/export/starexec/sandbox/solver/bin/starexec_run_ttt2 /export/starexec/sandbox/benchmark/theBenchmark.xml /export/starexec/sandbox/output/output_files


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YES

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

Proof:
 String Reversal Processor:
  b(a(x1)) -> a(b(x1))
  a(b(x1)) -> b(c(a(a(x1))))
  Bounds Processor:
   bound: 1
   enrichment: match
   automaton:
    final states: {4,1}
    transitions:
     b0(7) -> 4*
     b0(2) -> 3*
     b1(11) -> 12*
     b1(25) -> 26*
     c0(6) -> 7*
     a0(3) -> 1*
     a0(2) -> 5*
     a0(5) -> 6*
     f30() -> 2*
     c1(24) -> 25*
     c1(10) -> 11*
     a1(23) -> 24*
     a1(22) -> 23*
     a1(9) -> 10*
     a1(18) -> 19*
     a1(8) -> 9*
     19 -> 9*
     7 -> 22*
     12 -> 1*
     4 -> 9,5
     11 -> 18*
     2 -> 8*
     26 -> 10,6
     1 -> 3*
   problem:
    
   Qed