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Union_con_su_diferencia.thy
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Union_con_su_diferencia.thy
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(* Union_con_su_diferencia.thy
Unión con su diferencia
José A. Alonso Jiménez
Sevilla, 1 de marzo de 2024
---------------------------------------------------------------------
*)
(* ---------------------------------------------------------------------
Demostrar que
(s \ t) \<union> t = s \<union> t
------------------------------------------------------------------- *)
theory Union_con_su_diferencia
imports Main
begin
(* 1\<ordfeminine> demostración *)
lemma "(s - t) \<union> t = s \<union> t"
proof (rule equalityI)
show "(s - t) \<union> t \<subseteq> s \<union> t"
proof (rule subsetI)
fix x
assume "x \<in> (s - t) \<union> t"
then show "x \<in> s \<union> t"
proof (rule UnE)
assume "x \<in> s - t"
then have "x \<in> s"
by (simp only: DiffD1)
then show "x \<in> s \<union> t"
by (simp only: UnI1)
next
assume "x \<in> t"
then show "x \<in> s \<union> t"
by (simp only: UnI2)
qed
qed
next
show "s \<union> t \<subseteq> (s - t) \<union> t"
proof (rule subsetI)
fix x
assume "x \<in> s \<union> t"
then show "x \<in> (s - t) \<union> t"
proof (rule UnE)
assume "x \<in> s"
show "x \<in> (s - t) \<union> t"
proof (cases \<open>x \<in> t\<close>)
assume "x \<in> t"
then show "x \<in> (s - t) \<union> t"
by (simp only: UnI2)
next
assume "x \<notin> t"
with \<open>x \<in> s\<close> have "x \<in> s - t"
by (rule DiffI)
then show "x \<in> (s - t) \<union> t"
by (simp only: UnI1)
qed
next
assume "x \<in> t"
then show "x \<in> (s - t) \<union> t"
by (simp only: UnI2)
qed
qed
qed
(* 2\<ordfeminine> demostración *)
lemma "(s - t) \<union> t = s \<union> t"
proof
show "(s - t) \<union> t \<subseteq> s \<union> t"
proof
fix x
assume "x \<in> (s - t) \<union> t"
then show "x \<in> s \<union> t"
proof
assume "x \<in> s - t"
then have "x \<in> s"
by simp
then show "x \<in> s \<union> t"
by simp
next
assume "x \<in> t"
then show "x \<in> s \<union> t"
by simp
qed
qed
next
show "s \<union> t \<subseteq> (s - t) \<union> t"
proof
fix x
assume "x \<in> s \<union> t"
then show "x \<in> (s - t) \<union> t"
proof
assume "x \<in> s"
show "x \<in> (s - t) \<union> t"
proof
assume "x \<notin> t"
with \<open>x \<in> s\<close> show "x \<in> s - t"
by simp
qed
next
assume "x \<in> t"
then show "x \<in> (s - t) \<union> t"
by simp
qed
qed
qed
(* 3\<ordfeminine> demostración *)
lemma "(s - t) \<union> t = s \<union> t"
by (fact Un_Diff_cancel2)
(* 4\<ordfeminine> demostración *)
lemma "(s - t) \<union> t = s \<union> t"
by auto
end