{"id":518,"date":"2021-07-03T06:00:05","date_gmt":"2021-07-03T04:00:05","guid":{"rendered":"https:\/\/www.glc.us.es\/~jalonso\/calculemus\/?p=518"},"modified":"2021-06-24T17:05:58","modified_gmt":"2021-06-24T15:05:58","slug":"caracterizacion-de-producto-igual-al-primer-factor","status":"publish","type":"post","link":"https:\/\/www.glc.us.es\/~jalonso\/calculemus\/caracterizacion-de-producto-igual-al-primer-factor\/","title":{"rendered":"Caracterizaci\u00f3n de producto igual al primer factor"},"content":{"rendered":"<p>Un <a href=\"https:\/\/bit.ly\/3j9S0wt\">monoide cancelativo por la izquierda<\/a> es un <a href=\"https:\/\/bit.ly\/3h4notA\">monoide<\/a> M que cumple la propiedad cancelativa por la izquierda; es decir, para todo a, b \u2208 M<\/p>\n<pre lang=\"text\">\n   a * b = a * c \u2194 b = c.\n<\/pre>\n<p>En Lean la clase de los monoides cancelativos por la izquierda es left&#95;cancel&#95;monoid y la propiedad cancelativa por la izquierda es<\/p>\n<pre lang=\"text\">\n   mul_left_cancel_iff : a * b = a * c \u2194 b = c\n<\/pre>\n<p>Demostrar que si M es un monoide cancelativo por la izquierda y a, b \u2208 M, entonces<\/p>\n<pre lang=\"text\">\n   a * b = a \u2194 b = 1\n<\/pre>\n<p>Para ello, completar la siguiente teor\u00eda de Lean:<\/p>\n<pre lang=\"lean\">\nimport algebra.group.basic\n\nuniverse  u\nvariables {M : Type u} [left_cancel_monoid M]\nvariables {a b : M}\n\nexample : a * b = a \u2194 b = 1 :=\nsorry\n<\/pre>\n<p>[expand title=\u00bbSoluciones con Lean\u00bb]<\/p>\n<pre lang=\"lean\">\r\nimport algebra.group.basic\r\n\r\nuniverse  u\r\nvariables {M : Type u} [left_cancel_monoid M]\r\nvariables {a b : M}\r\n\r\n-- ?\u00aa demostraci\u00f3n\r\n-- ===============\r\n\r\nexample : a * b = a \u2194 b = 1 :=\r\nbegin\r\n  split,\r\n  { intro h,\r\n    rw \u2190 @mul_left_cancel_iff _ _ a b 1,\r\n    rw mul_one,\r\n    exact h, },\r\n  { intro h,\r\n    rw h,\r\n    exact mul_one a, },\r\nend\r\n\r\n-- ?\u00aa demostraci\u00f3n\r\n-- ===============\r\n\r\nexample : a * b = a \u2194 b = 1 :=\r\ncalc a * b = a \u2194 a * b = a * 1 : by rw mul_one\r\n           ... \u2194 b = 1         : mul_left_cancel_iff\r\n\r\n-- ?\u00aa demostraci\u00f3n\r\n-- ===============\r\n\r\nexample : a * b = a \u2194 b = 1 :=\r\nmul_right_eq_self\r\n\r\n-- ?\u00aa demostraci\u00f3n\r\n-- ===============\r\n\r\nexample : a * b = a \u2194 b = 1 :=\r\nby finish\r\n<\/pre>\n<p>Se puede interactuar con la prueba anterior en <a href=\"https:\/\/www.cs.us.es\/~jalonso\/lean-web-editor\/#url=https:\/\/raw.githubusercontent.com\/jaalonso\/Calculemus\/main\/src\/Caracterizacion_de_producto_igual_al_primer_factor.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;isar&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\ntheory Caracterizacion_de_producto_igual_al_primer_factor\r\nimports Main\r\nbegin\r\n\r\ncontext cancel_comm_monoid_add\r\nbegin\r\n\r\n(* 1\u00aa demostraci\u00f3n *)\r\n\r\nlemma \"a + b = a \u27f7 b = 0\"\r\nproof (rule iffI)\r\n  assume \"a + b = a\"\r\n  then have \"a + b = a + 0\"     by (simp only: add_0_right)\r\n  then show \"b = 0\"             by (simp only: add_left_cancel)\r\nnext\r\n  assume \"b = 0\"\r\n  have \"a + 0 = a\"              by (simp only: add_0_right)\r\n  with \u2039b = 0\u203a show \"a + b = a\" by (rule ssubst)\r\nqed\r\n\r\n(* 2\u00aa demostraci\u00f3n *)\r\n\r\nlemma \"a + b = a \u27f7 b = 0\"\r\nproof\r\n  assume \"a + b = a\"\r\n  then have \"a + b = a + 0\" by simp\r\n  then show \"b = 0\"         by simp\r\nnext\r\n  assume \"b = 0\"\r\n  have \"a + 0 = a\"          by simp\r\n  then show \"a + b = a\"     using \u2039b = 0\u203a by simp\r\nqed\r\n\r\n(* 3\u00aa demostraci\u00f3n *)\r\n\r\nlemma \"a + b = a \u27f7 b = 0\"\r\nproof -\r\n  have \"(a + b = a) \u27f7 (a + b = a + 0)\" by (simp only: add_0_right)\r\n  also have \"\u2026 \u27f7 (b = 0)\"              by (simp only: add_left_cancel)\r\n  finally show \"a + b = a \u27f7 b = 0\"     by this\r\nqed\r\n\r\n(* 4\u00aa demostraci\u00f3n *)\r\n\r\nlemma \"a + b = a \u27f7 b = 0\"\r\nproof -\r\n  have \"(a + b = a) \u27f7 (a + b = a + 0)\" by simp\r\n  also have \"\u2026 \u27f7 (b = 0)\"              by simp\r\n  finally show \"a + b = a \u27f7 b = 0\"     .\r\nqed\r\n\r\n(* 5\u00aa demostraci\u00f3n *)\r\n\r\nlemma \"a + b = a \u27f7 b = 0\"\r\n  by (simp only: add_cancel_left_right)\r\n\r\n(* 6\u00aa demostraci\u00f3n *)\r\n\r\nlemma \"a + b = a \u27f7 b = 0\"\r\n  by auto\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>Un monoide cancelativo por la izquierda es un monoide M que cumple la propiedad cancelativa por la izquierda; es decir, para todo a, b \u2208 M a * b = a * c \u2194 b = c. En Lean la clase de los monoides cancelativos por la izquierda es left&#95;cancel&#95;monoid y la propiedad cancelativa por la izquierda es mul_left_cancel_iff : a * b = a * c \u2194 b = c Demostrar que si M es un monoide cancelativo por la izquierda y a, b \u2208 M, entonces a * b = a \u2194 b = 1 Para ello, completar la siguiente teor\u00eda de Lean: import algebra.group.basic universe u variables {M : Type u} [left_cancel_monoid M] variables {a b : M} example : a *&#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\/518"}],"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=518"}],"version-history":[{"count":1,"href":"https:\/\/www.glc.us.es\/~jalonso\/calculemus\/wp-json\/wp\/v2\/posts\/518\/revisions"}],"predecessor-version":[{"id":519,"href":"https:\/\/www.glc.us.es\/~jalonso\/calculemus\/wp-json\/wp\/v2\/posts\/518\/revisions\/519"}],"wp:attachment":[{"href":"https:\/\/www.glc.us.es\/~jalonso\/calculemus\/wp-json\/wp\/v2\/media?parent=518"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.glc.us.es\/~jalonso\/calculemus\/wp-json\/wp\/v2\/categories?post=518"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.glc.us.es\/~jalonso\/calculemus\/wp-json\/wp\/v2\/tags?post=518"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}