{"id":706,"date":"2021-08-29T06:00:12","date_gmt":"2021-08-29T04:00:12","guid":{"rendered":"https:\/\/www.glc.us.es\/~jalonso\/calculemus\/?p=706"},"modified":"2021-08-23T16:56:27","modified_gmt":"2021-08-23T14:56:27","slug":"relacion-entre-los-indices-de-las-subsucesiones-y-los-de-la-sucesion","status":"publish","type":"post","link":"https:\/\/www.glc.us.es\/~jalonso\/calculemus\/relacion-entre-los-indices-de-las-subsucesiones-y-los-de-la-sucesion\/","title":{"rendered":"Relaci\u00f3n entre los \u00edndices de las subsucesiones y los de la sucesi\u00f3n"},"content":{"rendered":"<p>Para extraer una subsucesi\u00f3n se aplica una funci\u00f3n de extracci\u00f3n que conserva el orden; por ejemplo, la subsucesi\u00f3n<\/p>\n<pre lang=\"text\">\n   u\u2092, u\u2082, u\u2084, u\u2086, ...\n<\/pre>\n<p>se ha obtenido con la funci\u00f3n de extracci\u00f3n \u03c6 tal que \u03c6(n) = 2*n.<\/p>\n<p>En Lean, se puede definir que \u03c6 es una funci\u00f3n de extracci\u00f3n por<\/p>\n<pre lang=\"text\">\n   def extraccion (\u03c6 : \u2115 \u2192 \u2115) :=\n     \u2200 {n m}, n < m \u2192 \u03c6 n < \u03c6 m\n<\/pre>\n<p>Demostrar que si \u03c6 es una funci\u00f3n de extracci\u00f3n, entonces<\/p>\n<pre lang=\"text\">\n   \u2200 n, n \u2264 \u03c6 n\n<\/pre>\n<p>Para ello, completar la siguiente teor\u00eda de Lean:<\/p>\n<pre lang=\"lean\">\nimport tactic\nopen nat\n\nvariable {\u03c6 : \u2115 \u2192 \u2115}\n\ndef extraccion (\u03c6 : \u2115 \u2192 \u2115) :=\n  \u2200 {n m}, n < m \u2192 \u03c6 n < \u03c6 m\n\nexample :\n  extraccion \u03c6 \u2192 \u2200 n, n \u2264 \u03c6 n :=\nsorry\n<\/pre>\n<p>[expand title=\"Soluciones con Lean\"]<\/p>\n<pre lang=\"lean\">\r\nimport tactic\r\nopen nat\r\n\r\nvariable {\u03c6 : \u2115 \u2192 \u2115}\r\n\r\nset_option pp.structure_projections false\r\n\r\ndef extraccion (\u03c6 : \u2115 \u2192 \u2115) :=\r\n  \u2200 {n m}, n < m \u2192 \u03c6 n < \u03c6 m\r\n\r\n-- 1\u00aa demostraci\u00f3n\r\nexample :\r\n  extraccion \u03c6 \u2192 \u2200 n, n \u2264 \u03c6 n :=\r\nbegin\r\n  intros h n,\r\n  induction n with m HI,\r\n  { exact nat.zero_le (\u03c6 0), },\r\n  { apply nat.succ_le_of_lt,\r\n    have h1 : m < succ m := lt_add_one m,\r\n    calc m \u2264 \u03c6 m        : HI\r\n       ... < \u03c6 (succ m) : h h1, },\r\nend\r\n\r\n-- 2\u00aa demostraci\u00f3n\r\nexample :\r\n  extraccion \u03c6 \u2192 \u2200 n, n \u2264 \u03c6 n :=\r\nbegin\r\n  intros h n,\r\n  induction n with m HI,\r\n  { exact nat.zero_le (\u03c6 0), },\r\n  { apply nat.succ_le_of_lt,\r\n    calc m \u2264 \u03c6 m        : HI\r\n       ... < \u03c6 (succ m) : h (lt_add_one m), },\r\nend\r\n\r\n-- 3\u00aa demostraci\u00f3n\r\nexample :\r\n  extraccion \u03c6 \u2192 \u2200 n, n \u2264 \u03c6 n :=\r\nassume h : extraccion \u03c6,\r\nassume n,\r\nnat.rec_on n\r\n  ( show 0 \u2264 \u03c6 0,\r\n      from nat.zero_le (\u03c6 0) )\r\n  ( assume m,\r\n    assume HI : m \u2264 \u03c6 m,\r\n    have h1 : m < succ m,\r\n      from lt_add_one m,\r\n    have h2 : m < \u03c6 (succ m), from\r\n      calc m \u2264 \u03c6 m        : HI\r\n         ... < \u03c6 (succ m) : h h1,\r\n    show succ m \u2264 \u03c6 (succ m),\r\n      from nat.succ_le_of_lt h2)\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\/Relacion_entre_los_indices_de_las_subsucesiones_y_de_la_sucesion.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=\"Soluciones con Isabelle\/HOL\"]<\/p>\n<pre lang=\"isar\">\r\ntheory Relacion_entre_los_indices_de_las_subsucesiones_y_de_la_sucesion\r\nimports Main\r\nbegin\r\n\r\ndefinition extraccion :: \"(nat \u21d2 nat) \u21d2 bool\" where\r\n  \"extraccion \u03c6 \u27f7 (\u2200 n m. n < m \u27f6 \u03c6 n < \u03c6 m)\"\r\n\r\n(* En la demostraci\u00f3n se usar\u00e1 el siguiente lema *)\r\nlemma extraccionE:\r\n  assumes \"extraccion \u03c6\"\r\n          \"n < m\"\r\n  shows   \"\u03c6 n < \u03c6 m\"\r\nproof -\r\n  have \"\u2200 n m. n < m \u27f6 \u03c6 n < \u03c6 m\"\r\n    using assms(1) by (unfold extraccion_def)\r\n  then have \"n < m \u27f6 \u03c6 n < \u03c6 m\"\r\n    by (elim allE)\r\n  then show \"\u03c6 n < \u03c6 m\"\r\n    using assms(2) by (rule mp)\r\nqed\r\n\r\n(* 1\u00aa demostraci\u00f3n *)\r\nlemma\r\n  assumes \"extraccion \u03c6\"\r\n  shows   \"n \u2264 \u03c6 n\"\r\nproof (induct n)\r\n  show \"0 \u2264 \u03c6 0\"\r\n    by (rule le0)\r\nnext\r\n  fix n\r\n  assume \"n \u2264 \u03c6 n\"\r\n  also have \"\u03c6 n < \u03c6 (Suc n)\"\r\n  proof -\r\n    have \"n < Suc n\"\r\n      by (rule lessI)\r\n    with assms show \"\u03c6 n < \u03c6 (Suc n)\"\r\n      by (rule extraccionE)\r\n  qed\r\n  finally show \"Suc n \u2264 \u03c6 (Suc n)\"\r\n    by (rule Suc_leI)\r\nqed\r\n\r\n(* 2\u00aa demostraci\u00f3n *)\r\nlemma\r\n  assumes \"extraccion \u03c6\"\r\n  shows   \"n \u2264 \u03c6 n\"\r\nproof (induct n)\r\n  show \"0 \u2264 \u03c6 0\"\r\n    by (rule le0)\r\nnext\r\n  fix n\r\n  assume \"n \u2264 \u03c6 n\"\r\n  also have \"\u2026 < \u03c6 (Suc n)\"\r\n  using assms\r\n  proof (rule extraccionE)\r\n    show \"n < Suc n\"\r\n      by (rule lessI)\r\n  qed\r\n  finally show \"Suc n \u2264 \u03c6 (Suc n)\"\r\n    by (rule Suc_leI)\r\nqed\r\n\r\n(* 3\u00aa demostraci\u00f3n *)\r\nlemma\r\n  assumes \"extraccion \u03c6\"\r\n  shows   \"n \u2264 \u03c6 n\"\r\nproof (induct n)\r\n  show \"0 \u2264 \u03c6 0\"\r\n    by (rule le0)\r\nnext\r\n  fix n\r\n  assume \"n \u2264 \u03c6 n\"\r\n  also have \"\u2026 < \u03c6 (Suc n)\"\r\n    by (rule extraccionE [OF assms lessI])\r\n  finally show \"Suc n \u2264 \u03c6 (Suc n)\"\r\n    by (rule Suc_leI)\r\nqed\r\n\r\n(* 4\u00aa demostraci\u00f3n *)\r\nlemma\r\n  assumes \"extraccion \u03c6\"\r\n  shows   \"n \u2264 \u03c6 n\"\r\nproof (induct n)\r\n  show \"0 \u2264 \u03c6 0\"\r\n    by simp\r\nnext\r\n  fix n\r\n  assume HI : \"n \u2264 \u03c6 n\"\r\n  also have \"\u03c6 n < \u03c6 (Suc n)\"\r\n    using assms extraccion_def by blast\r\n  finally show \"Suc n \u2264 \u03c6 (Suc n)\"\r\n    by simp\r\nqed\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>Para extraer una subsucesi\u00f3n se aplica una funci\u00f3n de extracci\u00f3n que conserva el orden; por ejemplo, la subsucesi\u00f3n u\u2092, u\u2082, u\u2084, u\u2086, &#8230; se ha obtenido con la funci\u00f3n de extracci\u00f3n \u03c6 tal que \u03c6(n) = 2*n. En Lean, se puede definir que \u03c6 es una funci\u00f3n de extracci\u00f3n por def extraccion (\u03c6 : \u2115 \u2192 \u2115) := \u2200 {n m}, n < m \u2192 \u03c6 n < \u03c6 m Demostrar que si \u03c6 es una funci\u00f3n de extracci\u00f3n, entonces \u2200 n, n \u2264 \u03c6 n Para ello, completar la siguiente teor\u00eda de Lean: import tactic open nat variable {\u03c6 : \u2115 \u2192 \u2115} def extraccion (\u03c6 : \u2115 \u2192 \u2115) := \u2200 {n m}, n < m \u2192 \u03c6 n < \u03c6 m example...\n<\/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":[105,168],"tags":[54,92,175,177,178,179,176,51,100,180,114,49,169,63,173,174,171,170,172],"jetpack_featured_media_url":"","jetpack_sharing_enabled":true,"_links":{"self":[{"href":"https:\/\/www.glc.us.es\/~jalonso\/calculemus\/wp-json\/wp\/v2\/posts\/706"}],"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=706"}],"version-history":[{"count":1,"href":"https:\/\/www.glc.us.es\/~jalonso\/calculemus\/wp-json\/wp\/v2\/posts\/706\/revisions"}],"predecessor-version":[{"id":707,"href":"https:\/\/www.glc.us.es\/~jalonso\/calculemus\/wp-json\/wp\/v2\/posts\/706\/revisions\/707"}],"wp:attachment":[{"href":"https:\/\/www.glc.us.es\/~jalonso\/calculemus\/wp-json\/wp\/v2\/media?parent=706"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.glc.us.es\/~jalonso\/calculemus\/wp-json\/wp\/v2\/categories?post=706"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.glc.us.es\/~jalonso\/calculemus\/wp-json\/wp\/v2\/tags?post=706"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}