/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(a(x1)) -> b(x1)
 b(a(x1)) -> a(b(x1))
 b(b(c(x1))) -> c(a(x1))
 b(b(x1)) -> a(a(a(x1)))
 c(a(x1)) -> b(a(c(x1)))

Proof:
 Matrix Interpretation Processor: dim=1
  
  interpretation:
   [c](x0) = 3x0,
   
   [b](x0) = x0 + 8,
   
   [a](x0) = x0 + 4
  orientation:
   a(a(x1)) = x1 + 8 >= x1 + 8 = b(x1)
   
   b(a(x1)) = x1 + 12 >= x1 + 12 = a(b(x1))
   
   b(b(c(x1))) = 3x1 + 16 >= 3x1 + 12 = c(a(x1))
   
   b(b(x1)) = x1 + 16 >= x1 + 12 = a(a(a(x1)))
   
   c(a(x1)) = 3x1 + 12 >= 3x1 + 12 = b(a(c(x1)))
  problem:
   a(a(x1)) -> b(x1)
   b(a(x1)) -> a(b(x1))
   c(a(x1)) -> b(a(c(x1)))
  String Reversal Processor:
   a(a(x1)) -> b(x1)
   a(b(x1)) -> b(a(x1))
   a(c(x1)) -> c(a(b(x1)))
   Bounds Processor:
    bound: 1
    enrichment: match
    automaton:
     final states: {5,3,1}
     transitions:
      b1(8) -> 9*
      a1(7) -> 8*
      f30() -> 2*
      b0(2) -> 1*
      b0(4) -> 3*
      a0(2) -> 4*
      a0(1) -> 6*
      c0(6) -> 5*
      1 -> 8,4
      2 -> 7*
      3 -> 8,4
      5 -> 8,4
      9 -> 6*
    problem:
     
    Qed