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

Proof:
 String Reversal Processor:
  b(a(x1)) -> a(a(b(x1)))
  b(x1) -> c(a(c(x1)))
  a(a(x1)) -> a(c(a(x1)))
  Bounds Processor:
   bound: 2
   enrichment: match
   automaton:
    final states: {8,5,1}
    transitions:
     b0(2) -> 3*
     f30() -> 2*
     a1(16) -> 17*
     a1(37) -> 38*
     a1(21) -> 22*
     a1(19) -> 20*
     c2(39) -> 40*
     c2(28) -> 29*
     a2(40) -> 41*
     a2(47) -> 48*
     a2(27) -> 28*
     a2(29) -> 30*
     c1(17) -> 18*
     c1(20) -> 21*
     c1(15) -> 16*
     a0(3) -> 4*
     a0(4) -> 1*
     a0(6) -> 7*
     a0(2) -> 9*
     a0(10) -> 8*
     c0(2) -> 6*
     c0(7) -> 5*
     c0(9) -> 10*
     19 -> 47*
     48 -> 39*
     30 -> 48,39,20
     4 -> 37*
     2 -> 15*
     22 -> 38,17,1
     21 -> 27*
     38 -> 17*
     8 -> 9*
     1 -> 3,19
     20 -> 4*
     3 -> 19*
     41 -> 38,17
     18 -> 3*
   problem:
    
   Qed