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

Proof:
 Matrix Interpretation Processor: dim=4
  
  interpretation:
             [1 0 0 0]  
             [0 0 1 0]  
   [b](x0) = [1 0 0 0]x0
             [0 0 1 0]  ,
   
             [1 0 1 0]     [1]
             [0 0 0 0]     [0]
   [a](x0) = [0 1 0 1]x0 + [0]
             [0 0 0 0]     [0]
  orientation:
                    [2 0 2 0]     [3]    [2 0 2 0]     [2]                 
                    [0 0 0 0]     [0]    [0 0 0 0]     [0]                 
   a(b(b(a(x1)))) = [2 0 2 0]x1 + [2] >= [2 0 2 0]x1 + [2] = b(a(a(b(x1))))
                    [0 0 0 0]     [0]    [0 0 0 0]     [0]                 
   
                 [2 0 0 0]     [1]    [2 0 0 0]     [1]                 
                 [0 0 2 0]     [0]    [0 0 0 0]     [0]                 
   b(a(b(x1))) = [2 0 0 0]x1 + [1] >= [2 0 0 0]x1 + [0] = a(b(b(b(x1))))
                 [0 0 2 0]     [0]    [0 0 0 0]     [0]                 
  problem:
   b(a(b(x1))) -> a(b(b(b(x1))))
  String Reversal Processor:
   b(a(b(x1))) -> b(b(b(a(x1))))
   Bounds Processor:
    bound: 0
    enrichment: match
    automaton:
     final states: {1}
     transitions:
      f20() -> 2*
      a0(2) -> 3*
      b0(5) -> 1*
      b0(4) -> 5*
      b0(3) -> 4*
      1 -> 4*
    problem:
     
    Qed