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

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