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src/HOL/Library/Omega_Words_Fun.thy

author | wenzelm |

Wed, 08 Mar 2017 10:50:59 +0100 | |

changeset 65151 | a7394aa4d21c |

parent 64593 | 50c715579715 |

child 67443 | 3abf6a722518 |

permissions | -rw-r--r-- |

tuned proofs;

(* Author: Stefan Merz Author: Salomon Sickert Author: Julian Brunner Author: Peter Lammich *) section \<open>$\omega$-words\<close> theory Omega_Words_Fun imports Infinite_Set begin text \<open>Note: This theory is based on Stefan Merz's work.\<close> text \<open> Automata recognize languages, which are sets of words. For the theory of $\omega$-automata, we are mostly interested in $\omega$-words, but it is sometimes useful to reason about finite words, too. We are modeling finite words as lists; this lets us benefit from the existing library. Other formalizations could be investigated, such as representing words as functions whose domains are initial intervals of the natural numbers. \<close> subsection \<open>Type declaration and elementary operations\<close> text \<open> We represent $\omega$-words as functions from the natural numbers to the alphabet type. Other possible formalizations include a coinductive definition or a uniform encoding of finite and infinite words, as studied by M\"uller et al. \<close> type_synonym 'a word = "nat \<Rightarrow> 'a" text \<open> We can prefix a finite word to an $\omega$-word, and a way to obtain an $\omega$-word from a finite, non-empty word is by $\omega$-iteration. \<close> definition conc :: "['a list, 'a word] \<Rightarrow> 'a word" (infixr "\<frown>" 65) where "w \<frown> x == \<lambda>n. if n < length w then w!n else x (n - length w)" definition iter :: "'a list \<Rightarrow> 'a word" ("(_\<^sup>\<omega>)" [1000]) where "iter w == if w = [] then undefined else (\<lambda>n. w!(n mod (length w)))" lemma conc_empty[simp]: "[] \<frown> w = w" unfolding conc_def by auto lemma conc_fst[simp]: "n < length w \<Longrightarrow> (w \<frown> x) n = w!n" by (simp add: conc_def) lemma conc_snd[simp]: "\<not>(n < length w) \<Longrightarrow> (w \<frown> x) n = x (n - length w)" by (simp add: conc_def) lemma iter_nth [simp]: "0 < length w \<Longrightarrow> w\<^sup>\<omega> n = w!(n mod (length w))" by (simp add: iter_def) lemma conc_conc[simp]: "u \<frown> v \<frown> w = (u @ v) \<frown> w" (is "?lhs = ?rhs") proof fix n have u: "n < length u \<Longrightarrow> ?lhs n = ?rhs n" by (simp add: conc_def nth_append) have v: "\<lbrakk> \<not>(n < length u); n < length u + length v \<rbrakk> \<Longrightarrow> ?lhs n = ?rhs n" by (simp add: conc_def nth_append, arith) have w: "\<not>(n < length u + length v) \<Longrightarrow> ?lhs n = ?rhs n" by (simp add: conc_def nth_append, arith) from u v w show "?lhs n = ?rhs n" by blast qed lemma range_conc[simp]: "range (w\<^sub>1 \<frown> w\<^sub>2) = set w\<^sub>1 \<union> range w\<^sub>2" proof (intro equalityI subsetI) fix a assume "a \<in> range (w\<^sub>1 \<frown> w\<^sub>2)" then obtain i where 1: "a = (w\<^sub>1 \<frown> w\<^sub>2) i" by auto then show "a \<in> set w\<^sub>1 \<union> range w\<^sub>2" unfolding 1 by (cases "i < length w\<^sub>1") simp_all next fix a assume a: "a \<in> set w\<^sub>1 \<union> range w\<^sub>2" then show "a \<in> range (w\<^sub>1 \<frown> w\<^sub>2)" proof assume "a \<in> set w\<^sub>1" then obtain i where 1: "i < length w\<^sub>1" "a = w\<^sub>1 ! i" using in_set_conv_nth by metis show ?thesis proof show "a = (w\<^sub>1 \<frown> w\<^sub>2) i" using 1 by auto show "i \<in> UNIV" by rule qed next assume "a \<in> range w\<^sub>2" then obtain i where 1: "a = w\<^sub>2 i" by auto show ?thesis proof show "a = (w\<^sub>1 \<frown> w\<^sub>2) (length w\<^sub>1 + i)" using 1 by simp show "length w\<^sub>1 + i \<in> UNIV" by rule qed qed qed lemma iter_unroll: "0 < length w \<Longrightarrow> w\<^sup>\<omega> = w \<frown> w\<^sup>\<omega>" by (rule ext) (simp add: conc_def mod_geq) subsection \<open>Subsequence, Prefix, and Suffix\<close> definition suffix :: "[nat, 'a word] \<Rightarrow> 'a word" where "suffix k x \<equiv> \<lambda>n. x (k+n)" definition subsequence :: "'a word \<Rightarrow> nat \<Rightarrow> nat \<Rightarrow> 'a list" ("_ [_ \<rightarrow> _]" 900) where "subsequence w i j \<equiv> map w [i..<j]" abbreviation prefix :: "nat \<Rightarrow> 'a word \<Rightarrow> 'a list" where "prefix n w \<equiv> subsequence w 0 n" lemma suffix_nth [simp]: "(suffix k x) n = x (k+n)" by (simp add: suffix_def) lemma suffix_0 [simp]: "suffix 0 x = x" by (simp add: suffix_def) lemma suffix_suffix [simp]: "suffix m (suffix k x) = suffix (k+m) x" by (rule ext) (simp add: suffix_def add.assoc) lemma subsequence_append: "prefix (i + j) w = prefix i w @ (w [i \<rightarrow> i + j])" unfolding map_append[symmetric] upt_add_eq_append[OF le0] subsequence_def .. lemma subsequence_drop[simp]: "drop i (w [j \<rightarrow> k]) = w [j + i \<rightarrow> k]" by (simp add: subsequence_def drop_map) lemma subsequence_empty[simp]: "w [i \<rightarrow> j] = [] \<longleftrightarrow> j \<le> i" by (auto simp add: subsequence_def) lemma subsequence_length[simp]: "length (subsequence w i j) = j - i" by (simp add: subsequence_def) lemma subsequence_nth[simp]: "k < j - i \<Longrightarrow> (w [i \<rightarrow> j]) ! k = w (i + k)" unfolding subsequence_def by auto lemma subseq_to_zero[simp]: "w[i\<rightarrow>0] = []" by simp lemma subseq_to_smaller[simp]: "i\<ge>j \<Longrightarrow> w[i\<rightarrow>j] = []" by simp lemma subseq_to_Suc[simp]: "i\<le>j \<Longrightarrow> w [i \<rightarrow> Suc j] = w [ i \<rightarrow> j ] @ [w j]" by (auto simp: subsequence_def) lemma subsequence_singleton[simp]: "w [i \<rightarrow> Suc i] = [w i]" by (auto simp: subsequence_def) lemma subsequence_prefix_suffix: "prefix (j - i) (suffix i w) = w [i \<rightarrow> j]" proof (cases "i \<le> j") case True have "w [i \<rightarrow> j] = map w (map (\<lambda>n. n + i) [0..<j - i])" unfolding map_add_upt subsequence_def using le_add_diff_inverse2[OF True] by force also have "\<dots> = map (\<lambda>n. w (n + i)) [0..<j - i]" unfolding map_map comp_def by blast finally show ?thesis unfolding subsequence_def suffix_def add.commute[of i] by simp next case False then show ?thesis by (simp add: subsequence_def) qed lemma prefix_suffix: "x = prefix n x \<frown> (suffix n x)" by (rule ext) (simp add: subsequence_def conc_def) declare prefix_suffix[symmetric, simp] lemma word_split: obtains v\<^sub>1 v\<^sub>2 where "v = v\<^sub>1 \<frown> v\<^sub>2" "length v\<^sub>1 = k" proof show "v = prefix k v \<frown> suffix k v" by (rule prefix_suffix) show "length (prefix k v) = k" by simp qed lemma set_subsequence[simp]: "set (w[i\<rightarrow>j]) = w`{i..<j}" unfolding subsequence_def by auto lemma subsequence_take[simp]: "take i (w [j \<rightarrow> k]) = w [j \<rightarrow> min (j + i) k]" by (simp add: subsequence_def take_map min_def) lemma subsequence_shift[simp]: "(suffix i w) [j \<rightarrow> k] = w [i + j \<rightarrow> i + k]" by (metis add_diff_cancel_left subsequence_prefix_suffix suffix_suffix) lemma suffix_subseq_join[simp]: "i \<le> j \<Longrightarrow> v [i \<rightarrow> j] \<frown> suffix j v = suffix i v" by (metis (no_types, lifting) Nat.add_0_right le_add_diff_inverse prefix_suffix subsequence_shift suffix_suffix) lemma prefix_conc_fst[simp]: assumes "j \<le> length w" shows "prefix j (w \<frown> w') = take j w" proof - have "\<forall>i < j. (prefix j (w \<frown> w')) ! i = (take j w) ! i" using assms by (simp add: conc_fst subsequence_def) thus ?thesis by (simp add: assms list_eq_iff_nth_eq min.absorb2) qed lemma prefix_conc_snd[simp]: assumes "n \<ge> length u" shows "prefix n (u \<frown> v) = u @ prefix (n - length u) v" proof (intro nth_equalityI allI impI) show "length (prefix n (u \<frown> v)) = length (u @ prefix (n - length u) v)" using assms by simp fix i assume "i < length (prefix n (u \<frown> v))" then show "prefix n (u \<frown> v) ! i = (u @ prefix (n - length u) v) ! i" by (cases "i < length u") (auto simp: nth_append) qed lemma prefix_conc_length[simp]: "prefix (length w) (w \<frown> w') = w" by simp lemma suffix_conc_fst[simp]: assumes "n \<le> length u" shows "suffix n (u \<frown> v) = drop n u \<frown> v" proof show "suffix n (u \<frown> v) i = (drop n u \<frown> v) i" for i using assms by (cases "n + i < length u") (auto simp: algebra_simps) qed lemma suffix_conc_snd[simp]: assumes "n \<ge> length u" shows "suffix n (u \<frown> v) = suffix (n - length u) v" proof show "suffix n (u \<frown> v) i = suffix (n - length u) v i" for i using assms by simp qed lemma suffix_conc_length[simp]: "suffix (length w) (w \<frown> w') = w'" unfolding conc_def by force lemma concat_eq[iff]: assumes "length v\<^sub>1 = length v\<^sub>2" shows "v\<^sub>1 \<frown> u\<^sub>1 = v\<^sub>2 \<frown> u\<^sub>2 \<longleftrightarrow> v\<^sub>1 = v\<^sub>2 \<and> u\<^sub>1 = u\<^sub>2" (is "?lhs \<longleftrightarrow> ?rhs") proof assume ?lhs then have 1: "(v\<^sub>1 \<frown> u\<^sub>1) i = (v\<^sub>2 \<frown> u\<^sub>2) i" for i by auto show ?rhs proof (intro conjI ext nth_equalityI allI impI) show "length v\<^sub>1 = length v\<^sub>2" by (rule assms(1)) next fix i assume 2: "i < length v\<^sub>1" have 3: "i < length v\<^sub>2" using assms(1) 2 by simp show "v\<^sub>1 ! i = v\<^sub>2 ! i" using 1[of i] 2 3 by simp next show "u\<^sub>1 i = u\<^sub>2 i" for i using 1[of "length v\<^sub>1 + i"] assms(1) by simp qed next assume ?rhs then show ?lhs by simp qed lemma same_concat_eq[iff]: "u \<frown> v = u \<frown> w \<longleftrightarrow> v = w" by simp lemma comp_concat[simp]: "f \<circ> u \<frown> v = map f u \<frown> (f \<circ> v)" proof fix i show "(f \<circ> u \<frown> v) i = (map f u \<frown> (f \<circ> v)) i" by (cases "i < length u") simp_all qed subsection \<open>Prepending\<close> primrec build :: "'a \<Rightarrow> 'a word \<Rightarrow> 'a word" (infixr "##" 65) where "(a ## w) 0 = a" | "(a ## w) (Suc i) = w i" lemma build_eq[iff]: "a\<^sub>1 ## w\<^sub>1 = a\<^sub>2 ## w\<^sub>2 \<longleftrightarrow> a\<^sub>1 = a\<^sub>2 \<and> w\<^sub>1 = w\<^sub>2" proof assume 1: "a\<^sub>1 ## w\<^sub>1 = a\<^sub>2 ## w\<^sub>2" have 2: "(a\<^sub>1 ## w\<^sub>1) i = (a\<^sub>2 ## w\<^sub>2) i" for i using 1 by auto show "a\<^sub>1 = a\<^sub>2 \<and> w\<^sub>1 = w\<^sub>2" proof (intro conjI ext) show "a\<^sub>1 = a\<^sub>2" using 2[of "0"] by simp show "w\<^sub>1 i = w\<^sub>2 i" for i using 2[of "Suc i"] by simp qed next assume 1: "a\<^sub>1 = a\<^sub>2 \<and> w\<^sub>1 = w\<^sub>2" show "a\<^sub>1 ## w\<^sub>1 = a\<^sub>2 ## w\<^sub>2" using 1 by simp qed lemma build_cons[simp]: "(a # u) \<frown> v = a ## u \<frown> v" proof fix i show "((a # u) \<frown> v) i = (a ## u \<frown> v) i" proof (cases i) case 0 show ?thesis unfolding 0 by simp next case (Suc j) show ?thesis unfolding Suc by (cases "j < length u", simp+) qed qed lemma build_append[simp]: "(w @ a # u) \<frown> v = w \<frown> a ## u \<frown> v" unfolding conc_conc[symmetric] by simp lemma build_first[simp]: "w 0 ## suffix (Suc 0) w = w" proof show "(w 0 ## suffix (Suc 0) w) i = w i" for i by (cases i) simp_all qed lemma build_split[intro]: "w = w 0 ## suffix 1 w" by simp lemma build_range[simp]: "range (a ## w) = insert a (range w)" proof safe show "(a ## w) i \<notin> range w \<Longrightarrow> (a ## w) i = a" for i by (cases i) auto show "a \<in> range (a ## w)" proof (rule range_eqI) show "a = (a ## w) 0" by simp qed show "w i \<in> range (a ## w)" for i proof (rule range_eqI) show "w i = (a ## w) (Suc i)" by simp qed qed lemma suffix_singleton_suffix[simp]: "w i ## suffix (Suc i) w = suffix i w" using suffix_subseq_join[of i "Suc i" w] by simp text \<open>Find the first occurrence of a letter from a given set\<close> lemma word_first_split_set: assumes "A \<inter> range w \<noteq> {}" obtains u a v where "w = u \<frown> [a] \<frown> v" "A \<inter> set u = {}" "a \<in> A" proof - define i where "i = (LEAST i. w i \<in> A)" show ?thesis proof show "w = prefix i w \<frown> [w i] \<frown> suffix (Suc i) w" by simp show "A \<inter> set (prefix i w) = {}" apply safe subgoal premises prems for a proof - from prems obtain k where 3: "k < i" "w k = a" by auto have 4: "w k \<notin> A" using not_less_Least 3(1) unfolding i_def . show ?thesis using prems(1) 3(2) 4 by auto qed done show "w i \<in> A" using LeastI assms(1) unfolding i_def by fast qed qed subsection \<open>The limit set of an $\omega$-word\<close> text \<open> The limit set (also called infinity set) of an $\omega$-word is the set of letters that appear infinitely often in the word. This set plays an important role in defining acceptance conditions of $\omega$-automata. \<close> definition limit :: "'a word \<Rightarrow> 'a set" where "limit x \<equiv> {a . \<exists>\<^sub>\<infinity>n . x n = a}" lemma limit_iff_frequent: "a \<in> limit x \<longleftrightarrow> (\<exists>\<^sub>\<infinity>n . x n = a)" by (simp add: limit_def) text \<open> The following is a different way to define the limit, using the reverse image, making the laws about reverse image applicable to the limit set. (Might want to change the definition above?) \<close> lemma limit_vimage: "(a \<in> limit x) = infinite (x -` {a})" by (simp add: limit_def Inf_many_def vimage_def) lemma two_in_limit_iff: "({a, b} \<subseteq> limit x) = ((\<exists>n. x n =a ) \<and> (\<forall>n. x n = a \<longrightarrow> (\<exists>m>n. x m = b)) \<and> (\<forall>m. x m = b \<longrightarrow> (\<exists>n>m. x n = a)))" (is "?lhs = (?r1 \<and> ?r2 \<and> ?r3)") proof assume lhs: "?lhs" hence 1: "?r1" by (auto simp: limit_def elim: INFM_EX) from lhs have "\<forall>n. \<exists>m>n. x m = b" by (auto simp: limit_def INFM_nat) hence 2: "?r2" by simp from lhs have "\<forall>m. \<exists>n>m. x n = a" by (auto simp: limit_def INFM_nat) hence 3: "?r3" by simp from 1 2 3 show "?r1 \<and> ?r2 \<and> ?r3" by simp next assume "?r1 \<and> ?r2 \<and> ?r3" hence 1: "?r1" and 2: "?r2" and 3: "?r3" by simp+ have infa: "\<forall>m. \<exists>n\<ge>m. x n = a" proof fix m show "\<exists>n\<ge>m. x n = a" (is "?A m") proof (induct m) from 1 show "?A 0" by simp next fix m assume ih: "?A m" then obtain n where n: "n \<ge> m" "x n = a" by auto with 2 obtain k where k: "k>n" "x k = b" by auto with 3 obtain l where l: "l>k" "x l = a" by auto from n k l have "l \<ge> Suc m" by auto with l show "?A (Suc m)" by auto qed qed hence infa': "\<exists>\<^sub>\<infinity>n. x n = a" by (simp add: INFM_nat_le) have "\<forall>n. \<exists>m>n. x m = b" proof fix n from infa obtain k where k1: "k\<ge>n" and k2: "x k = a" by auto from 2 k2 obtain l where l1: "l>k" and l2: "x l = b" by auto from k1 l1 have "l > n" by auto with l2 show "\<exists>m>n. x m = b" by auto qed hence "\<exists>\<^sub>\<infinity>m. x m = b" by (simp add: INFM_nat) with infa' show "?lhs" by (auto simp: limit_def) qed text \<open> For $\omega$-words over a finite alphabet, the limit set is non-empty. Moreover, from some position onward, any such word contains only letters from its limit set. \<close> lemma limit_nonempty: assumes fin: "finite (range x)" shows "\<exists>a. a \<in> limit x" proof - from fin obtain a where "a \<in> range x \<and> infinite (x -` {a})" by (rule inf_img_fin_domE) auto hence "a \<in> limit x" by (auto simp add: limit_vimage) thus ?thesis .. qed lemmas limit_nonemptyE = limit_nonempty[THEN exE] lemma limit_inter_INF: assumes hyp: "limit w \<inter> S \<noteq> {}" shows "\<exists>\<^sub>\<infinity> n. w n \<in> S" proof - from hyp obtain x where "\<exists>\<^sub>\<infinity> n. w n = x" and "x \<in> S" by (auto simp add: limit_def) thus ?thesis by (auto elim: INFM_mono) qed text \<open> The reverse implication is true only if $S$ is finite. \<close> lemma INF_limit_inter: assumes hyp: "\<exists>\<^sub>\<infinity> n. w n \<in> S" and fin: "finite (S \<inter> range w)" shows "\<exists>a. a \<in> limit w \<inter> S" proof (rule ccontr) assume contra: "\<not>(\<exists>a. a \<in> limit w \<inter> S)" hence "\<forall>a\<in>S. finite {n. w n = a}" by (auto simp add: limit_def Inf_many_def) with fin have "finite (UN a:S \<inter> range w. {n. w n = a})" by auto moreover have "(UN a:S \<inter> range w. {n. w n = a}) = {n. w n \<in> S}" by auto moreover note hyp ultimately show "False" by (simp add: Inf_many_def) qed lemma fin_ex_inf_eq_limit: "finite A \<Longrightarrow> (\<exists>\<^sub>\<infinity>i. w i \<in> A) \<longleftrightarrow> limit w \<inter> A \<noteq> {}" by (metis INF_limit_inter equals0D finite_Int limit_inter_INF) lemma limit_in_range_suffix: "limit x \<subseteq> range (suffix k x)" proof fix a assume "a \<in> limit x" then obtain l where kl: "k < l" and xl: "x l = a" by (auto simp add: limit_def INFM_nat) from kl obtain m where "l = k+m" by (auto simp add: less_iff_Suc_add) with xl show "a \<in> range (suffix k x)" by auto qed lemma limit_in_range: "limit r \<subseteq> range r" using limit_in_range_suffix[of r 0] by simp lemmas limit_in_range_suffixD = limit_in_range_suffix[THEN subsetD] lemma limit_subset: "limit f \<subseteq> f ` {n..}" using limit_in_range_suffix[of f n] unfolding suffix_def by auto theorem limit_is_suffix: assumes fin: "finite (range x)" shows "\<exists>k. limit x = range (suffix k x)" proof - have "\<exists>k. range (suffix k x) \<subseteq> limit x" proof - \<comment> "The set of letters that are not in the limit is certainly finite." from fin have "finite (range x - limit x)" by simp \<comment> "Moreover, any such letter occurs only finitely often" moreover have "\<forall>a \<in> range x - limit x. finite (x -` {a})" by (auto simp add: limit_vimage) \<comment> "Thus, there are only finitely many occurrences of such letters." ultimately have "finite (UN a : range x - limit x. x -` {a})" by (blast intro: finite_UN_I) \<comment> "Therefore these occurrences are within some initial interval." then obtain k where "(UN a : range x - limit x. x -` {a}) \<subseteq> {..<k}" by (blast dest: finite_nat_bounded) \<comment> "This is just the bound we are looking for." hence "\<forall>m. k \<le> m \<longrightarrow> x m \<in> limit x" by (auto simp add: limit_vimage) hence "range (suffix k x) \<subseteq> limit x" by auto thus ?thesis .. qed then obtain k where "range (suffix k x) \<subseteq> limit x" .. with limit_in_range_suffix have "limit x = range (suffix k x)" by (rule subset_antisym) thus ?thesis .. qed lemmas limit_is_suffixE = limit_is_suffix[THEN exE] text \<open> The limit set enjoys some simple algebraic laws with respect to concatenation, suffixes, iteration, and renaming. \<close> theorem limit_conc [simp]: "limit (w \<frown> x) = limit x" proof (auto) fix a assume a: "a \<in> limit (w \<frown> x)" have "\<forall>m. \<exists>n. m<n \<and> x n = a" proof fix m from a obtain n where "m + length w < n \<and> (w \<frown> x) n = a" by (auto simp add: limit_def Inf_many_def infinite_nat_iff_unbounded) hence "m < n - length w \<and> x (n - length w) = a" by (auto simp add: conc_def) thus "\<exists>n. m<n \<and> x n = a" .. qed hence "infinite {n . x n = a}" by (simp add: infinite_nat_iff_unbounded) thus "a \<in> limit x" by (simp add: limit_def Inf_many_def) next fix a assume a: "a \<in> limit x" have "\<forall>m. length w < m \<longrightarrow> (\<exists>n. m<n \<and> (w \<frown> x) n = a)" proof (clarify) fix m assume m: "length w < m" with a obtain n where "m - length w < n \<and> x n = a" by (auto simp add: limit_def Inf_many_def infinite_nat_iff_unbounded) with m have "m < n + length w \<and> (w \<frown> x) (n + length w) = a" by (simp add: conc_def, arith) thus "\<exists>n. m<n \<and> (w \<frown> x) n = a" .. qed hence "infinite {n . (w \<frown> x) n = a}" by (simp add: unbounded_k_infinite) thus "a \<in> limit (w \<frown> x)" by (simp add: limit_def Inf_many_def) qed theorem limit_suffix [simp]: "limit (suffix n x) = limit x" proof - have "x = (prefix n x) \<frown> (suffix n x)" by (simp add: prefix_suffix) hence "limit x = limit (prefix n x \<frown> suffix n x)" by simp also have "\<dots> = limit (suffix n x)" by (rule limit_conc) finally show ?thesis by (rule sym) qed theorem limit_iter [simp]: assumes nempty: "0 < length w" shows "limit w\<^sup>\<omega> = set w" proof have "limit w\<^sup>\<omega> \<subseteq> range w\<^sup>\<omega>" by (auto simp add: limit_def dest: INFM_EX) also from nempty have "\<dots> \<subseteq> set w" by auto finally show "limit w\<^sup>\<omega> \<subseteq> set w" . next { fix a assume a: "a \<in> set w" then obtain k where k: "k < length w \<and> w!k = a" by (auto simp add: set_conv_nth) \<comment> "the following bound is terrible, but it simplifies the proof" from nempty k have "\<forall>m. w\<^sup>\<omega> ((Suc m)*(length w) + k) = a" by (simp add: mod_add_left_eq [symmetric]) moreover \<comment> "why is the following so hard to prove??" have "\<forall>m. m < (Suc m)*(length w) + k" proof fix m from nempty have "1 \<le> length w" by arith hence "m*1 \<le> m*length w" by simp hence "m \<le> m*length w" by simp with nempty have "m < length w + (m*length w) + k" by arith thus "m < (Suc m)*(length w) + k" by simp qed moreover note nempty ultimately have "a \<in> limit w\<^sup>\<omega>" by (auto simp add: limit_iff_frequent INFM_nat) } then show "set w \<subseteq> limit w\<^sup>\<omega>" by auto qed lemma limit_o [simp]: assumes a: "a \<in> limit w" shows "f a \<in> limit (f \<circ> w)" proof - from a have "\<exists>\<^sub>\<infinity>n. w n = a" by (simp add: limit_iff_frequent) hence "\<exists>\<^sub>\<infinity>n. f (w n) = f a" by (rule INFM_mono, simp) thus "f a \<in> limit (f \<circ> w)" by (simp add: limit_iff_frequent) qed text \<open> The converse relation is not true in general: $f(a)$ can be in the limit of $f \circ w$ even though $a$ is not in the limit of $w$. However, \<open>limit\<close> commutes with renaming if the function is injective. More generally, if $f(a)$ is the image of only finitely many elements, some of these must be in the limit of $w$. \<close> lemma limit_o_inv: assumes fin: "finite (f -` {x})" and x: "x \<in> limit (f \<circ> w)" shows "\<exists>a \<in> (f -` {x}). a \<in> limit w" proof (rule ccontr) assume contra: "\<not> ?thesis" \<comment> "hence, every element in the pre-image occurs only finitely often" then have "\<forall>a \<in> (f -` {x}). finite {n. w n = a}" by (simp add: limit_def Inf_many_def) \<comment> "so there are only finitely many occurrences of any such element" with fin have "finite (\<Union> a \<in> (f -` {x}). {n. w n = a})" by auto \<comment> \<open>these are precisely those positions where $x$ occurs in $f \circ w$\<close> moreover have "(\<Union> a \<in> (f -` {x}). {n. w n = a}) = {n. f(w n) = x}" by auto ultimately \<comment> "so $x$ can occur only finitely often in the translated word" have "finite {n. f(w n) = x}" by simp \<comment> \<open>\ldots\ which yields a contradiction\<close> with x show "False" by (simp add: limit_def Inf_many_def) qed theorem limit_inj [simp]: assumes inj: "inj f" shows "limit (f \<circ> w) = f ` (limit w)" proof show "f ` limit w \<subseteq> limit (f \<circ> w)" by auto show "limit (f \<circ> w) \<subseteq> f ` limit w" proof fix x assume x: "x \<in> limit (f \<circ> w)" from inj have "finite (f -` {x})" by (blast intro: finite_vimageI) with x obtain a where a: "a \<in> (f -` {x}) \<and> a \<in> limit w" by (blast dest: limit_o_inv) thus "x \<in> f ` (limit w)" by auto qed qed lemma limit_inter_empty: assumes fin: "finite (range w)" assumes hyp: "limit w \<inter> S = {}" shows "\<forall>\<^sub>\<infinity>n. w n \<notin> S" proof - from fin obtain k where k_def: "limit w = range (suffix k w)" using limit_is_suffix by blast have "w (k + k') \<notin> S" for k' using hyp unfolding k_def suffix_def image_def by blast thus ?thesis unfolding MOST_nat_le using le_Suc_ex by blast qed text \<open>If the limit is the suffix of the sequence's range, we may increase the suffix index arbitrarily\<close> lemma limit_range_suffix_incr: assumes "limit r = range (suffix i r)" assumes "j\<ge>i" shows "limit r = range (suffix j r)" (is "?lhs = ?rhs") proof - have "?lhs = range (suffix i r)" using assms by simp moreover have "\<dots> \<supseteq> ?rhs" using \<open>j\<ge>i\<close> by (metis (mono_tags, lifting) assms(2) image_subsetI le_Suc_ex range_eqI suffix_def suffix_suffix) moreover have "\<dots> \<supseteq> ?lhs" by (rule limit_in_range_suffix) ultimately show "?lhs = ?rhs" by (metis antisym_conv limit_in_range_suffix) qed text \<open>For two finite sequences, we can find a common suffix index such that the limits can be represented as these suffixes' ranges.\<close> lemma common_range_limit: assumes "finite (range x)" and "finite (range y)" obtains i where "limit x = range (suffix i x)" and "limit y = range (suffix i y)" proof - obtain i j where 1: "limit x = range (suffix i x)" and 2: "limit y = range (suffix j y)" using assms limit_is_suffix by metis have "limit x = range (suffix (max i j) x)" and "limit y = range (suffix (max i j) y)" using limit_range_suffix_incr[OF 1] limit_range_suffix_incr[OF 2] by auto thus ?thesis using that by metis qed subsection \<open>Index sequences and piecewise definitions\<close> text \<open> A word can be defined piecewise: given a sequence of words $w_0, w_1, \ldots$ and a strictly increasing sequence of integers $i_0, i_1, \ldots$ where $i_0=0$, a single word is obtained by concatenating subwords of the $w_n$ as given by the integers: the resulting word is \[ (w_0)_{i_0} \ldots (w_0)_{i_1-1} (w_1)_{i_1} \ldots (w_1)_{i_2-1} \ldots \] We prepare the field by proving some trivial facts about such sequences of indexes. \<close> definition idx_sequence :: "nat word \<Rightarrow> bool" where "idx_sequence idx \<equiv> (idx 0 = 0) \<and> (\<forall>n. idx n < idx (Suc n))" lemma idx_sequence_less: assumes iseq: "idx_sequence idx" shows "idx n < idx (Suc(n+k))" proof (induct k) from iseq show "idx n < idx (Suc (n + 0))" by (simp add: idx_sequence_def) next fix k assume ih: "idx n < idx (Suc(n+k))" from iseq have "idx (Suc(n+k)) < idx (Suc(n + Suc k))" by (simp add: idx_sequence_def) with ih show "idx n < idx (Suc(n + Suc k))" by (rule less_trans) qed lemma idx_sequence_inj: assumes iseq: "idx_sequence idx" and eq: "idx m = idx n" shows "m = n" proof (cases m n rule: linorder_cases) case greater then obtain k where "m = Suc(n+k)" by (auto simp add: less_iff_Suc_add) with iseq have "idx n < idx m" by (simp add: idx_sequence_less) with eq show ?thesis by simp next case less then obtain k where "n = Suc(m+k)" by (auto simp add: less_iff_Suc_add) with iseq have "idx m < idx n" by (simp add: idx_sequence_less) with eq show ?thesis by simp qed lemma idx_sequence_mono: assumes iseq: "idx_sequence idx" and m: "m \<le> n" shows "idx m \<le> idx n" proof (cases "m=n") case True thus ?thesis by simp next case False with m have "m < n" by simp then obtain k where "n = Suc(m+k)" by (auto simp add: less_iff_Suc_add) with iseq have "idx m < idx n" by (simp add: idx_sequence_less) thus ?thesis by simp qed text \<open> Given an index sequence, every natural number is contained in the interval defined by two adjacent indexes, and in fact this interval is determined uniquely. \<close> lemma idx_sequence_idx: assumes "idx_sequence idx" shows "idx k \<in> {idx k ..< idx (Suc k)}" using assms by (auto simp add: idx_sequence_def) lemma idx_sequence_interval: assumes iseq: "idx_sequence idx" shows "\<exists>k. n \<in> {idx k ..< idx (Suc k) }" (is "?P n" is "\<exists>k. ?in n k") proof (induct n) from iseq have "0 = idx 0" by (simp add: idx_sequence_def) moreover from iseq have "idx 0 \<in> {idx 0 ..< idx (Suc 0) }" by (rule idx_sequence_idx) ultimately show "?P 0" by auto next fix n assume "?P n" then obtain k where k: "?in n k" .. show "?P (Suc n)" proof (cases "Suc n < idx (Suc k)") case True with k have "?in (Suc n) k" by simp thus ?thesis .. next case False with k have "Suc n = idx (Suc k)" by auto with iseq have "?in (Suc n) (Suc k)" by (simp add: idx_sequence_def) thus ?thesis .. qed qed lemma idx_sequence_interval_unique: assumes iseq: "idx_sequence idx" and k: "n \<in> {idx k ..< idx (Suc k)}" and m: "n \<in> {idx m ..< idx (Suc m)}" shows "k = m" proof (cases k m rule: linorder_cases) case less hence "Suc k \<le> m" by simp with iseq have "idx (Suc k) \<le> idx m" by (rule idx_sequence_mono) with m have "idx (Suc k) \<le> n" by auto with k have "False" by simp thus ?thesis .. next case greater hence "Suc m \<le> k" by simp with iseq have "idx (Suc m) \<le> idx k" by (rule idx_sequence_mono) with k have "idx (Suc m) \<le> n" by auto with m have "False" by simp thus ?thesis .. qed lemma idx_sequence_unique_interval: assumes iseq: "idx_sequence idx" shows "\<exists>! k. n \<in> {idx k ..< idx (Suc k) }" proof (rule ex_ex1I) from iseq show "\<exists>k. n \<in> {idx k ..< idx (Suc k)}" by (rule idx_sequence_interval) next fix k y assume "n \<in> {idx k..<idx (Suc k)}" and "n \<in> {idx y..<idx (Suc y)}" with iseq show "k = y" by (auto elim: idx_sequence_interval_unique) qed text \<open> Now we can define the piecewise construction of a word using an index sequence. \<close> definition merge :: "'a word word \<Rightarrow> nat word \<Rightarrow> 'a word" where "merge ws idx \<equiv> \<lambda>n. let i = THE i. n \<in> {idx i ..< idx (Suc i) } in ws i n" lemma merge: assumes idx: "idx_sequence idx" and n: "n \<in> {idx i ..< idx (Suc i)}" shows "merge ws idx n = ws i n" proof - from n have "(THE k. n \<in> {idx k ..< idx (Suc k) }) = i" by (rule the_equality[OF _ sym[OF idx_sequence_interval_unique[OF idx n]]]) simp thus ?thesis by (simp add: merge_def Let_def) qed lemma merge0: assumes idx: "idx_sequence idx" shows "merge ws idx 0 = ws 0 0" proof (rule merge[OF idx]) from idx have "idx 0 < idx (Suc 0)" unfolding idx_sequence_def by blast with idx show "0 \<in> {idx 0 ..< idx (Suc 0)}" by (simp add: idx_sequence_def) qed lemma merge_Suc: assumes idx: "idx_sequence idx" and n: "n \<in> {idx i ..< idx (Suc i)}" shows "merge ws idx (Suc n) = (if Suc n = idx (Suc i) then ws (Suc i) else ws i) (Suc n)" proof auto assume eq: "Suc n = idx (Suc i)" from idx have "idx (Suc i) < idx (Suc(Suc i))" unfolding idx_sequence_def by blast with eq idx show "merge ws idx (idx (Suc i)) = ws (Suc i) (idx (Suc i))" by (simp add: merge) next assume neq: "Suc n \<noteq> idx (Suc i)" with n have "Suc n \<in> {idx i ..< idx (Suc i) }" by auto with idx show "merge ws idx (Suc n) = ws i (Suc n)" by (rule merge) qed end