{"id":7821,"date":"2022-10-23T17:14:06","date_gmt":"2022-10-23T15:14:06","guid":{"rendered":"https:\/\/www.glc.us.es\/~jalonso\/vestigium\/?p=7821"},"modified":"2022-10-23T17:14:06","modified_gmt":"2022-10-23T15:14:06","slug":"dao-la-semana-en-calculemus-21-de-octubre-de-2022","status":"publish","type":"post","link":"https:\/\/www.glc.us.es\/~jalonso\/vestigium\/dao-la-semana-en-calculemus-21-de-octubre-de-2022\/","title":{"rendered":"DAO: La semana en Calculemus (21 de octubre de 2022)"},"content":{"rendered":"<p>Esta semana he publicado en <a href=\"https:\/\/www.glc.us.es\/~jalonso\/calculemus\/\">Calculemus<\/a> las demostraciones con Lean de las siguientes propiedades:<\/p>\n<ul>\n<li><a href=\"#ej1\">1. Si R es un ret\u00edculo y x, y \u2208 R, entonces x \u2294 y = y \u2294 x<\/a><\/li>\n<li><a href=\"#ej2\">2. Si R es un ret\u00edculo y x, y, z \u2208 R, entonces (x \u2293 y) \u2293 z = x \u2293 (y \u2293 z)<\/a><\/li>\n<li><a href=\"#ej3\">3. Si R es un ret\u00edculo y x, y, z \u2208 R, entonces (x \u2294 y) \u2294 z = x \u2294 (y \u2294 z)<\/a><\/li>\n<li><a href=\"#ej4\">4. Si R es un ret\u00edculo y x, y \u2208 R, entonces x \u2293 (x \u2294 y) = x<\/a><\/li>\n<li><a href=\"#ej5\">5. Si R es un ret\u00edculo y x, y \u2208 R, entonces x \u2294 (x \u2293 y) = x<\/a><\/li>\n<\/ul>\n<p>A continuaci\u00f3n se muestran las soluciones.<br \/>\n<!--more--><br \/>\n<a name=\"ej1\"><\/a><\/p>\n<h3>1. Si R es un ret\u00edculo y x, y \u2208 R, entonces x \u2294 y = y \u2294 x<\/h3>\n<p>Demostrar que si R es un ret\u00edculo y x, y \u2208 R, entonces x \u2294 y = y \u2294 x.<\/p>\n<p>Para ello, completar la siguiente teor\u00eda de Lean:<\/p>\n<pre lang=\"lean\">\nimport order.lattice\n\nvariables {R : Type*} [lattice R]\nvariables x y : R\n\nexample : x \u2294 y = y \u2294 x :=\nsorry\n<\/pre>\n<p><b>Soluciones con Lean<\/b><\/p>\n<pre lang=\"lean\">\nimport order.lattice\n\nvariables {R : Type*} [lattice R]\nvariables x y : R\n\n-- 1\u00aa demostraci\u00f3n\n-- ===============\n\nlemma aux1 : x \u2294 y \u2264 y \u2294 x :=\nbegin\n  have h1 : x \u2264 y \u2294 x,\n    by exact le_sup_right,\n  have h2 : y \u2264 y \u2294 x,\n    by exact le_sup_left,\n  show x \u2294 y \u2264 y \u2294 x,\n    by exact sup_le h1 h2,\nend\n\nexample : x \u2294 y = y \u2294 x :=\nbegin\n  have h1 : x \u2294 y \u2264 y \u2294 x,\n    by exact aux1 x y,\n  have h2 : y \u2294 x \u2264 x \u2294 y,\n    by exact aux1 y x,\n  show x \u2294 y = y \u2294 x,\n    by exact le_antisymm h1 h2,\nend\n\n-- 2\u00aa demostraci\u00f3n\n-- ===============\n\nlemma aux2 : x \u2294 y \u2264 y \u2294 x :=\nsup_le le_sup_right le_sup_left\n\nexample : x \u2294 y = y \u2294 x :=\nle_antisymm (aux2 x y) (aux2 y x)\n\n-- 3\u00aa demostraci\u00f3n\n-- ===============\n\nlemma aux : x \u2294 y \u2264 y \u2294 x :=\nbegin\n  apply sup_le,\n  apply le_sup_right,\n  apply le_sup_left,\nend\n\nexample : x \u2294 y = y \u2294 x :=\nbegin\n  apply le_antisymm,\n  apply aux,\n  apply aux,\nend\n\n-- 4\u00aa demostraci\u00f3n\n-- ===============\n\nexample : x \u2294 y = y \u2294 x :=\nby apply le_antisymm; simp\n\n-- 5\u00aa demostraci\u00f3n\n-- ===============\n\nexample : x \u2294 y = y \u2294 x :=\n-- by library_search\nsup_comm\n\n-- 6\u00aa demostraci\u00f3n\n-- ===============\n\nexample : x \u2294 y = y \u2294 x :=\n-- by hint\nby finish\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\/Conmutatividad_del_supremo.lean\" rel=\"noopener noreferrer\" target=\"_blank\">esta sesi\u00f3n con Lean<\/a>.<\/p>\n<p><b>Referencias<\/b><\/p>\n<ul>\n<li>J. Avigad, K. Buzzard, R.Y. Lewis y P. Massot. <a href=\"https:\/\/bit.ly\/3U4UjBk\">Mathematics in Lean<\/a>, p. 22.<\/li>\n<\/ul>\n<p><a name=\"ej2\"><\/a><\/p>\n<h3>2. Si R es un ret\u00edculo y x, y, z \u2208 R, entonces (x \u2293 y) \u2293 z = x \u2293 (y \u2293 z)<\/h3>\n<p>Demostrar que si R es un ret\u00edculo y x, y, z \u2208 R, entonces<\/p>\n<pre lang=\"text\">\n(x \u2293 y) \u2293 z = x \u2293 (y \u2293 z)\n<\/pre>\n<p>Para ello, completar la siguiente teor\u00eda de Lean:<\/p>\n<pre lang=\"lean\">\nimport order.lattice\n\nvariables {R : Type*} [lattice R]\nvariables x y z : R\n\nexample : (x \u2293 y) \u2293 z = x \u2293 (y \u2293 z) :=\nsorry\n<\/pre>\n<p><b>Soluciones con Lean<\/b><\/p>\n<pre lang=\"lean\">\nimport order.lattice\n\nvariables {R : Type*} [lattice R]\nvariables x y z : R\n\n-- 1\u00aa demostraci\u00f3n\n-- ===============\n\nexample : (x \u2293 y) \u2293 z = x \u2293 (y \u2293 z) :=\nbegin\n  have h1 : (x \u2293 y) \u2293 z \u2264 x \u2293 (y \u2293 z),\n    { have h1a : (x \u2293 y) \u2293 z \u2264 x, calc\n        (x \u2293 y) \u2293 z \u2264 x \u2293 y : inf_le_left\n                ... \u2264 x     : inf_le_left,\n      have h1b : (x \u2293 y) \u2293 z \u2264 y \u2293 z,\n        { have h1b1 : (x \u2293 y) \u2293 z \u2264 y, calc\n            (x \u2293 y) \u2293 z \u2264 x \u2293 y : inf_le_left\n                    ... \u2264 y     : inf_le_right,\n          have h1b2 : (x \u2293 y) \u2293 z \u2264 z :=\n            inf_le_right,\n          show (x \u2293 y) \u2293 z \u2264 y \u2293 z,\n            by exact le_inf h1b1 h1b2, },\n      show (x \u2293 y) \u2293 z \u2264 x \u2293 (y \u2293 z),\n        by exact le_inf h1a h1b, },\n  have h2 : x \u2293 (y \u2293 z) \u2264 (x \u2293 y) \u2293 z,\n    { have h2a : x \u2293 (y \u2293 z) \u2264 x \u2293 y,\n        { have h2a1 : x \u2293 (y \u2293 z) \u2264 x,\n            by exact inf_le_left,\n          have h2a2 : x \u2293 (y \u2293 z) \u2264 y, calc\n            x \u2293 (y \u2293 z) \u2264 y \u2293 z : inf_le_right\n                    ... \u2264 y     : inf_le_left,\n          show x \u2293 (y \u2293 z) \u2264 x \u2293 y,\n            by exact le_inf h2a1 h2a2, },\n      have h2b : x \u2293 (y \u2293 z) \u2264 z, calc\n        x \u2293 (y \u2293 z) \u2264 y \u2293 z : inf_le_right\n                ... \u2264 z     : inf_le_right,\n      show x \u2293 (y \u2293 z) \u2264 (x \u2293 y) \u2293 z,\n        by exact le_inf h2a h2b, },\n  show (x \u2293 y) \u2293 z = x \u2293 (y \u2293 z),\n    by exact le_antisymm h1 h2,\nend\n\n-- 2\u00aa demostraci\u00f3n\n-- ===============\n\nexample : (x \u2293 y) \u2293 z = x \u2293 (y \u2293 z) :=\nbegin\n  apply le_antisymm,\n  { apply le_inf,\n    { apply inf_le_of_left_le inf_le_left, },\n    { apply le_inf (inf_le_of_left_le inf_le_right) inf_le_right}},\n  {apply le_inf,\n    { apply le_inf inf_le_left (inf_le_of_right_le inf_le_left), },\n    { apply inf_le_of_right_le inf_le_right, },},\nend\n\n-- 3\u00aa demostraci\u00f3n\n-- ===============\n\nexample : (x \u2293 y) \u2293 z = x \u2293 (y \u2293 z) :=\nle_antisymm\n  (le_inf\n    (inf_le_of_left_le inf_le_left)\n    (le_inf (inf_le_of_left_le inf_le_right) inf_le_right))\n  (le_inf\n    (le_inf inf_le_left (inf_le_of_right_le inf_le_left))\n    (inf_le_of_right_le inf_le_right))\n\n-- 4\u00aa demostraci\u00f3n\n-- ===============\n\nexample : (x \u2293 y) \u2293 z = x \u2293 (y \u2293 z) :=\n-- by library_search\ninf_assoc\n\n-- 5\u00aa demostraci\u00f3n\n-- ===============\n\nexample : (x \u2293 y) \u2293 z = x \u2293 (y \u2293 z) :=\n-- by hint\nby finish\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\/Asociatividad_del_infimo.lean\" rel=\"noopener noreferrer\" target=\"_blank\">esta sesi\u00f3n con Lean<\/a>.<\/p>\n<p><b>Referencias<\/b><\/p>\n<ul>\n<li>J. Avigad, K. Buzzard, R.Y. Lewis y P. Massot. <a href=\"https:\/\/bit.ly\/3U4UjBk\">Mathematics in Lean<\/a>, p. 22.<\/li>\n<\/ul>\n<p><a name=\"ej3\"><\/a><\/p>\n<h3>3. Si R es un ret\u00edculo y x, y, z \u2208 R, entonces (x \u2294 y) \u2294 z = x \u2294 (y \u2294 z)<\/h3>\n<p>Demostrar que si R es un ret\u00edculo y x, y, z \u2208 R, entonces<\/p>\n<pre lang=\"text\">\n(x \u2294 y) \u2294 z = x \u2294 (y \u2294 z)\n<\/pre>\n<p>Para ello, completar la siguiente teor\u00eda de Lean:<\/p>\n<pre lang=\"lean\">\nimport order.lattice\n\nvariables {R : Type*} [lattice R]\nvariables x y z : R\n\nexample : (x \u2294 y) \u2294 z = x \u2294 (y \u2294 z) :=\nsorry\n<\/pre>\n<p><b>Soluciones con Lean<\/b><\/p>\n<pre lang=\"lean\">\nimport order.lattice\n\nvariables {R : Type*} [lattice R]\nvariables x y z : R\n\n-- 1\u00aa demostraci\u00f3n\n-- ===============\n\nexample : (x \u2294 y) \u2294 z = x \u2294 (y \u2294 z) :=\nbegin\n  have h1 : (x \u2294 y) \u2294 z \u2264 x \u2294 (y \u2294 z),\n    { have h1a : x \u2294 y \u2264 x \u2294 (y \u2294 z), by finish,\n      have h1b : z \u2264 x \u2294 (y \u2294 z), by finish,\n      show (x \u2294 y) \u2294 z \u2264 x \u2294 (y \u2294 z),\n        by exact sup_le h1a h1b, },\n  have h2 : x \u2294 (y \u2294 z) \u2264 (x \u2294 y) \u2294 z,\n    { have h2a : x \u2264 (x \u2294 y) \u2294 z, by finish,\n      have h2b : y \u2294 z \u2264 (x \u2294 y) \u2294 z, by finish,\n      show x \u2294 (y \u2294 z) \u2264 (x \u2294 y) \u2294 z,\n        by exact sup_le h2a h2b, },\n  show (x \u2294 y) \u2294 z = x \u2294 (y \u2294 z),\n    by exact le_antisymm h1 h2,\nend\n\n-- 2\u00aa demostraci\u00f3n\n-- ===============\n\nexample : (x \u2294 y) \u2294 z = x \u2294 (y \u2294 z) :=\nbegin\n  have h1 : (x \u2294 y) \u2294 z \u2264 x \u2294 (y \u2294 z),\n    { have h1a : x \u2294 y \u2264 x \u2294 (y \u2294 z),\n        { have h1a1 : x \u2264 x \u2294 (y \u2294 z) :=\n            le_sup_left,\n          have h1a2 : y \u2264 x \u2294 (y \u2294 z), calc\n            y \u2264 y \u2294 z         : le_sup_left\n            ... \u2264 x \u2294 (y \u2294 z) : le_sup_right,\n          show x \u2294 y \u2264 x \u2294 (y \u2294 z),\n            by exact sup_le h1a1 h1a2, },\n      have h1b : z \u2264 x \u2294 (y \u2294 z), calc\n        z   \u2264 y \u2294 z       : le_sup_right\n        ... \u2264 x \u2294 (y \u2294 z) : le_sup_right,\n      show (x \u2294 y) \u2294 z \u2264 x \u2294 (y \u2294 z),\n        by exact sup_le h1a h1b, },\n  have h2 : x \u2294 (y \u2294 z) \u2264 (x \u2294 y) \u2294 z,\n    { have h2a : x \u2264 (x \u2294 y) \u2294 z, calc\n        x   \u2264 x \u2294 y       : le_sup_left\n        ... \u2264 (x \u2294 y) \u2294 z : le_sup_left,\n      have h2b : y \u2294 z \u2264 (x \u2294 y) \u2294 z,\n        { have h2b1 : y \u2264 (x \u2294 y) \u2294 z, calc\n            y   \u2264 x \u2294 y       : le_sup_right\n            ... \u2264 (x \u2294 y) \u2294 z : le_sup_left,\n          have h2b2 : z \u2264 (x \u2294 y) \u2294 z :=\n            le_sup_right,\n          show y \u2294 z \u2264 (x \u2294 y) \u2294 z,\n            by exact sup_le h2b1 h2b2, },\n      show x \u2294 (y \u2294 z) \u2264 (x \u2294 y) \u2294 z,\n        by exact sup_le h2a h2b, },\n  show (x \u2294 y) \u2294 z = x \u2294 (y \u2294 z),\n    by exact le_antisymm h1 h2,\nend\n\n-- 3\u00aa demostraci\u00f3n\n-- ===============\n\nexample : (x \u2294 y) \u2294 z = x \u2294 (y \u2294 z) :=\nbegin\n  apply le_antisymm,\n  { apply sup_le,\n    { apply sup_le le_sup_left (le_sup_of_le_right le_sup_left)},\n    { apply le_sup_of_le_right le_sup_right}},\n  { apply sup_le,\n    { apply le_sup_of_le_left le_sup_left},\n    { apply sup_le (le_sup_of_le_left le_sup_right) le_sup_right}},\nend\n\n-- 4\u00aa demostraci\u00f3n\n-- ===============\n\nexample : (x \u2294 y) \u2294 z = x \u2294 (y \u2294 z) :=\nle_antisymm\n  (sup_le\n    (sup_le le_sup_left (le_sup_of_le_right le_sup_left))\n    (le_sup_of_le_right le_sup_right))\n  (sup_le\n    (le_sup_of_le_left le_sup_left)\n    (sup_le (le_sup_of_le_left le_sup_right) le_sup_right))\n\n-- 5\u00aa demostraci\u00f3n\n-- ===============\n\nexample : x \u2294 y \u2294 z = x \u2294 (y \u2294 z) :=\n-- by library_search\nsup_assoc\n\n-- 6\u00aa demostraci\u00f3n\n-- ===============\n\nexample : x \u2294 y \u2294 z = x \u2294 (y \u2294 z) :=\n-- by hint\nby finish\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\/Asociatividad_del_supremo.lean\" rel=\"noopener noreferrer\" target=\"_blank\">esta sesi\u00f3n con Lean<\/a>.<\/p>\n<p><b>Referencias<\/b><\/p>\n<ul>\n<li>J. Avigad, K. Buzzard, R.Y. Lewis y P. Massot. <a href=\"https:\/\/bit.ly\/3U4UjBk\">Mathematics in Lean<\/a>, p. 22.<\/li>\n<\/ul>\n<p><a name=\"ej4\"><\/a><\/p>\n<h3>4. Si R es un ret\u00edculo y x, y \u2208 R, entonces x \u2293 (x \u2294 y) = x<\/h3>\n<p>Demostrar que si R es un ret\u00edculo y x, y \u2208 R, entonces<\/p>\n<pre lang=\"text\">\nx \u2293 (x \u2294 y) = x\n<\/pre>\n<p>Para ello, completar la siguiente teor\u00eda de Lean:<\/p>\n<pre lang=\"lean\">\nimport order.lattice\nvariables {R : Type*} [lattice R]\nvariables x y : R\n\nexample : x \u2293 (x \u2294 y) = x :=\nsorry\n<\/pre>\n<p><b>Soluciones con Lean<\/b><\/p>\n<pre lang=\"lean\">\nimport order.lattice\nvariables {R : Type*} [lattice R]\nvariables x y : R\n\n-- 1\u00aa demostraci\u00f3n\n-- ===============\n\nexample : x \u2293 (x \u2294 y) = x :=\nbegin\n  have h1 : x \u2293 (x \u2294 y) \u2264 x, finish,\n  have h2 : x \u2264 x \u2293 (x \u2294 y), finish,\n  show x \u2293 (x \u2294 y) = x,\n    by exact le_antisymm h1 h2,\nend\n\n-- 2\u00aa demostraci\u00f3n\n-- ===============\n\nexample : x \u2293 (x \u2294 y) = x :=\nbegin\n  have h1 : x \u2293 (x \u2294 y) \u2264 x := inf_le_left,\n  have h2 : x \u2264 x \u2293 (x \u2294 y),\n  { have h2a : x \u2264 x := rfl.ge,\n    have h2b : x \u2264 x \u2294 y := le_sup_left,\n    show x \u2264 x \u2293 (x \u2294 y),\n      by exact le_inf h2a h2b, },\n  show x \u2293 (x \u2294 y) = x,\n    by exact le_antisymm h1 h2,\nend\n\n-- 3\u00aa demostraci\u00f3n\n-- ===============\n\nexample : x \u2293 (x \u2294 y) = x :=\nbegin\n  apply le_antisymm,\n  { apply inf_le_left },\n  { apply le_inf,\n    { apply le_refl },\n    { apply le_sup_left }},\nend\n\n-- 4\u00aa demostraci\u00f3n\nexample : x \u2293 (x \u2294 y) = x :=\n-- by library_search\ninf_sup_self\n\n-- 5\u00aa demostraci\u00f3n\n-- ===============\n\nexample : x \u2293 (x \u2294 y) = x :=\n-- by hint\nby simp\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\/Ley_de_absorcion_1.lean\" rel=\"noopener noreferrer\" target=\"_blank\">esta sesi\u00f3n con Lean<\/a>.<\/p>\n<p><b>Referencias<\/b><\/p>\n<ul>\n<li>J. Avigad, K. Buzzard, R.Y. Lewis y P. Massot. <a href=\"https:\/\/bit.ly\/3U4UjBk\">Mathematics in Lean<\/a>, p. 22.<\/li>\n<\/ul>\n<p><a name=\"ej5\"><\/a><\/p>\n<h3>5. Si R es un ret\u00edculo y x, y \u2208 R, entonces x \u2294 (x \u2293 y) = x<\/h3>\n<p>Demostrar que si R es un ret\u00edculo y x, y \u2208 R, entonces<\/p>\n<pre lang=\"text\">\n   x \u2294 (x \u2293 y) = x\n<\/pre>\n<p>Para ello, completar la siguiente teor\u00eda de Lean:<\/p>\n<pre lang=\"lean\">\nimport order.lattice\nvariables {R : Type*} [lattice R]\nvariables x y : R\n\nexample : x \u2294 (x \u2293 y) = x :=\nsorry\n<\/pre>\n<p><b>Soluciones con Lean<\/b><\/p>\n<pre lang=\"lean\">\nimport order.lattice\nvariables {R : Type*} [lattice R]\nvariables x y : R\n\n-- 1\u00aa demostraci\u00f3n\n-- ===============\n\nexample : x \u2294 (x \u2293 y) = x :=\nbegin\n  have h1 : x \u2294 (x \u2293 y) \u2264 x, finish,\n  have h2 : x \u2264 x \u2294 (x \u2293 y), finish,\n  show x \u2294 (x \u2293 y) = x,\n    by exact le_antisymm h1 h2,\nend\n\n-- 2\u00aa demostraci\u00f3n\n-- ===============\n\nexample : x \u2294 (x \u2293 y) = x :=\nbegin\n  have h1 : x \u2294 (x \u2293 y) \u2264 x,\n  { have h1a : x \u2264 x := le_rfl,\n    have h1b : x \u2293 y \u2264 x := inf_le_left,\n    show x \u2294 (x \u2293 y) \u2264 x,\n      by exact sup_le h1a h1b,\n  },\n  have h2 : x \u2264 x \u2294 (x \u2293 y) := le_sup_left,\n  show x \u2294 (x \u2293 y) = x,\n    by exact le_antisymm h1 h2,\nend\n\n-- 3\u00aa demostraci\u00f3n\n-- ===============\n\nexample : x \u2294 (x \u2293 y) = x :=\nbegin\n  apply le_antisymm,\n  { apply sup_le,\n    { apply le_refl },\n    { apply inf_le_left }},\n  { apply le_sup_left },\nend\n\n-- 4\u00aa demostraci\u00f3n\n-- ===============\n\nexample : x \u2294 (x \u2293 y) = x :=\n-- by library_search\nsup_inf_self\n\n-- 4\u00aa demostraci\u00f3n\n-- ===============\n\nexample : x \u2294 (x \u2293 y) = x :=\n-- by hint\nby simp\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\/Ley_de_absorcion_2.lean\" rel=\"noopener noreferrer\" target=\"_blank\">esta sesi\u00f3n con Lean<\/a>.<\/p>\n<p><b>Referencias<\/b><\/p>\n<ul>\n<li>J. Avigad, K. Buzzard, R.Y. Lewis y P. Massot. <a href=\"https:\/\/bit.ly\/3U4UjBk\">Mathematics in Lean<\/a>, p. 22.<\/li>\n<\/ul>\n","protected":false},"excerpt":{"rendered":"<p>Esta semana he publicado en Calculemus las demostraciones con Lean de las siguientes propiedades: 1. Si R es un ret\u00edculo y x, y \u2208 R, entonces x \u2294 y = y \u2294 x 2. Si R es un ret\u00edculo y x, y, z \u2208 R, entonces (x \u2293 y) \u2293 z = x \u2293 (y&#8230;<\/p>\n","protected":false},"author":2,"featured_media":0,"comment_status":"closed","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,"footnotes":"","_jetpack_memberships_contains_paid_content":false},"categories":[335],"tags":[],"jetpack_featured_media_url":"","jetpack_sharing_enabled":true,"jetpack_likes_enabled":false,"_links":{"self":[{"href":"https:\/\/www.glc.us.es\/~jalonso\/vestigium\/wp-json\/wp\/v2\/posts\/7821"}],"collection":[{"href":"https:\/\/www.glc.us.es\/~jalonso\/vestigium\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/www.glc.us.es\/~jalonso\/vestigium\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/www.glc.us.es\/~jalonso\/vestigium\/wp-json\/wp\/v2\/users\/2"}],"replies":[{"embeddable":true,"href":"https:\/\/www.glc.us.es\/~jalonso\/vestigium\/wp-json\/wp\/v2\/comments?post=7821"}],"version-history":[{"count":1,"href":"https:\/\/www.glc.us.es\/~jalonso\/vestigium\/wp-json\/wp\/v2\/posts\/7821\/revisions"}],"predecessor-version":[{"id":7822,"href":"https:\/\/www.glc.us.es\/~jalonso\/vestigium\/wp-json\/wp\/v2\/posts\/7821\/revisions\/7822"}],"wp:attachment":[{"href":"https:\/\/www.glc.us.es\/~jalonso\/vestigium\/wp-json\/wp\/v2\/media?parent=7821"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.glc.us.es\/~jalonso\/vestigium\/wp-json\/wp\/v2\/categories?post=7821"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.glc.us.es\/~jalonso\/vestigium\/wp-json\/wp\/v2\/tags?post=7821"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}