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Imagen_de_la_interseccion_de_aplicaciones_inyectivas.thy
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Imagen_de_la_interseccion_de_aplicaciones_inyectivas.thy
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(* Imagen_de_la_interseccion_de_aplicaciones_inyectivas.thy
-- Si f es inyectiva, entonces f[s] \<inter> f[t] \<subseteq> f[s \<inter> t].
-- José A. Alonso Jiménez <https://jaalonso.github.io>
-- Sevilla, 15-abril-2024
-- ------------------------------------------------------------------ *)
(* ---------------------------------------------------------------------
-- Demostrar que si f es inyectiva, entonces
-- f ` s \<inter> f ` t \<subseteq> f ` (s \<inter> t)
-- ------------------------------------------------------------------ *)
theory Imagen_de_la_interseccion_de_aplicaciones_inyectivas
imports Main
begin
(* 1\<ordfeminine> demostración *)
lemma
assumes "inj f"
shows "f ` s \<inter> f ` t \<subseteq> f ` (s \<inter> t)"
proof (rule subsetI)
fix y
assume "y \<in> f ` s \<inter> f ` t"
then have "y \<in> f ` s"
by (rule IntD1)
then show "y \<in> f ` (s \<inter> t)"
proof (rule imageE)
fix x
assume "y = f x"
assume "x \<in> s"
have "x \<in> t"
proof -
have "y \<in> f ` t"
using \<open>y \<in> f ` s \<inter> f ` t\<close> by (rule IntD2)
then show "x \<in> t"
proof (rule imageE)
fix z
assume "y = f z"
assume "z \<in> t"
have "f x = f z"
using \<open>y = f x\<close> \<open>y = f z\<close> by (rule subst)
with \<open>inj f\<close> have "x = z"
by (simp only: inj_eq)
then show "x \<in> t"
using \<open>z \<in> t\<close> by (rule ssubst)
qed
qed
with \<open>x \<in> s\<close> have "x \<in> s \<inter> t"
by (rule IntI)
with \<open>y = f x\<close> show "y \<in> f ` (s \<inter> t)"
by (rule image_eqI)
qed
qed
(* 2\<ordfeminine> demostración *)
lemma
assumes "inj f"
shows "f ` s \<inter> f ` t \<subseteq> f ` (s \<inter> t)"
proof
fix y
assume "y \<in> f ` s \<inter> f ` t"
then have "y \<in> f ` s" by simp
then show "y \<in> f ` (s \<inter> t)"
proof
fix x
assume "y = f x"
assume "x \<in> s"
have "x \<in> t"
proof -
have "y \<in> f ` t" using \<open>y \<in> f ` s \<inter> f ` t\<close> by simp
then show "x \<in> t"
proof
fix z
assume "y = f z"
assume "z \<in> t"
have "f x = f z" using \<open>y = f x\<close> \<open>y = f z\<close> by simp
with \<open>inj f\<close> have "x = z" by (simp only: inj_eq)
then show "x \<in> t" using \<open>z \<in> t\<close> by simp
qed
qed
with \<open>x \<in> s\<close> have "x \<in> s \<inter> t" by simp
with \<open>y = f x\<close> show "y \<in> f ` (s \<inter> t)" by simp
qed
qed
(* 3\<ordfeminine> demostración *)
lemma
assumes "inj f"
shows "f ` s \<inter> f ` t \<subseteq> f ` (s \<inter> t)"
using assms
by (simp only: image_Int)
(* 4\<ordfeminine> demostración *)
lemma
assumes "inj f"
shows "f ` s \<inter> f ` t \<subseteq> f ` (s \<inter> t)"
using assms
unfolding inj_def
by auto
end