Theory Quotient_Type

theory Quotient_Type
imports Main
(*  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