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Imagen_de_la_interseccion_general_mediante_inyectiva.thy
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(* Imagen_de_la_interseccion_general_mediante_inyectiva.thy
-- Imagen de la interseccion general mediante inyectiva
-- José A. Alonso Jiménez <https://jaalonso.github.io>
-- Sevilla, 29-abril-2024
-- ------------------------------------------------------------------ *)
(* ---------------------------------------------------------------------
-- Demostrar que si f es inyectiva, entonces
-- \<Inter>ᵢf[Aᵢ] \<subseteq> f[\<Inter>ᵢAᵢ]
-- ------------------------------------------------------------------ *)
theory Imagen_de_la_interseccion_general_mediante_inyectiva
imports Main
begin
(* 1\<ordfeminine> demostración *)
lemma
assumes "i \<in> I"
"inj f"
shows "(\<Inter> i \<in> I. f ` A i) \<subseteq> f ` (\<Inter> i \<in> I. A i)"
proof (rule subsetI)
fix y
assume "y \<in> (\<Inter> i \<in> I. f ` A i)"
then have "y \<in> f ` A i"
using \<open>i \<in> I\<close> by (rule INT_D)
then show "y \<in> f ` (\<Inter> i \<in> I. A i)"
proof (rule imageE)
fix x
assume "y = f x"
assume "x \<in> A i"
have "x \<in> (\<Inter> i \<in> I. A i)"
proof (rule INT_I)
fix j
assume "j \<in> I"
show "x \<in> A j"
proof -
have "y \<in> f ` A j"
using \<open>y \<in> (\<Inter>i\<in>I. f ` A i)\<close> \<open>j \<in> I\<close> by (rule INT_D)
then show "x \<in> A j"
proof (rule imageE)
fix z
assume "y = f z"
assume "z \<in> A j"
have "f z = f x"
using \<open>y = f z\<close> \<open>y = f x\<close> by (rule subst)
with \<open>inj f\<close> have "z = x"
by (rule injD)
then show "x \<in> A j"
using \<open>z \<in> A j\<close> by (rule subst)
qed
qed
qed
then have "f x \<in> f ` (\<Inter> i \<in> I. A i)"
by (rule imageI)
with \<open>y = f x\<close> show "y \<in> f ` (\<Inter> i \<in> I. A i)"
by (rule ssubst)
qed
qed
(* 2\<ordfeminine> demostración *)
lemma
assumes "i \<in> I"
"inj f"
shows "(\<Inter> i \<in> I. f ` A i) \<subseteq> f ` (\<Inter> i \<in> I. A i)"
proof
fix y
assume "y \<in> (\<Inter> i \<in> I. f ` A i)"
then have "y \<in> f ` A i" using \<open>i \<in> I\<close> by simp
then show "y \<in> f ` (\<Inter> i \<in> I. A i)"
proof
fix x
assume "y = f x"
assume "x \<in> A i"
have "x \<in> (\<Inter> i \<in> I. A i)"
proof
fix j
assume "j \<in> I"
show "x \<in> A j"
proof -
have "y \<in> f ` A j"
using \<open>y \<in> (\<Inter>i\<in>I. f ` A i)\<close> \<open>j \<in> I\<close> by simp
then show "x \<in> A j"
proof
fix z
assume "y = f z"
assume "z \<in> A j"
have "f z = f x" using \<open>y = f z\<close> \<open>y = f x\<close> by simp
with \<open>inj f\<close> have "z = x" by (rule injD)
then show "x \<in> A j" using \<open>z \<in> A j\<close> by simp
qed
qed
qed
then have "f x \<in> f ` (\<Inter> i \<in> I. A i)" by simp
with \<open>y = f x\<close> show "y \<in> f ` (\<Inter> i \<in> I. A i)" by simp
qed
qed
(* 3\<ordfeminine> demostración *)
lemma
assumes "i \<in> I"
"inj f"
shows "(\<Inter> i \<in> I. f ` A i) \<subseteq> f ` (\<Inter> i \<in> I. A i)"
using assms
by (simp add: image_INT)
end