(* 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