src/HOL/Library/Quotient_Type.thy
 author Christian Sternagel Thu Aug 30 15:44:03 2012 +0900 (2012-08-30) changeset 49093 fdc301f592c4 parent 45694 4a8743618257 child 49834 b27bbb021df1 permissions -rw-r--r--
     1 (*  Title:      HOL/Library/Quotient_Type.thy

     2     Author:     Markus Wenzel, TU Muenchen

     3 *)

     4

     5 header {* Quotient types *}

     6

     7 theory Quotient_Type

     8 imports Main

     9 begin

    10

    11 text {*

    12  We introduce the notion of quotient types over equivalence relations

    13  via type classes.

    14 *}

    15

    16 subsection {* Equivalence relations and quotient types *}

    17

    18 text {*

    19  \medskip Type class @{text equiv} models equivalence relations @{text

    20  "\<sim> :: 'a => 'a => bool"}.

    21 *}

    22

    23 class eqv =

    24   fixes eqv :: "'a \<Rightarrow> 'a \<Rightarrow> bool"    (infixl "\<sim>" 50)

    25

    26 class equiv = eqv +

    27   assumes equiv_refl [intro]: "x \<sim> x"

    28   assumes equiv_trans [trans]: "x \<sim> y \<Longrightarrow> y \<sim> z \<Longrightarrow> x \<sim> z"

    29   assumes equiv_sym [sym]: "x \<sim> y \<Longrightarrow> y \<sim> x"

    30

    31 lemma equiv_not_sym [sym]: "\<not> (x \<sim> y) ==> \<not> (y \<sim> (x::'a::equiv))"

    32 proof -

    33   assume "\<not> (x \<sim> y)" then show "\<not> (y \<sim> x)"

    34     by (rule contrapos_nn) (rule equiv_sym)

    35 qed

    36

    37 lemma not_equiv_trans1 [trans]: "\<not> (x \<sim> y) ==> y \<sim> z ==> \<not> (x \<sim> (z::'a::equiv))"

    38 proof -

    39   assume "\<not> (x \<sim> y)" and "y \<sim> z"

    40   show "\<not> (x \<sim> z)"

    41   proof

    42     assume "x \<sim> z"

    43     also from y \<sim> z have "z \<sim> y" ..

    44     finally have "x \<sim> y" .

    45     with \<not> (x \<sim> y) show False by contradiction

    46   qed

    47 qed

    48

    49 lemma not_equiv_trans2 [trans]: "x \<sim> y ==> \<not> (y \<sim> z) ==> \<not> (x \<sim> (z::'a::equiv))"

    50 proof -

    51   assume "\<not> (y \<sim> z)" then have "\<not> (z \<sim> y)" ..

    52   also assume "x \<sim> y" then have "y \<sim> x" ..

    53   finally have "\<not> (z \<sim> x)" . then show "(\<not> x \<sim> z)" ..

    54 qed

    55

    56 text {*

    57  \medskip The quotient type @{text "'a quot"} consists of all

    58  \emph{equivalence classes} over elements of the base type @{typ 'a}.

    59 *}

    60

    61 definition "quot = {{x. a \<sim> x} | a::'a::eqv. True}"

    62

    63 typedef (open) 'a quot = "quot :: 'a::eqv set set"

    64   unfolding quot_def by blast

    65

    66 lemma quotI [intro]: "{x. a \<sim> x} \<in> quot"

    67   unfolding quot_def by blast

    68

    69 lemma quotE [elim]: "R \<in> quot ==> (!!a. R = {x. a \<sim> x} ==> C) ==> C"

    70   unfolding quot_def by blast

    71

    72 text {*

    73  \medskip Abstracted equivalence classes are the canonical

    74  representation of elements of a quotient type.

    75 *}

    76

    77 definition

    78   "class" :: "'a::equiv => 'a quot"  ("\<lfloor>_\<rfloor>") where

    79   "\<lfloor>a\<rfloor> = Abs_quot {x. a \<sim> x}"

    80

    81 theorem quot_exhaust: "\<exists>a. A = \<lfloor>a\<rfloor>"

    82 proof (cases A)

    83   fix R assume R: "A = Abs_quot R"

    84   assume "R \<in> quot" then have "\<exists>a. R = {x. a \<sim> x}" by blast

    85   with R have "\<exists>a. A = Abs_quot {x. a \<sim> x}" by blast

    86   then show ?thesis unfolding class_def .

    87 qed

    88

    89 lemma quot_cases [cases type: quot]: "(!!a. A = \<lfloor>a\<rfloor> ==> C) ==> C"

    90   using quot_exhaust by blast

    91

    92

    93 subsection {* Equality on quotients *}

    94

    95 text {*

    96  Equality of canonical quotient elements coincides with the original

    97  relation.

    98 *}

    99

   100 theorem quot_equality [iff?]: "(\<lfloor>a\<rfloor> = \<lfloor>b\<rfloor>) = (a \<sim> b)"

   101 proof

   102   assume eq: "\<lfloor>a\<rfloor> = \<lfloor>b\<rfloor>"

   103   show "a \<sim> b"

   104   proof -

   105     from eq have "{x. a \<sim> x} = {x. b \<sim> x}"

   106       by (simp only: class_def Abs_quot_inject quotI)

   107     moreover have "a \<sim> a" ..

   108     ultimately have "a \<in> {x. b \<sim> x}" by blast

   109     then have "b \<sim> a" by blast

   110     then show ?thesis ..

   111   qed

   112 next

   113   assume ab: "a \<sim> b"

   114   show "\<lfloor>a\<rfloor> = \<lfloor>b\<rfloor>"

   115   proof -

   116     have "{x. a \<sim> x} = {x. b \<sim> x}"

   117     proof (rule Collect_cong)

   118       fix x show "(a \<sim> x) = (b \<sim> x)"

   119       proof

   120         from ab have "b \<sim> a" ..

   121         also assume "a \<sim> x"

   122         finally show "b \<sim> x" .

   123       next

   124         note ab

   125         also assume "b \<sim> x"

   126         finally show "a \<sim> x" .

   127       qed

   128     qed

   129     then show ?thesis by (simp only: class_def)

   130   qed

   131 qed

   132

   133

   134 subsection {* Picking representing elements *}

   135

   136 definition

   137   pick :: "'a::equiv quot => 'a" where

   138   "pick A = (SOME a. A = \<lfloor>a\<rfloor>)"

   139

   140 theorem pick_equiv [intro]: "pick \<lfloor>a\<rfloor> \<sim> a"

   141 proof (unfold pick_def)

   142   show "(SOME x. \<lfloor>a\<rfloor> = \<lfloor>x\<rfloor>) \<sim> a"

   143   proof (rule someI2)

   144     show "\<lfloor>a\<rfloor> = \<lfloor>a\<rfloor>" ..

   145     fix x assume "\<lfloor>a\<rfloor> = \<lfloor>x\<rfloor>"

   146     then have "a \<sim> x" .. then show "x \<sim> a" ..

   147   qed

   148 qed

   149

   150 theorem pick_inverse [intro]: "\<lfloor>pick A\<rfloor> = A"

   151 proof (cases A)

   152   fix a assume a: "A = \<lfloor>a\<rfloor>"

   153   then have "pick A \<sim> a" by (simp only: pick_equiv)

   154   then have "\<lfloor>pick A\<rfloor> = \<lfloor>a\<rfloor>" ..

   155   with a show ?thesis by simp

   156 qed

   157

   158 text {*

   159  \medskip The following rules support canonical function definitions

   160  on quotient types (with up to two arguments).  Note that the

   161  stripped-down version without additional conditions is sufficient

   162  most of the time.

   163 *}

   164

   165 theorem quot_cond_function:

   166   assumes eq: "!!X Y. P X Y ==> f X Y == g (pick X) (pick Y)"

   167     and cong: "!!x x' y y'. \<lfloor>x\<rfloor> = \<lfloor>x'\<rfloor> ==> \<lfloor>y\<rfloor> = \<lfloor>y'\<rfloor>

   168       ==> P \<lfloor>x\<rfloor> \<lfloor>y\<rfloor> ==> P \<lfloor>x'\<rfloor> \<lfloor>y'\<rfloor> ==> g x y = g x' y'"

   169     and P: "P \<lfloor>a\<rfloor> \<lfloor>b\<rfloor>"

   170   shows "f \<lfloor>a\<rfloor> \<lfloor>b\<rfloor> = g a b"

   171 proof -

   172   from eq and P have "f \<lfloor>a\<rfloor> \<lfloor>b\<rfloor> = g (pick \<lfloor>a\<rfloor>) (pick \<lfloor>b\<rfloor>)" by (simp only:)

   173   also have "... = g a b"

   174   proof (rule cong)

   175     show "\<lfloor>pick \<lfloor>a\<rfloor>\<rfloor> = \<lfloor>a\<rfloor>" ..

   176     moreover

   177     show "\<lfloor>pick \<lfloor>b\<rfloor>\<rfloor> = \<lfloor>b\<rfloor>" ..

   178     moreover

   179     show "P \<lfloor>a\<rfloor> \<lfloor>b\<rfloor>" by (rule P)

   180     ultimately show "P \<lfloor>pick \<lfloor>a\<rfloor>\<rfloor> \<lfloor>pick \<lfloor>b\<rfloor>\<rfloor>" by (simp only:)

   181   qed

   182   finally show ?thesis .

   183 qed

   184

   185 theorem quot_function:

   186   assumes "!!X Y. f X Y == g (pick X) (pick Y)"

   187     and "!!x x' y y'. \<lfloor>x\<rfloor> = \<lfloor>x'\<rfloor> ==> \<lfloor>y\<rfloor> = \<lfloor>y'\<rfloor> ==> g x y = g x' y'"

   188   shows "f \<lfloor>a\<rfloor> \<lfloor>b\<rfloor> = g a b"

   189   using assms and TrueI

   190   by (rule quot_cond_function)

   191

   192 theorem quot_function':

   193   "(!!X Y. f X Y == g (pick X) (pick Y)) ==>

   194     (!!x x' y y'. x \<sim> x' ==> y \<sim> y' ==> g x y = g x' y') ==>

   195     f \<lfloor>a\<rfloor> \<lfloor>b\<rfloor> = g a b"

   196   by (rule quot_function) (simp_all only: quot_equality)

   197

   198 end