{"id":543,"date":"2021-07-09T06:00:17","date_gmt":"2021-07-09T04:00:17","guid":{"rendered":"https:\/\/www.glc.us.es\/~jalonso\/calculemus\/?p=543"},"modified":"2021-07-04T17:06:41","modified_gmt":"2021-07-04T15:06:41","slug":"potencias-de-potencias-en-monoides","status":"publish","type":"post","link":"https:\/\/www.glc.us.es\/~jalonso\/calculemus\/potencias-de-potencias-en-monoides\/","title":{"rendered":"Potencias de potencias en monoides"},"content":{"rendered":"<p>En los <a href=\"https:\/\/en.wikipedia.org\/wiki\/Monoid\">monoides<\/a> se define la potencia con exponentes naturales. En Lean la potencia x^n se caracteriza por los siguientes lemas:<\/p>\n<pre lang=\"text\">\n   pow_zero : x^0 = 1\n   pow_succ' : x^(succ n) = x * x^n\n<\/pre>\n<p>Demostrar que si M es un monoide, a \u2208 M y m, n \u2208 \u2115, entonces<\/p>\n<pre lang=\"text\">\n   a^(m * n) = (a^m)^n\n<\/pre>\n<p>Indicaci\u00f3n: Se puede usar el lema<\/p>\n<pre lang=\"text\">\n   pow_add : a^(m + n) = a^m * a^n\n<\/pre>\n<p>Para ello, completar la siguiente teor\u00eda de Lean:<\/p>\n<pre lang=\"lean\">\nimport algebra.group_power.basic\nopen monoid nat\n\nvariables {M : Type} [monoid M]\nvariable  a : M\nvariables (m n : \u2115)\n\nset_option pp.structure_projections false\n\nexample : a^(m * n) = (a^m)^n :=\nsorry\n<\/pre>\n<p>[expand title=\u00bbSoluciones con Lean\u00bb]<\/p>\n<pre lang=\"lean\">\r\nimport algebra.group_power.basic\r\nopen monoid nat\r\n\r\nvariables {M : Type} [monoid M]\r\nvariable  a : M\r\nvariables (m n : \u2115)\r\n\r\n-- Para que no use la notaci\u00f3n con puntos\r\nset_option pp.structure_projections false\r\n\r\n-- 1\u00aa demostraci\u00f3n\r\n-- ===============\r\n\r\nexample : a^(m * n) = (a^m)^n :=\r\nbegin\r\n  induction n with n HI,\r\n  { calc a^(m * 0)\r\n         = a^0             : congr_arg ((^) a) (nat.mul_zero m)\r\n     ... = 1               : pow_zero a\r\n     ... = (a^m)^0         : (pow_zero (a^m)).symm },\r\n  { calc a^(m * succ n)\r\n         = a^(m * n + m)   : congr_arg ((^) a) (nat.mul_succ m n)\r\n     ... = a^(m * n) * a^m : pow_add a (m * n) m\r\n     ... = (a^m)^n * a^m   : congr_arg (* a^m) HI\r\n     ... = (a^m)^(succ n)  : (pow_succ' (a^m) n).symm },\r\nend\r\n\r\n-- 2\u00aa demostraci\u00f3n\r\n-- ===============\r\n\r\nexample : a^(m * n) = (a^m)^n :=\r\nbegin\r\n  induction n with n HI,\r\n  { calc a^(m * 0)\r\n         = a^0             : by simp only [nat.mul_zero]\r\n     ... = 1               : by simp only [pow_zero]\r\n     ... = (a^m)^0         : by simp only [pow_zero] },\r\n  { calc a^(m * succ n)\r\n         = a^(m * n + m)   : by simp only [nat.mul_succ]\r\n     ... = a^(m * n) * a^m : by simp only [pow_add]\r\n     ... = (a^m)^n * a^m   : by simp only [HI]\r\n     ... = (a^m)^succ n    : by simp only [pow_succ'] },\r\nend\r\n\r\n-- 3\u00aa demostraci\u00f3n\r\n-- ===============\r\n\r\nexample : a^(m * n) = (a^m)^n :=\r\nbegin\r\n  induction n with n HI,\r\n  { calc a^(m * 0)\r\n         = a^0             : by simp [nat.mul_zero]\r\n     ... = 1               : by simp\r\n     ... = (a^m)^0         : by simp },\r\n  { calc a^(m * succ n)\r\n         = a^(m * n + m)   : by simp [nat.mul_succ]\r\n     ... = a^(m * n) * a^m : by simp [pow_add]\r\n     ... = (a^m)^n * a^m   : by simp [HI]\r\n     ... = (a^m)^succ n    : by simp [pow_succ'] },\r\nend\r\n\r\n-- 4\u00aa demostraci\u00f3n\r\n-- ===============\r\n\r\nexample : a^(m * n) = (a^m)^n :=\r\nbegin\r\n  induction n with n HI,\r\n  { by simp [nat.mul_zero] },\r\n  { by simp [nat.mul_succ,\r\n             pow_add,\r\n             HI,\r\n             pow_succ'] },\r\nend\r\n\r\n-- 5\u00aa demostraci\u00f3n\r\n-- ===============\r\n\r\nexample : a^(m * n) = (a^m)^n :=\r\nbegin\r\n  induction n with n HI,\r\n  { rw nat.mul_zero,\r\n    rw pow_zero,\r\n    rw pow_zero, },\r\n  { rw nat.mul_succ,\r\n    rw pow_add,\r\n    rw HI,\r\n    rw pow_succ', }\r\nend\r\n\r\n-- 6\u00aa demostraci\u00f3n\r\n-- ===============\r\n\r\nexample : a^(m * n) = (a^m)^n :=\r\nbegin\r\n  induction n with n HI,\r\n  { rw [nat.mul_zero, pow_zero, pow_zero] },\r\n  { rw [nat.mul_succ, pow_add, HI, pow_succ'] }\r\nend\r\n\r\n-- 7\u00aa demostraci\u00f3n\r\n-- ===============\r\n\r\nexample : a^(m * n) = (a^m)^n :=\r\npow_mul a m n\r\n<\/pre>\n<p>Se puede interactuar con la prueba anterior en <a href=\"https:\/\/leanprover-community.github.io\/lean-web-editor\/#url=https:\/\/raw.githubusercontent.com\/jaalonso\/Calculemus\/main\/src\/Potencias_de_potencias_en_monoides.lean\" rel=\"noopener noreferrer\" target=\"_blank\">esta sesi\u00f3n con Lean<\/a>.<\/p>\n<p>En los comentarios se pueden escribir otras soluciones, escribiendo el c\u00f3digo entre una l\u00ednea con &#60;pre lang=&quot;lean&quot;&#62; y otra con &#60;\/pre&#62;<br \/>\n[\/expand]<\/p>\n<p>[expand title=\u00bbSoluciones con Isabelle\/HOL\u00bb]<\/p>\n<pre lang=\"isar\">\r\n(* Potencias_de_potencias_en_monoides.thy\r\ntheory Potencias_de_potencias_en_monoides\r\nimports Main\r\nbegin\r\n\r\ncontext monoid_mult\r\nbegin\r\n\r\n(* 1\u00aa demostraci\u00f3n *)\r\n\r\nlemma  \"a^(m * n) = (a^m)^n\"\r\nproof (induct n)\r\n  have \"a ^ (m * 0) = a ^ 0\"\r\n    by (simp only: mult_0_right)\r\n  also have \"\u2026 = 1\"\r\n    by (simp only: power_0)\r\n  also have \"\u2026 = (a ^ m) ^ 0\"\r\n    by (simp only: power_0)\r\n  finally show \"a ^ (m * 0) = (a ^ m) ^ 0\"\r\n    by this\r\nnext\r\n  fix n\r\n  assume HI : \"a ^ (m * n) = (a ^ m) ^ n\"\r\n  have \"a ^ (m * Suc n) = a ^ (m + m * n)\"\r\n    by (simp only: mult_Suc_right)\r\n  also have \"\u2026 = a ^ m * a ^ (m * n)\"\r\n    by (simp only: power_add)\r\n  also have \"\u2026 = a ^ m * (a ^ m) ^ n\"\r\n    by (simp only: HI)\r\n  also have \"\u2026 = (a ^ m) ^ Suc n\"\r\n    by (simp only: power_Suc)\r\n  finally show \"a ^ (m * Suc n) = (a ^ m) ^ Suc n\"\r\n    by this\r\nqed\r\n\r\n(* 2\u00aa demostraci\u00f3n *)\r\n\r\nlemma  \"a^(m * n) = (a^m)^n\"\r\nproof (induct n)\r\n  have \"a ^ (m * 0) = a ^ 0\"               by simp\r\n  also have \"\u2026 = 1\"                        by simp\r\n  also have \"\u2026 = (a ^ m) ^ 0\"              by simp\r\n  finally show \"a ^ (m * 0) = (a ^ m) ^ 0\" .\r\nnext\r\n  fix n\r\n  assume HI : \"a ^ (m * n) = (a ^ m) ^ n\"\r\n  have \"a ^ (m * Suc n) = a ^ (m + m * n)\" by simp\r\n  also have \"\u2026 = a ^ m * a ^ (m * n)\"      by (simp add: power_add)\r\n  also have \"\u2026 = a ^ m * (a ^ m) ^ n\"      using HI by simp\r\n  also have \"\u2026 = (a ^ m) ^ Suc n\"          by simp\r\n  finally show \"a ^ (m * Suc n) =\r\n                (a ^ m) ^ Suc n\"           .\r\nqed\r\n\r\n(* 3\u00aa demostraci\u00f3n *)\r\n\r\nlemma  \"a^(m * n) = (a^m)^n\"\r\nproof (induct n)\r\n  case 0\r\n  then show ?case by simp\r\nnext\r\n  case (Suc n)\r\n  then show ?case by (simp add: power_add)\r\nqed\r\n\r\n(* 4\u00aa demostraci\u00f3n *)\r\n\r\nlemma  \"a^(m * n) = (a^m)^n\"\r\n  by (induct n) (simp_all add: power_add)\r\n\r\n(* 5\u00aa demostraci\u00f3n *)\r\n\r\nlemma \"a^(m * n) = (a^m)^n\"\r\n  by (simp only: power_mult)\r\n\r\nend\r\n\r\nend\r\n<\/pre>\n<p>En los comentarios se pueden escribir otras soluciones, escribiendo el c\u00f3digo entre una l\u00ednea con &#60;pre lang=&quot;isar&quot;&#62; y otra con &#60;\/pre&#62;<br \/>\n[\/expand]<\/p>\n","protected":false},"excerpt":{"rendered":"<p>En los monoides se define la potencia con exponentes naturales. En Lean la potencia x^n se caracteriza por los siguientes lemas: pow_zero : x^0 = 1 pow_succ&#8217; : x^(succ n) = x * x^n Demostrar que si M es un monoide, a \u2208 M y m, n \u2208 \u2115, entonces a^(m * n) = (a^m)^n Indicaci\u00f3n: Se puede usar el lema pow_add : a^(m + n) = a^m * a^n Para ello, completar la siguiente teor\u00eda de Lean: import algebra.group_power.basic open monoid nat variables {M : Type} [monoid M] variable a : M variables (m n : \u2115) set_option pp.structure_projections false example : a^(m * n) = (a^m)^n := sorry [expand title=\u00bbSoluciones con Lean\u00bb] import algebra.group_power.basic open monoid nat variables {M : Type} [monoid M]&#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":[9],"tags":[],"jetpack_featured_media_url":"","jetpack_sharing_enabled":true,"_links":{"self":[{"href":"https:\/\/www.glc.us.es\/~jalonso\/calculemus\/wp-json\/wp\/v2\/posts\/543"}],"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=543"}],"version-history":[{"count":3,"href":"https:\/\/www.glc.us.es\/~jalonso\/calculemus\/wp-json\/wp\/v2\/posts\/543\/revisions"}],"predecessor-version":[{"id":560,"href":"https:\/\/www.glc.us.es\/~jalonso\/calculemus\/wp-json\/wp\/v2\/posts\/543\/revisions\/560"}],"wp:attachment":[{"href":"https:\/\/www.glc.us.es\/~jalonso\/calculemus\/wp-json\/wp\/v2\/media?parent=543"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.glc.us.es\/~jalonso\/calculemus\/wp-json\/wp\/v2\/categories?post=543"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.glc.us.es\/~jalonso\/calculemus\/wp-json\/wp\/v2\/tags?post=543"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}