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src/HOL/Typedef.thy

author | haftmann |

Tue Jun 05 15:16:08 2007 +0200 (2007-06-05) | |

changeset 23247 | b99dce43d252 |

parent 22846 | fb79144af9a3 |

child 23433 | c2c10abd2a1e |

permissions | -rw-r--r-- |

merged Code_Generator.thy into HOL.thy

1 (* Title: HOL/Typedef.thy

2 ID: $Id$

3 Author: Markus Wenzel, TU Munich

4 *)

6 header {* HOL type definitions *}

8 theory Typedef

9 imports Set

10 uses

11 ("Tools/typedef_package.ML")

12 ("Tools/typecopy_package.ML")

13 ("Tools/typedef_codegen.ML")

14 begin

16 ML {*

17 structure HOL = struct val thy = theory "HOL" end;

18 *} -- "belongs to theory HOL"

20 locale type_definition =

21 fixes Rep and Abs and A

22 assumes Rep: "Rep x \<in> A"

23 and Rep_inverse: "Abs (Rep x) = x"

24 and Abs_inverse: "y \<in> A ==> Rep (Abs y) = y"

25 -- {* This will be axiomatized for each typedef! *}

26 begin

28 lemma Rep_inject:

29 "(Rep x = Rep y) = (x = y)"

30 proof

31 assume "Rep x = Rep y"

32 hence "Abs (Rep x) = Abs (Rep y)" by (simp only:)

33 also have "Abs (Rep x) = x" by (rule Rep_inverse)

34 also have "Abs (Rep y) = y" by (rule Rep_inverse)

35 finally show "x = y" .

36 next

37 assume "x = y"

38 thus "Rep x = Rep y" by (simp only:)

39 qed

41 lemma Abs_inject:

42 assumes x: "x \<in> A" and y: "y \<in> A"

43 shows "(Abs x = Abs y) = (x = y)"

44 proof

45 assume "Abs x = Abs y"

46 hence "Rep (Abs x) = Rep (Abs y)" by (simp only:)

47 also from x have "Rep (Abs x) = x" by (rule Abs_inverse)

48 also from y have "Rep (Abs y) = y" by (rule Abs_inverse)

49 finally show "x = y" .

50 next

51 assume "x = y"

52 thus "Abs x = Abs y" by (simp only:)

53 qed

55 lemma Rep_cases [cases set]:

56 assumes y: "y \<in> A"

57 and hyp: "!!x. y = Rep x ==> P"

58 shows P

59 proof (rule hyp)

60 from y have "Rep (Abs y) = y" by (rule Abs_inverse)

61 thus "y = Rep (Abs y)" ..

62 qed

64 lemma Abs_cases [cases type]:

65 assumes r: "!!y. x = Abs y ==> y \<in> A ==> P"

66 shows P

67 proof (rule r)

68 have "Abs (Rep x) = x" by (rule Rep_inverse)

69 thus "x = Abs (Rep x)" ..

70 show "Rep x \<in> A" by (rule Rep)

71 qed

73 lemma Rep_induct [induct set]:

74 assumes y: "y \<in> A"

75 and hyp: "!!x. P (Rep x)"

76 shows "P y"

77 proof -

78 have "P (Rep (Abs y))" by (rule hyp)

79 also from y have "Rep (Abs y) = y" by (rule Abs_inverse)

80 finally show "P y" .

81 qed

83 lemma Abs_induct [induct type]:

84 assumes r: "!!y. y \<in> A ==> P (Abs y)"

85 shows "P x"

86 proof -

87 have "Rep x \<in> A" by (rule Rep)

88 hence "P (Abs (Rep x))" by (rule r)

89 also have "Abs (Rep x) = x" by (rule Rep_inverse)

90 finally show "P x" .

91 qed

93 end

95 use "Tools/typedef_package.ML"

96 use "Tools/typecopy_package.ML"

97 use "Tools/typedef_codegen.ML"

99 setup {*

100 TypecopyPackage.setup

101 #> TypedefCodegen.setup

102 *}

104 end