--- a/src/HOL/Typedef.thy Thu Dec 12 17:07:09 2024 +0100
+++ b/src/HOL/Typedef.thy Thu Dec 12 17:07:17 2024 +0100
@@ -23,12 +23,11 @@
proof
assume "Rep x = Rep y"
then have "Abs (Rep x) = Abs (Rep y)" by (simp only:)
- moreover have "Abs (Rep x) = x" by (rule Rep_inverse)
- moreover have "Abs (Rep y) = y" by (rule Rep_inverse)
- ultimately show "x = y" by simp
+ also have "Abs (Rep x) = x" by (rule Rep_inverse)
+ also have "Abs (Rep y) = y" by (rule Rep_inverse)
+ finally show "x = y" .
next
- assume "x = y"
- then show "Rep x = Rep y" by (simp only:)
+ show "x = y \<Longrightarrow> Rep x = Rep y" by (simp only:)
qed
lemma Abs_inject:
@@ -37,12 +36,11 @@
proof
assume "Abs x = Abs y"
then have "Rep (Abs x) = Rep (Abs y)" by (simp only:)
- moreover from \<open>x \<in> A\<close> have "Rep (Abs x) = x" by (rule Abs_inverse)
- moreover from \<open>y \<in> A\<close> have "Rep (Abs y) = y" by (rule Abs_inverse)
- ultimately show "x = y" by simp
+ also from \<open>x \<in> A\<close> have "Rep (Abs x) = x" by (rule Abs_inverse)
+ also from \<open>y \<in> A\<close> have "Rep (Abs y) = y" by (rule Abs_inverse)
+ finally show "x = y" .
next
- assume "x = y"
- then show "Abs x = Abs y" by (simp only:)
+ show "x = y \<Longrightarrow> Abs x = Abs y" by (simp only:)
qed
lemma Rep_cases [cases set]:
@@ -69,8 +67,8 @@
shows "P y"
proof -
have "P (Rep (Abs y))" by (rule hyp)
- moreover from y have "Rep (Abs y) = y" by (rule Abs_inverse)
- ultimately show "P y" by simp
+ also from y have "Rep (Abs y) = y" by (rule Abs_inverse)
+ finally show "P y" .
qed
lemma Abs_induct [induct type]:
@@ -79,8 +77,8 @@
proof -
have "Rep x \<in> A" by (rule Rep)
then have "P (Abs (Rep x))" by (rule r)
- moreover have "Abs (Rep x) = x" by (rule Rep_inverse)
- ultimately show "P x" by simp
+ also have "Abs (Rep x) = x" by (rule Rep_inverse)
+ finally show "P x" .
qed
lemma Rep_range: "range Rep = A"
@@ -99,10 +97,11 @@
show "Abs ` A \<subseteq> UNIV" by (rule subset_UNIV)
show "UNIV \<subseteq> Abs ` A"
proof
- fix x
- have "x = Abs (Rep x)" by (rule Rep_inverse [symmetric])
- moreover have "Rep x \<in> A" by (rule Rep)
- ultimately show "x \<in> Abs ` A" by (rule image_eqI)
+ show "x \<in> Abs ` A" for x
+ proof (rule image_eqI)
+ show "x = Abs (Rep x)" by (rule Rep_inverse [symmetric])
+ show "Rep x \<in> A" by (rule Rep)
+ qed
qed
qed