/export/starexec/sandbox/solver/bin/starexec_run_ttt2-1.17+nonreach /export/starexec/sandbox/benchmark/theBenchmark.xml /export/starexec/sandbox/output/output_files


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YES

Problem:
 a(b(x1)) -> b(a(a(x1)))
 b(x1) -> c(a(c(x1)))
 a(a(x1)) -> a(c(a(x1)))

Proof:
 Matrix Interpretation Processor: dim=1
  
  interpretation:
   [c](x0) = x0,
   
   [a](x0) = x0,
   
   [b](x0) = x0 + 1
  orientation:
   a(b(x1)) = x1 + 1 >= x1 + 1 = b(a(a(x1)))
   
   b(x1) = x1 + 1 >= x1 = c(a(c(x1)))
   
   a(a(x1)) = x1 >= x1 = a(c(a(x1)))
  problem:
   a(b(x1)) -> b(a(a(x1)))
   a(a(x1)) -> a(c(a(x1)))
  String Reversal Processor:
   b(a(x1)) -> a(a(b(x1)))
   a(a(x1)) -> a(c(a(x1)))
   Bounds Processor:
    bound: 2
    enrichment: match
    automaton:
     final states: {5,1}
     transitions:
      f30() -> 2*
      a0(7) -> 5*
      a0(2) -> 6*
      a0(4) -> 1*
      a0(3) -> 4*
      b0(2) -> 3*
      c0(6) -> 7*
      a1(12) -> 13*
      a1(24) -> 25*
      a1(14) -> 15*
      c1(13) -> 14*
      a2(20) -> 21*
      a2(22) -> 23*
      a2(29) -> 30*
      c2(21) -> 22*
      c2(28) -> 29*
      1 -> 3,12
      3 -> 12*
      4 -> 24*
      5 -> 6*
      14 -> 20*
      15 -> 25,1
      23 -> 28,4,13
      25 -> 21*
      30 -> 25*
    problem:
     
    Qed