wenzelm@35100: (* Title: HOL/Library/Quotient_Type.thy wenzelm@10483: Author: Markus Wenzel, TU Muenchen wenzelm@10250: *) wenzelm@10250: wenzelm@14706: header {* Quotient types *} wenzelm@10250: wenzelm@35100: theory Quotient_Type haftmann@30738: imports Main nipkow@15131: begin wenzelm@10250: wenzelm@10250: text {* wenzelm@10285: We introduce the notion of quotient types over equivalence relations haftmann@22390: via type classes. wenzelm@10250: *} wenzelm@10250: wenzelm@10285: subsection {* Equivalence relations and quotient types *} wenzelm@10250: wenzelm@10250: text {* wenzelm@10390: \medskip Type class @{text equiv} models equivalence relations @{text wenzelm@10390: "\ :: 'a => 'a => bool"}. wenzelm@10250: *} wenzelm@10250: haftmann@29608: class eqv = haftmann@25062: fixes eqv :: "'a \ 'a \ bool" (infixl "\" 50) wenzelm@10250: haftmann@22390: class equiv = eqv + haftmann@25062: assumes equiv_refl [intro]: "x \ x" haftmann@25062: assumes equiv_trans [trans]: "x \ y \ y \ z \ x \ z" haftmann@25062: assumes equiv_sym [sym]: "x \ y \ y \ x" wenzelm@10250: wenzelm@12371: lemma equiv_not_sym [sym]: "\ (x \ y) ==> \ (y \ (x::'a::equiv))" wenzelm@10477: proof - wenzelm@23373: assume "\ (x \ y)" then show "\ (y \ x)" wenzelm@10477: by (rule contrapos_nn) (rule equiv_sym) wenzelm@10477: qed wenzelm@10477: wenzelm@10477: lemma not_equiv_trans1 [trans]: "\ (x \ y) ==> y \ z ==> \ (x \ (z::'a::equiv))" wenzelm@10477: proof - wenzelm@23373: assume "\ (x \ y)" and "y \ z" wenzelm@10477: show "\ (x \ z)" wenzelm@10477: proof wenzelm@10477: assume "x \ z" wenzelm@23373: also from `y \ z` have "z \ y" .. wenzelm@10477: finally have "x \ y" . wenzelm@23373: with `\ (x \ y)` show False by contradiction wenzelm@10477: qed wenzelm@10477: qed wenzelm@10477: wenzelm@10477: lemma not_equiv_trans2 [trans]: "x \ y ==> \ (y \ z) ==> \ (x \ (z::'a::equiv))" wenzelm@10477: proof - wenzelm@23373: assume "\ (y \ z)" then have "\ (z \ y)" .. wenzelm@23373: also assume "x \ y" then have "y \ x" .. wenzelm@23373: finally have "\ (z \ x)" . then show "(\ x \ z)" .. wenzelm@10477: qed wenzelm@10477: wenzelm@10250: text {* wenzelm@10285: \medskip The quotient type @{text "'a quot"} consists of all wenzelm@10285: \emph{equivalence classes} over elements of the base type @{typ 'a}. wenzelm@10250: *} wenzelm@10250: wenzelm@45694: definition "quot = {{x. a \ x} | a::'a::eqv. True}" wenzelm@45694: wenzelm@49834: typedef 'a quot = "quot :: 'a::eqv set set" wenzelm@45694: unfolding quot_def by blast wenzelm@10250: wenzelm@10250: lemma quotI [intro]: "{x. a \ x} \ quot" wenzelm@18730: unfolding quot_def by blast wenzelm@10250: wenzelm@10250: lemma quotE [elim]: "R \ quot ==> (!!a. R = {x. a \ x} ==> C) ==> C" wenzelm@18730: unfolding quot_def by blast wenzelm@10250: wenzelm@10250: text {* wenzelm@10250: \medskip Abstracted equivalence classes are the canonical wenzelm@10250: representation of elements of a quotient type. wenzelm@10250: *} wenzelm@10250: wenzelm@19086: definition wenzelm@21404: "class" :: "'a::equiv => 'a quot" ("\_\") where wenzelm@19086: "\a\ = Abs_quot {x. a \ x}" wenzelm@10250: wenzelm@10311: theorem quot_exhaust: "\a. A = \a\" wenzelm@10278: proof (cases A) wenzelm@10278: fix R assume R: "A = Abs_quot R" wenzelm@23373: assume "R \ quot" then have "\a. R = {x. a \ x}" by blast wenzelm@10278: with R have "\a. A = Abs_quot {x. a \ x}" by blast wenzelm@23373: then show ?thesis unfolding class_def . wenzelm@10250: qed wenzelm@10250: wenzelm@10311: lemma quot_cases [cases type: quot]: "(!!a. A = \a\ ==> C) ==> C" wenzelm@18730: using quot_exhaust by blast wenzelm@10250: wenzelm@10250: wenzelm@10285: subsection {* Equality on quotients *} wenzelm@10250: wenzelm@10250: text {* wenzelm@10286: Equality of canonical quotient elements coincides with the original wenzelm@10286: relation. wenzelm@10250: *} wenzelm@10250: wenzelm@12371: theorem quot_equality [iff?]: "(\a\ = \b\) = (a \ b)" wenzelm@10285: proof wenzelm@10285: assume eq: "\a\ = \b\" wenzelm@10285: show "a \ b" wenzelm@10285: proof - wenzelm@10285: from eq have "{x. a \ x} = {x. b \ x}" wenzelm@10551: by (simp only: class_def Abs_quot_inject quotI) wenzelm@10285: moreover have "a \ a" .. wenzelm@10285: ultimately have "a \ {x. b \ x}" by blast wenzelm@23373: then have "b \ a" by blast wenzelm@23373: then show ?thesis .. wenzelm@10285: qed wenzelm@10285: next wenzelm@10250: assume ab: "a \ b" wenzelm@10285: show "\a\ = \b\" wenzelm@10285: proof - wenzelm@10285: have "{x. a \ x} = {x. b \ x}" wenzelm@10285: proof (rule Collect_cong) wenzelm@10285: fix x show "(a \ x) = (b \ x)" wenzelm@10285: proof wenzelm@10285: from ab have "b \ a" .. wenzelm@10285: also assume "a \ x" wenzelm@10285: finally show "b \ x" . wenzelm@10285: next wenzelm@10285: note ab wenzelm@10285: also assume "b \ x" wenzelm@10285: finally show "a \ x" . wenzelm@10285: qed wenzelm@10250: qed wenzelm@23373: then show ?thesis by (simp only: class_def) wenzelm@10250: qed wenzelm@10250: qed wenzelm@10250: wenzelm@10250: wenzelm@10285: subsection {* Picking representing elements *} wenzelm@10250: wenzelm@19086: definition wenzelm@21404: pick :: "'a::equiv quot => 'a" where wenzelm@19086: "pick A = (SOME a. A = \a\)" wenzelm@10250: wenzelm@10285: theorem pick_equiv [intro]: "pick \a\ \ a" wenzelm@10250: proof (unfold pick_def) wenzelm@10250: show "(SOME x. \a\ = \x\) \ a" wenzelm@10250: proof (rule someI2) wenzelm@10250: show "\a\ = \a\" .. wenzelm@10250: fix x assume "\a\ = \x\" wenzelm@23373: then have "a \ x" .. then show "x \ a" .. wenzelm@10250: qed wenzelm@10250: qed wenzelm@10250: wenzelm@10483: theorem pick_inverse [intro]: "\pick A\ = A" wenzelm@10250: proof (cases A) wenzelm@10250: fix a assume a: "A = \a\" wenzelm@23373: then have "pick A \ a" by (simp only: pick_equiv) wenzelm@23373: then have "\pick A\ = \a\" .. wenzelm@10250: with a show ?thesis by simp wenzelm@10250: qed wenzelm@10250: wenzelm@10285: text {* wenzelm@10285: \medskip The following rules support canonical function definitions wenzelm@10483: on quotient types (with up to two arguments). Note that the wenzelm@10483: stripped-down version without additional conditions is sufficient wenzelm@10483: most of the time. wenzelm@10285: *} wenzelm@10285: wenzelm@10483: theorem quot_cond_function: wenzelm@18372: assumes eq: "!!X Y. P X Y ==> f X Y == g (pick X) (pick Y)" wenzelm@18372: and cong: "!!x x' y y'. \x\ = \x'\ ==> \y\ = \y'\ wenzelm@18372: ==> P \x\ \y\ ==> P \x'\ \y'\ ==> g x y = g x' y'" wenzelm@18372: and P: "P \a\ \b\" wenzelm@18372: shows "f \a\ \b\ = g a b" wenzelm@10473: proof - wenzelm@18372: from eq and P have "f \a\ \b\ = g (pick \a\) (pick \b\)" by (simp only:) wenzelm@10505: also have "... = g a b" wenzelm@10491: proof (rule cong) wenzelm@10483: show "\pick \a\\ = \a\" .. wenzelm@10483: moreover wenzelm@10483: show "\pick \b\\ = \b\" .. wenzelm@10491: moreover wenzelm@23373: show "P \a\ \b\" by (rule P) wenzelm@10491: ultimately show "P \pick \a\\ \pick \b\\" by (simp only:) wenzelm@10285: qed wenzelm@10285: finally show ?thesis . wenzelm@10285: qed wenzelm@10285: wenzelm@10483: theorem quot_function: wenzelm@18372: assumes "!!X Y. f X Y == g (pick X) (pick Y)" wenzelm@18372: and "!!x x' y y'. \x\ = \x'\ ==> \y\ = \y'\ ==> g x y = g x' y'" wenzelm@18372: shows "f \a\ \b\ = g a b" wenzelm@23394: using assms and TrueI wenzelm@18372: by (rule quot_cond_function) wenzelm@10285: bauerg@10499: theorem quot_function': bauerg@10499: "(!!X Y. f X Y == g (pick X) (pick Y)) ==> bauerg@10499: (!!x x' y y'. x \ x' ==> y \ y' ==> g x y = g x' y') ==> bauerg@10499: f \a\ \b\ = g a b" wenzelm@18372: by (rule quot_function) (simp_all only: quot_equality) bauerg@10499: wenzelm@10250: end