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Quelle  MergeAnomaly.thy

  Sprache: Isabelle
 

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 * Project : HOL-CSP - A Shallow Embedding of CSP in Isabelle/HOL
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chapter Annex: A Note on a Classical Example: The ``Merge Anomaly''

theory MergeAnomaly
  imports "HOL-CSP"
begin

textManfred Broy proposed in his 'Habilitationsschrift' (🪙Published in
 `A Theory for Nondeterminism, Parallelism, Communication, and Concurrency. TCS (1986)''
)
  stream processing language AMPL, which is in many respect similar to 🍋HOL-CSP.
 , the underlying Powerset-construction contained a error that made it possible
  in an interleaving between a stream 1\ and a stream 1.0\ a 0 can get ahead of the
  1 which contradicts the intuition that interleaving should preserve causality in
  inputs.

  we believe in the importance of a 🪙formally verified semantics of a language for
 , we take this anecdotical reference to the Merge-Anomaly as an example to
  in the process algebra 🍋HOL-CSP, so in a setting with trace, failure and
 /divergence semantics.

 , we define the three processes corresponding to the notation 1\, 0\ and 1.0\:
 


definition ones      :: "nat process" where ones = (μ X. 1 X)
definition zeros     :: "nat process" where zeros = (μ X. 0 X)
definition oneszeros :: "nat process" where oneszeros = (μ X. x{0,1} X)

text... and derive the more handy recursive stream-equations:

lemma ones_rec: "ones = 1 ones"  by (subst cont_process_rec[OF ones_def]) simp_all

lemma zeros_rec: "zeros = 0 zeros" by (subst cont_process_rec[OF zeros_def]) simp_all

lemma oneszeros_rec: "oneszeros = x{0,1} oneszeros" 
      by (subst cont_process_rec[OF oneszeros_def]) simp_all

textNow, we can establish that ones ||| zeros and oneszeros are indeed equal
  the failure/divergence semantics in 🍋HOL-CSP. This is formally proven as follows:


lemma ones_Inter_zeros_eq_oneszeros : ones ||| zeros = oneszeros
proof (rule FD_antisym)
  show oneszeros FD ones ||| zeros
  proof (unfold oneszeros_def, induct rule: cont_fix_ind)
    fix X assume X FD ones ||| zeros
    have ones ||| zeros = (1 (ones ||| 0 zeros)) (0 (1 ones ||| zeros))
      by (subst ones_rec, subst zeros_rec) (simp add: write0_Inter_write0)
    also have  = (1 (ones ||| zeros)) (0 (ones ||| zeros))
      by (simp del: One_nat_def flip: ones_rec zeros_rec)
    also have  = a{0, 1} (ones ||| zeros)
      by (metis Mprefix_Un_distrib Mprefix_singl Un_insert_right sup_bot.right_neutral)
    also have a{0, 1} X FD
      by (simp add: X FD ones ||| zeros mono_Mprefix_FD)
    finally show a{0, 1} X FD ones ||| zeros .
  qed simp_all
next
  show ones ||| zeros FD oneszeros
    ― With stronger theoretical footprint, this could be skipped since constoneszeros
 is a deterministic process (therefore maximal for term(FD)).

  proof (unfold ones_def zeros_def, induct rule: parallel_fix_ind_inc)
    fix X Y
    assume hyps : (Λ X. 1 X)X ||| Y FD oneszeros
                  X ||| (Λ Y. 0 Y)Y FD oneszeros
    have (Λ X. 1 X)X ||| (Λ Y. 0 Y)Y = (1 X) ||| (0 Y) by simp
    also have  = (1 (X ||| 0 Y)) (0 (1 X ||| Y))
      by (simp add: write0_Inter_write0)
    also have  = a {0, 1} (if a = 0 then 1 X ||| Y else X ||| 0 Y)
      by (auto simp add: write0_def Mprefix_Det_Mprefix intro: mono_Mprefix_eq)
    also have  FD oneszeros
      by (subst oneszeros_rec)
        (use hyps in auto intro: mono_Mprefix_FD)
    finally show (Λ X. 1 X)X ||| (Λ Y. 0 Y)Y FD oneszeros .
  qed simp_all
qed

textAs a consequence, in the trace model, we can establish that there will be no ``Anomaly''
  🍋HOL-CSP, so ones ||| (1 zeros) will be equal to 1 oneszeros in the
  intended trace-projection. The proof proceeds indirectly over induction over the traces;
  is the recommended proof strategy if arguments over trace sets have to be established. In
 , an argument over the termlfp-operator, which seems natural at first sight, is amazingly
 . We have:


lemma T (ones ||| (1 zeros)) = T (1 oneszeros) (is ?lhs = ?rhs)
proof (rule set_eqI)
  show t ?lhs t ?rhs for t
  proof (induct t)
    case Nil show ?case by simp
  next
    case (Cons e t)
    show ?case
    proof (rule iffI)
      assume e # t T (1 oneszeros)
      hence e = ev 1 t T oneszeros
        by (simp_all add: T_write0)
      thus e # t T (ones ||| 1 zeros)
        apply (subst ones_rec)
        apply (simp del: One_nat_def add: write0_Inter_write0 T_Det T_write0)
        apply (fold ones_rec ones_Inter_zeros_eq_oneszeros)
        ..
    next
      show e # t T (ones ||| 1 zeros) ==> e # t T (1 oneszeros)
        apply (subst (asm) ones_rec)
        apply (simp del: One_nat_def add: write0_Inter_write0 T_Det T_write0)
        apply (fold ones_rec)
        apply (unfold Cons.hyps)
        apply (unfold ones_Inter_zeros_eq_oneszeros)
        apply (subst (asm) (2) oneszeros_rec, subst oneszeros_rec)
        by (auto simp add: T_Mprefix T_write0)
    qed
  qed
qed


textHowever, ones ||| (1 zeros) will be 🪙not equal to 1 oneszeros in the failure
  and therefore not in the failure/divergence model. The deeper reason is the interleave
  🪙can neither refuse 0 or 1; it is designed to be sensitive to the process
  and let pass any possible interleaving admitted by the context which is reflected
  the rule @{thm Read_Write_CSP_Laws.write0_Inter_write0}.



lemma ones ||| (1 zeros) 1 oneszeros (is ?P1 ?P2
proof -
  have ([ev 1], {ev 0}) F ?P2
    by (subst oneszeros_rec)
      (simp add: Process_eq_spec F_write0 F_Mprefix)
  moreover have ([ev 1], {ev 0}) F ?P1
    by (subst ones_rec, subst ones_rec)
      (simp add: write0_Inter_write0 Process_eq_spec F_Det F_write0)
  ultimately show ?P1 ?P2 by metis
qed

textOne might ask what happens if we would have defined the process constoneszeros via the
 -deterministic choice, so using the following definition:


definition oneszeros' :: "nat process" where oneszeros' = (μ X. x{0,1} X)

lemma oneszeros'_rec : oneszeros' = x {0, 1} oneszeros'
  by (subst cont_process_rec[OF oneszeros'_def]) simp_all

textBut this means that already the first step in the above argument breaks down:
  interleaving 🪙can neither refuse 0 or 1, while the deterministic choice of
 constoneszeros' does, we can easily show:


lemma ones ||| zeros oneszeros'
proof (rule notI)
  assume ones ||| zeros = oneszeros'
  with ones_Inter_zeros_eq_oneszeros have oneszeros' = oneszeros by metis
  moreover have ([], {ev 0}) F oneszeros'
    by (subst oneszeros'_rec)
      (simp add: F_Mndetprefix')
  moreover have ([], {ev 0}) F oneszeros
    by (subst oneszeros_rec) (simp add: F_Mprefix)
  ultimately show False by simp
qed



end

Messung V0.5 in Prozent
C=69 H=95 G=82

¤ Dauer der Verarbeitung: 0.10 Sekunden  (vorverarbeitet am  2026-06-10) ¤

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