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

Proof:
 Matrix Interpretation Processor: dim=1
  
  interpretation:
   [b](x0) = x0,
   
   [a](x0) = 2x0 + 1
  orientation:
   a(b(a(a(x1)))) = 8x1 + 7 >= 2x1 + 1 = a(b(b(b(x1))))
   
   b(b(a(a(x1)))) = 4x1 + 3 >= 4x1 + 3 = a(b(a(b(x1))))
   
   b(a(a(b(x1)))) = 4x1 + 3 >= 4x1 + 3 = b(a(b(a(x1))))
  problem:
   b(b(a(a(x1)))) -> a(b(a(b(x1))))
   b(a(a(b(x1)))) -> b(a(b(a(x1))))
  String Reversal Processor:
   a(a(b(b(x1)))) -> b(a(b(a(x1))))
   b(a(a(b(x1)))) -> a(b(a(b(x1))))
   Matrix Interpretation Processor: dim=3
    
    interpretation:
               [1 0 0]     [0]
     [b](x0) = [0 0 1]x0 + [1]
               [0 1 0]     [0],
     
               [1 0 1]  
     [a](x0) = [1 0 0]x0
               [1 0 0]  
    orientation:
                      [2 0 1]     [1]    [2 0 1]     [0]                 
     a(a(b(b(x1)))) = [1 0 1]x1 + [1] >= [1 0 1]x1 + [1] = b(a(b(a(x1))))
                      [1 0 1]     [1]    [1 0 1]     [0]                 
     
                      [2 1 0]     [0]    [2 1 0]                   
     b(a(a(b(x1)))) = [1 1 0]x1 + [1] >= [1 1 0]x1 = a(b(a(b(x1))))
                      [1 1 0]     [0]    [1 1 0]                   
    problem:
     b(a(a(b(x1)))) -> a(b(a(b(x1))))
    String Reversal Processor:
     b(a(a(b(x1)))) -> b(a(b(a(x1))))
     KBO Processor:
      weight function:
       w0 = 1
       w(b) = 1
       w(a) = 0
      precedence:
       a > b
      problem:
       
      Qed