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src/HOL/Library/Quotient.thy

author | wenzelm |

Fri, 03 Nov 2000 21:35:59 +0100 | |

changeset 10390 | 1d54567bed24 |

parent 10333 | f12ff6a4bc7b |

child 10392 | f27afee8475d |

permissions | -rw-r--r-- |

tuned;

(* Title: HOL/Library/Quotient.thy ID: $Id$ Author: Gertrud Bauer and Markus Wenzel, TU Muenchen *) header {* \title{Quotients} \author{Gertrud Bauer and 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" 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 equivalence_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 equivalence_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 equivalence_class_eq [iff?]: "(\<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: equivalence_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: equivalence_class_def) qed qed 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: "\<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. *} theorem cong_definition1: "(!!X. f X == g (pick X)) ==> (!!x x'. x \<sim> x' ==> g x = g x') ==> f \<lfloor>a\<rfloor> = g a" proof - assume cong: "!!x x'. x \<sim> x' ==> g x = g x'" assume "!!X. f X == g (pick X)" hence "f \<lfloor>a\<rfloor> = g (pick \<lfloor>a\<rfloor>)" by (simp only:) also have "\<dots> = g a" proof (rule cong) show "pick \<lfloor>a\<rfloor> \<sim> a" .. qed finally show ?thesis . qed theorem cong_definition2: "(!!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" proof - assume cong: "!!x x' y y'. x \<sim> x' ==> y \<sim> y' ==> g x y = g x' y'" assume "!!X Y. f X Y == g (pick X) (pick Y)" hence "f \<lfloor>a\<rfloor> \<lfloor>b\<rfloor> = g (pick \<lfloor>a\<rfloor>) (pick \<lfloor>b\<rfloor>)" by (simp only:) also have "\<dots> = g a b" proof (rule cong) show "pick \<lfloor>a\<rfloor> \<sim> a" .. show "pick \<lfloor>b\<rfloor> \<sim> b" .. qed finally show ?thesis . qed end