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

Proof:
 String Reversal Processor:
  a(a(b(a(x1)))) -> a(b(b(a(x1))))
  b(a(b(a(x1)))) -> a(b(b(a(x1))))
  a(a(b(b(x1)))) -> a(a(b(a(x1))))
  b(b(b(b(x1)))) -> a(a(a(a(x1))))
  Matrix Interpretation Processor: dim=3
   
   interpretation:
              [1 0 1]     [0]
    [b](x0) = [1 0 1]x0 + [1]
              [0 1 0]     [0],
    
              [1 1 0]  
    [a](x0) = [0 0 0]x0
              [0 0 0]  
   orientation:
                     [2 2 0]     [1]    [2 2 0]     [1]                 
    a(a(b(a(x1)))) = [0 0 0]x1 + [0] >= [0 0 0]x1 + [0] = a(b(b(a(x1))))
                     [0 0 0]     [0]    [0 0 0]     [0]                 
    
                     [2 2 0]     [1]    [2 2 0]     [1]                 
    b(a(b(a(x1)))) = [2 2 0]x1 + [2] >= [0 0 0]x1 + [0] = a(b(b(a(x1))))
                     [0 0 0]     [0]    [0 0 0]     [0]                 
    
                     [2 2 2]     [1]    [2 2 0]     [1]                 
    a(a(b(b(x1)))) = [0 0 0]x1 + [0] >= [0 0 0]x1 + [0] = a(a(b(a(x1))))
                     [0 0 0]     [0]    [0 0 0]     [0]                 
    
                     [3 2 3]     [2]    [1 1 0]                   
    b(b(b(b(x1)))) = [3 2 3]x1 + [3] >= [0 0 0]x1 = a(a(a(a(x1))))
                     [2 1 2]     [2]    [0 0 0]                   
   problem:
    a(a(b(a(x1)))) -> a(b(b(a(x1))))
    b(a(b(a(x1)))) -> a(b(b(a(x1))))
    a(a(b(b(x1)))) -> a(a(b(a(x1))))
   String Reversal Processor:
    a(b(a(a(x1)))) -> a(b(b(a(x1))))
    a(b(a(b(x1)))) -> a(b(b(a(x1))))
    b(b(a(a(x1)))) -> a(b(a(a(x1))))
    Matrix Interpretation Processor: dim=3
     
     interpretation:
                [1 0 1]     [1]
      [b](x0) = [1 0 2]x0 + [0]
                [0 1 0]     [0],
      
                [1 0 0]     [1]
      [a](x0) = [0 0 0]x0 + [0]
                [1 0 0]     [0]
     orientation:
                       [2 0 0]     [5]    [2 0 0]     [4]                 
      a(b(a(a(x1)))) = [0 0 0]x1 + [0] >= [0 0 0]x1 + [0] = a(b(b(a(x1))))
                       [2 0 0]     [4]    [2 0 0]     [3]                 
      
                       [2 0 2]     [5]    [2 0 0]     [4]                 
      a(b(a(b(x1)))) = [0 0 0]x1 + [0] >= [0 0 0]x1 + [0] = a(b(b(a(x1))))
                       [2 0 2]     [4]    [2 0 0]     [3]                 
      
                       [2 0 0]     [5]    [2 0 0]     [5]                 
      b(b(a(a(x1)))) = [2 0 0]x1 + [4] >= [0 0 0]x1 + [0] = a(b(a(a(x1))))
                       [3 0 0]     [4]    [2 0 0]     [4]                 
     problem:
      b(b(a(a(x1)))) -> a(b(a(a(x1))))
     KBO Processor:
      weight function:
       w0 = 1
       w(b) = w(a) = 1
      precedence:
       b > a
      problem:
       
      Qed