Seq.thy
changeset 368 2d6beddb6fa6
equal deleted inserted replaced
367:79401279ba21 368:2d6beddb6fa6
       
     1 (*  Title:       Infinite Sequences
       
     2     Author:      Christian Sternagel <c-sterna@jaist.ac.jp>
       
     3                  René Thiemann       <rene.thiemann@uibk.ac.at>
       
     4     Maintainer:  Christian Sternagel and René Thiemann
       
     5     License:     LGPL
       
     6 *)
       
     7 
       
     8 (*
       
     9 Copyright 2012 Christian Sternagel, René Thiemann
       
    10 
       
    11 This file is part of IsaFoR/CeTA.
       
    12 
       
    13 IsaFoR/CeTA is free software: you can redistribute it and/or modify it under the
       
    14 terms of the GNU Lesser General Public License as published by the Free Software
       
    15 Foundation, either version 3 of the License, or (at your option) any later
       
    16 version.
       
    17 
       
    18 IsaFoR/CeTA is distributed in the hope that it will be useful, but WITHOUT ANY
       
    19 WARRANTY; without even the implied warranty of MERCHANTABILITY or FITNESS FOR A
       
    20 PARTICULAR PURPOSE.  See the GNU Lesser General Public License for more details.
       
    21 
       
    22 You should have received a copy of the GNU Lesser General Public License along
       
    23 with IsaFoR/CeTA. If not, see <http://www.gnu.org/licenses/>.
       
    24 *)
       
    25 header {* Infinite Sequences *}
       
    26 theory Seq
       
    27 imports
       
    28   Main
       
    29   "~~/src/HOL/Library/Infinite_Set"
       
    30 begin
       
    31 
       
    32 text {*Infinite sequences are represented by functions of type @{typ "nat \<Rightarrow> 'a"}.*}
       
    33 type_synonym 'a seq = "nat \<Rightarrow> 'a"
       
    34 
       
    35 subsection {*Operations on Infinite Sequences*}
       
    36 
       
    37 text {*Adding a new element at the front.*}
       
    38 abbreviation cons :: "'a \<Rightarrow> 'a seq \<Rightarrow> 'a seq" (infixr "#s" 65) where (*FIXME: find better infix symbol*)
       
    39   "x #s S \<equiv> (\<lambda>i. case i of 0 \<Rightarrow> x | Suc n \<Rightarrow> S n)"
       
    40 
       
    41 text {*An infinite sequence is \emph{linked} by a binary predicate @{term P} if every two
       
    42 consecutive elements satisfy it. Such a sequence is called a \emph{@{term P}-chain}. *}
       
    43 abbreviation (input) chainp :: "('a \<Rightarrow> 'a \<Rightarrow> bool) \<Rightarrow>'a seq \<Rightarrow> bool" where
       
    44   "chainp P S \<equiv> \<forall>i. P (S i) (S (Suc i))"
       
    45 
       
    46 text {*Special version for relations.*}
       
    47 abbreviation (input) chain :: "'a rel \<Rightarrow> 'a seq \<Rightarrow> bool" where
       
    48   "chain r S \<equiv> chainp (\<lambda>x y. (x, y) \<in> r) S"
       
    49 
       
    50 text {*Extending a chain at the front.*}
       
    51 lemma cons_chainp:
       
    52   assumes "P x (S 0)" and "chainp P S"
       
    53   shows "chainp P (x #s S)" (is "chainp P ?S")
       
    54 proof
       
    55   fix i show "P (?S i) (?S (Suc i))" using assms by (cases i) simp_all
       
    56 qed
       
    57 
       
    58 text {*Special version for relations.*}
       
    59 lemma cons_chain:
       
    60   assumes "(x, S 0) \<in> r" and "chain r S" shows "chain r (x #s S)"
       
    61   using cons_chainp[of "\<lambda>x y. (x, y) \<in> r", OF assms] .
       
    62 
       
    63 text {*A chain admits arbitrary transitive steps.*}
       
    64 lemma chainp_imp_relpowp:
       
    65   assumes "chainp P S" shows "(P^^j) (S i) (S (i + j))"
       
    66 proof (induct "i + j" arbitrary: j)
       
    67   case (Suc n) thus ?case using assms by (cases j) auto
       
    68 qed simp
       
    69 
       
    70 lemma chain_imp_relpow:
       
    71   assumes "chain r S" shows "(S i, S (i + j)) \<in> r^^j"
       
    72 proof (induct "i + j" arbitrary: j)
       
    73   case (Suc n) thus ?case using assms by (cases j) auto
       
    74 qed simp
       
    75 
       
    76 lemma chainp_imp_tranclp:
       
    77   assumes "chainp P S" and "i < j" shows "P^++ (S i) (S j)"
       
    78 proof -
       
    79   from less_imp_Suc_add[OF assms(2)] obtain n where "j = i + Suc n" by auto
       
    80   with chainp_imp_relpowp[of P S "Suc n" i, OF assms(1)]
       
    81     show ?thesis
       
    82       unfolding trancl_power[of "(S i, S j)", to_pred]
       
    83       by force
       
    84 qed
       
    85 
       
    86 lemma chain_imp_trancl:
       
    87   assumes "chain r S" and "i < j" shows "(S i, S j) \<in> r^+"
       
    88 proof -
       
    89   from less_imp_Suc_add[OF assms(2)] obtain n where "j = i + Suc n" by auto
       
    90   with chain_imp_relpow[OF assms(1), of i "Suc n"]
       
    91     show ?thesis unfolding trancl_power by force
       
    92 qed
       
    93 
       
    94 text {*A chain admits arbitrary reflexive and transitive steps.*}
       
    95 lemma chainp_imp_rtranclp:
       
    96   assumes "chainp P S" and "i \<le> j" shows "P^** (S i) (S j)"
       
    97 proof -
       
    98   from assms(2) obtain n where "j = i + n" by (induct "j - i" arbitrary: j) force+
       
    99   with chainp_imp_relpowp[of P S, OF assms(1), of n i] show ?thesis
       
   100     by (simp add: relpow_imp_rtrancl[of "(S i, S (i + n))", to_pred])
       
   101 qed
       
   102 
       
   103 lemma chain_imp_rtrancl:
       
   104   assumes "chain r S" and "i \<le> j" shows "(S i, S j) \<in> r^*"
       
   105 proof -
       
   106   from assms(2) obtain n where "j = i + n" by (induct "j - i" arbitrary: j) force+
       
   107   with chain_imp_relpow[OF assms(1), of i n] show ?thesis by (simp add: relpow_imp_rtrancl)
       
   108 qed
       
   109 
       
   110 text {*If for every @{term i} there is a later index @{term "f i"} such that the
       
   111 corresponding elements satisfy the predicate @{term P}, then there is a @{term P}-chain.*}
       
   112 lemma stepfun_imp_chainp:
       
   113   assumes "\<forall>i\<ge>n::nat. f i > i \<and> P (S i) (S (f i))"
       
   114   shows "chainp P (\<lambda>i. S ((f ^^ i) n))" (is "chainp P ?T")
       
   115 proof
       
   116   fix i
       
   117   from assms have "(f ^^ i) n \<ge> n" by (induct i) auto
       
   118   with assms[THEN spec[of _ "(f ^^ i) n"]]
       
   119     show "P (?T i) (?T (Suc i))" by simp
       
   120 qed
       
   121 
       
   122 text {*If for every @{term i} there is a later index @{term j} such that the
       
   123 corresponding elements satisfy the predicate @{term P}, then there is a @{term P}-chain.*}
       
   124 lemma steps_imp_chainp:
       
   125   assumes "\<forall>i\<ge>n::nat. \<exists>j>i. P (S i) (S j)" shows "\<exists>T. chainp P T"
       
   126 proof -
       
   127   from assms have "\<forall>i\<in>{i. i \<ge> n}. \<exists>j>i. P (S i) (S j)" by auto
       
   128   from bchoice[OF this]
       
   129     obtain f where seq: "\<forall>i\<ge>n. f i > i \<and> P (S i) (S (f i))" by auto
       
   130   from stepfun_imp_chainp[of n f P S, OF this] show ?thesis by fast
       
   131 qed
       
   132 
       
   133 
       
   134 subsection {* Predicates on Natural Numbers *}
       
   135 
       
   136 text {*If some property holds for infinitely many natural numbers, obtain
       
   137 an index function that points to these numbers in increasing order.*}
       
   138 
       
   139 locale infinitely_many =
       
   140   fixes p :: "nat \<Rightarrow> bool"
       
   141   assumes infinite: "INFM j. p j"
       
   142 begin
       
   143 
       
   144 lemma inf: "\<exists>j\<ge>i. p j" using infinite[unfolded INFM_nat_le] by auto
       
   145 
       
   146 fun index :: "nat seq" where
       
   147   "index 0 = (LEAST n. p n)"
       
   148 | "index (Suc n) = (LEAST k. p k \<and> k > index n)"
       
   149 
       
   150 lemma index_p: "p (index n)"
       
   151 proof (induct n)
       
   152   case 0
       
   153   from inf obtain j where "p j" by auto
       
   154   with LeastI[of p j] show ?case by auto
       
   155 next
       
   156   case (Suc n)
       
   157   from inf obtain k where "k \<ge> Suc (index n) \<and> p k" by auto
       
   158   with LeastI[of "\<lambda> k. p k \<and> k > index n" k] show ?case by auto
       
   159 qed
       
   160 
       
   161 lemma index_ordered: "index n < index (Suc n)"
       
   162 proof -
       
   163   from inf obtain k where "k \<ge> Suc (index n) \<and> p k" by auto
       
   164   with LeastI[of "\<lambda> k. p k \<and> k > index n" k] show ?thesis by auto
       
   165 qed
       
   166 
       
   167 lemma index_not_p_between:
       
   168   assumes i1: "index n < i"
       
   169     and i2: "i < index (Suc n)"
       
   170   shows "\<not> p i"
       
   171 proof -
       
   172   from not_less_Least[OF i2[simplified]] i1 show ?thesis by auto
       
   173 qed
       
   174 
       
   175 lemma index_ordered_le:
       
   176   assumes "i \<le> j" shows "index i \<le> index j"
       
   177 proof - 
       
   178   from assms have "j = i + (j - i)" by auto
       
   179   then obtain k where j: "j = i + k" by auto
       
   180   have "index i \<le> index (i + k)"
       
   181   proof (induct k)
       
   182     case (Suc k)
       
   183     with index_ordered[of "i + k"]
       
   184     show ?case by auto
       
   185   qed simp
       
   186   thus ?thesis unfolding j .
       
   187 qed
       
   188 
       
   189 lemma index_surj:
       
   190   assumes "k \<ge> index l"
       
   191   shows "\<exists>i j. k = index i + j \<and> index i + j < index (Suc i)"
       
   192 proof -
       
   193   from assms have "k = index l + (k - index l)" by auto
       
   194   then obtain u where k: "k = index l + u" by auto
       
   195   show ?thesis unfolding k
       
   196   proof (induct u)
       
   197     case 0
       
   198     show ?case
       
   199       by (intro exI conjI, rule refl, insert index_ordered[of l], simp)
       
   200   next
       
   201     case (Suc u)
       
   202     then obtain i j
       
   203       where lu: "index l + u = index i + j" and lt: "index i + j < index (Suc i)" by auto
       
   204     hence "index l + u < index (Suc i)" by auto
       
   205     show ?case
       
   206     proof (cases "index l + (Suc u) = index (Suc i)")
       
   207       case False
       
   208       show ?thesis
       
   209         by (rule exI[of _ i], rule exI[of _ "Suc j"], insert lu lt False, auto)
       
   210     next
       
   211       case True
       
   212       show ?thesis
       
   213         by (rule exI[of _ "Suc i"], rule exI[of _ 0], insert True index_ordered[of "Suc i"], auto)
       
   214     qed
       
   215   qed
       
   216 qed
       
   217 
       
   218 lemma index_ordered_less:
       
   219   assumes "i < j" shows "index i < index j"
       
   220 proof - 
       
   221   from assms have "Suc i \<le> j" by auto
       
   222   from index_ordered_le[OF this]
       
   223   have "index (Suc i) \<le> index j" .
       
   224   with index_ordered[of i] show ?thesis by auto
       
   225 qed
       
   226 
       
   227 lemma index_not_p_start: assumes i: "i < index 0" shows "\<not> p i"
       
   228 proof -
       
   229   from i[simplified index.simps] have "i < Least p" .
       
   230   from not_less_Least[OF this] show ?thesis .
       
   231 qed
       
   232 
       
   233 end
       
   234 
       
   235 
       
   236 subsection {* Assembling Infinite Words from Finite Words *}
       
   237 
       
   238 text {*Concatenate infinitely many non-empty words to an infinite word.*}
       
   239 
       
   240 fun inf_concat_simple :: "(nat \<Rightarrow> nat) \<Rightarrow> nat \<Rightarrow> (nat \<times> nat)" where
       
   241   "inf_concat_simple f 0 = (0, 0)"
       
   242 | "inf_concat_simple f (Suc n) = (
       
   243     let (i, j) = inf_concat_simple f n in 
       
   244     if Suc j < f i then (i, Suc j)
       
   245     else (Suc i, 0))"
       
   246 
       
   247 lemma inf_concat_simple_add:
       
   248   assumes ck: "inf_concat_simple f k = (i, j)"
       
   249     and jl: "j + l < f i"
       
   250   shows "inf_concat_simple f (k + l) = (i,j + l)"
       
   251 using jl
       
   252 proof (induct l)
       
   253   case 0
       
   254   thus ?case using ck by simp
       
   255 next
       
   256   case (Suc l)
       
   257   hence c: "inf_concat_simple f (k + l) = (i, j+ l)" by auto
       
   258   show ?case 
       
   259     by (simp add: c, insert Suc(2), auto)
       
   260 qed
       
   261 
       
   262 lemma inf_concat_simple_surj_zero: "\<exists> k. inf_concat_simple f k = (i,0)"
       
   263 proof (induct i)
       
   264   case 0
       
   265   show ?case 
       
   266     by (rule exI[of _ 0], simp)
       
   267 next
       
   268   case (Suc i)
       
   269   then obtain k where ck: "inf_concat_simple f k = (i,0)" by auto
       
   270   show ?case
       
   271   proof (cases "f i")
       
   272     case 0
       
   273     show ?thesis
       
   274       by (rule exI[of _ "Suc k"], simp add: ck 0)
       
   275   next
       
   276     case (Suc n)
       
   277     hence "0 + n < f i" by auto
       
   278     from inf_concat_simple_add[OF ck, OF this] Suc
       
   279     show ?thesis
       
   280       by (intro exI[of _ "k + Suc n"], auto)
       
   281   qed
       
   282 qed
       
   283 
       
   284 lemma inf_concat_simple_surj:
       
   285   assumes "j < f i"
       
   286   shows "\<exists> k. inf_concat_simple f k = (i,j)"
       
   287 proof -
       
   288   from assms have j: "0 + j < f i" by auto
       
   289   from inf_concat_simple_surj_zero obtain k where "inf_concat_simple f k = (i,0)" by auto
       
   290   from inf_concat_simple_add[OF this, OF j] show ?thesis by auto
       
   291 qed
       
   292 
       
   293 lemma inf_concat_simple_mono:
       
   294   assumes "k \<le> k'" shows "fst (inf_concat_simple f k) \<le> fst (inf_concat_simple f k')"
       
   295 proof -
       
   296   from assms have "k' = k + (k' - k)" by auto
       
   297   then obtain l where k': "k' = k + l" by auto
       
   298   show ?thesis  unfolding k'
       
   299   proof (induct l)
       
   300     case (Suc l)
       
   301     obtain i j where ckl: "inf_concat_simple f (k+l) = (i,j)" by (cases "inf_concat_simple f (k+l)", auto)
       
   302     with Suc have "fst (inf_concat_simple f k) \<le> i" by auto
       
   303     also have "... \<le> fst (inf_concat_simple f (k + Suc l))"
       
   304       by (simp add: ckl)
       
   305     finally show ?case .
       
   306   qed simp
       
   307 qed
       
   308 
       
   309 
       
   310 (* inf_concat assembles infinitely many (possibly empty) words to an infinite word *)
       
   311 fun inf_concat :: "(nat \<Rightarrow> nat) \<Rightarrow> nat \<Rightarrow> nat \<times> nat" where
       
   312   "inf_concat n 0 = (LEAST j. n j > 0, 0)"
       
   313 | "inf_concat n (Suc k) = (let (i, j) = inf_concat n k in (if Suc j < n i then (i, Suc j) else (LEAST i'. i' > i \<and> n i' > 0, 0)))"
       
   314 
       
   315 lemma inf_concat_bounds:
       
   316   assumes inf: "INFM i. n i > 0"
       
   317     and res: "inf_concat n k = (i,j)"
       
   318   shows "j < n i"
       
   319 proof (cases k)
       
   320   case 0
       
   321   with res have i: "i = (LEAST i. n i > 0)" and j: "j = 0" by auto
       
   322   from inf[unfolded INFM_nat_le] obtain i' where i': "0 < n i'" by auto
       
   323   have "0 < n (LEAST i. n i > 0)" 
       
   324     by (rule LeastI, rule i')
       
   325   with i j show ?thesis by auto
       
   326 next
       
   327   case (Suc k')
       
   328   obtain i' j' where res': "inf_concat n k' = (i',j')" by force
       
   329   note res = res[unfolded Suc inf_concat.simps res' Let_def split]
       
   330   show ?thesis 
       
   331   proof (cases "Suc j' < n i'")
       
   332     case True
       
   333     with res show ?thesis by auto
       
   334   next
       
   335     case False
       
   336     with res have i: "i = (LEAST f. i' < f \<and> 0 < n f)" and j: "j = 0" by auto
       
   337     from inf[unfolded INFM_nat] obtain f where f: "i' < f \<and> 0 < n f" by auto
       
   338     have "0 < n (LEAST f. i' < f \<and> 0 < n f)"
       
   339       using LeastI[of "\<lambda> f. i' < f \<and> 0 < n f", OF f]
       
   340       by auto
       
   341     with i j show ?thesis by auto
       
   342   qed
       
   343 qed
       
   344 
       
   345 lemma inf_concat_add:
       
   346   assumes res: "inf_concat n k = (i,j)"
       
   347     and j: "j + m < n i"
       
   348   shows "inf_concat n (k + m) = (i,j+m)"
       
   349   using j
       
   350 proof (induct m)
       
   351   case 0 show ?case using res by auto
       
   352 next
       
   353   case (Suc m)
       
   354   hence "inf_concat n (k + m) = (i, j+m)" by auto
       
   355   with Suc(2)
       
   356   show ?case by auto
       
   357 qed
       
   358 
       
   359 lemma inf_concat_step:
       
   360   assumes res: "inf_concat n k = (i,j)"
       
   361     and j: "Suc (j + m) = n i"
       
   362   shows "inf_concat n (k + Suc m) = (LEAST i'. i' > i \<and> 0 < n i', 0)"
       
   363 proof -
       
   364   from j have "j + m < n i" by auto
       
   365   note res = inf_concat_add[OF res, OF this]
       
   366   show ?thesis by (simp add: res j)
       
   367 qed
       
   368 
       
   369 lemma inf_concat_surj_zero:
       
   370   assumes "0 < n i"
       
   371   shows "\<exists>k. inf_concat n k = (i, 0)"
       
   372 proof -
       
   373   {
       
   374     fix l
       
   375     have "\<forall> j. j < l \<and> 0 < n j \<longrightarrow> (\<exists> k. inf_concat n k = (j,0))"
       
   376     proof (induct l)
       
   377       case 0
       
   378       thus ?case by auto
       
   379     next
       
   380       case (Suc l)
       
   381       show ?case
       
   382       proof (intro allI impI, elim conjE)
       
   383         fix j
       
   384         assume j: "j < Suc l" and nj: "0 < n j"
       
   385         show "\<exists> k. inf_concat n k = (j, 0)"
       
   386         proof (cases "j < l")
       
   387           case True
       
   388           from Suc[THEN spec[of _ j]] True nj show ?thesis by auto
       
   389         next
       
   390           case False
       
   391           with j have j: "j = l" by auto
       
   392           show ?thesis
       
   393           proof (cases "\<exists> j'. j' < l \<and> 0 < n j'")
       
   394             case False
       
   395             have l: "(LEAST i. 0 < n i) = l"
       
   396             proof (rule Least_equality, rule nj[unfolded j])
       
   397               fix l'
       
   398               assume "0 < n l'"
       
   399               with False have "\<not> l' < l" by auto
       
   400               thus "l \<le> l'" by auto
       
   401             qed
       
   402             show ?thesis
       
   403               by (rule exI[of _ 0], simp add: l j)
       
   404           next
       
   405             case True
       
   406             then obtain lll where lll: "lll < l" and nlll: "0 < n lll" by auto 
       
   407             then obtain ll where l: "l = Suc ll" by (cases l, auto)   
       
   408             from lll l have lll: "lll = ll - (ll - lll)" by auto
       
   409             let ?l' = "LEAST d. 0 < n (ll - d)"
       
   410             have nl': "0 < n (ll - ?l')"
       
   411             proof (rule LeastI)
       
   412               show "0 < n (ll - (ll - lll))" using lll nlll by auto
       
   413             qed
       
   414             with Suc[THEN spec[of _ "ll - ?l'"]] obtain k where k:
       
   415               "inf_concat n k = (ll - ?l',0)" unfolding l by auto
       
   416             from nl' obtain off where off: "Suc (0 + off) = n (ll - ?l')" by (cases "n (ll - ?l')", auto)
       
   417             from inf_concat_step[OF k, OF off]
       
   418             have id: "inf_concat n (k + Suc off) = (LEAST i'. ll - ?l' < i' \<and> 0 < n i',0)" (is "_ = (?l,0)") .
       
   419             have ll: "?l = l" unfolding l
       
   420             proof (rule Least_equality)
       
   421               show "ll - ?l' < Suc ll \<and> 0 < n (Suc ll)" using nj[unfolded j l] by simp
       
   422             next
       
   423               fix l'
       
   424               assume ass: "ll - ?l' < l' \<and> 0 < n l'"
       
   425               show "Suc ll \<le> l'" 
       
   426               proof (rule ccontr)
       
   427                 assume not: "\<not> ?thesis"
       
   428                 hence "l' \<le> ll" by auto
       
   429                 hence "ll = l' + (ll - l')" by auto
       
   430                 then obtain k where ll: "ll = l' + k" by auto
       
   431                 from ass have "l' + k - ?l' < l'" unfolding ll by auto
       
   432                 hence kl': "k < ?l'" by auto
       
   433                 have "0 < n (ll - k)" using ass unfolding ll by simp
       
   434                 from Least_le[of "\<lambda> k. 0 < n (ll - k)", OF this] kl'
       
   435                 show False by auto
       
   436               qed
       
   437             qed            
       
   438             show ?thesis unfolding j
       
   439               by (rule exI[of _ "k + Suc off"], unfold id ll, simp)
       
   440           qed
       
   441         qed
       
   442       qed
       
   443     qed
       
   444   }
       
   445   with assms show ?thesis by auto
       
   446 qed
       
   447 
       
   448 lemma inf_concat_surj:
       
   449   assumes j: "j < n i"
       
   450   shows "\<exists>k. inf_concat n k = (i, j)"
       
   451 proof -
       
   452   from j have "0 < n i" by auto
       
   453   from inf_concat_surj_zero[of n, OF this]
       
   454   obtain k where "inf_concat n k = (i,0)" by auto
       
   455   from inf_concat_add[OF this, of j] j
       
   456   show ?thesis by auto
       
   457 qed
       
   458 
       
   459 lemma inf_concat_mono:
       
   460   assumes inf: "INFM i. n i > 0"
       
   461     and resk: "inf_concat n k = (i, j)"
       
   462     and reskp: "inf_concat n k' = (i', j')"
       
   463     and lt: "i < i'"
       
   464   shows "k < k'"
       
   465 proof -
       
   466   note bounds = inf_concat_bounds[OF inf]
       
   467   {
       
   468     assume "k' \<le> k"
       
   469     hence "k = k' + (k - k')" by auto
       
   470     then obtain l where k: "k = k' + l" by auto
       
   471     have "i' \<le> fst (inf_concat n (k' + l))" 
       
   472     proof (induct l)
       
   473       case 0
       
   474       with reskp show ?case by auto
       
   475     next      
       
   476       case (Suc l)
       
   477       obtain i'' j'' where l: "inf_concat n (k' + l) = (i'',j'')" by force
       
   478       with Suc have one: "i' \<le> i''" by auto
       
   479       from bounds[OF l] have j'': "j'' < n i''" by auto
       
   480       show ?case 
       
   481       proof (cases "Suc j'' < n i''")
       
   482         case True
       
   483         show ?thesis by (simp add: l True one)
       
   484       next
       
   485         case False
       
   486         let ?i = "LEAST i'. i'' < i' \<and> 0 < n i'"
       
   487         from inf[unfolded INFM_nat] obtain k where "i'' < k \<and> 0 < n k" by auto
       
   488         from LeastI[of "\<lambda> k. i'' < k \<and> 0 < n k", OF this]
       
   489         have "i'' < ?i" by auto
       
   490         with one show ?thesis by (simp add: l False)
       
   491       qed
       
   492     qed      
       
   493     with resk k lt have False by auto
       
   494   }
       
   495   thus ?thesis by arith
       
   496 qed
       
   497 
       
   498 lemma inf_concat_Suc:
       
   499   assumes inf: "INFM i. n i > 0"
       
   500     and f: "\<And> i. f i (n i) = f (Suc i) 0"
       
   501     and resk: "inf_concat n k = (i, j)"
       
   502     and ressk: "inf_concat n (Suc k) = (i', j')"
       
   503   shows "f i' j' = f i (Suc j)"
       
   504 proof - 
       
   505   note bounds = inf_concat_bounds[OF inf]
       
   506   from bounds[OF resk] have j: "j < n i" .
       
   507   show ?thesis
       
   508   proof (cases "Suc j < n i")
       
   509     case True
       
   510     with ressk resk
       
   511     show ?thesis by simp
       
   512   next
       
   513     case False
       
   514     let ?p = "\<lambda> i'. i < i' \<and> 0 < n i'"
       
   515     let ?i' = "LEAST i'. ?p i'"
       
   516     from False j have id: "Suc (j + 0) = n i" by auto
       
   517     from inf_concat_step[OF resk, OF id] ressk
       
   518     have i': "i' = ?i'" and j': "j' = 0" by auto
       
   519     from id have j: "Suc j = n i" by simp
       
   520     from inf[unfolded INFM_nat] obtain k where "?p k" by auto
       
   521     from LeastI[of ?p, OF this] have "?p ?i'" .
       
   522     hence "?i' = Suc i + (?i' - Suc i)" by simp
       
   523     then obtain d where ii': "?i' = Suc i + d" by auto
       
   524     from not_less_Least[of _ ?p, unfolded ii'] have d': "\<And> d'. d' < d \<Longrightarrow> n (Suc i + d') = 0" by auto
       
   525     have "f (Suc i) 0 = f ?i' 0" unfolding ii' using d'
       
   526     proof (induct d)
       
   527       case 0
       
   528       show ?case by simp
       
   529     next
       
   530       case (Suc d)
       
   531       hence "f (Suc i) 0 = f (Suc i + d) 0" by auto
       
   532       also have "... = f (Suc (Suc i + d)) 0"
       
   533         unfolding f[symmetric]
       
   534         using Suc(2)[of d] by simp
       
   535       finally show ?case by simp
       
   536     qed
       
   537     thus ?thesis unfolding i' j' j f by simp
       
   538   qed
       
   539 qed
       
   540 
       
   541 end