Theory Omega_Words_Fun

(*
    Author:     Stefan Merz
    Author:     Salomon Sickert
    Author:     Julian Brunner
    Author:     Peter Lammich
*)

section $\omega$-words

theory Omega_Words_Fun

imports Infinite_Set
begin

text Note: This theory is based on Stefan Merz's work.

text 
  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.



subsection Type declaration and elementary operations

text 
  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.


type_synonym
  'a word = "nat  'a"

text 
  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.


definition
  conc :: "['a list, 'a word]  'a word"  (infixr  65)
  where "w  x == λn. if n < length w then w!n else x (n - length w)"

definition
  iter :: "'a list  'a word"  ((_ω) [1000])
  where "iter w == if w = [] then undefined else (λn. w!(n mod (length w)))"

lemma conc_empty[simp]: "[]  w = w"
  unfolding conc_def by auto

lemma conc_fst[simp]: "n < length w  (w  x) n = w!n"
  by (simp add: conc_def)

lemma conc_snd[simp]: "¬(n < length w)  (w  x) n = x (n - length w)"
  by (simp add: conc_def)

lemma iter_nth [simp]: "0 < length w  wω n = w!(n mod (length w))"
  by (simp add: iter_def)

lemma conc_conc[simp]: "u  v  w = (u @ v)  w" (is "?lhs = ?rhs")
proof
  fix n
  have u: "n < length u  ?lhs n = ?rhs n"
    by (simp add: conc_def nth_append)
  have v: " ¬(n < length u); n < length u + length v   ?lhs n = ?rhs n"
    by (simp add: conc_def nth_append, arith)
  have w: "¬(n < length u + length v)  ?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 (w1  w2) = set w1  range w2"
proof (intro equalityI subsetI)
  fix a
  assume "a  range (w1  w2)"
  then obtain i where 1: "a = (w1  w2) i" by auto
  then show "a  set w1  range w2"
    unfolding 1 by (cases "i < length w1") simp_all
next
  fix a
  assume a: "a  set w1  range w2"
  then show "a  range (w1  w2)"
  proof
    assume "a  set w1"
    then obtain i where 1: "i < length w1" "a = w1 ! i"
      using in_set_conv_nth by metis
    show ?thesis
    proof
      show "a = (w1  w2) i" using 1 by auto
      show "i  UNIV" by rule
    qed
  next
    assume "a  range w2"
    then obtain i where 1: "a = w2 i" by auto
    show ?thesis
    proof
      show "a = (w1  w2) (length w1 + i)" using 1 by simp
      show "length w1 + i  UNIV" by rule
    qed
  qed
qed


lemma iter_unroll: "0 < length w  wω = w  wω"
  by (rule ext) (simp add: conc_def mod_geq)


subsection Subsequence, Prefix, and Suffix

definition suffix :: "[nat, 'a word]  'a word"
  where "suffix k x  λn. x (k+n)"

definition subsequence :: "'a word  nat  nat  'a list"  (_ [_  _] 900)
  where "subsequence w i j  map w [i..<j]"

abbreviation prefix :: "nat  'a word  'a list"
  where "prefix n w  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  i + j])"
  unfolding map_append[symmetric] upt_add_eq_append[OF le0] subsequence_def ..

lemma subsequence_drop[simp]: "drop i (w [j  k]) = w [j + i  k]"
  by (simp add: subsequence_def drop_map)

lemma subsequence_empty[simp]: "w [i  j] = []  j  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  (w [i  j]) ! k = w (i + k)"
  unfolding subsequence_def
  by auto

lemma subseq_to_zero[simp]: "w[i0] = []"
  by simp

lemma subseq_to_smaller[simp]: "ij  w[ij] = []"
  by simp

lemma subseq_to_Suc[simp]: "ij  w [i  Suc j] = w [ i  j ] @ [w j]"
  by (auto simp: subsequence_def)

lemma subsequence_singleton[simp]: "w [i  Suc i] = [w i]"
  by (auto simp: subsequence_def)


lemma subsequence_prefix_suffix: "prefix (j - i) (suffix i w) = w [i  j]"
proof (cases "i  j")
  case True
  have "w [i  j] = map w (map (λn. n + i) [0..<j - i])"
    unfolding map_add_upt subsequence_def
    using le_add_diff_inverse2[OF True] by force
  also
  have " = map (λn. 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  (suffix n x)"
  by (rule ext) (simp add: subsequence_def conc_def)

declare prefix_suffix[symmetric, simp]


lemma word_split: obtains v1 v2 where "v = v1  v2" "length v1 = k"
proof
  show "v = prefix k v  suffix k v"
    by (rule prefix_suffix)
  show "length (prefix k v) = k"
    by simp
qed


lemma set_subsequence[simp]: "set (w[ij]) = w`{i..<j}"
  unfolding subsequence_def by auto

lemma subsequence_take[simp]: "take i (w [j  k]) = w [j  min (j + i) k]"
  by (simp add: subsequence_def take_map min_def)

lemma subsequence_shift[simp]: "(suffix i w) [j  k] = w [i + j  i + k]"
  by (metis add_diff_cancel_left subsequence_prefix_suffix suffix_suffix)

lemma suffix_subseq_join[simp]: "i  j  v [i  j]  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  length w"
  shows "prefix j (w  w') = take j w"
proof -
  have "i < j. (prefix j (w  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  length u"
  shows "prefix n (u  v) = u @ prefix (n - length u) v"
proof (intro nth_equalityI)
  show "length (prefix n (u  v)) = length (u @ prefix (n - length u) v)"
    using assms by simp
  fix i
  assume "i < length (prefix n (u  v))"
  then show "prefix n (u  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  w') = w"
  by simp

lemma suffix_conc_fst[simp]:
  assumes "n  length u"
  shows "suffix n (u  v) = drop n u  v"
proof
  show "suffix n (u  v) i = (drop n u  v) i" for i
    using assms by (cases "n + i < length u") (auto simp: algebra_simps)
qed

lemma suffix_conc_snd[simp]:
  assumes "n  length u"
  shows "suffix n (u  v) = suffix (n - length u) v"
proof
  show "suffix n (u  v) i = suffix (n - length u) v i" for i
    using assms by simp
qed

lemma suffix_conc_length[simp]: "suffix (length w) (w  w') = w'"
  unfolding conc_def by force

lemma concat_eq[iff]:
  assumes "length v1 = length v2"
  shows "v1  u1 = v2  u2  v1 = v2  u1 = u2"
  (is "?lhs  ?rhs")
proof
  assume ?lhs
  then have 1: "(v1  u1) i = (v2  u2) i" for i by auto
  show ?rhs
  proof (intro conjI ext nth_equalityI)
    show "length v1 = length v2" by (rule assms(1))
  next
    fix i
    assume 2: "i < length v1"
    have 3: "i < length v2" using assms(1) 2 by simp
    show "v1 ! i = v2 ! i" using 1[of i] 2 3 by simp
  next
    show "u1 i = u2 i" for i
      using 1[of "length v1 + i"] assms(1) by simp
  qed
next
  assume ?rhs
  then show ?lhs by simp
qed

lemma same_concat_eq[iff]: "u  v = u  w  v = w"
  by simp

lemma comp_concat[simp]: "f  u  v = map f u  (f  v)"
proof
  fix i
  show "(f  u  v) i = (map f u  (f  v)) i"
    by (cases "i < length u") simp_all
qed


subsection Prepending

primrec build :: "'a  'a word  'a word"  (infixr ## 65)
  where "(a ## w) 0 = a" | "(a ## w) (Suc i) = w i"

lemma build_eq[iff]: "a1 ## w1 = a2 ## w2  a1 = a2  w1 = w2"
proof
  assume 1: "a1 ## w1 = a2 ## w2"
  have 2: "(a1 ## w1) i = (a2 ## w2) i" for i
    using 1 by auto
  show "a1 = a2  w1 = w2"
  proof (intro conjI ext)
    show "a1 = a2"
      using 2[of "0"] by simp
    show "w1 i = w2 i" for i
      using 2[of "Suc i"] by simp
  qed
next
  assume 1: "a1 = a2  w1 = w2"
  show "a1 ## w1 = a2 ## w2" using 1 by simp
qed

lemma build_cons[simp]: "(a # u)  v = a ## u  v"
proof
  fix i
  show "((a # u)  v) i = (a ## u  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)  v = w  a ## u  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  range w  (a ## w) i = a" for i
    by (cases i) auto
  show "a  range (a ## w)"
  proof (rule range_eqI)
    show "a = (a ## w) 0" by simp
  qed
  show "w i  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 Find the first occurrence of a letter from a given set
lemma word_first_split_set:
  assumes "A  range w  {}"
  obtains u a v where "w = u  [a]  v" "A  set u = {}" "a  A"
proof -
  define i where "i = (LEAST i. w i  A)"
  show ?thesis
  proof
    show "w = prefix i w  [w i]  suffix (Suc i) w"
      by simp
    show "A  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  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  A"
      using LeastI assms(1) unfolding i_def by fast
  qed
qed


subsection The limit set of an $\omega$-word

text 
  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.


definition limit :: "'a word  'a set"
  where "limit x  {a . n . x n = a}"

lemma limit_iff_frequent: "a  limit x  (n . x n = a)"
  by (simp add: limit_def)

text 
  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?)


lemma limit_vimage: "(a  limit x) = infinite (x -` {a})"
  by (simp add: limit_def Inf_many_def vimage_def)

lemma two_in_limit_iff:
  "({a, b}  limit x) =
    ((n. x n =a )  (n. x n = a  (m>n. x m = b))  (m. x m = b  (n>m. x n = a)))"
  (is "?lhs = (?r1  ?r2  ?r3)")
proof
  assume lhs: "?lhs"
  hence 1: "?r1" by (auto simp: limit_def elim: INFM_EX)
  from lhs have "n. m>n. x m = b" by (auto simp: limit_def INFM_nat)
  hence 2: "?r2" by simp
  from lhs have "m. n>m. x n = a" by (auto simp: limit_def INFM_nat)
  hence 3: "?r3" by simp
  from 1 2 3 show "?r1  ?r2  ?r3" by simp
next
  assume "?r1  ?r2  ?r3"
  hence 1: "?r1" and 2: "?r2" and 3: "?r3" by simp+
  have infa: "m. nm. x n = a"
  proof
    fix m
    show "nm. 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  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  Suc m" by auto
      with l show "?A (Suc m)" by auto
    qed
  qed
  hence infa': "n. x n = a" by (simp add: INFM_nat_le)
  have "n. m>n. x m = b"
  proof
    fix n
    from infa obtain k where k1: "kn" 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 "m>n. x m = b" by auto
  qed
  hence "m. x m = b" by (simp add: INFM_nat)
  with infa' show "?lhs" by (auto simp: limit_def)
qed

text 
  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.


lemma limit_nonempty:
  assumes fin: "finite (range x)"
  shows "a. a  limit x"
proof -
  from fin obtain a where "a  range x  infinite (x -` {a})"
    by (rule inf_img_fin_domE) auto
  hence "a  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  S  {}"
  shows " n. w n  S"
proof -
  from hyp obtain x where " n. w n = x" and "x  S"
    by (auto simp add: limit_def)
  thus ?thesis
    by (auto elim: INFM_mono)
qed

text 
  The reverse implication is true only if $S$ is finite.


lemma INF_limit_inter:
  assumes hyp: " n. w n   S"
    and fin: "finite (S  range w)"
  shows  "a. a  limit w  S"
proof (rule ccontr)
  assume contra: "¬(a. a  limit w  S)"
  hence "aS. finite {n. w n = a}"
    by (auto simp add: limit_def Inf_many_def)
  with fin have "finite (UN a:S  range w. {n. w n = a})"
    by auto
  moreover
  have "(UN a:S  range w. {n. w n = a}) = {n. w n  S}"
    by auto
  moreover
  note hyp
  ultimately show "False"
    by (simp add: Inf_many_def)
qed

lemma fin_ex_inf_eq_limit: "finite A  (i. w i  A)  limit w  A  {}"
  by (metis INF_limit_inter equals0D finite_Int limit_inter_INF)

lemma limit_in_range_suffix: "limit x  range (suffix k x)"
proof
  fix a
  assume "a  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  range (suffix k x)"
    by auto
qed

lemma limit_in_range: "limit r  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  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 "k. limit x = range (suffix k x)"
proof -
  have "k. range (suffix k x)  limit x"
  proof -
    ― ‹The set of letters that are not in the limit is certainly finite.
    from fin have "finite (range x - limit x)"
      by simp
    ― ‹Moreover, any such letter occurs only finitely often
    moreover
    have "a  range x - limit x. finite (x -` {a})"
      by (auto simp add: limit_vimage)
    ― ‹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)
    ― ‹Therefore these occurrences are within some initial interval.
    then obtain k where "(UN a : range x - limit x. x -` {a})  {..<k}"
      by (blast dest: finite_nat_bounded)
    ― ‹This is just the bound we are looking for.
    hence "m. k  m  x m  limit x"
      by (auto simp add: limit_vimage)
    hence "range (suffix k x)  limit x"
      by auto
    thus ?thesis ..
  qed
  then obtain k where "range (suffix k x)  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 
  The limit set enjoys some simple algebraic laws with respect
  to concatenation, suffixes, iteration, and renaming.


theorem limit_conc [simp]: "limit (w  x) = limit x"
proof (auto)
  fix a assume a: "a  limit (w  x)"
  have "m. n. m<n  x n = a"
  proof
    fix m
    from a obtain n where "m + length w < n  (w  x) n = a"
      by (auto simp add: limit_def Inf_many_def infinite_nat_iff_unbounded)
    hence "m < n - length w  x (n - length w) = a"
      by (auto simp add: conc_def)
    thus "n. m<n  x n = a" ..
  qed
  hence "infinite {n . x n = a}"
    by (simp add: infinite_nat_iff_unbounded)
  thus "a  limit x"
    by (simp add: limit_def Inf_many_def)
next
  fix a assume a: "a  limit x"
  have "m. length w < m  (n. m<n  (w  x) n = a)"
  proof (clarify)
    fix m
    assume m: "length w < m"
    with a obtain n where "m - length w < n  x n = a"
      by (auto simp add: limit_def Inf_many_def infinite_nat_iff_unbounded)
    with m have "m < n + length w  (w  x) (n + length w) = a"
      by (simp add: conc_def, arith)
    thus "n. m<n  (w  x) n = a" ..
  qed
  hence "infinite {n . (w  x) n = a}"
    by (simp add: unbounded_k_infinite)
  thus "a  limit (w  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)  (suffix n x)"
    by (simp add: prefix_suffix)
  hence "limit x = limit (prefix n x  suffix n x)"
    by simp
  also have " = 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 wω = set w"
proof
  have "limit wω  range wω"
    by (auto simp add: limit_def dest: INFM_EX)
  also from nempty have "  set w"
    by auto
  finally show "limit wω  set w" .
next
  {
    fix a assume a: "a  set w"
    then obtain k where k: "k < length w  w!k = a"
      by (auto simp add: set_conv_nth)
    ― ‹the following bound is terrible, but it simplifies the proof
    from nempty k have "m. wω ((Suc m)*(length w) + k) = a"
      by (simp add: mod_add_left_eq [symmetric])
    moreover
    ― ‹why is the following so hard to prove??
    have "m. m < (Suc m)*(length w) + k"
    proof
      fix m
      from nempty have "1  length w" by arith
      hence "m*1  m*length w" by simp
      hence "m  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  limit wω"
      by (auto simp add: limit_iff_frequent INFM_nat)
  }
  then show "set w  limit wω" by auto
qed

lemma limit_o [simp]:
  assumes a: "a  limit w"
  shows "f a  limit (f  w)"
proof -
  from a
  have "n. w n = a"
    by (simp add: limit_iff_frequent)
  hence "n. f (w n) = f a"
    by (rule INFM_mono, simp)
  thus "f a  limit (f  w)"
    by (simp add: limit_iff_frequent)
qed

text 
  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, limit› 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 $w$.


lemma limit_o_inv:
  assumes fin: "finite (f -` {x})"
    and x: "x  limit (f  w)"
  shows "a  (f -` {x}). a  limit w"
proof (rule ccontr)
  assume contra: "¬ ?thesis"
  ― ‹hence, every element in the pre-image occurs only finitely often
  then have "a  (f -` {x}). finite {n. w n = a}"
    by (simp add: limit_def Inf_many_def)
  ― ‹so there are only finitely many occurrences of any such element
  with fin have "finite ( a  (f -` {x}). {n. w n = a})"
    by auto
  ― ‹these are precisely those positions where $x$ occurs in $f \circ w$
  moreover
  have "( a  (f -` {x}). {n. w n = a}) = {n. f(w n) = x}"
    by auto
  ultimately
  ― ‹so $x$ can occur only finitely often in the translated word
  have "finite {n. f(w n) = x}"
    by simp
  ― ‹\ldots\ which yields a contradiction
  with x show "False"
    by (simp add: limit_def Inf_many_def)
qed

theorem limit_inj [simp]:
  assumes inj: "inj f"
  shows "limit (f  w) = f ` (limit w)"
proof
  show "f ` limit w  limit (f  w)"
    by auto
  show "limit (f  w)  f ` limit w"
  proof
    fix x
    assume x: "x  limit (f  w)"
    from inj have "finite (f -` {x})"
      by (blast intro: finite_vimageI)
    with x obtain a where a: "a  (f -` {x})  a  limit w"
      by (blast dest: limit_o_inv)
    thus "x  f ` (limit w)"
      by auto
  qed
qed

lemma limit_inter_empty:
  assumes fin: "finite (range w)"
  assumes hyp: "limit w  S = {}"
  shows "n. w n  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')  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 If the limit is the suffix of the sequence's range,
  we may increase the suffix index arbitrarily
lemma limit_range_suffix_incr:
  assumes "limit r = range (suffix i r)"
  assumes "ji"
  shows "limit r = range (suffix j r)"
    (is "?lhs = ?rhs")
proof -
  have "?lhs = range (suffix i r)"
    using assms by simp
  moreover
  have "  ?rhs" using ji
    by (metis (mono_tags, lifting) assms(2)
        image_subsetI le_Suc_ex range_eqI suffix_def suffix_suffix)
  moreover
  have "  ?lhs" by (rule limit_in_range_suffix)
  ultimately
  show "?lhs = ?rhs"
    by (metis antisym_conv limit_in_range_suffix)
qed

text For two finite sequences, we can find a common suffix index such
  that the limits can be represented as these suffixes' ranges.
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 Index sequences and piecewise definitions

text 
  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.


definition idx_sequence :: "nat word  bool"
  where "idx_sequence idx  (idx 0 = 0)  (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  n"
  shows "idx m  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 
  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.


lemma idx_sequence_idx:
  assumes "idx_sequence idx"
  shows "idx k  {idx k ..< idx (Suc k)}"
using assms by (auto simp add: idx_sequence_def)

lemma idx_sequence_interval:
  assumes iseq: "idx_sequence idx"
  shows "k. n  {idx k ..< idx (Suc k) }"
    (is "?P n" is "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  {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  {idx k ..< idx (Suc k)}"
    and m: "n  {idx m ..< idx (Suc m)}"
  shows "k = m"
proof (cases k m rule: linorder_cases)
  case less
  hence "Suc k  m" by simp
  with iseq have "idx (Suc k)  idx m"
    by (rule idx_sequence_mono)
  with m have "idx (Suc k)  n"
    by auto
  with k have "False"
    by simp
  thus ?thesis ..
next
  case greater
  hence "Suc m  k" by simp
  with iseq have "idx (Suc m)  idx k"
    by (rule idx_sequence_mono)
  with k have "idx (Suc m)  n"
    by auto
  with m have "False"
    by simp
  thus ?thesis ..
qed

lemma idx_sequence_unique_interval:
  assumes iseq: "idx_sequence idx"
  shows "∃! k. n  {idx k ..< idx (Suc k) }"
proof (rule ex_ex1I)
  from iseq show "k. n  {idx k ..< idx (Suc k)}"
    by (rule idx_sequence_interval)
next
  fix k y
  assume "n  {idx k..<idx (Suc k)}" and "n  {idx y..<idx (Suc y)}"
  with iseq show "k = y" by (auto elim: idx_sequence_interval_unique)
qed

text 
  Now we can define the piecewise construction of a word using
  an index sequence.


definition merge :: "'a word word  nat word  'a word"
  where "merge ws idx  λn. let i = THE i. n  {idx i ..< idx (Suc i) } in ws i n"

lemma merge:
  assumes idx: "idx_sequence idx"
    and n: "n  {idx i ..< idx (Suc i)}"
  shows "merge ws idx n = ws i n"
proof -
  from n have "(THE k. n  {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  {idx 0 ..< idx (Suc 0)}"
    by (simp add: idx_sequence_def)
qed

lemma merge_Suc:
  assumes idx: "idx_sequence idx"
    and n: "n  {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  idx (Suc i)"
  with n have "Suc n  {idx i ..< idx (Suc i) }"
    by auto
  with idx show "merge ws idx (Suc n) = ws i (Suc n)"
    by (rule merge)
qed

end