(* Title: HOL/Library/Quotient_Type.thy
Author: Markus Wenzel, TU Muenchen
*)
header {* Quotient types *}
theory Quotient_Type
imports Main
begin
text {*
We introduce the notion of quotient types over equivalence relations
via type classes.
*}
subsection {* Equivalence relations and quotient types *}
text {*
\medskip Type class @{text equiv} models equivalence relations @{text
"\ :: 'a => 'a => bool"}.
*}
class eqv =
fixes eqv :: "'a \ 'a \ bool" (infixl "\" 50)
class equiv = eqv +
assumes equiv_refl [intro]: "x \ x"
assumes equiv_trans [trans]: "x \ y \ y \ z \ x \ z"
assumes equiv_sym [sym]: "x \ y \ y \ x"
lemma equiv_not_sym [sym]: "\ (x \ y) ==> \ (y \ (x::'a::equiv))"
proof -
assume "\ (x \ y)" then show "\ (y \ x)"
by (rule contrapos_nn) (rule equiv_sym)
qed
lemma not_equiv_trans1 [trans]: "\ (x \ y) ==> y \ z ==> \ (x \ (z::'a::equiv))"
proof -
assume "\ (x \ y)" and "y \ z"
show "\ (x \ z)"
proof
assume "x \ z"
also from `y \ z` have "z \ y" ..
finally have "x \ y" .
with `\ (x \ y)` show False by contradiction
qed
qed
lemma not_equiv_trans2 [trans]: "x \ y ==> \ (y \ z) ==> \ (x \ (z::'a::equiv))"
proof -
assume "\ (y \ z)" then have "\ (z \ y)" ..
also assume "x \ y" then have "y \ x" ..
finally have "\ (z \ x)" . then show "(\ x \ z)" ..
qed
text {*
\medskip The quotient type @{text "'a quot"} consists of all
\emph{equivalence classes} over elements of the base type @{typ 'a}.
*}
definition "quot = {{x. a \ x} | a::'a::eqv. True}"
typedef 'a quot = "quot :: 'a::eqv set set"
unfolding quot_def by blast
lemma quotI [intro]: "{x. a \ x} \ quot"
unfolding quot_def by blast
lemma quotE [elim]: "R \ quot ==> (!!a. R = {x. a \ x} ==> C) ==> C"
unfolding quot_def by blast
text {*
\medskip Abstracted equivalence classes are the canonical
representation of elements of a quotient type.
*}
definition
"class" :: "'a::equiv => 'a quot" ("\_\") where
"\a\ = Abs_quot {x. a \ x}"
theorem quot_exhaust: "\a. A = \a\"
proof (cases A)
fix R assume R: "A = Abs_quot R"
assume "R \ quot" then have "\a. R = {x. a \ x}" by blast
with R have "\a. A = Abs_quot {x. a \ x}" by blast
then show ?thesis unfolding class_def .
qed
lemma quot_cases [cases type: quot]: "(!!a. A = \a\ ==> C) ==> C"
using quot_exhaust by blast
subsection {* Equality on quotients *}
text {*
Equality of canonical quotient elements coincides with the original
relation.
*}
theorem quot_equality [iff?]: "(\a\ = \b\) = (a \ b)"
proof
assume eq: "\a\ = \b\"
show "a \ b"
proof -
from eq have "{x. a \ x} = {x. b \ x}"
by (simp only: class_def Abs_quot_inject quotI)
moreover have "a \ a" ..
ultimately have "a \ {x. b \ x}" by blast
then have "b \ a" by blast
then show ?thesis ..
qed
next
assume ab: "a \ b"
show "\a\ = \b\"
proof -
have "{x. a \ x} = {x. b \ x}"
proof (rule Collect_cong)
fix x show "(a \ x) = (b \ x)"
proof
from ab have "b \ a" ..
also assume "a \ x"
finally show "b \ x" .
next
note ab
also assume "b \ x"
finally show "a \ x" .
qed
qed
then show ?thesis by (simp only: class_def)
qed
qed
subsection {* Picking representing elements *}
definition
pick :: "'a::equiv quot => 'a" where
"pick A = (SOME a. A = \a\)"
theorem pick_equiv [intro]: "pick \a\ \ a"
proof (unfold pick_def)
show "(SOME x. \a\ = \x\) \ a"
proof (rule someI2)
show "\a\ = \a\" ..
fix x assume "\a\ = \x\"
then have "a \ x" .. then show "x \ a" ..
qed
qed
theorem pick_inverse [intro]: "\pick A\ = A"
proof (cases A)
fix a assume a: "A = \a\"
then have "pick A \ a" by (simp only: pick_equiv)
then have "\pick A\ = \a\" ..
with a show ?thesis by simp
qed
text {*
\medskip The following rules support canonical function definitions
on quotient types (with up to two arguments). Note that the
stripped-down version without additional conditions is sufficient
most of the time.
*}
theorem quot_cond_function:
assumes eq: "!!X Y. P X Y ==> f X Y == g (pick X) (pick Y)"
and cong: "!!x x' y y'. \x\ = \x'\ ==> \y\ = \y'\
==> P \x\ \y\ ==> P \x'\ \y'\ ==> g x y = g x' y'"
and P: "P \a\ \b\"
shows "f \a\ \b\ = g a b"
proof -
from eq and P have "f \a\ \b\ = g (pick \a\) (pick \b\)" by (simp only:)
also have "... = g a b"
proof (rule cong)
show "\pick \a\\ = \a\" ..
moreover
show "\pick \b\\ = \b\" ..
moreover
show "P \a\ \b\" by (rule P)
ultimately show "P \pick \a\\ \pick \b\\" by (simp only:)
qed
finally show ?thesis .
qed
theorem quot_function:
assumes "!!X Y. f X Y == g (pick X) (pick Y)"
and "!!x x' y y'. \x\ = \x'\ ==> \y\ = \y'\ ==> g x y = g x' y'"
shows "f \a\ \b\ = g a b"
using assms and TrueI
by (rule quot_cond_function)
theorem quot_function':
"(!!X Y. f X Y == g (pick X) (pick Y)) ==>
(!!x x' y y'. x \ x' ==> y \ y' ==> g x y = g x' y') ==>
f \a\ \b\ = g a b"
by (rule quot_function) (simp_all only: quot_equality)
end