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
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```     2     ID:         \$Id\$
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```     3     Author:     Markus Wenzel, TU Munich
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```     4 *)
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```     5
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```     6 header {* HOL type definitions *}
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```     7
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```     8 theory Typedef = Set
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```     9 files ("Tools/typedef_package.ML"):
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```    10
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```    11 locale type_definition =
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```    12   fixes Rep and Abs and A
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```    13   assumes Rep: "Rep x \<in> A"
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```    14     and Rep_inverse: "Abs (Rep x) = x"
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```    15     and Abs_inverse: "y \<in> A ==> Rep (Abs y) = y"
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```    16   -- {* This will be axiomatized for each typedef! *}
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```    17
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```    18 lemmas type_definitionI [intro] =
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```    19   type_definition.intro [OF type_definition_axioms.intro]
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```    20
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```    21 lemma (in type_definition) Rep_inject:
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```    22   "(Rep x = Rep y) = (x = y)"
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```    23 proof
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```    24   assume "Rep x = Rep y"
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```    25   hence "Abs (Rep x) = Abs (Rep y)" by (simp only:)
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```    26   also have "Abs (Rep x) = x" by (rule Rep_inverse)
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```    27   also have "Abs (Rep y) = y" by (rule Rep_inverse)
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```    28   finally show "x = y" .
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```    29 next
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```    30   assume "x = y"
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```    31   thus "Rep x = Rep y" by (simp only:)
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```    32 qed
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```    33
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```    34 lemma (in type_definition) Abs_inject:
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```    35   assumes x: "x \<in> A" and y: "y \<in> A"
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```    36   shows "(Abs x = Abs y) = (x = y)"
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```    37 proof
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```    38   assume "Abs x = Abs y"
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```    39   hence "Rep (Abs x) = Rep (Abs y)" by (simp only:)
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```    40   also from x have "Rep (Abs x) = x" by (rule Abs_inverse)
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```    41   also from y have "Rep (Abs y) = y" by (rule Abs_inverse)
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```    42   finally show "x = y" .
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```    43 next
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```    44   assume "x = y"
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```    45   thus "Abs x = Abs y" by (simp only:)
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```    46 qed
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```    47
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```    48 lemma (in type_definition) Rep_cases [cases set]:
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```    49   assumes y: "y \<in> A"
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```    50     and hyp: "!!x. y = Rep x ==> P"
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```    51   shows P
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```    52 proof (rule hyp)
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```    53   from y have "Rep (Abs y) = y" by (rule Abs_inverse)
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```    54   thus "y = Rep (Abs y)" ..
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```    55 qed
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```    56
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```    57 lemma (in type_definition) Abs_cases [cases type]:
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```    58   assumes r: "!!y. x = Abs y ==> y \<in> A ==> P"
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```    59   shows P
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```    60 proof (rule r)
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```    61   have "Abs (Rep x) = x" by (rule Rep_inverse)
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```    62   thus "x = Abs (Rep x)" ..
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```    63   show "Rep x \<in> A" by (rule Rep)
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```    64 qed
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```    65
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```    66 lemma (in type_definition) Rep_induct [induct set]:
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```    67   assumes y: "y \<in> A"
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```    68     and hyp: "!!x. P (Rep x)"
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```    69   shows "P y"
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```    70 proof -
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```    71   have "P (Rep (Abs y))" by (rule hyp)
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```    72   also from y have "Rep (Abs y) = y" by (rule Abs_inverse)
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```    73   finally show "P y" .
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```    74 qed
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```    75
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```    76 lemma (in type_definition) Abs_induct [induct type]:
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```    77   assumes r: "!!y. y \<in> A ==> P (Abs y)"
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```    78   shows "P x"
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```    79 proof -
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```    80   have "Rep x \<in> A" by (rule Rep)
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```    81   hence "P (Abs (Rep x))" by (rule r)
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```    82   also have "Abs (Rep x) = x" by (rule Rep_inverse)
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```    83   finally show "P x" .
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```    84 qed
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```    85
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```    86 use "Tools/typedef_package.ML"
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```    87
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```    88 end
```