src/HOL/Typedef.thy
 author wenzelm Wed Jul 24 22:15:55 2002 +0200 (2002-07-24) changeset 13421 8fcdf4a26468 parent 13412 666137b488a4 child 15131 c69542757a4d permissions -rw-r--r--
simplified locale predicates;
```     1 (*  Title:      HOL/Typedef.thy
```
```     2     ID:         \$Id\$
```
```     3     Author:     Markus Wenzel, TU Munich
```
```     4 *)
```
```     5
```
```     6 header {* HOL type definitions *}
```
```     7
```
```     8 theory Typedef = Set
```
```     9 files ("Tools/typedef_package.ML"):
```
```    10
```
```    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! *}
```
```    17
```
```    18 lemma (in type_definition) Rep_inject:
```
```    19   "(Rep x = Rep y) = (x = y)"
```
```    20 proof
```
```    21   assume "Rep x = Rep y"
```
```    22   hence "Abs (Rep x) = Abs (Rep y)" by (simp only:)
```
```    23   also have "Abs (Rep x) = x" by (rule Rep_inverse)
```
```    24   also have "Abs (Rep y) = y" by (rule Rep_inverse)
```
```    25   finally show "x = y" .
```
```    26 next
```
```    27   assume "x = y"
```
```    28   thus "Rep x = Rep y" by (simp only:)
```
```    29 qed
```
```    30
```
```    31 lemma (in type_definition) Abs_inject:
```
```    32   assumes x: "x \<in> A" and y: "y \<in> A"
```
```    33   shows "(Abs x = Abs y) = (x = y)"
```
```    34 proof
```
```    35   assume "Abs x = Abs y"
```
```    36   hence "Rep (Abs x) = Rep (Abs y)" by (simp only:)
```
```    37   also from x have "Rep (Abs x) = x" by (rule Abs_inverse)
```
```    38   also from y have "Rep (Abs y) = y" by (rule Abs_inverse)
```
```    39   finally show "x = y" .
```
```    40 next
```
```    41   assume "x = y"
```
```    42   thus "Abs x = Abs y" by (simp only:)
```
```    43 qed
```
```    44
```
```    45 lemma (in type_definition) Rep_cases [cases set]:
```
```    46   assumes y: "y \<in> A"
```
```    47     and hyp: "!!x. y = Rep x ==> P"
```
```    48   shows P
```
```    49 proof (rule hyp)
```
```    50   from y have "Rep (Abs y) = y" by (rule Abs_inverse)
```
```    51   thus "y = Rep (Abs y)" ..
```
```    52 qed
```
```    53
```
```    54 lemma (in type_definition) Abs_cases [cases type]:
```
```    55   assumes r: "!!y. x = Abs y ==> y \<in> A ==> P"
```
```    56   shows P
```
```    57 proof (rule r)
```
```    58   have "Abs (Rep x) = x" by (rule Rep_inverse)
```
```    59   thus "x = Abs (Rep x)" ..
```
```    60   show "Rep x \<in> A" by (rule Rep)
```
```    61 qed
```
```    62
```
```    63 lemma (in type_definition) Rep_induct [induct set]:
```
```    64   assumes y: "y \<in> A"
```
```    65     and hyp: "!!x. P (Rep x)"
```
```    66   shows "P y"
```
```    67 proof -
```
```    68   have "P (Rep (Abs y))" by (rule hyp)
```
```    69   also from y have "Rep (Abs y) = y" by (rule Abs_inverse)
```
```    70   finally show "P y" .
```
```    71 qed
```
```    72
```
```    73 lemma (in type_definition) Abs_induct [induct type]:
```
```    74   assumes r: "!!y. y \<in> A ==> P (Abs y)"
```
```    75   shows "P x"
```
```    76 proof -
```
```    77   have "Rep x \<in> A" by (rule Rep)
```
```    78   hence "P (Abs (Rep x))" by (rule r)
```
```    79   also have "Abs (Rep x) = x" by (rule Rep_inverse)
```
```    80   finally show "P x" .
```
```    81 qed
```
```    82
```
```    83 use "Tools/typedef_package.ML"
```
```    84
```
```    85 end
```