/export/starexec/sandbox2/solver/bin/starexec_run_ttt2 /export/starexec/sandbox2/benchmark/theBenchmark.xml /export/starexec/sandbox2/output/output_files


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YES

Problem:
 A(b(x1)) -> b(a(B(A(x1))))
 B(a(x1)) -> a(b(A(B(x1))))
 A(a(x1)) -> x1
 B(b(x1)) -> x1

Proof:
 String Reversal Processor:
  b(A(x1)) -> A(B(a(b(x1))))
  a(B(x1)) -> B(A(b(a(x1))))
  a(A(x1)) -> x1
  b(B(x1)) -> x1
  Bounds Processor:
   bound: 0
   enrichment: match
   automaton:
    final states: {2,6,1}
    transitions:
     A0(5) -> 1*
     A0(8) -> 9*
     a0(3) -> 4*
     a0(2) -> 7*
     b0(2) -> 3*
     b0(7) -> 8*
     f40() -> 2*
     B0(4) -> 5*
     B0(9) -> 6*
     2 -> 8,4,3,7
     6 -> 4,7
     1 -> 8,3
     5 -> 4*
     9 -> 8*
   problem:
    
   Qed