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

Proof:
 String Reversal Processor:
  b(a(b(b(x1)))) -> a(b(b(b(x1))))
  a(a(b(b(x1)))) -> b(b(a(a(x1))))
  a(a(a(b(x1)))) -> a(b(b(b(x1))))
  Matrix Interpretation Processor: dim=1
   
   interpretation:
    [a](x0) = x0 + 2,
    
    [b](x0) = x0
   orientation:
    b(a(b(b(x1)))) = x1 + 2 >= x1 + 2 = a(b(b(b(x1))))
    
    a(a(b(b(x1)))) = x1 + 4 >= x1 + 4 = b(b(a(a(x1))))
    
    a(a(a(b(x1)))) = x1 + 6 >= x1 + 2 = a(b(b(b(x1))))
   problem:
    b(a(b(b(x1)))) -> a(b(b(b(x1))))
    a(a(b(b(x1)))) -> b(b(a(a(x1))))
   Matrix Interpretation Processor: dim=3
    
    interpretation:
               [1 0 0]  
     [a](x0) = [0 0 1]x0
               [0 1 0]  ,
     
               [1 0 1]     [0]
     [b](x0) = [0 0 1]x0 + [1]
               [0 0 1]     [0]
    orientation:
                      [1 0 3]     [1]    [1 0 3]     [0]                 
     b(a(b(b(x1)))) = [0 0 1]x1 + [2] >= [0 0 1]x1 + [0] = a(b(b(b(x1))))
                      [0 0 1]     [1]    [0 0 1]     [1]                 
     
                      [1 0 2]     [0]    [1 0 2]     [0]                 
     a(a(b(b(x1)))) = [0 0 1]x1 + [1] >= [0 0 1]x1 + [1] = b(b(a(a(x1))))
                      [0 0 1]     [0]    [0 0 1]     [0]                 
    problem:
     a(a(b(b(x1)))) -> b(b(a(a(x1))))
    KBO Processor:
     weight function:
      w0 = 1
      w(b) = 1
      w(a) = 0
     precedence:
      a > b
     problem:
      
     Qed