imports Main QuotScript QuotList

uses ("quotient.ML")

begin

locale QUOT_TYPE =

  fixes R :: "'a \<Rightarrow> 'a \<Rightarrow> bool"

  and   Abs :: "('a \<Rightarrow> bool) \<Rightarrow> 'b"

  and   Rep :: "'b \<Rightarrow> ('a \<Rightarrow> bool)"

  assumes equiv: "EQUIV R"

  and     rep_prop: "\<And>y. \<exists>x. Rep y = R x"

  and     rep_inverse: "\<And>x. Abs (Rep x) = x"

  and     abs_inverse: "\<And>x. (Rep (Abs (R x))) = (R x)"

  and     rep_inject: "\<And>x y. (Rep x = Rep y) = (x = y)"

definition

  "ABS x \<equiv> Abs (R x)"

definition

  "REP a = Eps (Rep a)"

lemma lem9:

  shows "R (Eps (R x)) = R x"

proof -

  have a: "R x x" using equiv by (simp add: EQUIV_REFL_SYM_TRANS REFL_def)

  then have "R x (Eps (R x))" by (rule someI)

  then show "R (Eps (R x)) = R x"

    using equiv unfolding EQUIV_def by simp

qed

theorem thm10:

  shows "ABS (REP a) \<equiv> a"

  apply  (rule eq_reflection)

  unfolding ABS_def REP_def

proof -

  from rep_prop

  obtain x where eq: "Rep a = R x" by auto

  have "Abs (R (Eps (Rep a))) = Abs (R (Eps (R x)))" using eq by simp

  also have "\<dots> = Abs (R x)" using lem9 by simp

  also have "\<dots> = Abs (Rep a)" using eq by simp

  also have "\<dots> = a" using rep_inverse by simp

  finally

  show "Abs (R (Eps (Rep a))) = a" by simp

qed

lemma REP_refl:

  shows "R (REP a) (REP a)"

unfolding REP_def

by (simp add: equiv[simplified EQUIV_def])

lemma lem7:

  shows "(R x = R y) = (Abs (R x) = Abs (R y))"

apply(rule iffI)

apply(simp)

apply(drule rep_inject[THEN iffD2])

apply(simp add: abs_inverse)

done

theorem thm11:

  shows "R r r' = (ABS r = ABS r')"

unfolding ABS_def

by (simp only: equiv[simplified EQUIV_def] lem7)

lemma REP_ABS_rsp:

  shows "R f (REP (ABS g)) = R f g"

  and   "R (REP (ABS g)) f = R g f"

by (simp_all add: thm10 thm11)

lemma QUOTIENT:

  "QUOTIENT R ABS REP"

apply(unfold QUOTIENT_def)

apply(simp add: thm10)

apply(simp add: REP_refl)

apply(subst thm11[symmetric])

apply(simp add: equiv[simplified EQUIV_def])

done

lemma R_trans:

  assumes ab: "R a b"

  and     bc: "R b c"

  shows "R a c"

proof -

  have tr: "TRANS R" using equiv EQUIV_REFL_SYM_TRANS[of R] by simp

  moreover have ab: "R a b" by fact

  moreover have bc: "R b c" by fact

  ultimately show "R a c" unfolding TRANS_def by blast

qed

lemma R_sym:

  assumes ab: "R a b"

  shows "R b a"

proof -

  have re: "SYM R" using equiv EQUIV_REFL_SYM_TRANS[of R] by simp

  then show "R b a" using ab unfolding SYM_def by blast

qed

lemma R_trans2:

  assumes ac: "R a c"

  and     bd: "R b d"

  shows "R a b = R c d"

proof

  assume "R a b"

  then have "R b a" using R_sym by blast

  then have "R b c" using ac R_trans by blast

  then have "R c b" using R_sym by blast

  then show "R c d" using bd R_trans by blast

next

  assume "R c d"

  then have "R a d" using ac R_trans by blast

  then have "R d a" using R_sym by blast

  then have "R b a" using bd R_trans by blast

  then show "R a b" using R_sym by blast

qed

lemma REPS_same:

  shows "R (REP a) (REP b) \<equiv> (a = b)"

proof -

  have "R (REP a) (REP b) = (a = b)"

  proof

    assume as: "R (REP a) (REP b)"

    from rep_prop

    obtain x y

      where eqs: "Rep a = R x" "Rep b = R y" by blast

    from eqs have "R (Eps (R x)) (Eps (R y))" using as unfolding REP_def by simp

    then have "R x (Eps (R y))" using lem9 by simp

    then have "R (Eps (R y)) x" using R_sym by blast

    then have "R y x" using lem9 by simp

    then have "R x y" using R_sym by blast

    then have "ABS x = ABS y" using thm11 by simp

    then have "Abs (Rep a) = Abs (Rep b)" using eqs unfolding ABS_def by simp

    then show "a = b" using rep_inverse by simp

  next

    assume ab: "a = b"

    have "REFL R" using equiv EQUIV_REFL_SYM_TRANS[of R] by simp

    then show "R (REP a) (REP b)" unfolding REFL_def using ab by auto

  qed

  then show "R (REP a) (REP b) \<equiv> (a = b)" by simp

qed

end

use "quotient.ML"