src/HOL/Typedef.thy

predicate defs via locales;

(*  Title:      HOL/Typedef.thy
    ID:         $Id$
    Author:     Markus Wenzel, TU Munich
*)

header {* HOL type definitions *}

theory Typedef = Set
files ("Tools/typedef_package.ML"):

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! *}

lemmas type_definitionI [intro] =
  type_definition.intro [OF type_definition_axioms.intro]

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

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

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

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

lemma (in type_definition) Rep_induct [induct set]:
  assumes y: "y \<in> A"
    and hyp: "!!x. P (Rep x)"
  shows "P y"
proof -
  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

lemma (in type_definition) Abs_induct [induct type]:
  assumes r: "!!y. y \<in> A ==> P (Abs y)"
  shows "P x"
proof -
  have "Rep x \<in> A" by (rule Rep)
  hence "P (Abs (Rep x))" by (rule r)
  also have "Abs (Rep x) = x" by (rule Rep_inverse)
  finally show "P x" .
qed

use "Tools/typedef_package.ML"

end