Theory Typedef (Isabelle2005: October 2005)

# Theory Typedef

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theory Typedef
imports Set
uses (Tools/typedef_package.ML)
begin

    (*  Title:      HOL/Typedef.thy
        ID:         $Id: Typedef.thy,v 1.14 2005/06/17 14:13:05 haftmann Exp $
        Author:     Markus Wenzel, TU Munich
    *)

    header {* HOL type definitions *}

    theory Typedef
    imports Set
    uses ("Tools/typedef_package.ML")
    begin

    locale type_definition =
      fixes Rep and Abs and A
      assumes Rep: "Rep x ∈ A"
        and Rep_inverse: "Abs (Rep x) = x"
        and Abs_inverse: "y ∈ A ==> Rep (Abs y) = y"
      -- {* This will be axiomatized for each typedef! *}

    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 ∈ A" and y: "y ∈ 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 ∈ 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 ∈ A ==> P"
      shows P
    proof (rule r)
      have "Abs (Rep x) = x" by (rule Rep_inverse)
      thus "x = Abs (Rep x)" ..
      show "Rep x ∈ A" by (rule Rep)
    qed

    lemma (in type_definition) Rep_induct [induct set]:
      assumes y: "y ∈ 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 ∈ A ==> P (Abs y)"
      shows "P x"
    proof -
      have "Rep x ∈ 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"

    setup TypedefPackage.setup

    end

lemma Rep_inject:

      type_definition Rep Abs A ==> (Rep x = Rep y) = (x = y)

lemma Abs_inject:

      [| type_definition Rep Abs A; x ∈ A; y ∈ A |] ==> (Abs x = Abs y) = (x = y)

lemma Rep_cases:

      [| type_definition Rep Abs A; y ∈ A; !!x. y = Rep x ==> P |] ==> P

lemma Abs_cases:

      [| type_definition Rep Abs A; !!y. [| x = Abs y; y ∈ A |] ==> P |] ==> P

lemma Rep_induct:

      [| type_definition Rep Abs A; y ∈ A; !!x. P (Rep x) |] ==> P y

lemma Abs_induct:

      [| type_definition Rep Abs A; !!y. y ∈ A ==> P (Abs y) |] ==> P x