Prove.thy
author Christian Urban <urbanc@in.tum.de>
Mon, 12 Oct 2009 23:06:14 +0200
changeset 79 c0c41fefeb06
parent 57 13be92f5b638
child 83 e8f352546ad8
permissions -rw-r--r--
added quotient command (you need to update isar-keywords-prove.el)

theory Prove
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)"
begin

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"

ML {*
val r = ref (NONE:(unit -> term) option)
*}

ML {*
let
  fun after_qed thm_name thms lthy =
       LocalTheory.note Thm.theoremK (thm_name, (flat thms)) lthy |> snd
    
  fun setup_proof (name_spec, (txt, pos)) lthy =
  let
    val trm = ML_Context.evaluate lthy true ("r", r) txt
  in
    Proof.theorem_i NONE (after_qed name_spec) [[(trm,[])]] lthy
  end

  val parser = SpecParse.opt_thm_name ":" -- OuterParse.ML_source
in
  OuterSyntax.local_theory_to_proof "prove" "proving a proposition" 
    OuterKeyword.thy_goal (parser >> setup_proof)
end
*}

end