--- a/src/HOL/Typedef.thy Wed Jul 24 00:09:44 2002 +0200
+++ b/src/HOL/Typedef.thy Wed Jul 24 00:10:52 2002 +0200
@@ -8,105 +8,79 @@
theory Typedef = Set
files ("Tools/typedef_package.ML"):
-constdefs
- type_definition :: "('a => 'b) => ('b => 'a) => 'b set => bool"
- "type_definition Rep Abs A ==
- (\<forall>x. Rep x \<in> A) \<and>
- (\<forall>x. Abs (Rep x) = x) \<and>
- (\<forall>y \<in> A. Rep (Abs y) = y)"
- -- {* This will be stated as an axiom for each typedef! *}
+locale type_definition =
+ 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"
+ -- {* This will be axiomatized for each typedef! *}
-lemma type_definitionI [intro]:
- "(!!x. Rep x \<in> A) ==>
- (!!x. Abs (Rep x) = x) ==>
- (!!y. y \<in> A ==> Rep (Abs y) = y) ==>
- type_definition Rep Abs A"
- by (unfold type_definition_def) blast
-
-theorem Rep: "type_definition Rep Abs A ==> Rep x \<in> A"
- by (unfold type_definition_def) blast
+lemmas type_definitionI [intro] =
+ type_definition.intro [OF type_definition_axioms.intro]
-theorem Rep_inverse: "type_definition Rep Abs A ==> Abs (Rep x) = x"
- by (unfold type_definition_def) blast
-
-theorem Abs_inverse: "type_definition Rep Abs A ==> y \<in> A ==> Rep (Abs y) = y"
- by (unfold type_definition_def) blast
+lemma (in type_definition) Rep_inject:
+ "(Rep x = Rep y) = (x = y)"
+proof
+ assume "Rep x = Rep y"
+ hence "Abs (Rep x) = Abs (Rep y)" by (simp only:)
+ 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"
+ thus "Rep x = Rep y" by (simp only:)
+qed
-theorem Rep_inject: "type_definition Rep Abs A ==> (Rep x = Rep y) = (x = y)"
-proof -
- assume tydef: "type_definition Rep Abs A"
- show ?thesis
- proof
- assume "Rep x = Rep y"
- hence "Abs (Rep x) = Abs (Rep y)" by (simp only:)
- thus "x = y" by (simp only: Rep_inverse [OF tydef])
- next
- assume "x = y"
- thus "Rep x = Rep y" by simp
- qed
+lemma (in type_definition) Abs_inject:
+ assumes x: "x \<in> A" and y: "y \<in> A"
+ shows "(Abs x = Abs y) = (x = y)"
+proof
+ assume "Abs x = Abs y"
+ hence "Rep (Abs x) = Rep (Abs y)" by (simp only:)
+ also from x have "Rep (Abs x) = x" by (rule Abs_inverse)
+ also from y have "Rep (Abs y) = y" by (rule Abs_inverse)
+ finally show "x = y" .
+next
+ assume "x = y"
+ thus "Abs x = Abs y" by (simp only:)
qed
-theorem Abs_inject:
- "type_definition Rep Abs A ==> x \<in> A ==> y \<in> A ==> (Abs x = Abs y) = (x = y)"
-proof -
- assume tydef: "type_definition Rep Abs A"
- assume x: "x \<in> A" and y: "y \<in> A"
- show ?thesis
- proof
- assume "Abs x = Abs y"
- hence "Rep (Abs x) = Rep (Abs y)" by simp
- moreover from x have "Rep (Abs x) = x" by (rule Abs_inverse [OF tydef])
- moreover from y have "Rep (Abs y) = y" by (rule Abs_inverse [OF tydef])
- ultimately show "x = y" by (simp only:)
- next
- assume "x = y"
- thus "Abs x = Abs y" by simp
- qed
+lemma (in type_definition) Rep_cases [cases set]:
+ assumes y: "y \<in> A"
+ and hyp: "!!x. y = Rep x ==> P"
+ shows P
+proof (rule hyp)
+ from y have "Rep (Abs y) = y" by (rule Abs_inverse)
+ thus "y = Rep (Abs y)" ..
qed
-theorem Rep_cases:
- "type_definition Rep Abs A ==> y \<in> A ==> (!!x. y = Rep x ==> P) ==> P"
-proof -
- assume tydef: "type_definition Rep Abs A"
- assume y: "y \<in> A" and r: "(!!x. y = Rep x ==> P)"
- show P
- proof (rule r)
- from y have "Rep (Abs y) = y" by (rule Abs_inverse [OF tydef])
- thus "y = Rep (Abs y)" ..
- qed
+lemma (in type_definition) Abs_cases [cases type]:
+ assumes r: "!!y. x = Abs y ==> y \<in> A ==> P"
+ shows P
+proof (rule r)
+ have "Abs (Rep x) = x" by (rule Rep_inverse)
+ thus "x = Abs (Rep x)" ..
+ show "Rep x \<in> A" by (rule Rep)
qed
-theorem Abs_cases:
- "type_definition Rep Abs A ==> (!!y. x = Abs y ==> y \<in> A ==> P) ==> P"
+lemma (in type_definition) Rep_induct [induct set]:
+ assumes y: "y \<in> A"
+ and hyp: "!!x. P (Rep x)"
+ shows "P y"
proof -
- assume tydef: "type_definition Rep Abs A"
- assume r: "!!y. x = Abs y ==> y \<in> A ==> P"
- show P
- proof (rule r)
- have "Abs (Rep x) = x" by (rule Rep_inverse [OF tydef])
- thus "x = Abs (Rep x)" ..
- show "Rep x \<in> A" by (rule Rep [OF tydef])
- qed
+ have "P (Rep (Abs y))" by (rule hyp)
+ also from y have "Rep (Abs y) = y" by (rule Abs_inverse)
+ finally show "P y" .
qed
-theorem Rep_induct:
- "type_definition Rep Abs A ==> y \<in> A ==> (!!x. P (Rep x)) ==> P y"
+lemma (in type_definition) Abs_induct [induct type]:
+ assumes r: "!!y. y \<in> A ==> P (Abs y)"
+ shows "P x"
proof -
- assume tydef: "type_definition Rep Abs A"
- assume "!!x. P (Rep x)" hence "P (Rep (Abs y))" .
- moreover assume "y \<in> A" hence "Rep (Abs y) = y" by (rule Abs_inverse [OF tydef])
- ultimately show "P y" by (simp only:)
-qed
-
-theorem Abs_induct:
- "type_definition Rep Abs A ==> (!!y. y \<in> A ==> P (Abs y)) ==> P x"
-proof -
- assume tydef: "type_definition Rep Abs A"
- assume r: "!!y. y \<in> A ==> P (Abs y)"
- have "Rep x \<in> A" by (rule Rep [OF tydef])
+ have "Rep x \<in> A" by (rule Rep)
hence "P (Abs (Rep x))" by (rule r)
- moreover have "Abs (Rep x) = x" by (rule Rep_inverse [OF tydef])
- ultimately show "P x" by (simp only:)
+ also have "Abs (Rep x) = x" by (rule Rep_inverse)
+ finally show "P x" .
qed
use "Tools/typedef_package.ML"