     todos (\u03bbx. x>(2::nat)) [2,6,4] = False\n  ------------------------------------------------------------------ *}\n\nfun todos :: \"('a \u21d2 bool) \u21d2 'a list \u21d2 bool\" where\n  \"todos p []     = True\"\n| \"todos p (x#xs) = (p x \u2227 todos p xs)\"\n\nvalue \"todos (\u03bbx. x>(1::nat)) [2,6,4]\" -- \"= True\"\nvalue \"todos (\u03bbx. x>(2::nat)) [2,6,4]\" -- \"= False\"\n\ntext {* --------------------------------------------------------------- \n  Ejercicio 3.2. Demostrar detalladamente que todos los elementos de\n  (copia n x) son iguales a x. \n  ------------------------------------------------------------------- *}\n\n-- \"La demostraci\u00f3n detallada es\"\nlemma \"todos (\u03bby. y=x) (copia n x)\"\nproof (induct n)\n  show \"todos (\u03bby. y = x) (copia 0 x)\" by simp\nnext\n  fix n\n  assume HI: \"todos (\u03bby. y = x) (copia n x)\" \n  have \"todos (\u03bby. y = x) (copia (Suc n) x) = \n        todos (\u03bby. y = x) (x # copia n x)\" by simp\n  also have \"... = (x = x \u2227 todos (\u03bby. y = x) (copia n x))\" \n    by simp\n  also have \"... = True\" using HI by simp\n  finally show \"todos (\u03bby. y = x) (copia (Suc n) x)\" by simp\nqed\n\n-- \"La demostraci\u00f3n autom\u00e1tica es\"\nlemma \"todos (\u03bby. y=x) (copia n x)\"\nby (induct n) auto\n\ntext {* --------------------------------------------------------------- \n  Ejercicio 4.1. Definir la funci\u00f3n\n    factR :: nat \u21d2 nat\n  tal que (factR n) es el factorial de n. Por ejemplo,\n    factR 4 = 24\n  ------------------------------------------------------------------ *}\n\nfun factR :: \"nat \u21d2 nat\" where\n  \"factR 0       = 1\"\n| \"factR (Suc n) = Suc n * factR n\"\n\ntext {* --------------------------------------------------------------- \n  Ejercicio 4.2. Se considera la siguiente definici\u00f3n iterativa de la\n  funci\u00f3n factorial \n     factI :: \"nat \u21d2 nat\" where\n     factI n = factI' n 1\n     \n     factI' :: nat \u21d2 nat \u21d2 nat\" where\n     factI' 0       x = x\n     factI' (Suc n) x = factI' n (Suc n)*x\n  Demostrar que, para todo n y todo x, se tiene \n     factI' n x = x * factR n\n  Indicaci\u00f3n: La propiedad mult_Suc es \n     (Suc m) * n = n + m * n\n  Puede que se necesite desactivarla en un paso con \n     (simp del: mult_Suc)\n  ------------------------------------------------------------------- *}\n\nfun factI' :: \"nat \u21d2 nat \u21d2 nat\" where\n  \"factI' 0       x = x\"\n| \"factI' (Suc n) x = factI' n (x * Suc n)\"\n\nfun factI :: \"nat \u21d2 nat\" where\n  \"factI n = factI' n 1\"\n\n-- \"La demostraci\u00f3n detallada es\"     \nlemma fact: \"factI' n x = x * factR n\"\nproof (induct n arbitrary: x)\n  show \"\u22c0x. factI' 0 x = x * factR 0\" by simp\nnext\n  fix n\n  assume HI: \"\u22c0x. factI' n x = x * factR n\"\n  show \"\u22c0x. factI' (Suc n) x = x * factR (Suc n)\"\n  proof -\n    fix x\n    have \"factI' (Suc n) x = factI' n (x * Suc n)\" by simp\n    also have \"... = (x * Suc n) * factR n\" using HI by simp\n    also have \"... = x * (Suc n * factR n)\" by (simp del: mult_Suc)\n    also have \"... = x * factR (Suc n)\" by simp\n    finally show \"factI' (Suc n) x = x * factR (Suc n)\" by simp\n  qed\nqed\n\n-- \"La demostraci\u00f3n autom\u00e1tica es\"     \nlemma \"factI' n x = x * factR n\"\nby (induct n arbitrary: x) \n   (auto simp del: mult_Suc)\n\ntext {* --------------------------------------------------------------- \n  Ejercicio 4.3. Escribir la demostraci\u00f3n detallada de\n     factI n = factR n\n  ------------------------------------------------------------------- *}\n\ncorollary \"factI n = factR n\"\nby (simp add: fact)\n\ntext {* --------------------------------------------------------------- \n  Ejercicio 5.1. Definir, recursivamente y sin usar (@), la funci\u00f3n\n     amplia :: 'a list \u21d2 'a \u21d2 'a list\n  tal que (amplia xs y) es la lista obtenida a\u00f1adiendo el elemento y al\n  final de la lista xs. Por ejemplo,\n     amplia [d,a] t = [d,a,t]\n  ------------------------------------------------------------------ *}\n\nfun amplia :: \"'a list \u21d2 'a \u21d2 'a list\" where\n  \"amplia []     y = [y]\"\n| \"amplia (x#xs) y = x # amplia xs y\"\n\ntext {* --------------------------------------------------------------- \n  Ejercicio 5.2. Escribir la demostraci\u00f3n detallada de\n     amplia xs y = xs @ [y]\n  ------------------------------------------------------------------- *}\n\n-- \"La demostraci\u00f3n detallada es\"\nlemma \"amplia xs y = xs @ [y]\"\nproof (induct xs)\n  show \"amplia [] y = [] @ [y]\" by simp\nnext\n  fix x xs\n  assume HI: \"amplia xs y = xs @ [y]\"\n  have \"amplia (x # xs) y = x # amplia xs y\" by simp\n  also have \"... = x # (xs @ [y])\" using HI by simp\n  also have \"... = (x # xs) @ [y]\" by simp\n  finally show \"amplia (x # xs) y = (x # xs) @ [y]\" by simp\nqed\n\n-- \"La demostraci\u00f3n autom\u00e1tica es\"\nlemma \"amplia xs y = xs @ [y]\"\nby (induct xs) auto\n\nend\n<\/pre>\n","protected":false},"excerpt":{"rendered":"<p>En la primera parte de la clase de hoy del curso de Razonamiento autom\u00e1tico se han comentado las soluciones de la 3\u00aa relaci\u00f3n de ejercicios sobre razonamiento detallado sobre programas en Isabelle\/HOL La teor\u00eda con las soluciones de los ejercicios es la siguiente<\/p>\n","protected":false},"author":2,"featured_media":0,"comment_status":"closed","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"jetpack_post_was_ever_published":false,"_kad_post_transparent":"","_kad_post_title":"","_kad_post_layout":"","_kad_post_sidebar_id":"","_kad_post_content_style":"","_kad_post_vertical_padding":"","_kad_post_feature":"","_kad_post_feature_position":"","_kad_post_header":false,"_kad_post_footer":false,"_jetpack_newsletter_access":"","_jetpack_dont_email_post_to_subs":false,"_jetpack_newsletter_tier_id":0,"_jetpack_memberships_contains_paywalled_content":false,"footnotes":"","_jetpack_memberships_contains_paid_content":false},"categories":[266],"tags":[144,317],"jetpack_featured_media_url":"","jetpack_sharing_enabled":true,"jetpack_likes_enabled":false,"_links":{"self":[{"href":"https:\/\/www.glc.us.es\/~jalonso\/vestigium\/wp-json\/wp\/v2\/posts\/5856"}],"collection":[{"href":"https:\/\/www.glc.us.es\/~jalonso\/vestigium\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/www.glc.us.es\/~jalonso\/vestigium\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/www.glc.us.es\/~jalonso\/vestigium\/wp-json\/wp\/v2\/users\/2"}],"replies":[{"embeddable":true,"href":"https:\/\/www.glc.us.es\/~jalonso\/vestigium\/wp-json\/wp\/v2\/comments?post=5856"}],"version-history":[{"count":1,"href":"https:\/\/www.glc.us.es\/~jalonso\/vestigium\/wp-json\/wp\/v2\/posts\/5856\/revisions"}],"predecessor-version":[{"id":5857,"href":"https:\/\/www.glc.us.es\/~jalonso\/vestigium\/wp-json\/wp\/v2\/posts\/5856\/revisions\/5857"}],"wp:attachment":[{"href":"https:\/\/www.glc.us.es\/~jalonso\/vestigium\/wp-json\/wp\/v2\/media?parent=5856"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.glc.us.es\/~jalonso\/vestigium\/wp-json\/wp\/v2\/categories?post=5856"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.glc.us.es\/~jalonso\/vestigium\/wp-json\/wp\/v2\/tags?post=5856"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}