src/HOL/Library/Quotient.thy
 author wenzelm Thu, 30 Nov 2000 20:05:34 +0100 changeset 10551 ec9fab41b3a0 parent 10505 b89e6cc963e3 child 10681 ec76e17f73c5 permissions -rw-r--r--
renamed "equivalence_class" to "class";

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

\title{Quotient types}
\author{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"

lemma not_equiv_sym [elim?]: "\<not> (x \<sim> y) ==> \<not> (y \<sim> (x::'a::equiv))"
proof -
assume "\<not> (x \<sim> y)" thus "\<not> (y \<sim> x)"
by (rule contrapos_nn) (rule equiv_sym)
qed

lemma not_equiv_trans1 [trans]: "\<not> (x \<sim> y) ==> y \<sim> z ==> \<not> (x \<sim> (z::'a::equiv))"
proof -
assume "\<not> (x \<sim> y)" and yz: "y \<sim> z"
show "\<not> (x \<sim> z)"
proof
assume "x \<sim> z"
also from yz have "z \<sim> y" ..
finally have "x \<sim> y" .
qed
qed

lemma not_equiv_trans2 [trans]: "x \<sim> y ==> \<not> (y \<sim> z) ==> \<not> (x \<sim> (z::'a::equiv))"
proof -
assume "\<not> (y \<sim> z)" hence "\<not> (z \<sim> y)" ..
also assume "x \<sim> y" hence "y \<sim> x" ..
finally have "\<not> (z \<sim> x)" . thus "(\<not> x \<sim> z)" ..
qed

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
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 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 quot_equality: "(\<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: 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: class_def)
qed
qed

lemma quot_equalI [intro?]: "a \<sim> b ==> \<lfloor>a\<rfloor> = \<lfloor>b\<rfloor>"
by (simp only: quot_equality)

lemma quot_equalD [dest?]: "\<lfloor>a\<rfloor> = \<lfloor>b\<rfloor> ==> a \<sim> b"
by (simp only: quot_equality)

lemma quot_not_equalI [intro?]: "\<not> (a \<sim> b) ==> \<lfloor>a\<rfloor> \<noteq> \<lfloor>b\<rfloor>"

lemma quot_not_equalD [dest?]: "\<lfloor>a\<rfloor> \<noteq> \<lfloor>b\<rfloor> ==> \<not> (a \<sim> b)"

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 [intro]: "\<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 (with up to two arguments).  Note that the
stripped-down version without additional conditions is sufficient
most of the time.
*}

theorem quot_cond_function:
"(!!X Y. P X Y ==> f X Y == g (pick X) (pick Y)) ==>
(!!x x' y y'. \<lfloor>x\<rfloor> = \<lfloor>x'\<rfloor> ==> \<lfloor>y\<rfloor> = \<lfloor>y'\<rfloor>
==> P \<lfloor>x\<rfloor> \<lfloor>y\<rfloor> ==> P \<lfloor>x'\<rfloor> \<lfloor>y'\<rfloor> ==> g x y = g x' y') ==>
P \<lfloor>a\<rfloor> \<lfloor>b\<rfloor> ==> f \<lfloor>a\<rfloor> \<lfloor>b\<rfloor> = g a b"
(is "PROP ?eq ==> PROP ?cong ==> _ ==> _")
proof -
assume cong: "PROP ?cong"
assume "PROP ?eq" and "P \<lfloor>a\<rfloor> \<lfloor>b\<rfloor>"
hence "f \<lfloor>a\<rfloor> \<lfloor>b\<rfloor> = g (pick \<lfloor>a\<rfloor>) (pick \<lfloor>b\<rfloor>)" by (simp only:)
also have "... = g a b"
proof (rule cong)
show "\<lfloor>pick \<lfloor>a\<rfloor>\<rfloor> = \<lfloor>a\<rfloor>" ..
moreover
show "\<lfloor>pick \<lfloor>b\<rfloor>\<rfloor> = \<lfloor>b\<rfloor>" ..
moreover
show "P \<lfloor>a\<rfloor> \<lfloor>b\<rfloor>" .
ultimately show "P \<lfloor>pick \<lfloor>a\<rfloor>\<rfloor> \<lfloor>pick \<lfloor>b\<rfloor>\<rfloor>" by (simp only:)
qed
finally show ?thesis .
qed

theorem quot_function:
"(!!X Y. f X Y == g (pick X) (pick Y)) ==>
(!!x x' y y'. \<lfloor>x\<rfloor> = \<lfloor>x'\<rfloor> ==> \<lfloor>y\<rfloor> = \<lfloor>y'\<rfloor> ==> g x y = g x' y') ==>
f \<lfloor>a\<rfloor> \<lfloor>b\<rfloor> = g a b"
proof -
case antecedent from this TrueI
show ?thesis by (rule quot_cond_function)
qed

theorem quot_function':
"(!!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"
by  (rule quot_function) (simp only: quot_equality)+

end