diff -r 7b915ca69af1 -r 0ec9fd8d8119 src/HOL/Typedef.thy
--- a/src/HOL/Typedef.thy	Thu Sep 03 16:41:43 2015 +0200
+++ b/src/HOL/Typedef.thy	Thu Sep 03 17:14:57 2015 +0200
@@ -13,12 +13,11 @@
   fixes Rep and Abs and A
   assumes Rep: "Rep x \<in> A"
     and Rep_inverse: "Abs (Rep x) = x"
-    and Abs_inverse: "y \<in> A ==> Rep (Abs y) = y"
+    and Abs_inverse: "y \<in> A \<Longrightarrow> Rep (Abs y) = y"
   -- \<open>This will be axiomatized for each typedef!\<close>
 begin
 
-lemma Rep_inject:
-  "(Rep x = Rep y) = (x = y)"
+lemma Rep_inject: "Rep x = Rep y \<longleftrightarrow> x = y"
 proof
   assume "Rep x = Rep y"
   then have "Abs (Rep x) = Abs (Rep y)" by (simp only:)
@@ -27,44 +26,44 @@
   ultimately show "x = y" by simp
 next
   assume "x = y"
-  thus "Rep x = Rep y" by (simp only:)
+  then show "Rep x = Rep y" by (simp only:)
 qed
 
 lemma Abs_inject:
-  assumes x: "x \<in> A" and y: "y \<in> A"
-  shows "(Abs x = Abs y) = (x = y)"
+  assumes "x \<in> A" and "y \<in> A"
+  shows "Abs x = Abs y \<longleftrightarrow> x = y"
 proof
   assume "Abs x = Abs y"
   then have "Rep (Abs x) = Rep (Abs y)" by (simp only:)
-  moreover from x have "Rep (Abs x) = x" by (rule Abs_inverse)
-  moreover from y have "Rep (Abs y) = y" by (rule Abs_inverse)
+  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
 next
   assume "x = y"
-  thus "Abs x = Abs y" by (simp only:)
+  then show "Abs x = Abs y" by (simp only:)
 qed
 
 lemma Rep_cases [cases set]:
-  assumes y: "y \<in> A"
-    and hyp: "!!x. y = Rep x ==> P"
+  assumes "y \<in> A"
+    and hyp: "\<And>x. y = Rep x \<Longrightarrow> P"
   shows P
 proof (rule hyp)
-  from y have "Rep (Abs y) = y" by (rule Abs_inverse)
-  thus "y = Rep (Abs y)" ..
+  from \<open>y \<in> A\<close> have "Rep (Abs y) = y" by (rule Abs_inverse)
+  then show "y = Rep (Abs y)" ..
 qed
 
 lemma Abs_cases [cases type]:
-  assumes r: "!!y. x = Abs y ==> y \<in> A ==> P"
+  assumes r: "\<And>y. x = Abs y \<Longrightarrow> y \<in> A \<Longrightarrow> P"
   shows P
 proof (rule r)
   have "Abs (Rep x) = x" by (rule Rep_inverse)
-  thus "x = Abs (Rep x)" ..
+  then show "x = Abs (Rep x)" ..
   show "Rep x \<in> A" by (rule Rep)
 qed
 
 lemma Rep_induct [induct set]:
   assumes y: "y \<in> A"
-    and hyp: "!!x. P (Rep x)"
+    and hyp: "\<And>x. P (Rep x)"
   shows "P y"
 proof -
   have "P (Rep (Abs y))" by (rule hyp)
@@ -73,7 +72,7 @@
 qed
 
 lemma Abs_induct [induct type]:
-  assumes r: "!!y. y \<in> A ==> P (Abs y)"
+  assumes r: "\<And>y. y \<in> A \<Longrightarrow> P (Abs y)"
   shows "P x"
 proof -
   have "Rep x \<in> A" by (rule Rep)
@@ -84,25 +83,24 @@
 
 lemma Rep_range: "range Rep = A"
 proof
-  show "range Rep <= A" using Rep by (auto simp add: image_def)
-  show "A <= range Rep"
+  show "range Rep \<subseteq> A" using Rep by (auto simp add: image_def)
+  show "A \<subseteq> range Rep"
   proof
-    fix x assume "x : A"
-    hence "x = Rep (Abs x)" by (rule Abs_inverse [symmetric])
-    thus "x : range Rep" by (rule range_eqI)
+    fix x assume "x \<in> A"
+    then have "x = Rep (Abs x)" by (rule Abs_inverse [symmetric])
+    then show "x \<in> range Rep" by (rule range_eqI)
   qed
 qed
 
 lemma Abs_image: "Abs ` A = UNIV"
 proof
-  show "Abs ` A <= UNIV" by (rule subset_UNIV)
-next
-  show "UNIV <= Abs ` A"
+  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 : A" by (rule Rep)
-    ultimately show "x : Abs ` A" by (rule image_eqI)
+    moreover have "Rep x \<in> A" by (rule Rep)
+    ultimately show "x \<in> Abs ` A" by (rule image_eqI)
   qed
 qed