src/HOL/Typedef.thy
 author wenzelm Wed Jul 24 00:10:52 2002 +0200 (2002-07-24) changeset 13412 666137b488a4 parent 12023 d982f98e0f0d child 13421 8fcdf4a26468 permissions -rw-r--r--
predicate defs via locales;
1 (*  Title:      HOL/Typedef.thy
2     ID:         \$Id\$
3     Author:     Markus Wenzel, TU Munich
4 *)
6 header {* HOL type definitions *}
8 theory Typedef = Set
9 files ("Tools/typedef_package.ML"):
11 locale type_definition =
12   fixes Rep and Abs and A
13   assumes Rep: "Rep x \<in> A"
14     and Rep_inverse: "Abs (Rep x) = x"
15     and Abs_inverse: "y \<in> A ==> Rep (Abs y) = y"
16   -- {* This will be axiomatized for each typedef! *}
18 lemmas type_definitionI [intro] =
19   type_definition.intro [OF type_definition_axioms.intro]
21 lemma (in type_definition) Rep_inject:
22   "(Rep x = Rep y) = (x = y)"
23 proof
24   assume "Rep x = Rep y"
25   hence "Abs (Rep x) = Abs (Rep y)" by (simp only:)
26   also have "Abs (Rep x) = x" by (rule Rep_inverse)
27   also have "Abs (Rep y) = y" by (rule Rep_inverse)
28   finally show "x = y" .
29 next
30   assume "x = y"
31   thus "Rep x = Rep y" by (simp only:)
32 qed
34 lemma (in type_definition) Abs_inject:
35   assumes x: "x \<in> A" and y: "y \<in> A"
36   shows "(Abs x = Abs y) = (x = y)"
37 proof
38   assume "Abs x = Abs y"
39   hence "Rep (Abs x) = Rep (Abs y)" by (simp only:)
40   also from x have "Rep (Abs x) = x" by (rule Abs_inverse)
41   also from y have "Rep (Abs y) = y" by (rule Abs_inverse)
42   finally show "x = y" .
43 next
44   assume "x = y"
45   thus "Abs x = Abs y" by (simp only:)
46 qed
48 lemma (in type_definition) Rep_cases [cases set]:
49   assumes y: "y \<in> A"
50     and hyp: "!!x. y = Rep x ==> P"
51   shows P
52 proof (rule hyp)
53   from y have "Rep (Abs y) = y" by (rule Abs_inverse)
54   thus "y = Rep (Abs y)" ..
55 qed
57 lemma (in type_definition) Abs_cases [cases type]:
58   assumes r: "!!y. x = Abs y ==> y \<in> A ==> P"
59   shows P
60 proof (rule r)
61   have "Abs (Rep x) = x" by (rule Rep_inverse)
62   thus "x = Abs (Rep x)" ..
63   show "Rep x \<in> A" by (rule Rep)
64 qed
66 lemma (in type_definition) Rep_induct [induct set]:
67   assumes y: "y \<in> A"
68     and hyp: "!!x. P (Rep x)"
69   shows "P y"
70 proof -
71   have "P (Rep (Abs y))" by (rule hyp)
72   also from y have "Rep (Abs y) = y" by (rule Abs_inverse)
73   finally show "P y" .
74 qed
76 lemma (in type_definition) Abs_induct [induct type]:
77   assumes r: "!!y. y \<in> A ==> P (Abs y)"
78   shows "P x"
79 proof -
80   have "Rep x \<in> A" by (rule Rep)
81   hence "P (Abs (Rep x))" by (rule r)
82   also have "Abs (Rep x) = x" by (rule Rep_inverse)
83   finally show "P x" .
84 qed
86 use "Tools/typedef_package.ML"
88 end