{"id":8048,"date":"2023-11-11T13:18:08","date_gmt":"2023-11-11T12:18:08","guid":{"rendered":"https:\/\/www.glc.us.es\/~jalonso\/vestigium\/?p=8048"},"modified":"2023-11-11T13:21:05","modified_gmt":"2023-11-11T12:21:05","slug":"11-nov-23","status":"publish","type":"post","link":"https:\/\/www.glc.us.es\/~jalonso\/vestigium\/11-nov-23\/","title":{"rendered":"La semana en Calculemus (11 de noviembre de 2023)"},"content":{"rendered":"<p><br \/>\nEsta semana he publicado en <a href=\"https:\/\/www.glc.us.es\/~jalonso\/calculemus\/\">Calculemus<\/a> las demostraciones con Lean4 de las siguientes propiedades:<\/p>\n<ul>\n<li><a href=\"#ej1\">1. Transitividad de la divisibilidad<\/a><\/li>\n<li><a href=\"#ej2\">2. Si a divide a b y a c, entonces divide a b+c<\/a><\/li>\n<li><a href=\"#ej3\">3. La funci\u00f3n (x \u21a6 x + c) es suprayectiva<\/a><\/li>\n<li><a href=\"#ej4\">4. Si c \u2260 0, entonces la funci\u00f3n (x \u21a6 cx) es suprayectiva<\/a><\/li>\n<li><a href=\"#ej5\">5. Si c \u2260 0, entonces la funci\u00f3n (x \u21a6 cx + d) es suprayectiva<\/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. Transitividad de la divisibilidad<\/h3>\n<p>Demostrar con Lean4 la transitividad de la divisibilidad.<\/p>\n<p>Para ello, completar la siguiente teor\u00eda de Lean4:<\/p>\n<pre lang=\"lean\">\nimport Mathlib.Tactic\n\nvariable {a b c : \u2115}\n\nexample\n  (divab : a \u2223 b)\n  (divbc : b \u2223 c) :\n  a \u2223 c :=\nby sorry\n<\/pre>\n<p><b>Demostraci\u00f3n en lenguaje natural<\/b><\/p>\n<p>Supongamos que &#92;(a | b&#92;) y &#92;(b | c&#92;). Entonces, existen &#92;(d&#92;) y &#92;(e&#92;) tales que<br \/>\n&#92;begin{align}<br \/>\n   b &amp;= ad &#92;tag{1} &#92;&#92;<br \/>\n   c &amp;= be &#92;tag{2}<br \/>\n&#92;end{align}<br \/>\nPor tanto,<br \/>\n&#92;begin{align}<br \/>\n   c &amp;= be       &amp;&amp;&#92;text{[por (2)]} &#92;&#92;<br \/>\n     &amp;= (ad)e    &amp;&amp;&#92;text{[por (1)]} &#92;&#92;<br \/>\n     &amp;= a(de)<br \/>\n&#92;end{align}<br \/>\nPor consiguiente, &#92;(a | c&#92;).<\/p>\n<p><b>Demostraciones con Lean4<\/b><\/p>\n<pre lang=\"lean\">\nimport Mathlib.Tactic\n\nvariable {a b c : \u2115}\n\n-- 1\u00aa demostraci\u00f3n\nexample\n  (divab : a \u2223 b)\n  (divbc : b \u2223 c) :\n  a \u2223 c :=\nby\n  rcases divab with \u27e8d, beq : b = a * d\u27e9\n  rcases divbc with \u27e8e, ceq : c = b * e\u27e9\n  have h1 : c = a * (d * e) :=\n    calc c = b * e      := ceq\n        _ = (a * d) * e := congrArg (. * e) beq\n        _ = a * (d * e) := mul_assoc a d e\n  show a \u2223 c\n  exact Dvd.intro (d * e) h1.symm\n\n-- 2\u00aa demostraci\u00f3n\nexample\n  (divab : a \u2223 b)\n  (divbc : b \u2223 c) :\n  a \u2223 c :=\nby\n  rcases divab with \u27e8d, beq : b = a * d\u27e9\n  rcases divbc with \u27e8e, ceq : c = b * e\u27e9\n  use (d * e)\n  -- \u22a2 c = a * (d * e)\n  rw [ceq, beq]\n  -- \u22a2 (a * d) * e = a * (d * e)\n  exact mul_assoc a d e\n\n-- 3\u00aa demostraci\u00f3n\nexample\n  (divab : a \u2223 b)\n  (divbc : b \u2223 c) :\n  a \u2223 c :=\nby\n  rcases divbc with \u27e8e, rfl\u27e9\n  -- \u22a2 a \u2223 b * e\n  rcases divab with \u27e8d, rfl\u27e9\n  -- \u22a2 a \u2223 a * d * e\n  use (d * e)\n  -- \u22a2 a * d * e = a * (d * e)\n  ring\n\n-- 4\u00aa demostraci\u00f3n\nexample\n  (divab : a \u2223 b)\n  (divbc : b \u2223 c) :\n  a \u2223 c :=\nby\n  cases' divab with d beq\n  -- d : \u2115\n  -- beq : b = a * d\n  cases' divbc with e ceq\n  -- e : \u2115\n  -- ceq : c = b * e\n  rw [ceq, beq]\n  -- \u22a2 a \u2223 a * d * e\n  use (d * e)\n  -- \u22a2 (a * d) * e = a * (d * e)\n  exact mul_assoc a d e\n\n-- 5\u00aa demostraci\u00f3n\nexample\n  (divab : a \u2223 b)\n  (divbc : b \u2223 c) :\n  a \u2223 c :=\nby exact dvd_trans divab divbc\n\n-- Lemas usados\n-- ============\n\n-- #check (mul_assoc a b c : (a * b) * c = a * (b * c))\n-- #check (Dvd.intro c : a * c = b \u2192 a \u2223 b)\n-- #check (dvd_trans : a \u2223 b \u2192 b \u2223 c \u2192 a \u2223 c)\n<\/pre>\n<p><b>Demostraciones interactivas<\/b><\/p>\n<p>Se puede interactuar con las demostraciones anteriores en <a href=\"https:\/\/live.lean-lang.org\/#url=https:\/\/raw.githubusercontent.com\/jaalonso\/Calculemus2\/main\/src\/Transitividad_de_la_divisibilidad.lean\" rel=\"noopener noreferrer\" target=\"_blank\">Lean 4 Web<\/a>.<\/p>\n<p><b>Referencias<\/b><\/p>\n<ul>\n<li> J. Avigad y P. Massot. <a href=\"https:\/\/bit.ly\/3U4UjBk\">Mathematics in Lean<\/a>, p. 30.<\/li>\n<\/ul>\n<p><a name=\"ej2\"><\/a><\/p>\n<h3>2. Si a divide a b y a c, entonces divide a b+c<\/h3>\n<p>Demostrar con Lean4 que si &#92;(a&#92;) divide a &#92;(b&#92;) y a &#92;(c&#92;), entonces divide a &#92;(b+c&#92;).<\/p>\n<p>Para ello, completar la siguiente teor\u00eda de Lean4:<\/p>\n<pre lang=\"lean\">\nimport Mathlib.Tactic\n\nvariable {a b c : \u2115}\n\nexample\n  (h1 : a \u2223 b)\n  (h2 : a \u2223 c)\n  : a \u2223 (b + c) :=\nby sorry\n<\/pre>\n<p><b>Demostraci\u00f3n en lenguaje natural<\/b><\/p>\n<p>Puesto que &#92;(a&#92;) divide a &#92;(b&#92;) y a &#92;(c&#92;), existen &#92;(d&#92;) y &#92;(e&#92;) tales que<br \/>\n&#92;begin{align}<br \/>\n   b &amp;= ad &#92;tag{1} &#92;&#92;<br \/>\n   c &amp;= ae &#92;tag{2}<br \/>\n&#92;end{align}<br \/>\nPor tanto,<br \/>\n&#92;begin{align}<br \/>\n   b + c &amp;= ad + c     &amp;&amp;&#92;text{[por (1)]} &#92;&#92;<br \/>\n         &amp;= ad + ae    &amp;&amp;&#92;text{[por (2)]} &#92;&#92;<br \/>\n         &amp;= a(d + e)   &amp;&amp;&#92;text{[por la distributiva]}<br \/>\n&#92;end{align}<br \/>\nPor consiguiente, &#92;(a&#92;) divide a &#92;(b + c&#92;).<\/p>\n<p><b>Demostraciones con Lean4<\/b><\/p>\n<pre lang=\"lean\">\nimport Mathlib.Tactic\n\nvariable {a b c : \u2115}\n\n-- 1\u00aa demostraci\u00f3n\nexample\n  (h1 : a \u2223 b)\n  (h2 : a \u2223 c)\n  : a \u2223 (b + c) :=\nby\n  rcases h1 with \u27e8d, beq : b = a * d\u27e9\n  rcases h2 with \u27e8e, ceq: c = a * e\u27e9\n  have h1 : b + c = a * (d + e) :=\n    calc b + c\n         = (a * d) + c       := congrArg (. + c) beq\n       _ = (a * d) + (a * e) := congrArg ((a * d) + .) ceq\n       _ = a * (d + e)       := by rw [\u2190 mul_add]\n  show a \u2223 (b + c)\n  exact Dvd.intro (d + e) h1.symm\n\n-- 2\u00aa demostraci\u00f3n\nexample\n  (h1 : a \u2223 b)\n  (h2 : a \u2223 c)\n  : a \u2223 (b + c) :=\nby\n  rcases h1 with \u27e8d, beq : b = a * d\u27e9\n  rcases h2 with \u27e8e, ceq: c = a * e\u27e9\n  have h1 : b + c = a * (d + e) := by linarith\n  show a \u2223 (b + c)\n  exact Dvd.intro (d + e) h1.symm\n\n-- 3\u00aa demostraci\u00f3n\nexample\n  (h1 : a \u2223 b)\n  (h2 : a \u2223 c)\n  : a \u2223 (b + c) :=\nby\n  rcases h1 with \u27e8d, beq : b = a * d\u27e9\n  rcases h2 with \u27e8e, ceq: c = a * e\u27e9\n  show a \u2223 (b + c)\n  exact Dvd.intro (d + e) (by linarith)\n\n-- 4\u00aa demostraci\u00f3n\nexample\n  (h1 : a \u2223 b)\n  (h2 : a \u2223 c)\n  : a \u2223 (b + c) :=\nby\n  cases' h1 with d beq\n  -- d : \u2115\n  -- beq : b = a * d\n  cases' h2 with e ceq\n  -- e : \u2115\n  -- ceq : c = a * e\n  rw [ceq, beq]\n  -- \u22a2 a \u2223 a * d + a * e\n  use (d + e)\n  -- \u22a2 a * d + a * e = a * (d + e)\n  ring\n\n-- 5\u00aa demostraci\u00f3n\nexample\n  (h1 : a \u2223 b)\n  (h2 : a \u2223 c)\n  : a \u2223 (b + c) :=\nby\n  rcases h1 with \u27e8d, rfl\u27e9\n  -- \u22a2 a \u2223 a * d + c\n  rcases h2 with \u27e8e, rfl\u27e9\n  -- \u22a2 a \u2223 a * d + a * e\n  use (d + e)\n  -- \u22a2 a * d + a * e = a * (d + e)\n  ring\n\n-- 6\u00aa demostraci\u00f3n\nexample\n  (h1 : a \u2223 b)\n  (h2 : a \u2223 c)\n  : a \u2223 (b + c) :=\ndvd_add h1 h2\n\n-- Lemas usados\n-- ============\n\n-- #check (Dvd.intro c : a * c = b \u2192 a \u2223 b)\n-- #check (dvd_add : a \u2223 b \u2192  a \u2223 c \u2192 a \u2223 (b + c))\n-- #check (mul_add a b c : a * (b + c) = a * b + a * c)\n<\/pre>\n<p><b>Demostraciones interactivas<\/b><\/p>\n<p>Se puede interactuar con las demostraciones anteriores en <a href=\"https:\/\/live.lean-lang.org\/#url=https:\/\/raw.githubusercontent.com\/jaalonso\/Calculemus2\/main\/src\/Suma_divisible.lean\" rel=\"noopener noreferrer\" target=\"_blank\">Lean 4 Web<\/a>.<\/p>\n<p><b>Referencias<\/b><\/p>\n<ul>\n<li> J. Avigad y P. Massot. <a href=\"https:\/\/bit.ly\/3U4UjBk\">Mathematics in Lean<\/a>, p. 30.<\/li>\n<\/ul>\n<p><a name=\"ej3\"><\/a><\/p>\n<h3>3. La funci\u00f3n (x \u21a6 x + c) es suprayectiva<\/h3>\n<p>Demostrar con Lean4 que la funci\u00f3n &#92;(x \u21a6 x + c&#92;) es suprayectiva.<\/p>\n<p>Para ello, completar la siguiente teor\u00eda de Lean4:<\/p>\n<pre lang=\"lean\">\nimport Mathlib.Data.Real.Basic\nvariable {c : \u211d}\nopen Function\n\nexample : Surjective (fun x \u21a6 x + c) :=\nby sorry\n<\/pre>\n<p><b>Demostraci\u00f3n en lenguaje natural<\/b><\/p>\n<p>Tenemos que demostrar que<br \/>\n&#92;[ (\u2200 x \u2208 \u211d)(\u2203 y \u2208 \u211d)[y+c = x] &#92;]<br \/>\nSea &#92;(x \u2208 \u211d&#92;). Entonces, &#92;(y = x-c \u2208 \u211d&#92;) y<br \/>\n&#92;begin{align}<br \/>\n   y + c &amp;= (x &#8211; c) + c &#92;&#92;<br \/>\n         &amp;= x<br \/>\n&#92;end{align}<\/p>\n<p><b>Demostraciones con Lean4<\/b><\/p>\n<pre lang=\"lean\">\nimport Mathlib.Data.Real.Basic\n\nvariable {c : \u211d}\n\nopen Function\n\n-- 1\u00aa demostraci\u00f3n\nexample : Surjective (fun x \u21a6 x + c) :=\nby\n  intro x\n  -- x : \u211d\n  -- \u22a2 \u2203 a, (fun x => x + c) a = x\n  use x - c\n  -- \u22a2 (fun x => x + c) (x - c) = x\n  dsimp\n  -- \u22a2 (x - c) + c = x\n  exact sub_add_cancel x c\n\n-- 2\u00aa demostraci\u00f3n\nexample : Surjective (fun x \u21a6 x + c) :=\nby\n  intro x\n  -- x : \u211d\n  -- \u22a2 \u2203 a, (fun x => x + c) a = x\n  use x - c\n  -- \u22a2 (fun x => x + c) (x - c) = x\n  change (x - c) + c = x\n  -- \u22a2 (x - c) + c = x\n  exact sub_add_cancel x c\n\n-- 3\u00aa demostraci\u00f3n\nexample : Surjective (fun x \u21a6 x + c) :=\nby\n  intro x\n  -- x : \u211d\n  -- \u22a2 \u2203 a, (fun x => x + c) a = x\n  use x - c\n  -- \u22a2 (fun x => x + c) (x - c) = x\n  exact sub_add_cancel x c\n\n-- 4\u00aa demostraci\u00f3n\nexample : Surjective (fun x \u21a6 x + c) :=\nfun x \u21a6 \u27e8x - c, sub_add_cancel x c\u27e9\n\n-- 5\u00aa demostraci\u00f3n\nexample : Surjective (fun x \u21a6 x + c) :=\nfun x \u21a6 \u27e8x - c, by ring\u27e9\n\n-- 6\u00aa demostraci\u00f3n\nexample : Surjective (fun x \u21a6 x + c) :=\nadd_right_surjective c\n\n-- Lemas usados\n-- ============\n\n-- variable (a b : \u211d)\n-- #check (sub_add_cancel a b : (a - b) + b = a)\n-- #check (add_right_surjective c : Surjective (fun x \u21a6 x + c))\n<\/pre>\n<p><b>Demostraciones interactivas<\/b><\/p>\n<p>Se puede interactuar con las demostraciones anteriores en <a href=\"https:\/\/live.lean-lang.org\/#url=https:\/\/raw.githubusercontent.com\/jaalonso\/Calculemus2\/main\/src\/Suma_constante_es_suprayectiva.lean\" rel=\"noopener noreferrer\" target=\"_blank\">Lean 4 Web<\/a>.<\/p>\n<p><b>Referencias<\/b><\/p>\n<ul>\n<li> J. Avigad y P. Massot. <a href=\"https:\/\/bit.ly\/3U4UjBk\">Mathematics in Lean<\/a>, p. 31.<\/li>\n<\/ul>\n<p><a name=\"ej4\"><\/a><\/p>\n<h3>4. Si c \u2260 0, entonces la funci\u00f3n (x \u21a6 cx) es suprayectiva<\/h3>\n<p>Demostrar con Lean4 que si &#92;(c \u2260 0&#92;), entonces la funci\u00f3n &#92;(x \u21a6 cx&#92;) es suprayectiva.<\/p>\n<p>Para ello, completar la siguiente teor\u00eda de Lean4:<\/p>\n<pre lang=\"lean\">\nimport Mathlib.Data.Real.Basic\nvariable {c : \u211d}\nopen Function\n\nexample\n  (h : c \u2260 0)\n  : Surjective (fun x \u21a6 c * x) :=\nby sorry\n<\/pre>\n<p><b>Demostraci\u00f3n en lenguaje natural<\/b><\/p>\n<p>Hay que demostrar que<br \/>\n&#92;[ (\u2200 x \u2208 \u211d)(\u2203 y \u2208 \u211d)[cy = x] &#92;]<br \/>\nSea &#92;(x \u2208 \u211d&#92;). Entonces, &#92;(y = x\/c \u2208 R&#92;) y<br \/>\n&#92;begin{align}<br \/>\n   cy &amp;= c(x\/c) &#92;&#92;<br \/>\n      &amp;= y<br \/>\n&#92;end{align}<\/p>\n<p><b>Demostraciones con Lean4<\/b><\/p>\n<pre lang=\"lean\">\nimport Mathlib.Data.Real.Basic\nvariable {c : \u211d}\nopen Function\n\n-- 1\u00aa demostraci\u00f3n\nexample\n  (h : c \u2260 0)\n  : Surjective (fun x \u21a6 c * x) :=\nby\n  intro x\n  -- x : \u211d\n  -- \u22a2 \u2203 a, (fun x => c * x) a = x\n  use (x \/ c)\n  -- \u22a2 (fun x => c * x) (x \/ c) = x\n  dsimp\n  -- \u22a2 c * (x \/ c) = x\n  rw [mul_comm]\n  -- \u22a2 (x \/ c) * c = x\n  exact div_mul_cancel x h\n\n-- 2\u00aa demostraci\u00f3n\nexample\n  (h : c \u2260 0)\n  : Surjective (fun x \u21a6 c * x) :=\nby\n  intro x\n  -- x : \u211d\n  -- \u22a2 \u2203 a, (fun x => c * x) a = x\n  use (x \/ c)\n  -- \u22a2 (fun x => c * x) (x \/ c) = x\n  exact mul_div_cancel' x h\n\n-- 3\u00aa demostraci\u00f3n\nexample\n  (h : c \u2260 0)\n  : Surjective (fun x \u21a6 c * x) :=\nfun x \u21a6 \u27e8x \/ c, mul_div_cancel' x h\u27e9\n\n-- 4\u00aa demostraci\u00f3n\nexample\n  (h : c \u2260 0)\n  : Surjective (fun x \u21a6 c * x) :=\nmul_left_surjective\u2080 h\n\n-- Lemas usados\n-- ============\n\n-- variable (a b : \u211d)\n-- #check (div_mul_cancel a : b \u2260 0 \u2192 (a \/ b) * b = a)\n-- #check (mul_comm a b : a * b = b * a)\n-- #check (mul_div_cancel' a : b \u2260 0 \u2192 b * (a \/ b) = a)\n-- #check (mul_left_surjective\u2080 : c \u2260 0 \u2192 Surjective (fun x \u21a6 c * x))\n<\/pre>\n<p><b>Demostraciones interactivas<\/b><\/p>\n<p>Se puede interactuar con las demostraciones anteriores en <a href=\"https:\/\/live.lean-lang.org\/#url=https:\/\/raw.githubusercontent.com\/jaalonso\/Calculemus2\/main\/src\/Producto_por_no_nula_es_suprayectiva.lean\" rel=\"noopener noreferrer\" target=\"_blank\">Lean 4 Web<\/a>.<\/p>\n<p><b>Referencias<\/b><\/p>\n<ul>\n<li> J. Avigad y P. Massot. <a href=\"https:\/\/bit.ly\/3U4UjBk\">Mathematics in Lean<\/a>, p. 31.<\/li>\n<\/ul>\n<p><a name=\"ej5\"><\/a><\/p>\n<h3>5. Si c \u2260 0, entonces la funci\u00f3n (x \u21a6 cx + d) es suprayectiva<\/h3>\n<p>Demostrar con Lean4 que si &#92;(c \u2260 0&#92;), entonces la funci\u00f3n &#92;(x \u21a6 cx + d&#92;) es suprayectiva.<\/p>\n<p>Para ello, completar la siguiente teor\u00eda de Lean4:<\/p>\n<pre lang=\"lean\">\nimport Mathlib.Data.Real.Basic\nvariable {c d : \u211d}\nopen Function\n\nexample\n  (h : c \u2260 0)\n  : Surjective (fun x \u21a6 c * x + d) :=\nby sorry\n<\/pre>\n<p><b>Demostraci\u00f3n en lenguaje natural<\/b><\/p>\n<p>Hay que demostrar que<br \/>\n&#92;[ (\u2200x \u2208 \u211d)(\u2203y \u2208 \u211d)[cy+d = x] &#92;]<br \/>\nPara &#92;(x \u2208 \u211d&#92;), sea &#92;(y = &#92;dfrac{x-d}{c}&#92;). Entonces,<br \/>\n&#92;begin{align}<br \/>\n   cy &amp;= c&#92;left(&#92;frac{x-d}{c}&#92;right)+d &#92;&#92;<br \/>\n      &amp;= (x-d)+d      &#92;&#92;<br \/>\n      &amp;= x<br \/>\n&#92;end{align}<\/p>\n<p><b>Demostraciones con Lean4<\/b><\/p>\n<pre lang=\"lean\">\nimport Mathlib.Data.Real.Basic\nvariable {c d : \u211d}\nopen Function\n\n-- 1\u00aa demostraci\u00f3n\n-- ===============\n\nexample\n  (h : c \u2260 0)\n  : Surjective (fun x \u21a6 c * x + d) :=\nby\n  intro x\n  -- x : \u211d\n  -- \u22a2 \u2203 a, (fun x => c * x + d) a = x\n  use ((x - d) \/ c)\n  -- \u22a2 (fun x => c * x + d) ((x - d) \/ c) = x\n  dsimp\n  -- \u22a2 c * ((x - d) \/ c) + d = x\n  show c * ((x - d) \/ c) + d = x\n  calc c * ((x - d) \/ c) + d\n         = (x - d) + d := congrArg (. + d) (mul_div_cancel' (x - d) h)\n       _ = x           := sub_add_cancel x d\n\n-- 2\u00aa demostraci\u00f3n\n-- ===============\n\nexample\n  (h : c \u2260 0)\n  : Surjective (fun x \u21a6 c * x + d) :=\nby\n  intro x\n  -- x : \u211d\n  -- \u22a2 \u2203 a, (fun x => c * x + d) a = x\n  use ((x - d) \/ c)\n  -- \u22a2 (fun x => c * x + d) ((x - d) \/ c) = x\n  dsimp\n  -- \u22a2 c * ((x - d) \/ c) + d = x\n  simp [mul_div_cancel', h]\n\n-- 3\u00aa demostraci\u00f3n\n-- ===============\n\nexample\n  (h : c \u2260 0)\n  : Surjective (fun x \u21a6 c * x + d) :=\nby\n  intro x\n  -- x : \u211d\n  -- \u22a2 \u2203 a, (fun x => c * x + d) a = x\n  use ((x - d) \/ c)\n  -- \u22a2 (fun x => c * x + d) ((x - d) \/ c) = x\n  simp [mul_div_cancel', h]\n\n-- 4\u00aa demostraci\u00f3n\n-- ===============\n\nexample\n  (h : c \u2260 0)\n  : Surjective (fun x \u21a6 c * x + d) :=\nfun x \u21a6 \u27e8(x - d) \/ c, by simp [mul_div_cancel', h]\u27e9\n\n-- Lemas usados\n-- ============\n\n-- variable (a b : \u211d)\n-- #check (mul_div_cancel' a : b \u2260 0 \u2192 b * (a \/ b) = a)\n-- #check (sub_add_cancel a b : a - b + b = a)\n<\/pre>\n<p><b>Demostraciones interactivas<\/b><\/p>\n<p>Se puede interactuar con las demostraciones anteriores en <a href=\"https:\/\/live.lean-lang.org\/#url=https:\/\/raw.githubusercontent.com\/jaalonso\/Calculemus2\/main\/src\/Producto_por_no_nula_y_suma_es_suprayectiva.lean\" rel=\"noopener noreferrer\" target=\"_blank\">Lean 4 Web<\/a>.<\/p>\n<p><b>Referencias<\/b><\/p>\n<ul>\n<li> J. Avigad y P. Massot. <a href=\"https:\/\/bit.ly\/3U4UjBk\">Mathematics in Lean<\/a>, p. 32.<\/li>\n<\/ul>\n","protected":false},"excerpt":{"rendered":"<p>Esta semana he publicado en Calculemus las demostraciones con Lean4 de las siguientes propiedades: 1. Transitividad de la divisibilidad 2. Si a divide a b y a c, entonces divide a b+c 3. La funci\u00f3n (x \u21a6 x + c) es suprayectiva 4. Si c \u2260 0, entonces la funci\u00f3n (x \u21a6 cx) es suprayectiva&#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\/8048"}],"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=8048"}],"version-history":[{"count":1,"href":"https:\/\/www.glc.us.es\/~jalonso\/vestigium\/wp-json\/wp\/v2\/posts\/8048\/revisions"}],"predecessor-version":[{"id":8049,"href":"https:\/\/www.glc.us.es\/~jalonso\/vestigium\/wp-json\/wp\/v2\/posts\/8048\/revisions\/8049"}],"wp:attachment":[{"href":"https:\/\/www.glc.us.es\/~jalonso\/vestigium\/wp-json\/wp\/v2\/media?parent=8048"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.glc.us.es\/~jalonso\/vestigium\/wp-json\/wp\/v2\/categories?post=8048"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.glc.us.es\/~jalonso\/vestigium\/wp-json\/wp\/v2\/tags?post=8048"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}