1 theory turing_basic

     2 imports Main

     3 begin

     4 

     5 section {* Basic definitions of Turing machine *}

     6 

     7 (* Title: Turing machine's definition and its charater

     8    Author: Xu Jian <xujian817@hotmail.com>

     9    Maintainer: Xu Jian

    10 *)

    11 

    12 text {*

    13   Actions of Turing machine (Abbreviated TM in the following* ).

    14 *}

    15 

    16 datatype taction = 

    17   -- {* Write zero *}

    18   W0 | 

    19   -- {* Write one *}

    20   W1 | 

    21   -- {* Move left *}

    22   L | 

    23   -- {* Move right *}

    24   R | 

    25   -- {* Do nothing *}

    26   Nop

    27 

    28 text {*

    29   Tape contents in every block.

    30 *}

    31 

    32 datatype block = 

    33   -- {* Blank *}

    34   Bk | 

    35   -- {* Occupied *}

    36   Oc

    37 

    38 text {*

    39   Tape is represented as a pair of lists $(L_{left}, L_{right})$,

    40   where $L_left$, named {\em left list}, is used to represent

    41   the tape to the left of RW-head and

    42   $L_{right}$, named {\em right list}, is used to represent the tape

    43   under and to the right of RW-head.

    44 *}

    45 

    46 type_synonym tape = "block list \<times> block list"

    47 

    48 text {* The state of turing machine.*}

    49 type_synonym tstate = nat

    50 

    51 text {*

    52   Turing machine instruction is represented as a 

    53   pair @{text "(action, next_state)"},

    54   where @{text "action"} is the action to take at the current state 

    55   and @{text "next_state"} is the next state the machine is getting into

    56   after the action.

    57 *}

    58 type_synonym tinst = "taction \<times> tstate"

    59 

    60 text {*

    61   Program of Turing machine is represented as a list of Turing instructions

    62   and the execution of the program starts from the head of the list.

    63   *}

    64 type_synonym tprog = "tinst list"

    65 

    66 

    67 text {*

    68   Turing machine configuration, which consists of the current state 

    69   and the tape.

    70 *}

    71 type_synonym t_conf = "tstate \<times> tape"

    72 

    73 fun nth_of ::  "'a list \<Rightarrow> nat \<Rightarrow> 'a option"

    74   where

    75   "nth_of xs n = (if n < length xs then Some (xs!n)

    76                   else None)"

    77 

    78 text {*

    79   The function used to fetech instruction out of Turing program.

    80   *}

    81 

    82 fun fetch :: "tprog \<Rightarrow> tstate \<Rightarrow> block \<Rightarrow> tinst"

    83   where

    84   "fetch p s b = (if s = 0 then (Nop, 0) else

    85                   case b of 

    86                      Bk \<Rightarrow> case nth_of p (2 * (s - 1)) of

    87                           Some i \<Rightarrow> i

    88                         | None \<Rightarrow> (Nop, 0) 

    89                    | Oc \<Rightarrow> case nth_of p (2 * (s - 1) +1) of

    90                           Some i \<Rightarrow> i

    91                         | None \<Rightarrow> (Nop, 0))"

    92 

    93 

    94 fun new_tape :: "taction \<Rightarrow> tape \<Rightarrow> tape"

    95 where 

    96    "new_tape action (leftn, rightn) = (case action of

    97                                          W0 \<Rightarrow> (leftn, Bk#(tl rightn)) |

    98                                          W1 \<Rightarrow> (leftn, Oc#(tl rightn)) |

    99                                          L  \<Rightarrow>  (if leftn = [] then (tl leftn, Bk#rightn)

   100                                                else (tl leftn, (hd leftn) # rightn)) |

   101                                          R  \<Rightarrow> if rightn = [] then (Bk#leftn,tl rightn) 

   102                                                else ((hd rightn)#leftn, tl rightn) |

   103                                          Nop \<Rightarrow> (leftn, rightn)

   104                                        )"

   105 

   106 text {*

   107   The one step function used to transfer Turing machine configuration.

   108 *}

   109 fun tstep :: "t_conf \<Rightarrow> tprog \<Rightarrow> t_conf"

   110   where

   111   "tstep c p = (let (s, l, r) = c in 

   112                      let (ac, ns) = (fetch p s (case r of [] \<Rightarrow> Bk |     

   113                                                                x # xs \<Rightarrow> x)) in

   114                        (ns, new_tape ac (l, r)))"

   115 

   116 text {*

   117   The many-step function.

   118 *}

   119 fun steps :: "t_conf \<Rightarrow> tprog \<Rightarrow> nat \<Rightarrow> t_conf"

   120   where

   121   "steps c p 0 = c" |

   122   "steps c p (Suc n) = steps (tstep c p) p n"

   123 

   124 lemma tstep_red: "steps c p (Suc n) = tstep (steps c p n) p"

   125 proof(induct n arbitrary: c)

   126   fix c

   127   show "steps c p (Suc 0) = tstep (steps c p 0) p" by(simp add: steps.simps)

   128 next

   129   fix n c

   130   assume ind: "\<And> c. steps c p (Suc n) = tstep (steps c p n) p"

   131   have "steps (tstep c p) p (Suc n) = tstep (steps (tstep c p) p n) p"

   132     by(rule ind)

   133   thus "steps c p (Suc (Suc n)) = tstep (steps c p (Suc n)) p" by(simp add: steps.simps)

   134 qed

   135 

   136 declare Let_def[simp] option.split[split]

   137 

   138 definition 

   139   "iseven n \<equiv> \<exists> x. n = 2 * x"

   140 

   141 

   142 text {*

   143   The following @{text "t_correct"} function is used to specify the wellformedness of Turing

   144   machine.

   145 *}

   146 fun t_correct :: "tprog \<Rightarrow> bool"

   147   where

   148   "t_correct p = (length p \<ge> 2 \<and> iseven (length p) \<and> 

   149                    list_all (\<lambda> (acn, s). s \<le> length p div 2) p)"

   150 

   151 declare t_correct.simps[simp del]

   152 

   153 lemma allimp: "\<lbrakk>\<forall>x. P x \<longrightarrow> Q x; \<forall>x. P x\<rbrakk> \<Longrightarrow> \<forall>x. Q x"

   154 by(auto elim: allE)

   155 

   156 lemma halt_lemma: "\<lbrakk>wf LE; \<forall> n. (\<not> P (f n) \<longrightarrow> (f (Suc n), (f n)) \<in> LE)\<rbrakk> \<Longrightarrow> \<exists> n. P (f n)"

   157 apply(rule exCI, drule allimp, auto)

   158 apply(drule_tac f = f  in wf_inv_image, simp add: inv_image_def)

   159 apply(erule wf_induct, auto)

   160 done

   161 

   162 lemma steps_add: "steps c t (x + y) = steps (steps c t x) t y"

   163 by(induct x arbitrary: c, auto simp: steps.simps tstep_red)

   164 

   165 lemma listall_set: "list_all p t \<Longrightarrow> \<forall> a \<in> set t. p a"

   166 by(induct t, auto)

   167 

   168 lemma fetch_ex: "\<exists>b a. fetch T aa ab = (b, a)"

   169 by(simp add: fetch.simps)

   170 definition exponent :: "'a \<Rightarrow> nat \<Rightarrow> 'a list" ("_\<^bsup>_\<^esup>" [0, 0]100)

   171   where "exponent x n = replicate n x"

   172 

   173 text {* 

   174   @{text "tinres l1 l2"} means left list @{text "l1"} is congruent with

   175   @{text "l2"} with respect to the execution of Turing machine. 

   176   Appending Blank to the right of eigther one does not affect the 

   177   outcome of excution. 

   178 *}

   179 

   180 definition tinres :: "block list \<Rightarrow> block list \<Rightarrow> bool"

   181   where

   182   "tinres bx by = (\<exists> n. bx = by@Bk\<^bsup>n\<^esup> \<or> by = bx @ Bk\<^bsup>n\<^esup>)"

   183 

   184 lemma exp_zero: "a\<^bsup>0\<^esup> = []"

   185 by(simp add: exponent_def)

   186 lemma exp_ind_def: "a\<^bsup>Suc x \<^esup> = a # a\<^bsup>x\<^esup>"

   187 by(simp add: exponent_def)

   188 

   189 text {*

   190   The following lemma shows the meaning of @{text "tinres"} with respect to 

   191   one step execution.

   192   *}

   193 lemma tinres_step: 

   194   "\<lbrakk>tinres l l'; tstep (ss, l, r) t = (sa, la, ra); tstep (ss, l', r) t = (sb, lb, rb)\<rbrakk>

   195     \<Longrightarrow> tinres la lb \<and> ra = rb \<and> sa = sb"

   196 apply(auto simp: tstep.simps fetch.simps new_tape.simps 

   197         split: if_splits taction.splits list.splits

   198                  block.splits)

   199 apply(case_tac [!] "t ! (2 * (ss - Suc 0))", 

   200      auto simp: exponent_def tinres_def split: if_splits taction.splits list.splits

   201                  block.splits)

   202 apply(case_tac [!] "t ! (2 * (ss - Suc 0) + Suc 0)", 

   203      auto simp: exponent_def tinres_def split: if_splits taction.splits list.splits

   204                  block.splits)

   205 done

   206 

   207 declare tstep.simps[simp del] steps.simps[simp del]

   208 

   209 text {*

   210   The following lemma shows the meaning of @{text "tinres"} with respect to 

   211   many step execution.

   212   *}

   213 lemma tinres_steps: 

   214   "\<lbrakk>tinres l l'; steps (ss, l, r) t stp = (sa, la, ra); steps (ss, l', r) t stp = (sb, lb, rb)\<rbrakk>

   215     \<Longrightarrow> tinres la lb \<and> ra = rb \<and> sa = sb"

   216 apply(induct stp arbitrary: sa la ra sb lb rb, simp add: steps.simps)

   217 apply(simp add: tstep_red)

   218 apply(case_tac "(steps (ss, l, r) t stp)")

   219 apply(case_tac "(steps (ss, l', r) t stp)")

   220 proof -

   221   fix stp sa la ra sb lb rb a b c aa ba ca

   222   assume ind: "\<And>sa la ra sb lb rb. \<lbrakk>steps (ss, l, r) t stp = (sa, la, ra); 

   223           steps (ss, l', r) t stp = (sb, lb, rb)\<rbrakk> \<Longrightarrow> tinres la lb \<and> ra = rb \<and> sa = sb"

   224   and h: " tinres l l'" "tstep (steps (ss, l, r) t stp) t = (sa, la, ra)"

   225          "tstep (steps (ss, l', r) t stp) t = (sb, lb, rb)" "steps (ss, l, r) t stp = (a, b, c)" 

   226          "steps (ss, l', r) t stp = (aa, ba, ca)"

   227   have "tinres b ba \<and> c = ca \<and> a = aa"

   228     apply(rule_tac ind, simp_all add: h)

   229     done

   230   thus "tinres la lb \<and> ra = rb \<and> sa = sb"

   231     apply(rule_tac l = b and l' = ba and r = c  and ss = a   

   232             and t = t in tinres_step)

   233     using h

   234     apply(simp, simp, simp)

   235     done

   236 qed

   237 

   238 text {*

   239   The following function @{text "tshift tp n"} is used to shift Turing programs 

   240   @{text "tp"} by @{text "n"} when it is going to be combined with others.

   241   *}

   242 

   243 fun tshift :: "tprog \<Rightarrow> nat \<Rightarrow> tprog"

   244   where

   245   "tshift tp off = (map (\<lambda> (action, state). (action, (if state = 0 then 0

   246                                                       else state + off))) tp)"

   247 

   248 text {*

   249   When two Turing programs are combined, the end state (state @{text "0"}) of the one 

   250   at the prefix position needs to be connected to the start state 

   251   of the one at postfix position. If @{text "tp"} is the Turing program

   252   to be at the prefix, @{text "change_termi_state tp"} is the transformed Turing program.

   253   *}

   254 fun change_termi_state :: "tprog \<Rightarrow> tprog"

   255   where

   256   "change_termi_state t = 

   257        (map (\<lambda> (acn, ns). if ns = 0 then (acn, Suc ((length t) div 2)) else (acn, ns)) t)"

   258 

   259 text {*

   260   @{text "t_add tp1 tp2"} is the combined Truing program.

   261 *}

   262 

   263 fun t_add :: "tprog \<Rightarrow> tprog \<Rightarrow> tprog" ("_ |+| _" [0, 0] 100)

   264   where

   265   "t_add t1 t2 = ((change_termi_state t1) @ (tshift t2 ((length t1) div 2)))"

   266 

   267 text {*

   268   Tests whether the current configuration is at state @{text "0"}.

   269 *}

   270 definition isS0 :: "t_conf \<Rightarrow> bool"

   271   where

   272   "isS0 c = (let (s, l, r) = c in s = 0)"

   273 

   274 declare tstep.simps[simp del] steps.simps[simp del] 

   275         t_add.simps[simp del] fetch.simps[simp del]

   276         new_tape.simps[simp del]

   277 

   278 

   279 text {*

   280   Single step execution starting from state @{text "0"} will not make any progress.

   281 *}

   282 lemma tstep_0: "tstep (0, tp) p = (0, tp)"

   283 apply(simp add: tstep.simps fetch.simps new_tape.simps)

   284 done

   285 

   286 

   287 text {*

   288   Many step executions starting from state @{text "0"} will not make any progress.

   289 *}

   290 

   291 lemma steps_0: "steps (0, tp) p stp = (0, tp)"

   292 apply(induct stp)

   293 apply(simp add: steps.simps)

   294 apply(simp add: tstep_red tstep_0)

   295 done

   296 

   297 lemma s_keep_step: "\<lbrakk>a \<le> length A div 2; tstep (a, b, c) A = (s, l, r); t_correct A\<rbrakk>

   298   \<Longrightarrow> s \<le> length A div 2"

   299 apply(simp add: tstep.simps fetch.simps t_correct.simps iseven_def 

   300   split: if_splits block.splits list.splits)

   301 apply(case_tac [!] a, auto simp: list_all_length)

   302 apply(erule_tac x = "2 * nat" in allE, auto)

   303 apply(erule_tac x = "2 * nat" in allE, auto)

   304 apply(erule_tac x = "Suc (2 * nat)" in allE, auto)

   305 done

   306 

   307 lemma s_keep: "\<lbrakk>steps (Suc 0, tp) A stp = (s, l, r);  t_correct A\<rbrakk> \<Longrightarrow> s \<le> length A div 2"

   308 proof(induct stp arbitrary: s l r)

   309   case 0 thus "?case" by(auto simp: t_correct.simps steps.simps)

   310 next

   311   fix stp s l r

   312   assume ind: "\<And>s l r. \<lbrakk>steps (Suc 0, tp) A stp = (s, l, r); t_correct A\<rbrakk> \<Longrightarrow> s \<le> length A div 2"

   313   and h1: "steps (Suc 0, tp) A (Suc stp) = (s, l, r)"

   314   and h2: "t_correct A"

   315   from h1 h2 show "s \<le> length A div 2"

   316   proof(simp add: tstep_red, cases "(steps (Suc 0, tp) A stp)", simp)

   317     fix a b c 

   318     assume h3: "tstep (a, b, c) A = (s, l, r)"

   319     and h4: "steps (Suc 0, tp) A stp = (a, b, c)"

   320     have "a \<le> length A div 2"

   321       using h2 h4

   322       by(rule_tac l = b and r = c in ind, auto)

   323     thus "?thesis"

   324       using h3 h2

   325       by(simp add: s_keep_step)

   326   qed

   327 qed

   328 

   329 lemma t_merge_fetch_pre:

   330   "\<lbrakk>fetch A s b = (ac, ns); s \<le> length A div 2; t_correct A; s \<noteq> 0\<rbrakk> \<Longrightarrow> 

   331   fetch (A |+| B) s b = (ac, if ns = 0 then Suc (length A div 2) else ns)"

   332 apply(subgoal_tac "2 * (s - Suc 0) < length A \<and> Suc (2 * (s - Suc 0)) < length A")

   333 apply(auto simp: fetch.simps t_add.simps split: if_splits block.splits)

   334 apply(simp_all add: nth_append change_termi_state.simps)

   335 done

   336 

   337 lemma [simp]:  "\<lbrakk>\<not> a \<le> length A div 2; t_correct A\<rbrakk> \<Longrightarrow> fetch A a b = (Nop, 0)"

   338 apply(auto simp: fetch.simps del: nth_of.simps split: block.splits)

   339 apply(case_tac [!] a, auto simp: t_correct.simps iseven_def)

   340 done

   341 

   342 lemma  [elim]: "\<lbrakk>t_correct A; \<not> isS0 (tstep (a, b, c) A)\<rbrakk> \<Longrightarrow> a \<le> length A div 2"

   343 apply(rule_tac classical, auto simp: tstep.simps new_tape.simps isS0_def)

   344 done

   345 

   346 lemma [elim]: "\<lbrakk>t_correct A; \<not> isS0 (tstep (a, b, c) A)\<rbrakk> \<Longrightarrow> 0 < a"

   347 apply(rule_tac classical, simp add: tstep_0 isS0_def)

   348 done

   349 

   350 

   351 lemma t_merge_pre_eq_step: "\<lbrakk>tstep (a, b, c) A = cf; t_correct A; \<not> isS0 cf\<rbrakk> 

   352         \<Longrightarrow> tstep (a, b, c) (A |+| B) = cf"

   353 apply(subgoal_tac "a \<le> length A div 2 \<and> a \<noteq> 0")

   354 apply(simp add: tstep.simps)

   355 apply(case_tac "fetch A a (case c of [] \<Rightarrow> Bk | x # xs \<Rightarrow> x)", simp)

   356 apply(drule_tac B = B in t_merge_fetch_pre, simp, simp, simp, simp add: isS0_def, auto)

   357 done

   358 

   359 lemma t_merge_pre_eq:  "\<lbrakk>steps (Suc 0, tp) A stp = cf; \<not> isS0 cf; t_correct A\<rbrakk>

   360     \<Longrightarrow> steps (Suc 0, tp) (A |+| B) stp = cf"

   361 proof(induct stp arbitrary: cf)

   362   case 0 thus "?case" by(simp add: steps.simps)

   363 next

   364   fix stp cf

   365   assume ind: "\<And>cf. \<lbrakk>steps (Suc 0, tp) A stp = cf; \<not> isS0 cf; t_correct A\<rbrakk> 

   366                  \<Longrightarrow> steps (Suc 0, tp) (A |+| B) stp = cf"

   367   and h1: "steps (Suc 0, tp) A (Suc stp) = cf"

   368   and h2: "\<not> isS0 cf"

   369   and h3: "t_correct A"

   370   from h1 h2 h3 show "steps (Suc 0, tp) (A |+| B) (Suc stp) = cf"

   371   proof(simp add: tstep_red, cases "steps (Suc 0, tp) (A) stp", simp)

   372     fix a b c

   373     assume h4: "tstep (a, b, c) A = cf"

   374     and h5: "steps (Suc 0, tp) A stp = (a, b, c)"

   375     have "steps (Suc 0, tp) (A |+| B) stp = (a, b, c)"

   376     proof(cases a)

   377       case 0 thus "?thesis"

   378         using h4 h2

   379         apply(simp add: tstep_0, cases cf, simp add: isS0_def)

   380         done

   381     next

   382       case (Suc n) thus "?thesis"

   383         using h5 h3

   384         apply(rule_tac ind, auto simp: isS0_def)

   385         done

   386     qed

   387     thus "tstep (steps (Suc 0, tp) (A |+| B) stp) (A |+| B) = cf"

   388       using h4 h5 h3 h2

   389       apply(simp)

   390       apply(rule t_merge_pre_eq_step, auto)

   391       done

   392   qed

   393 qed

   394 

   395 declare nth.simps[simp del] tshift.simps[simp del] change_termi_state.simps[simp del]

   396 

   397 lemma [simp]: "length (change_termi_state A) = length A"

   398 by(simp add: change_termi_state.simps)

   399 

   400 lemma first_halt_point: "steps (Suc 0, tp) A stp = (0, tp')

   401  \<Longrightarrow> \<exists>stp. \<not> isS0 (steps (Suc 0, tp) A stp) \<and> steps (Suc 0, tp) A (Suc stp) = (0, tp')"

   402 proof(induct stp)

   403   case 0  thus "?case" by(simp add: steps.simps)

   404 next

   405   case (Suc n) 

   406   fix stp

   407   assume ind: "steps (Suc 0, tp) A stp = (0, tp') \<Longrightarrow> 

   408        \<exists>stp. \<not> isS0 (steps (Suc 0, tp) A stp) \<and> steps (Suc 0, tp) A (Suc stp) = (0, tp')"

   409     and h: "steps (Suc 0, tp) A (Suc stp) = (0, tp')"

   410   from h show "\<exists>stp. \<not> isS0 (steps (Suc 0, tp) A stp) \<and> steps (Suc 0, tp) A (Suc stp) = (0, tp')"

   411   proof(simp add: tstep_red, cases "steps (Suc 0, tp) A stp", simp, case_tac a)

   412     fix a b c

   413     assume g1: "a = (0::nat)"

   414     and g2: "tstep (a, b, c) A = (0, tp')"

   415     and g3: "steps (Suc 0, tp) A stp = (a, b, c)"

   416     have "steps (Suc 0, tp) A stp = (0, tp')"

   417       using g2 g1 g3

   418       by(simp add: tstep_0)

   419     hence "\<exists> stp. \<not> isS0 (steps (Suc 0, tp) A stp) \<and> steps (Suc 0, tp) A (Suc stp) = (0, tp')"

   420       by(rule ind)

   421     thus "\<exists>stp. \<not> isS0 (steps (Suc 0, tp) A stp) \<and> tstep (steps (Suc 0, tp) A stp) A = (0, tp')" 

   422       apply(simp add: tstep_red)

   423       done

   424   next

   425     fix a b c nat

   426     assume g1: "steps (Suc 0, tp) A stp = (a, b, c)"

   427     and g2: "steps (Suc 0, tp) A (Suc stp) = (0, tp')" "a= Suc nat"

   428     thus "\<exists>stp. \<not> isS0 (steps (Suc 0, tp) A stp) \<and> tstep (steps (Suc 0, tp) A stp) A = (0, tp')"

   429       apply(rule_tac x = stp in exI)

   430       apply(simp add: isS0_def tstep_red)

   431       done

   432   qed

   433 qed 

   434    

   435 lemma t_merge_pre_halt_same': 

   436   "\<lbrakk>\<not> isS0 (steps (Suc 0, tp) A stp) ; steps (Suc 0, tp) A (Suc stp) = (0, tp'); t_correct A\<rbrakk>

   437   \<Longrightarrow> steps (Suc 0, tp) (A |+| B) (Suc stp) = (Suc (length A div 2), tp')"    

   438 proof(simp add: tstep_red, cases "steps (Suc 0, tp) A stp", simp)

   439   fix a b c 

   440   assume h1: "\<not> isS0 (a, b, c)"

   441   and h2: "tstep (a, b, c) A = (0, tp')"

   442   and h3: "t_correct A"

   443   and h4: "steps (Suc 0, tp) A stp = (a, b, c)"

   444   have "steps (Suc 0, tp) (A |+| B) stp = (a, b, c)"

   445     using h1 h4 h3

   446     apply(rule_tac  t_merge_pre_eq, auto)

   447     done

   448   moreover have "tstep (a, b, c) (A |+| B) = (Suc (length A div 2), tp')"

   449     using h2 h3 h1 h4 

   450     apply(simp add: tstep.simps)

   451     apply(case_tac " fetch A a (case c of [] \<Rightarrow> Bk | x # xs \<Rightarrow> x)", simp)

   452     apply(drule_tac B = B in t_merge_fetch_pre, auto simp: isS0_def intro: s_keep)

   453     done

   454   ultimately show "tstep (steps (Suc 0, tp) (A |+| B) stp) (A |+| B) = (Suc (length A div 2), tp')"

   455     by(simp)

   456 qed

   457 

   458 text {*

   459   When Turing machine @{text "A"} and @{text "B"} are combined and the execution

   460   of @{text "A"} can termination within @{text "stp"} steps, 

   461   the combined machine @{text "A |+| B"} will eventually get into the starting 

   462   state of machine @{text "B"}.

   463 *}

   464 lemma t_merge_pre_halt_same: "

   465   \<lbrakk>steps (Suc 0, tp) A stp = (0, tp'); t_correct A; t_correct B\<rbrakk>

   466      \<Longrightarrow> \<exists> stp. steps (Suc 0, tp) (A |+| B) stp = (Suc (length A div 2), tp')"

   467 proof -

   468   assume a_wf: "t_correct A"

   469   and b_wf: "t_correct B"

   470   and a_ht: "steps (Suc 0, tp) A stp = (0, tp')"

   471   have halt_point: "\<exists> stp. \<not> isS0 (steps (Suc 0, tp) A stp) \<and> steps (Suc 0, tp) A (Suc stp) = (0, tp')"

   472     using a_ht

   473     by(erule_tac first_halt_point)

   474   then obtain stp' where "\<not> isS0 (steps (Suc 0, tp) A stp') \<and> steps (Suc 0, tp) A (Suc stp') = (0, tp')"..

   475   hence "steps (Suc 0, tp) (A |+| B) (Suc stp') = (Suc (length A div 2), tp')"

   476     using a_wf

   477     apply(rule_tac t_merge_pre_halt_same', auto)

   478     done

   479   thus "?thesis" ..

   480 qed

   481 

   482 lemma fetch_0: "fetch p 0 b = (Nop, 0)"

   483 by(simp add: fetch.simps)

   484 

   485 lemma [simp]: "length (tshift B x) = length B"

   486 by(simp add: tshift.simps)

   487 

   488 lemma [simp]: "t_correct A \<Longrightarrow> 2 * (length A div 2) = length A"

   489 apply(simp add: t_correct.simps iseven_def, auto)

   490 done

   491 

   492 lemma t_merge_fetch_snd: 

   493   "\<lbrakk>fetch B a b = (ac, ns); t_correct A; t_correct B; a > 0 \<rbrakk>

   494   \<Longrightarrow> fetch (A |+| B) (a + length A div 2) b

   495   = (ac, if ns = 0 then 0 else ns + length A div 2)"

   496 apply(auto simp: fetch.simps t_add.simps split: if_splits block.splits)

   497 apply(case_tac [!] a, simp_all)

   498 apply(simp_all add: nth_append change_termi_state.simps tshift.simps)

   499 done

   500 

   501 lemma t_merge_snd_eq_step: 

   502   "\<lbrakk>tstep (s, l, r) B = (s', l', r'); t_correct A; t_correct B; s > 0\<rbrakk>

   503     \<Longrightarrow> tstep (s + length A div 2, l, r) (A |+| B) = 

   504        (if s' = 0 then 0 else s' + length A div 2, l' ,r') "

   505 apply(simp add: tstep.simps)

   506 apply(cases "fetch B s (case r of [] \<Rightarrow> Bk | x # xs \<Rightarrow> x)")

   507 apply(auto simp: t_merge_fetch_snd)

   508 apply(frule_tac [!] t_merge_fetch_snd, auto)

   509 done 

   510 

   511 text {*

   512   Relates the executions of TM @{text "B"}, one is when @{text "B"} is executed alone,

   513   the other is the execution when @{text "B"} is in the combined TM.

   514 *}

   515 lemma t_merge_snd_eq_steps: 

   516   "\<lbrakk>t_correct A; t_correct B; steps (s, l, r) B stp = (s', l', r'); s > 0\<rbrakk>

   517   \<Longrightarrow> steps (s + length A div 2, l, r) (A |+| B) stp = 

   518       (if s' = 0 then 0 else s' + length A div 2, l', r')"

   519 proof(induct stp arbitrary: s' l' r')

   520   case 0 thus "?case" 

   521     by(simp add: steps.simps)

   522 next

   523   fix stp s' l' r'

   524   assume ind: "\<And>s' l' r'. \<lbrakk>t_correct A; t_correct B; steps (s, l, r) B stp = (s', l', r'); 0 < s\<rbrakk>

   525                    \<Longrightarrow> steps (s + length A div 2, l, r) (A |+| B) stp = 

   526                           (if s' = 0 then 0 else s' + length A div 2, l', r')"

   527   and h1: "steps (s, l, r) B (Suc stp) = (s', l', r')"

   528   and h2: "t_correct A"

   529   and h3: "t_correct B"

   530   and h4: "0 < s"

   531   from h1 show "steps (s + length A div 2, l, r) (A |+| B) (Suc stp) 

   532             = (if s' = 0 then 0 else s' + length A div 2, l', r')"

   533   proof(simp only: tstep_red, cases "steps (s, l, r) B stp")

   534     fix a b c 

   535     assume h5: "steps (s, l, r) B stp = (a, b, c)" "tstep (steps (s, l, r) B stp) B = (s', l', r')"

   536     hence h6: "(steps (s + length A div 2, l, r) (A |+| B) stp) = 

   537                 ((if a = 0 then 0 else a + length A div 2, b, c))"

   538       using h2 h3 h4

   539       by(rule_tac ind, auto)

   540     thus "tstep (steps (s + length A div 2, l, r) (A |+| B) stp) (A |+| B) = 

   541        (if s' = 0 then 0 else s'+ length A div 2, l', r')"

   542       using h5

   543     proof(auto)

   544       assume "tstep (0, b, c) B = (0, l', r')" thus "tstep (0, b, c) (A |+| B) = (0, l', r')"

   545         by(simp add: tstep_0)

   546     next

   547       assume "tstep (0, b, c) B = (s', l', r')" "0 < s'"

   548       thus "tstep (0, b, c) (A |+| B) = (s' + length A div 2, l', r')"

   549         by(simp add: tstep_0)

   550     next

   551       assume "tstep (a, b, c) B = (0, l', r')" "0 < a"

   552       thus "tstep (a + length A div 2, b, c) (A |+| B) = (0, l', r')"

   553         using h2 h3

   554         by(drule_tac t_merge_snd_eq_step, auto)

   555     next

   556       assume "tstep (a, b, c) B = (s', l', r')" "0 < a" "0 < s'"

   557       thus "tstep (a + length A div 2, b, c) (A |+| B) = (s' + length A div 2, l', r')"

   558         using h2 h3

   559         by(drule_tac t_merge_snd_eq_step, auto)

   560     qed

   561   qed

   562 qed

   563 

   564 lemma t_merge_snd_halt_eq: 

   565   "\<lbrakk>steps (Suc 0, tp) B stp = (0, tp'); t_correct A; t_correct B\<rbrakk>

   566   \<Longrightarrow> \<exists>stp. steps (Suc (length A div 2), tp) (A |+| B) stp = (0, tp')"

   567 apply(case_tac tp, cases tp', simp)

   568 apply(drule_tac  s = "Suc 0" in t_merge_snd_eq_steps, auto)

   569 done

   570 

   571 lemma t_inj: "\<lbrakk>steps (Suc 0, tp) A stpa = (0, tp1); steps (Suc 0, tp) A stpb = (0, tp2)\<rbrakk> 

   572       \<Longrightarrow> tp1 = tp2"

   573 proof -

   574   assume h1: "steps (Suc 0, tp) A stpa = (0, tp1)" 

   575   and h2: "steps (Suc 0, tp) A stpb = (0, tp2)"

   576   thus "?thesis"

   577   proof(cases "stpa < stpb")

   578     case True thus "?thesis"

   579       using h1 h2

   580       apply(drule_tac less_imp_Suc_add, auto)

   581       apply(simp del: add_Suc_right add_Suc add: add_Suc_right[THEN sym] steps_add steps_0)

   582       done

   583   next

   584     case False thus "?thesis"

   585       using h1 h2

   586       apply(drule_tac leI)

   587       apply(case_tac "stpb = stpa", auto)

   588       apply(subgoal_tac "stpb < stpa")

   589       apply(drule_tac less_imp_Suc_add, auto)

   590       apply(simp del: add_Suc_right add_Suc add: add_Suc_right[THEN sym] steps_add steps_0)

   591       done

   592   qed

   593 qed

   594 

   595 type_synonym t_assert = "tape \<Rightarrow> bool"

   596 

   597 definition t_imply :: "t_assert \<Rightarrow> t_assert \<Rightarrow> bool" ("_ \<turnstile>-> _" [0, 0] 100)

   598   where

   599   "t_imply a1 a2 = (\<forall> tp. a1 tp \<longrightarrow> a2 tp)"

   600 

   601 

   602 locale turing_merge =

   603   fixes A :: "tprog" and B :: "tprog" and P1 :: "t_assert"

   604   and P2 :: "t_assert"

   605   and P3 :: "t_assert"

   606   and P4 :: "t_assert"

   607   and Q1:: "t_assert"

   608   and Q2 :: "t_assert"

   609   assumes 

   610   A_wf : "t_correct A"

   611   and B_wf : "t_correct B"

   612   and A_halt : "P1 tp \<Longrightarrow> \<exists> stp. let (s, tp') = steps (Suc 0, tp) A stp in s = 0 \<and> Q1 tp'"

   613   and B_halt : "P2 tp \<Longrightarrow> \<exists> stp. let (s, tp') = steps (Suc 0, tp) B stp in s = 0 \<and> Q2 tp'"

   614   and A_uhalt : "P3 tp \<Longrightarrow> \<not> (\<exists> stp. isS0 (steps (Suc 0, tp) A stp))"

   615   and B_uhalt: "P4 tp \<Longrightarrow> \<not> (\<exists> stp. isS0 (steps (Suc 0, tp) B stp))"

   616 begin

   617 

   618 

   619 text {*

   620   The following lemma tries to derive the Hoare logic rule for sequentially combined TMs.

   621   It deals with the situtation when both @{text "A"} and @{text "B"} are terminated.

   622 *}

   623 

   624 lemma t_merge_halt: 

   625   assumes aimpb: "Q1 \<turnstile>-> P2"

   626   shows "P1 \<turnstile>->  \<lambda> tp. (\<exists> stp tp'. steps (Suc 0, tp) (A |+| B)  stp = (0, tp') \<and> Q2 tp')"

   627 proof(simp add: t_imply_def, auto)

   628   fix a b

   629   assume h: "P1 (a, b)"

   630   hence "\<exists> stp. let (s, tp') = steps (Suc 0, a, b) A stp in s = 0 \<and> Q1 tp'"

   631     using A_halt by simp

   632   from this obtain stp1 where "let (s, tp') = steps (Suc 0, a, b) A stp1 in s = 0 \<and> Q1 tp'" ..

   633   thus "\<exists>stp aa ba. steps (Suc 0, a, b) (A |+| B) stp = (0, aa, ba) \<and> Q2 (aa, ba)"

   634   proof(case_tac "steps (Suc 0, a, b) A stp1", simp, erule_tac conjE)

   635     fix aa ba c

   636     assume g1: "Q1 (ba, c)" 

   637       and g2: "steps (Suc 0, a, b) A stp1 = (0, ba, c)"

   638     hence "P2 (ba, c)"

   639       using aimpb apply(simp add: t_imply_def)

   640       done

   641     hence "\<exists> stp. let (s, tp') = steps (Suc 0, ba, c) B stp in s = 0 \<and> Q2 tp'"

   642       using B_halt by simp

   643     from this obtain stp2 where "let (s, tp') = steps (Suc 0, ba, c) B stp2 in s = 0 \<and> Q2 tp'" ..

   644     thus "?thesis"

   645     proof(case_tac "steps (Suc 0, ba, c) B stp2", simp, erule_tac conjE)

   646       fix aa bb ca

   647       assume g3: " Q2 (bb, ca)" "steps (Suc 0, ba, c) B stp2 = (0, bb, ca)"

   648       have "\<exists> stp. steps (Suc 0, a, b) (A |+| B) stp = (Suc (length A div 2), ba , c)"

   649         using g2 A_wf B_wf

   650         by(rule_tac t_merge_pre_halt_same, auto)

   651       moreover have "\<exists> stp. steps (Suc (length A div 2), ba, c) (A |+| B) stp = (0, bb, ca)"

   652         using g3 A_wf B_wf

   653         apply(rule_tac t_merge_snd_halt_eq, auto)

   654         done

   655       ultimately show "\<exists>stp aa ba. steps (Suc 0, a, b) (A |+| B) stp = (0, aa, ba) \<and> Q2 (aa, ba)"

   656         apply(erule_tac exE, erule_tac exE)

   657         apply(rule_tac x = "stp + stpa" in exI, simp add: steps_add)

   658         using g3 by simp

   659     qed

   660   qed

   661 qed

   662 

   663 lemma  t_merge_uhalt_tmp:

   664   assumes B_uh: "\<forall>stp. \<not> isS0 (steps (Suc 0, b, c) B stp)"

   665   and merge_ah: "steps (Suc 0, tp) (A |+| B) stpa = (Suc (length A div 2), b, c)" 

   666   shows "\<forall> stp. \<not> isS0 (steps (Suc 0, tp) (A |+| B) stp)"

   667   using B_uh merge_ah

   668 apply(rule_tac allI)

   669 apply(case_tac "stp > stpa")

   670 apply(erule_tac x = "stp - stpa" in allE)

   671 apply(case_tac "(steps (Suc 0, b, c) B (stp - stpa))", simp)

   672 proof -

   673   fix stp a ba ca 

   674   assume h1: "\<not> isS0 (a, ba, ca)" "stpa < stp"

   675   and h2: "steps (Suc 0, b, c) B (stp - stpa) = (a, ba, ca)"

   676   have "steps (Suc 0 + length A div 2, b, c) (A |+| B) (stp - stpa) = 

   677       (if a = 0 then 0 else a + length A div 2, ba, ca)"

   678     using A_wf B_wf h2

   679     by(rule_tac t_merge_snd_eq_steps, auto)

   680   moreover have "a > 0" using h1 by(simp add: isS0_def)

   681   moreover have "\<exists> stpb. stp = stpa + stpb" 

   682     using h1 by(rule_tac x = "stp - stpa" in exI, simp)

   683   ultimately show "\<not> isS0 (steps (Suc 0, tp) (A |+| B) stp)"

   684     using merge_ah

   685     by(auto simp: steps_add isS0_def)

   686 next

   687   fix stp

   688   assume h: "steps (Suc 0, tp) (A |+| B) stpa = (Suc (length A div 2), b, c)" "\<not> stpa < stp"

   689   hence "\<exists> stpb. stpa = stp + stpb" apply(rule_tac x = "stpa - stp" in exI, auto) done

   690   thus "\<not> isS0 (steps (Suc 0, tp) (A |+| B) stp)"

   691     using h

   692     apply(auto)

   693     apply(cases "steps (Suc 0, tp) (A |+| B) stp", simp add: steps_add isS0_def steps_0)

   694     done

   695 qed

   696 

   697 text {*

   698   The following lemma deals with the situation when TM @{text "B"} can not terminate.

   699   *}

   700 

   701 lemma t_merge_uhalt: 

   702   assumes aimpb: "Q1 \<turnstile>-> P4"

   703   shows "P1 \<turnstile>-> \<lambda> tp. \<not> (\<exists> stp. isS0 (steps (Suc 0, tp) (A |+| B) stp))"

   704 proof(simp only: t_imply_def, rule_tac allI, rule_tac impI)

   705   fix tp 

   706   assume init_asst: "P1 tp"

   707   show "\<not> (\<exists>stp. isS0 (steps (Suc 0, tp) (A |+| B) stp))"

   708   proof -

   709     have "\<exists> stp. let (s, tp') = steps (Suc 0, tp) A stp in s = 0 \<and> Q1 tp'"

   710       using A_halt[of tp] init_asst

   711       by(simp)

   712     from this obtain stpx where "let (s, tp') = steps (Suc 0, tp) A stpx in s = 0 \<and> Q1 tp'" ..

   713     thus "?thesis"

   714     proof(cases "steps (Suc 0, tp) A stpx", simp, erule_tac conjE)

   715       fix a b c

   716       assume "Q1 (b, c)"

   717         and h3: "steps (Suc 0, tp) A stpx = (0, b, c)"

   718       hence h2: "P4 (b, c)"  using aimpb

   719         by(simp add: t_imply_def)

   720       have "\<exists> stp. steps (Suc 0, tp) (A |+| B) stp = (Suc (length A div 2), b, c)"

   721         using h3 A_wf B_wf

   722         apply(rule_tac stp = stpx in t_merge_pre_halt_same, auto)

   723         done

   724       from this obtain stpa where h4:"steps (Suc 0, tp) (A |+| B) stpa = (Suc (length A div 2), b, c)" ..

   725       have " \<not> (\<exists> stp. isS0 (steps (Suc 0, b, c) B stp))"

   726         using B_uhalt [of "(b, c)"] h2 apply simp

   727         done

   728       from this and h4 show "\<forall>stp. \<not> isS0 (steps (Suc 0, tp) (A |+| B) stp)"

   729         by(rule_tac t_merge_uhalt_tmp, auto)

   730     qed

   731   qed

   732 qed

   733 end

   734 

   735 end

   736