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src/HOL/Library/Quotient_Type.thy

author | wenzelm |

Sun, 28 Dec 2014 12:37:03 +0100 | |

changeset 59192 | a1d6d6db781b |

parent 58881 | b9556a055632 |

child 61260 | e6f03fae14d5 |

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

modernized historic example;

(* Title: HOL/Library/Quotient_Type.thy Author: Markus Wenzel, TU Muenchen *) section \<open>Quotient types\<close> theory Quotient_Type imports Main begin text \<open>We introduce the notion of quotient types over equivalence relations via type classes.\<close> subsection \<open>Equivalence relations and quotient types\<close> text \<open>Type class @{text equiv} models equivalence relations @{text "\<sim> :: 'a \<Rightarrow> 'a \<Rightarrow> bool"}.\<close> class eqv = fixes eqv :: "'a \<Rightarrow> 'a \<Rightarrow> bool" (infixl "\<sim>" 50) class equiv = eqv + assumes equiv_refl [intro]: "x \<sim> x" and equiv_trans [trans]: "x \<sim> y \<Longrightarrow> y \<sim> z \<Longrightarrow> x \<sim> z" and equiv_sym [sym]: "x \<sim> y \<Longrightarrow> y \<sim> x" begin lemma equiv_not_sym [sym]: "\<not> x \<sim> y \<Longrightarrow> \<not> y \<sim> x" proof - assume "\<not> x \<sim> y" then show "\<not> y \<sim> x" by (rule contrapos_nn) (rule equiv_sym) qed lemma not_equiv_trans1 [trans]: "\<not> x \<sim> y \<Longrightarrow> y \<sim> z \<Longrightarrow> \<not> x \<sim> z" proof - assume "\<not> x \<sim> y" and "y \<sim> z" show "\<not> x \<sim> z" proof assume "x \<sim> z" also from \<open>y \<sim> z\<close> have "z \<sim> y" .. finally have "x \<sim> y" . with \<open>\<not> x \<sim> y\<close> show False by contradiction qed qed lemma not_equiv_trans2 [trans]: "x \<sim> y \<Longrightarrow> \<not> y \<sim> z \<Longrightarrow> \<not> x \<sim> z" proof - assume "\<not> y \<sim> z" then have "\<not> z \<sim> y" .. also assume "x \<sim> y" then have "y \<sim> x" .. finally have "\<not> z \<sim> x" . then show "\<not> x \<sim> z" .. qed end text \<open>The quotient type @{text "'a quot"} consists of all \emph{equivalence classes} over elements of the base type @{typ 'a}.\<close> definition (in eqv) "quot = {{x. a \<sim> x} | a. True}" typedef 'a quot = "quot :: 'a::eqv set set" unfolding quot_def by blast lemma quotI [intro]: "{x. a \<sim> x} \<in> quot" unfolding quot_def by blast lemma quotE [elim]: assumes "R \<in> quot" obtains a where "R = {x. a \<sim> x}" using assms unfolding quot_def by blast text \<open>Abstracted equivalence classes are the canonical representation of elements of a quotient type.\<close> definition "class" :: "'a::equiv \<Rightarrow> 'a quot" ("\<lfloor>_\<rfloor>") where "\<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" then have "\<exists>a. R = {x. a \<sim> x}" by blast with R have "\<exists>a. A = Abs_quot {x. a \<sim> x}" by blast then show ?thesis unfolding class_def . qed lemma quot_cases [cases type: quot]: obtains a where "A = \<lfloor>a\<rfloor>" using quot_exhaust by blast subsection \<open>Equality on quotients\<close> text \<open>Equality of canonical quotient elements coincides with the original relation.\<close> theorem quot_equality [iff?]: "\<lfloor>a\<rfloor> = \<lfloor>b\<rfloor> \<longleftrightarrow> 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: class_def Abs_quot_inject quotI) moreover have "a \<sim> a" .. ultimately have "a \<in> {x. b \<sim> x}" by blast then have "b \<sim> a" by blast then show ?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 then show ?thesis by (simp only: class_def) qed qed subsection \<open>Picking representing elements\<close> definition pick :: "'a::equiv quot \<Rightarrow> 'a" where "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>" then have "a \<sim> x" .. then show "x \<sim> a" .. qed qed theorem pick_inverse [intro]: "\<lfloor>pick A\<rfloor> = A" proof (cases A) fix a assume a: "A = \<lfloor>a\<rfloor>" then have "pick A \<sim> a" by (simp only: pick_equiv) then have "\<lfloor>pick A\<rfloor> = \<lfloor>a\<rfloor>" .. with a show ?thesis by simp qed text \<open>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.\<close> theorem quot_cond_function: assumes eq: "\<And>X Y. P X Y \<Longrightarrow> f X Y \<equiv> g (pick X) (pick Y)" and cong: "\<And>x x' y y'. \<lfloor>x\<rfloor> = \<lfloor>x'\<rfloor> \<Longrightarrow> \<lfloor>y\<rfloor> = \<lfloor>y'\<rfloor> \<Longrightarrow> P \<lfloor>x\<rfloor> \<lfloor>y\<rfloor> \<Longrightarrow> P \<lfloor>x'\<rfloor> \<lfloor>y'\<rfloor> \<Longrightarrow> g x y = g x' y'" and P: "P \<lfloor>a\<rfloor> \<lfloor>b\<rfloor>" shows "f \<lfloor>a\<rfloor> \<lfloor>b\<rfloor> = g a b" proof - from eq and P have "f \<lfloor>a\<rfloor> \<lfloor>b\<rfloor> = g (pick \<lfloor>a\<rfloor>) (pick \<lfloor>b\<rfloor>)" by (simp only:) also have "... = g a b" proof (rule cong) show "\<lfloor>pick \<lfloor>a\<rfloor>\<rfloor> = \<lfloor>a\<rfloor>" .. moreover show "\<lfloor>pick \<lfloor>b\<rfloor>\<rfloor> = \<lfloor>b\<rfloor>" .. moreover show "P \<lfloor>a\<rfloor> \<lfloor>b\<rfloor>" by (rule P) ultimately show "P \<lfloor>pick \<lfloor>a\<rfloor>\<rfloor> \<lfloor>pick \<lfloor>b\<rfloor>\<rfloor>" by (simp only:) qed finally show ?thesis . qed theorem quot_function: assumes "\<And>X Y. f X Y \<equiv> g (pick X) (pick Y)" and "\<And>x x' y y'. \<lfloor>x\<rfloor> = \<lfloor>x'\<rfloor> \<Longrightarrow> \<lfloor>y\<rfloor> = \<lfloor>y'\<rfloor> \<Longrightarrow> g x y = g x' y'" shows "f \<lfloor>a\<rfloor> \<lfloor>b\<rfloor> = g a b" using assms and TrueI by (rule quot_cond_function) theorem quot_function': "(\<And>X Y. f X Y \<equiv> g (pick X) (pick Y)) \<Longrightarrow> (\<And>x x' y y'. x \<sim> x' \<Longrightarrow> y \<sim> y' \<Longrightarrow> g x y = g x' y') \<Longrightarrow> f \<lfloor>a\<rfloor> \<lfloor>b\<rfloor> = g a b" by (rule quot_function) (simp_all only: quot_equality) end