{"id":685,"date":"2021-08-26T06:00:33","date_gmt":"2021-08-26T04:00:33","guid":{"rendered":"https:\/\/www.glc.us.es\/~jalonso\/calculemus\/?p=685"},"modified":"2021-08-21T12:46:11","modified_gmt":"2021-08-21T10:46:11","slug":"las-familias-de-conjuntos-definen-relaciones-simetricas","status":"publish","type":"post","link":"https:\/\/www.glc.us.es\/~jalonso\/calculemus\/las-familias-de-conjuntos-definen-relaciones-simetricas\/","title":{"rendered":"Las familias de conjuntos definen relaciones sim\u00e9tricas"},"content":{"rendered":"<p>Cada familia de conjuntos P define una relaci\u00f3n de forma que dos elementos est\u00e1n relacionados si alg\u00fan conjunto de P contiene a ambos elementos. Se puede definir en Lean por<\/p>\n<pre lang=\"text\">\n   def relacion (P : set (set X)) (x y : X) :=\n     \u2203 A \u2208 P, x \u2208 A \u2227 y \u2208 A\n<\/pre>\n<p>Demostrar que si P es una familia de subconjunt\u2759os de X, entonces la relaci\u00f3n definida por P es sim\u00e9trica.<\/p>\n<p>Para ello, completar la siguiente teor\u00eda de Lean:<\/p>\n<pre lang=\"lean\">\nimport tactic\n\nvariable {X : Type}\nvariable (P : set (set X))\n\ndef relacion (P : set (set X)) (x y : X) :=\n  \u2203 A \u2208 P, x \u2208 A \u2227 y \u2208 A\n\nexample : symmetric (relacion P) :=\nsorry\n<\/pre>\n<p>[expand title=\u00bbSoluciones con Lean\u00bb]<\/p>\n<pre lang=\"lean\">\r\nimport tactic\r\n\r\nvariable {X : Type}\r\nvariable (P : set (set X))\r\n\r\ndef relacion (P : set (set X)) (x y : X) :=\r\n  \u2203 A \u2208 P, x \u2208 A \u2227 y \u2208 A\r\n\r\n-- 1\u00aa demostraci\u00f3n\r\nexample : symmetric (relacion P) :=\r\nbegin\r\n  unfold symmetric,\r\n  intros x y hxy,\r\n  unfold relacion at *,\r\n  rcases hxy with \u27e8B, hBP, \u27e8hxB, hyB\u27e9\u27e9,\r\n  use B,\r\n  repeat { split },\r\n  { exact hBP, },\r\n  { exact hyB, },\r\n  { exact hxB, },\r\nend\r\n\r\n-- 2\u00aa demostraci\u00f3n\r\nexample : symmetric (relacion P) :=\r\nbegin\r\n  intros x y hxy,\r\n  rcases hxy with \u27e8B, hBP, \u27e8hxB, hyB\u27e9\u27e9,\r\n  use B,\r\n  repeat { split } ;\r\n  assumption,\r\nend\r\n\r\n-- 3\u00aa demostraci\u00f3n\r\nexample : symmetric (relacion P) :=\r\nbegin\r\n  intros x y hxy,\r\n  rcases hxy with \u27e8B, hBP, \u27e8hxB, hyB\u27e9\u27e9,\r\n  use [B, \u27e8hBP, hyB, hxB\u27e9],\r\nend\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\/Las_familias_de_conjuntos_definen_relaciones_simetricas.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\ntheory Las_familias_de_conjuntos_definen_relaciones_simetricas\r\nimports Main\r\nbegin\r\n\r\ndefinition relacion :: \"('a set) set \u21d2 'a \u21d2 'a \u21d2 bool\" where\r\n  \"relacion P x y \u27f7 (\u2203A\u2208P. x \u2208 A \u2227 y \u2208 A)\"\r\n\r\n(* 1\u00aa demostraci\u00f3n *)\r\nlemma \"symp (relacion P)\"\r\nproof (rule sympI)\r\n  fix x y\r\n  assume \"relacion P x y\"\r\n  then have \"\u2203A\u2208P. x \u2208 A \u2227 y \u2208 A\"\r\n    by (unfold relacion_def)\r\n  then have \"\u2203A\u2208P. y \u2208 A \u2227 x \u2208 A\"\r\n  proof (rule bexE)\r\n    fix A\r\n    assume hA1 : \"A \u2208 P\" and hA2 : \"x \u2208 A \u2227 y \u2208 A\"\r\n    have \"y \u2208 A \u2227 x \u2208 A\"\r\n      using hA2 by (simp only: conj_commute)\r\n    then show \"\u2203A\u2208P. y \u2208 A \u2227 x \u2208 A\"\r\n      using hA1 by (rule bexI)\r\n  qed\r\n  then show \"relacion P y x\"\r\n    by (unfold relacion_def)\r\nqed\r\n\r\n(* 2\u00aa demostraci\u00f3n *)\r\nlemma \"symp (relacion P)\"\r\nproof (rule sympI)\r\n  fix x y\r\n  assume \"relacion P x y\"\r\n  then obtain A where \"A \u2208 P \u2227 x \u2208 A \u2227 y \u2208 A\"\r\n    using relacion_def\r\n    by metis\r\n  then show \"relacion P y x\"\r\n    using relacion_def\r\n    by metis\r\nqed\r\n\r\n(* 3\u00aa demostraci\u00f3n *)\r\nlemma \"symp (relacion P)\"\r\n  using relacion_def\r\n  by (metis sympI)\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>Cada familia de conjuntos P define una relaci\u00f3n de forma que dos elementos est\u00e1n relacionados si alg\u00fan conjunto de P contiene a ambos elementos. Se puede definir en Lean por def relacion (P : set (set X)) (x y : X) := \u2203 A \u2208 P, x \u2208 A \u2227 y \u2208 A Demostrar que si P es una familia de subconjunt\u2759os de X, entonces la relaci\u00f3n definida por P es sim\u00e9trica. Para ello, completar la siguiente teor\u00eda de Lean: import tactic variable {X : Type} variable (P : set (set X)) def relacion (P : set (set X)) (x y : X) := \u2203 A \u2208 P, x \u2208 A \u2227 y \u2208 A example : symmetric (relacion P) := sorry [expand title=\u00bbSoluciones con&#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":[29],"tags":[66,59,67,60,51,100,65,62,64,50,49,63,45,47,48,46],"jetpack_featured_media_url":"","jetpack_sharing_enabled":true,"_links":{"self":[{"href":"https:\/\/www.glc.us.es\/~jalonso\/calculemus\/wp-json\/wp\/v2\/posts\/685"}],"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=685"}],"version-history":[{"count":1,"href":"https:\/\/www.glc.us.es\/~jalonso\/calculemus\/wp-json\/wp\/v2\/posts\/685\/revisions"}],"predecessor-version":[{"id":686,"href":"https:\/\/www.glc.us.es\/~jalonso\/calculemus\/wp-json\/wp\/v2\/posts\/685\/revisions\/686"}],"wp:attachment":[{"href":"https:\/\/www.glc.us.es\/~jalonso\/calculemus\/wp-json\/wp\/v2\/media?parent=685"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.glc.us.es\/~jalonso\/calculemus\/wp-json\/wp\/v2\/categories?post=685"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.glc.us.es\/~jalonso\/calculemus\/wp-json\/wp\/v2\/tags?post=685"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}