{"id":165,"date":"2020-01-24T12:54:59","date_gmt":"2020-01-24T10:54:59","guid":{"rendered":"https:\/\/www.glc.us.es\/~jalonso\/calculemus\/?p=165"},"modified":"2021-08-21T12:58:52","modified_gmt":"2021-08-21T10:58:52","slug":"reto-f-f-f-b-f-b","status":"publish","type":"post","link":"https:\/\/www.glc.us.es\/~jalonso\/calculemus\/reto-f-f-f-b-f-b\/","title":{"rendered":"Reto f (f (f b)) = f b"},"content":{"rendered":"<p>El problema de hoy est\u00e1 basado en el reto <a href=\"https:\/\/dev.to\/thepracticaldev\/daily-challenge-168-code-golf-f-f-f-b-f-b-1nd9\">Daily Challenge #168<\/a> que se plante\u00f3 ayer. El reto consiste en demostrar que si <code>f<\/code> es una funci\u00f3n de los booleanos en los booleanos, entonces para todo <code>b<\/code> se verifica que <code>f (f (f b)) = f b<\/code>.<\/p>\n<p>Concretamente, el reto consiste en completar la siguiente demostraci\u00f3n<\/p>\n<pre lang=\"isar\">\nlemma \n  fixes f :: \"bool \u21d2 bool\"\n  shows \"f (f (f b)) = f b\"\n  sorry\n<\/pre>\n<h4>Soluciones con Isabelle\/HOL<\/h4>\n<pre lang=\"isar\">\ntheory reto\n\nimports Main\nbegin\n\n(* 1\u00aa soluci\u00f3n *)\n\nlemma \"\u2200(f :: bool \u21d2 bool). f (f (f b)) = f b\"\n  by smt\n\n(* 2\u00aa soluci\u00f3n *)\n\nlemma \n  fixes f :: \"bool \u21d2 bool\"\n  shows \"f (f (f b)) = f b\"\n  by smt\n\n(* 3\u00aa soluci\u00f3n *)\n\nlemma\n  fixes f :: \"bool \u21d2 bool\"\n  shows \"f (f (f b)) = f b\"\n  by (cases b; cases \"f True\"; cases \"f False\"; simp)\n\n(* 4\u00aa soluci\u00f3n *)\n\nlemma\n  fixes f :: \"bool \u21d2 bool\"\n  shows \"f (f (f b)) = f b\"\nproof (cases \"f True\"; cases \"f False\"; cases b)\n  assume \"f True\" \"f False\" \"b\"\n  then have \"f b = True\" by simp\n  then show \"f (f (f b)) = f b\" using \u2039f True\u203a by simp \nnext\n  assume \"f True\" \"f False\" \"\u00ac b\"\n  then have \"f b = True\" by simp\n  then show \"f (f (f b)) = f b\" using \u2039f True\u203a \u2039f False\u203a \u2039\u00ac b\u203a by simp \nnext\n  assume \"f True\" \"\u00ac f False\" \"b\"\n  then have \"f b = True\" by simp\n  then show \"f (f (f b)) = f b\" using \u2039f True\u203a \u2039\u00ac f False\u203a \u2039b\u203a by simp \nnext\n  assume \"f True\" \"\u00ac f False\" \"\u00ac b\"\n  then have \"f b = False\" by simp\n  then show \"f (f (f b)) = f b\" using \u2039f True\u203a \u2039\u00ac f False\u203a \u2039\u00ac b\u203a by simp \nnext \n  assume \"\u00ac f True\" \"f False\" \"b\"\n  then have \"f b = False\" by simp\n  then show \"f (f (f b)) = f b\" using \u2039\u00ac f True\u203a \u2039f False\u203a \u2039b\u203a by simp \nnext\n  assume \"\u00ac f True\" \"f False\" \"\u00ac b\"\n  then have \"f b = True\" by simp\n  then show \"f (f (f b)) = f b\" using \u2039\u00ac f True\u203a \u2039f False\u203a \u2039\u00ac b\u203a by simp \nnext\n  assume \"\u00ac f True\" \"\u00ac f False\" \"b\"\n  then have \"f b = False\" by simp\n  then show \"f (f (f b)) = f b\" using \u2039\u00ac f True\u203a \u2039\u00ac f False\u203a \u2039b\u203a by simp \nnext\n  assume \"\u00ac f True\" \"\u00ac f False\" \"\u00ac b\"\n  then have \"f b = False\" by simp\n  then show \"f (f (f b)) = f b\" using \u2039\u00ac f True\u203a \u2039\u00ac f False\u203a \u2039\u00acb\u203a by simp \nqed\n\n(* 5\u00aa soluci\u00f3n *)\n\nlemma\n  fixes f :: \"bool \u21d2 bool\"\n  shows \"f (f (f b)) = f b\"\nproof (cases \"f True\"; cases \"f False\"; cases b)\n  assume \"f True\" \"f False\" \"b\"\n  have \"f (f (f b)) = f (f (f True))\" using \u2039b\u203a by simp\n  also have \"\u2026 = f (f True)\" using \u2039f True\u203a by simp\n  also have \"\u2026 = f True\" using \u2039f True\u203a by simp\n  also have \"\u2026 = f b\" using \u2039b\u203a by simp\n  finally show \"f (f (f b)) = f b\" by this \nnext\n  assume \"f True\" \"f False\" \"\u00ac b\"\n  have \"f (f (f b)) = f (f (f False))\" using \u2039\u00acb\u203a by simp\n  also have \"\u2026 = f (f True)\" using \u2039f False\u203a by simp\n  also have \"\u2026 = f True\" using \u2039f True\u203a by simp\n  also have \"\u2026 = f False\" using \u2039f True\u203a \u2039f False\u203a by simp\n  also have \"\u2026 = f b\" using \u2039\u00ac b\u203a by simp\n  finally show \"f (f (f b)) = f b\" by this \nnext\n  assume \"f True\" \"\u00ac f False\" \"b\"\n  have \"f (f (f b)) = f (f (f True))\" using \u2039b\u203a by simp\n  also have \"\u2026 = f (f True)\" using \u2039f True\u203a by simp\n  also have \"\u2026 = f True\" using \u2039f True\u203a by simp\n  also have \"\u2026 = f b\" using \u2039b\u203a by simp\n  finally show \"f (f (f b)) = f b\" by this \nnext\n  assume \"f True\" \"\u00ac f False\" \"\u00ac b\"\n  have \"f (f (f b)) = f (f (f False))\" using \u2039\u00acb\u203a by simp\n  also have \"\u2026 = f (f False)\" using \u2039\u00ac f False\u203a by simp\n  also have \"\u2026 = f False\" using \u2039\u00ac f False\u203a by simp\n  also have \"\u2026 = f b\" using \u2039\u00ac b\u203a by simp\n  finally show \"f (f (f b)) = f b\" by this \nnext \n  assume \"\u00ac f True\" \"f False\" \"b\"\n  have \"f (f (f b)) = f (f (f True))\" using \u2039b\u203a by simp\n  also have \"\u2026 = f (f False)\" using \u2039\u00ac f True\u203a by simp\n  also have \"\u2026 = f True\" using \u2039f False\u203a by simp\n  also have \"\u2026 = f b\" using \u2039b\u203a by simp\n  finally show \"f (f (f b)) = f b\" by this \nnext\n  assume \"\u00ac f True\" \"f False\" \"\u00ac b\"\n  have \"f (f (f b)) = f (f (f False))\" using \u2039\u00acb\u203a by simp\n  also have \"\u2026 = f (f True)\" using \u2039f False\u203a by simp\n  also have \"\u2026 = f False\" using \u2039\u00ac f True\u203a by simp\n  also have \"\u2026 = f b\" using \u2039\u00ac b\u203a by simp\n  finally show \"f (f (f b)) = f b\" by this \nnext\n  assume \"\u00ac f True\" \"\u00ac f False\" \"b\"\n  have \"f (f (f b)) = f (f (f True))\" using \u2039b\u203a by simp\n  also have \"\u2026 = f (f False)\" using \u2039\u00ac f True\u203a by simp\n  also have \"\u2026 = f False\" using \u2039\u00ac f False\u203a by simp\n  also have \"\u2026 = False\" using \u2039\u00ac f False\u203a by simp\n  also have \"\u2026 = f True\" using \u2039\u00ac f True\u203a by simp\n  also have \"\u2026 = f b\" using \u2039b\u203a by simp\n  finally show \"f (f (f b)) = f b\" by this \nnext\n  assume \"\u00ac f True\" \"\u00ac f False\" \"\u00ac b\"\n  have \"f (f (f b)) = f (f (f False))\" using \u2039\u00acb\u203a by simp\n  also have \"\u2026 = f (f False)\" using \u2039\u00ac f False\u203a by simp\n  also have \"\u2026 = f False\" using \u2039\u00ac f False\u203a by simp\n  also have \"\u2026 = f b\" using \u2039\u00ac b\u203a by simp\n  finally show \"f (f (f b)) = f b\" by this \nqed\n\n(* 6\u00aa soluci\u00f3n *)\n\ntheorem\n  fixes f :: \"bool \u21d2 bool\"\n  shows \"f (f (f b)) = f b\" \nproof (cases b)\n  assume b: b\n  show ?thesis\n  proof (cases \"f True\")\n    assume ft: \"f True\"\n    show ?thesis\n      using b ft by auto\n  next\n    assume ft: \"\u00ac f True\"\n    show ?thesis\n    proof (cases \"f False\")\n      assume ff: \"f False\"\n      show ?thesis\n        using b ft ff by auto\n    next\n      assume ff: \"\u00ac f False\"\n      show ?thesis\n        using b ft ff by auto\n    qed\n  qed\nnext\n  assume b: \"\u00ac b\"\n  show ?thesis\n  proof (cases \"f True\")\n    assume ft: \"f True\"\n    show ?thesis\n    proof (cases \"f False\")\n      assume ff: \"f False\"\n      show ?thesis\n        using b ft ff by auto\n    next\n      assume ff: \"\u00ac f False\"\n      show ?thesis\n        using b ft ff by auto\n    qed\n  next\n    assume ft: \"\u00ac f True\"\n    show ?thesis\n    proof (cases \"f False\")\n      assume ff: \"f False\"\n      show ?thesis\n        using b ft ff by auto\n    next\n      assume ff: \"\u00ac f False\"\n      show ?thesis\n        using b ft ff by auto\n    qed\n  qed\nqed\n\n(* 7\u00aa soluci\u00f3n *)\n\ntheorem\n  fixes f :: \"bool \u21d2 bool\"\n  shows \"f (f (f b)) = f b\"\n  by (cases b \"f True\" \"f False\"\n      rule: bool.exhaust [ case_product bool.exhaust\n                         , case_product bool.exhaust])\n    auto\n\nend\n<\/pre>\n<h4>Otras soluciones<\/h4>\n<ul>\n<li>Se pueden escribir otras soluciones en los comentarios.\n<li>El c\u00f3digo se debe escribir entre una l\u00ednea con &#60;pre lang=&quot;isar&quot;&#62; y otra con &#60;\/pre&#62;\n<\/ul>\n","protected":false},"excerpt":{"rendered":"<p>El problema de hoy est\u00e1 basado en el reto Daily Challenge #168 que se plante\u00f3 ayer. El reto consiste en demostrar que si f es una funci\u00f3n de los booleanos en los booleanos, entonces para todo b se verifica que f (f (f b)) = f b. Concretamente, el reto consiste en completar la siguiente demostraci\u00f3n lemma fixes f :: \u00abbool \u21d2 bool\u00bb shows \u00abf (f (f b)) = f b\u00bb sorry Soluciones con Isabelle\/HOL theory reto imports Main begin (* 1\u00aa soluci\u00f3n *) lemma \u00ab\u2200(f :: bool \u21d2 bool). f (f (f b)) = f b\u00bb by smt (* 2\u00aa soluci\u00f3n *) lemma fixes f :: \u00abbool \u21d2 bool\u00bb shows \u00abf (f (f b)) = f b\u00bb by smt (* 3\u00aa soluci\u00f3n *) lemma&#8230;<\/p>\n","protected":false},"author":1,"featured_media":0,"comment_status":"open","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,"_jetpack_memberships_contains_paid_content":false,"footnotes":""},"categories":[106],"tags":[],"jetpack_featured_media_url":"","jetpack_sharing_enabled":true,"_links":{"self":[{"href":"https:\/\/www.glc.us.es\/~jalonso\/calculemus\/wp-json\/wp\/v2\/posts\/165"}],"collection":[{"href":"https:\/\/www.glc.us.es\/~jalonso\/calculemus\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/www.glc.us.es\/~jalonso\/calculemus\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/www.glc.us.es\/~jalonso\/calculemus\/wp-json\/wp\/v2\/users\/1"}],"replies":[{"embeddable":true,"href":"https:\/\/www.glc.us.es\/~jalonso\/calculemus\/wp-json\/wp\/v2\/comments?post=165"}],"version-history":[{"count":18,"href":"https:\/\/www.glc.us.es\/~jalonso\/calculemus\/wp-json\/wp\/v2\/posts\/165\/revisions"}],"predecessor-version":[{"id":273,"href":"https:\/\/www.glc.us.es\/~jalonso\/calculemus\/wp-json\/wp\/v2\/posts\/165\/revisions\/273"}],"wp:attachment":[{"href":"https:\/\/www.glc.us.es\/~jalonso\/calculemus\/wp-json\/wp\/v2\/media?parent=165"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.glc.us.es\/~jalonso\/calculemus\/wp-json\/wp\/v2\/categories?post=165"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.glc.us.es\/~jalonso\/calculemus\/wp-json\/wp\/v2\/tags?post=165"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}