src/HOL/Library/Quotient.thy
 author wenzelm Fri, 03 Nov 2000 21:35:59 +0100 changeset 10390 1d54567bed24 parent 10333 f12ff6a4bc7b child 10392 f27afee8475d permissions -rw-r--r--
tuned;

(*  Title:      HOL/Library/Quotient.thy
ID:         $Id$
Author:     Gertrud Bauer and Markus Wenzel, TU Muenchen
*)

\title{Quotients}
\author{Gertrud Bauer and Markus Wenzel}
*}

theory Quotient = Main:

text {*
We introduce the notion of quotient types over equivalence relations
via axiomatic type classes.
*}

subsection {* Equivalence relations and quotient types *}

text {*
\medskip Type class @{text equiv} models equivalence relations @{text
"\<sim> :: 'a => 'a => bool"}.
*}

axclass eqv < "term"
consts
eqv :: "('a::eqv) => 'a => bool"    (infixl "\<sim>" 50)

axclass equiv < eqv
equiv_refl [intro]: "x \<sim> x"
equiv_trans [trans]: "x \<sim> y ==> y \<sim> z ==> x \<sim> z"
equiv_sym [elim?]: "x \<sim> y ==> y \<sim> x"

text {*
\medskip The quotient type @{text "'a quot"} consists of all
\emph{equivalence classes} over elements of the base type @{typ 'a}.
*}

typedef 'a quot = "{{x. a \<sim> x}| a::'a::eqv. True}"
by blast

lemma quotI [intro]: "{x. a \<sim> x} \<in> quot"
by (unfold quot_def) blast

lemma quotE [elim]: "R \<in> quot ==> (!!a. R = {x. a \<sim> x} ==> C) ==> C"
by (unfold quot_def) blast

text {*
\medskip Abstracted equivalence classes are the canonical
representation of elements of a quotient type.
*}

constdefs
equivalence_class :: "'a::equiv => 'a quot"    ("\<lfloor>_\<rfloor>")
"\<lfloor>a\<rfloor> == Abs_quot {x. a \<sim> x}"

theorem quot_exhaust: "\<exists>a. A = \<lfloor>a\<rfloor>"
proof (cases A)
fix R assume R: "A = Abs_quot R"
assume "R \<in> quot" hence "\<exists>a. R = {x. a \<sim> x}" by blast
with R have "\<exists>a. A = Abs_quot {x. a \<sim> x}" by blast
thus ?thesis by (unfold equivalence_class_def)
qed

lemma quot_cases [cases type: quot]: "(!!a. A = \<lfloor>a\<rfloor> ==> C) ==> C"
by (insert quot_exhaust) blast

subsection {* Equality on quotients *}

text {*
Equality of canonical quotient elements coincides with the original
relation.
*}

theorem equivalence_class_eq [iff?]: "(\<lfloor>a\<rfloor> = \<lfloor>b\<rfloor>) = (a \<sim> b)"
proof
assume eq: "\<lfloor>a\<rfloor> = \<lfloor>b\<rfloor>"
show "a \<sim> b"
proof -
from eq have "{x. a \<sim> x} = {x. b \<sim> x}"
by (simp only: equivalence_class_def Abs_quot_inject quotI)
moreover have "a \<sim> a" ..
ultimately have "a \<in> {x. b \<sim> x}" by blast
hence "b \<sim> a" by blast
thus ?thesis ..
qed
next
assume ab: "a \<sim> b"
show "\<lfloor>a\<rfloor> = \<lfloor>b\<rfloor>"
proof -
have "{x. a \<sim> x} = {x. b \<sim> x}"
proof (rule Collect_cong)
fix x show "(a \<sim> x) = (b \<sim> x)"
proof
from ab have "b \<sim> a" ..
also assume "a \<sim> x"
finally show "b \<sim> x" .
next
note ab
also assume "b \<sim> x"
finally show "a \<sim> x" .
qed
qed
thus ?thesis by (simp only: equivalence_class_def)
qed
qed

subsection {* Picking representing elements *}

constdefs
pick :: "'a::equiv quot => 'a"
"pick A == SOME a. A = \<lfloor>a\<rfloor>"

theorem pick_equiv [intro]: "pick \<lfloor>a\<rfloor> \<sim> a"
proof (unfold pick_def)
show "(SOME x. \<lfloor>a\<rfloor> = \<lfloor>x\<rfloor>) \<sim> a"
proof (rule someI2)
show "\<lfloor>a\<rfloor> = \<lfloor>a\<rfloor>" ..
fix x assume "\<lfloor>a\<rfloor> = \<lfloor>x\<rfloor>"
hence "a \<sim> x" .. thus "x \<sim> a" ..
qed
qed

theorem pick_inverse: "\<lfloor>pick A\<rfloor> = A"
proof (cases A)
fix a assume a: "A = \<lfloor>a\<rfloor>"
hence "pick A \<sim> a" by (simp only: pick_equiv)
hence "\<lfloor>pick A\<rfloor> = \<lfloor>a\<rfloor>" ..
with a show ?thesis by simp
qed

text {*
\medskip The following rules support canonical function definitions
on quotient types.
*}

theorem cong_definition1:
"(!!X. f X == g (pick X)) ==>
(!!x x'. x \<sim> x' ==> g x = g x') ==>
f \<lfloor>a\<rfloor> = g a"
proof -
assume cong: "!!x x'. x \<sim> x' ==> g x = g x'"
assume "!!X. f X == g (pick X)"
hence "f \<lfloor>a\<rfloor> = g (pick \<lfloor>a\<rfloor>)" by (simp only:)
also have "\<dots> = g a"
proof (rule cong)
show "pick \<lfloor>a\<rfloor> \<sim> a" ..
qed
finally show ?thesis .
qed

theorem cong_definition2:
"(!!X Y. f X Y == g (pick X) (pick Y)) ==>
(!!x x' y y'. x \<sim> x' ==> y \<sim> y' ==> g x y = g x' y') ==>
f \<lfloor>a\<rfloor> \<lfloor>b\<rfloor> = g a b"
proof -
assume cong: "!!x x' y y'. x \<sim> x' ==> y \<sim> y' ==> g x y = g x' y'"
assume "!!X Y. f X Y == g (pick X) (pick Y)"
hence "f \<lfloor>a\<rfloor> \<lfloor>b\<rfloor> = g (pick \<lfloor>a\<rfloor>) (pick \<lfloor>b\<rfloor>)" by (simp only:)
also have "\<dots> = g a b"
proof (rule cong)
show "pick \<lfloor>a\<rfloor> \<sim> a" ..
show "pick \<lfloor>b\<rfloor> \<sim> b" ..
qed
finally show ?thesis .
qed

end