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Datei: Lift_Fun.thy Sprache: Isabelle

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(*  Title:      HOL/Quotient_Examples/Lift_Fun.thy
    Author:     Ondrej Kuncar
*)

section \<open>Example of lifting definitions with contravariant or co/contravariant type variables\<close>

theory Lift_Fun
imports Main "HOL-Library.Quotient_Syntax"
begin

text \<open>This file is meant as a test case.
  It contains examples of lifting definitions with quotients that have contravariant
  type variables or type variables which are covariant and contravariant in the same time.\<close>

subsection \<open>Contravariant type variables\<close>

text \<open>'a is a contravariant type variable and we are able to map over this variable
  in the following four definitions. This example is based on HOL/Fun.thy.\<close>

quotient_type
('a, 'b) fun' (infixr "\" 55) = "'a \ 'b" / "(=)"
  by (simp add: identity_equivp)

quotient_definition "comp' :: ('b \ 'c) \ ('a \ 'b) \ 'a \ 'c" is
  "comp :: ('b \ 'c) \ ('a \ 'b) \ 'a \ 'c" done

quotient_definition "fcomp' :: ('a \ 'b) \ ('b \ 'c) \ 'a \ 'c" is
  fcomp done

quotient_definition "map_fun' :: ('c \ 'a) \ ('b \ 'd) \ ('a \ 'b) \ 'c \ 'd"
  is "map_fun::('c \ 'a) \ ('b \ 'd) \ ('a \ 'b) \ 'c \ 'd" done

quotient_definition "inj_on' :: ('a \ 'b) \ 'a set \ bool" is inj_on done

quotient_definition "bij_betw' :: ('a \ 'b) \ 'a set \ 'b set \ bool" is bij_betw done

subsection \<open>Co/Contravariant type variables\<close>

text \<open>'a is a covariant and contravariant type variable in the same time.
  The following example is a bit artificial. We haven't had a natural one yet.\

quotient_type 'a endofun = "'a \<Rightarrow> 'a" / "(=)" by (simp add: identity_equivp)

definition map_endofun' :: "('a \<Rightarrow> 'b) \<Rightarrow> ('b \<Rightarrow> 'a) \<Rightarrow> ('a => 'a) \<Rightarrow> ('b => 'b)"
  where "map_endofun' f g e = map_fun g f e"

quotient_definition "map_endofun :: ('a \ 'b) \ ('b \ 'a) \ 'a endofun \ 'b endofun" is
  map_endofun' done

text \<open>Registration of the map function for 'a endofun.\<close>

functor map_endofun : map_endofun
proof -
  have "\ x. abs_endofun (rep_endofun x) = x" using Quotient3_endofun by (auto simp: Quotient3_def)
  then show "map_endofun id id = id"
    by (auto simp: map_endofun_def map_endofun'_def map_fun_def fun_eq_iff)

  have a:"\ x. rep_endofun (abs_endofun x) = x" using Quotient3_endofun
    Quotient3_rep_abs[of "(=)" abs_endofun rep_endofun] by blast
  show "\f g h i. map_endofun f g \ map_endofun h i = map_endofun (f \ h) (i \ g)"
    by (auto simp: map_endofun_def map_endofun'_def map_fun_def fun_eq_iff) (simp add: a o_assoc)
qed

text \<open>Relator for 'a endofun.\<close>

definition
  rel_endofun' :: "('a \<Rightarrow> 'b \<Rightarrow> bool) \<Rightarrow> ('a \<Rightarrow> 'a) \<Rightarrow> ('b \<Rightarrow> 'b) \<Rightarrow> bool"
where
  "rel_endofun' R = (\f g. (R ===> R) f g)"

quotient_definition "rel_endofun :: ('a \ 'b \ bool) \ 'a endofun \ 'b endofun \ bool" is
  rel_endofun' done

lemma endofun_quotient:
assumes a: "Quotient3 R Abs Rep"
shows "Quotient3 (rel_endofun R) (map_endofun Abs Rep) (map_endofun Rep Abs)"
proof (intro Quotient3I)
  show "\a. map_endofun Abs Rep (map_endofun Rep Abs a) = a"
    by (metis (hide_lams, no_types) a abs_o_rep id_apply map_endofun.comp map_endofun.id o_eq_dest_lhs)
next
  show "\a. rel_endofun R (map_endofun Rep Abs a) (map_endofun Rep Abs a)"
  using fun_quotient3[OF a a, THEN Quotient3_rep_reflp]
  unfolding rel_endofun_def map_endofun_def map_fun_def o_def map_endofun'_def rel_endofun'_def id_def
    by (metis (mono_tags) Quotient3_endofun rep_abs_rsp)
next
  have abs_to_eq: "\ x y. abs_endofun x = abs_endofun y \ x = y"
  by (drule arg_cong[where f=rep_endofun]) (simp add: Quotient3_rep_abs[OF Quotient3_endofun])
  fix r s
  show "rel_endofun R r s =
          (rel_endofun R r r \<and>
           rel_endofun R s s \<and> map_endofun Abs Rep r = map_endofun Abs Rep s)"
    apply(auto simp add: rel_endofun_def rel_endofun'_def map_endofun_def map_endofun'_def)
    using fun_quotient3[OF a a,THEN Quotient3_refl1]
    apply metis
    using fun_quotient3[OF a a,THEN Quotient3_refl2]
    apply metis
    using fun_quotient3[OF a a, THEN Quotient3_rel]
    apply metis
    by (auto intro: fun_quotient3[OF a a, THEN Quotient3_rel, THEN iffD1] simp add: abs_to_eq)
qed

declare [[mapQ3 endofun = (rel_endofun, endofun_quotient)]]

quotient_definition "endofun_id_id :: ('a endofun) endofun" is "id :: ('a \ 'a) \ ('a \ 'a)" done

term  endofun_id_id
thm  endofun_id_id_def

quotient_type 'a endofun' = "'a endofun" / "(=)" by (simp add: identity_equivp)

text \<open>We have to map "'a endofun" to "('a endofun') endofun", i.e., mapping (lifting)
  over a type variable which is a covariant and contravariant type variable.\<close>

quotient_definition "endofun'_id_id :: ('a endofun') endofun'" is endofun_id_id done

term  endofun'_id_id
thm  endofun'_id_id_def

end

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