Closures2.thy
author urbanc
Tue, 07 Feb 2012 00:50:23 +0000
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parent 233 e2dc11e12e0b
child 348 bea94f1e6771
permissions -rw-r--r--
more on paper
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theory Closures2
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imports 
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  Closures
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  Higman
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  (* "~~/src/HOL/Proofs/Extraction/Higman" *)
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begin
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section {* Closure under @{text SUBSEQ} and @{text SUPSEQ} *}
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(* compatibility with Higman theory by Stefan *)
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notation
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  emb ("_ \<preceq> _")
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declare  emb0 [intro]
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declare  emb1 [intro]
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declare  emb2 [intro]
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lemma letter_UNIV:
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  shows "UNIV = {A, B::letter}"
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apply(auto)
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apply(case_tac x)
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apply(auto)
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done
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instance letter :: finite
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apply(default)
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apply(simp add: letter_UNIV)
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done
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hide_const A
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hide_const B
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hide_const R
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(*
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inductive 
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  emb :: "'a list \<Rightarrow> 'a list \<Rightarrow> bool" ("_ \<preceq> _")
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where
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   emb0 [intro]: "emb [] y"
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 | emb1 [intro]: "emb x y \<Longrightarrow> emb x (c # y)"
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 | emb2 [intro]: "emb x y \<Longrightarrow> emb (c # x) (c # y)"
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*)
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lemma emb_refl [intro]:
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  shows "x \<preceq> x"
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by (induct x) (auto)
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lemma emb_right [intro]:
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  assumes a: "x \<preceq> y"
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  shows "x \<preceq> y @ y'"
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using a 
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by (induct arbitrary: y') (auto)
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lemma emb_left [intro]:
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  assumes a: "x \<preceq> y"
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  shows "x \<preceq> y' @ y"
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using a by (induct y') (auto)
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lemma emb_appendI [intro]:
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  assumes a: "x \<preceq> x'"
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  and     b: "y \<preceq> y'"
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  shows "x @ y \<preceq> x' @ y'"
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using a b by (induct) (auto)
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lemma emb_cons_leftD:
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  assumes "a # x \<preceq> y"
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  shows "\<exists>y1 y2. y = y1 @ [a] @ y2 \<and> x \<preceq> y2"
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using assms
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apply(induct x\<equiv>"a # x" y\<equiv>"y" arbitrary: a x y)
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apply(auto)
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apply(metis append_Cons)
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done
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lemma emb_append_leftD:
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  assumes "x1 @ x2 \<preceq> y"
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  shows "\<exists>y1 y2. y = y1 @ y2 \<and> x1 \<preceq> y1 \<and> x2 \<preceq> y2"
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using assms
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apply(induct x1 arbitrary: x2 y)
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apply(auto)
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apply(drule emb_cons_leftD)
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apply(auto)
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apply(drule_tac x="x2" in meta_spec)
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apply(drule_tac x="y2" in meta_spec)
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apply(auto)
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apply(rule_tac x="y1 @ (a # y1a)" in exI)
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apply(rule_tac x="y2a" in exI)
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apply(auto)
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done
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lemma emb_trans:
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  assumes a: "x1 \<preceq> x2"
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  and     b: "x2 \<preceq> x3"
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  shows "x1 \<preceq> x3"
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using a b
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apply(induct arbitrary: x3)
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apply(metis emb0)
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apply(metis emb_cons_leftD emb_left)
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apply(drule_tac emb_cons_leftD)
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apply(auto)
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done
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lemma emb_strict_length:
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  assumes a: "x \<preceq> y" "x \<noteq> y" 
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  shows "length x < length y"
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using a
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by (induct) (auto simp add: less_Suc_eq)
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lemma emb_antisym:
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  assumes a: "x \<preceq> y" "y \<preceq> x" 
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  shows "x = y"
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using a emb_strict_length
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by (metis not_less_iff_gr_or_eq)
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lemma emb_wf:
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  shows "wf {(x, y). x \<preceq> y \<and> x \<noteq> y}"
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proof -
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  have "wf (measure length)" by simp
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  moreover
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  have "{(x, y). x \<preceq> y \<and> x \<noteq> y} \<subseteq> measure length"
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    unfolding measure_def by (auto simp add: emb_strict_length)
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  ultimately 
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  show "wf {(x, y). x \<preceq> y \<and> x \<noteq> y}" by (rule wf_subset)
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qed
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subsection {* Sub- and Supsequences *}
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definition
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 "SUBSEQ A \<equiv> {x. \<exists>y \<in> A. x \<preceq> y}"
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definition
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 "SUPSEQ A \<equiv> {x. \<exists>y \<in> A. y \<preceq> x}"
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lemma SUPSEQ_simps [simp]:
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  shows "SUPSEQ {} = {}"
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  and   "SUPSEQ {[]} = UNIV"
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unfolding SUPSEQ_def by auto
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lemma SUPSEQ_atom [simp]:
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  shows "SUPSEQ {[c]} = UNIV \<cdot> {[c]} \<cdot> UNIV"
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unfolding SUPSEQ_def conc_def
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by (auto) (metis append_Cons emb_cons_leftD)
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lemma SUPSEQ_union [simp]:
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  shows "SUPSEQ (A \<union> B) = SUPSEQ A \<union> SUPSEQ B"
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unfolding SUPSEQ_def by auto
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lemma SUPSEQ_conc [simp]:
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  shows "SUPSEQ (A \<cdot> B) = SUPSEQ A \<cdot> SUPSEQ B"
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unfolding SUPSEQ_def conc_def
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by (auto) (metis emb_append_leftD)
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lemma SUPSEQ_star [simp]:
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  shows "SUPSEQ (A\<star>) = UNIV"
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apply(subst star_unfold_left)
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apply(simp only: SUPSEQ_union) 
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apply(simp)
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done
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definition
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  Allreg :: "'a::finite rexp"
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where
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  "Allreg \<equiv> \<Uplus>(Atom ` UNIV)"
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lemma Allreg_lang [simp]:
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  shows "lang Allreg = (\<Union>a. {[a]})"
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unfolding Allreg_def by auto
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lemma [simp]:
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  shows "(\<Union>a. {[a]})\<star> = UNIV"
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apply(auto)
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apply(induct_tac x rule: list.induct)
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apply(auto)
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apply(subgoal_tac "[a] @ list \<in> (\<Union>a. {[a]})\<star>")
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apply(simp)
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apply(rule append_in_starI)
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apply(auto)
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done
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lemma Star_Allreg_lang [simp]:
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  shows "lang (Star Allreg) = UNIV"
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by simp
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221
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fun 
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  UP :: "'a::finite rexp \<Rightarrow> 'a rexp"
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where
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  "UP (Zero) = Zero"
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| "UP (One) = Star Allreg"
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| "UP (Atom c) = Times (Star Allreg) (Times (Atom c) (Star Allreg))"   
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| "UP (Plus r1 r2) = Plus (UP r1) (UP r2)"
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| "UP (Times r1 r2) = Times (UP r1) (UP r2)"
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| "UP (Star r) = Star Allreg"
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221
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lemma lang_UP:
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  fixes r::"letter rexp"
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  shows "lang (UP r) = SUPSEQ (lang r)"
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by (induct r) (simp_all)
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lemma regular_SUPSEQ: 
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  fixes A::"letter lang"
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  assumes "regular A"
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  shows "regular (SUPSEQ A)"
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proof -
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  from assms obtain r::"letter rexp" where "lang r = A" by auto
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  then have "lang (UP r) = SUPSEQ A" by (simp add: lang_UP)
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  then show "regular (SUPSEQ A)" by auto
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qed
221
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lemma SUPSEQ_subset:
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  shows "A \<subseteq> SUPSEQ A"
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unfolding SUPSEQ_def by auto
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223
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lemma SUBSEQ_complement:
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  shows "- (SUBSEQ A) = SUPSEQ (- (SUBSEQ A))"
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   213
proof -
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  have "- (SUBSEQ A) \<subseteq> SUPSEQ (- (SUBSEQ A))"
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    by (rule SUPSEQ_subset)
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   216
  moreover 
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  have "SUPSEQ (- (SUBSEQ A)) \<subseteq> - (SUBSEQ A)"
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  proof (rule ccontr)
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    assume "\<not> (SUPSEQ (- (SUBSEQ A)) \<subseteq> - (SUBSEQ A))"
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    then obtain x where 
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       a: "x \<in> SUPSEQ (- (SUBSEQ A))" and 
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       b: "x \<notin> - (SUBSEQ A)" by auto
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    from a obtain y where c: "y \<in> - (SUBSEQ A)" and d: "y \<preceq> x"
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      by (auto simp add: SUPSEQ_def)
221
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223
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    from b have "x \<in> SUBSEQ A" by simp
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    then obtain x' where f: "x' \<in> A" and e: "x \<preceq> x'"
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      by (auto simp add: SUBSEQ_def)
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    from d e have "y \<preceq> x'" by (rule emb_trans)
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    then have "y \<in> SUBSEQ A" using f
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      by (auto simp add: SUBSEQ_def)
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    with c show "False" by simp
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  qed
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  ultimately show "- (SUBSEQ A) = SUPSEQ (- (SUBSEQ A))" by simp
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qed
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   238
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lemma Higman_antichains:
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  assumes a: "\<forall>x \<in> A. \<forall>y \<in> A. x \<noteq> y \<longrightarrow> \<not>(x \<preceq> y) \<and> \<not>(y \<preceq> x)"
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  shows "finite A"
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proof (rule ccontr)
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  assume "infinite A"
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  then obtain f::"nat \<Rightarrow> letter list" where b: "inj f" and c: "range f \<subseteq> A"
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    by (auto simp add: infinite_iff_countable_subset)
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  from higman_idx obtain i j where d: "i < j" and e: "f i \<preceq> f j" by blast
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  have "f i \<noteq> f j" using b d by (auto simp add: inj_on_def)
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  moreover
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  have "f i \<in> A" using c by auto
223
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  moreover
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  have "f j \<in> A" using c by auto
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  ultimately have "\<not>(f i \<preceq> f j)" using a by simp
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  with e show "False" by simp
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qed
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68e28debe995 solved the SUBSEQ/SUPSEQ problem
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definition
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  minimal :: "letter list \<Rightarrow> letter lang \<Rightarrow> bool"
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where
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  "minimal x A \<equiv> (\<forall>y \<in> A. y \<preceq> x \<longrightarrow> x \<preceq> y)"
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221
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lemma main_lemma:
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  shows "\<exists>M. finite M \<and> SUPSEQ A = SUPSEQ M"
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   263
proof -
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  def M \<equiv> "{x \<in> A. minimal x A}"
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  have "finite M"
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    unfolding M_def minimal_def
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    by (rule Higman_antichains) (auto simp add: emb_antisym)
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  moreover
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  have "SUPSEQ A \<subseteq> SUPSEQ M"
223
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  proof
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    fix x
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    assume "x \<in> SUPSEQ A"
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    then obtain y where "y \<in> A" and "y \<preceq> x" by (auto simp add: SUPSEQ_def)
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    then have a: "y \<in> {y' \<in> A. y' \<preceq> x}" by simp
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    obtain z where b: "z \<in> A" "z \<preceq> x" and c: "\<forall>y. y \<preceq> z \<and> y \<noteq> z \<longrightarrow> y \<notin> {y' \<in> A. y' \<preceq> x}"
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      using wfE_min[OF emb_wf a] by auto
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   277
    then have "z \<in> M"
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      unfolding M_def minimal_def
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      by (auto intro: emb_trans)
233
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    with b(2) show "x \<in> SUPSEQ M"
223
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      by (auto simp add: SUPSEQ_def)
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  qed
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   283
  moreover
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   284
  have "SUPSEQ M \<subseteq> SUPSEQ A"
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   285
    by (auto simp add: SUPSEQ_def M_def)
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   286
  ultimately
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   287
  show "\<exists>M. finite M \<and> SUPSEQ A = SUPSEQ M" by blast
221
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qed
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   289
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lemma closure_SUPSEQ:
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  fixes A::"letter lang" 
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  shows "regular (SUPSEQ A)"
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   293
proof -
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   294
  obtain M where a: "finite M" and b: "SUPSEQ A = SUPSEQ M"
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   295
    using main_lemma by blast
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   296
  have "regular M" using a by (rule finite_regular)
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   297
  then have "regular (SUPSEQ M)" by (rule regular_SUPSEQ)
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   298
  then show "regular (SUPSEQ A)" using b by simp
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   299
qed
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   300
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lemma closure_SUBSEQ:
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  fixes A::"letter lang"
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   303
  shows "regular (SUBSEQ A)"
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   304
proof -
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   305
  have "regular (SUPSEQ (- SUBSEQ A))" by (rule closure_SUPSEQ)
223
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   306
  then have "regular (- SUBSEQ A)" by (subst SUBSEQ_complement) (simp)
221
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parents: 220
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   307
  then have "regular (- (- (SUBSEQ A)))" by (rule closure_complement)
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   308
  then show "regular (SUBSEQ A)" by simp
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   309
qed
68e28debe995 solved the SUBSEQ/SUPSEQ problem
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220
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end