(* 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