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

Proof:
 String Reversal Processor:
  b(a(a(x1))) -> a(a(b(b(x1))))
  b(a(b(x1))) -> a(a(a(a(x1))))
  DP Processor:
   DPs:
    b#(a(a(x1))) -> b#(x1)
    b#(a(a(x1))) -> b#(b(x1))
   TRS:
    b(a(a(x1))) -> a(a(b(b(x1))))
    b(a(b(x1))) -> a(a(a(a(x1))))
   Arctic Interpretation Processor:
    dimension: 2
    usable rules:
     b(a(a(x1))) -> a(a(b(b(x1))))
     b(a(b(x1))) -> a(a(a(a(x1))))
    interpretation:
     [b#](x0) = [1 0]x0,
     
               [-& 1 ]     [0 ]
     [a](x0) = [0  -&]x0 + [-&],
     
               [0  -&]     [-&]
     [b](x0) = [2  0 ]x0 + [0 ]
    orientation:
     b#(a(a(x1))) = [2 1]x1 + [1] >= [1 0]x1 = b#(x1)
     
     b#(a(a(x1))) = [2 1]x1 + [1] >= [2 0]x1 + [0] = b#(b(x1))
     
                   [1  -&]     [0]    [1  -&]     [0]                 
     b(a(a(x1))) = [3  1 ]x1 + [2] >= [3  1 ]x1 + [1] = a(a(b(b(x1))))
     
                   [3 1]     [1]    [2  -&]     [1]                 
     b(a(b(x1))) = [5 3]x1 + [3] >= [-& 2 ]x1 + [1] = a(a(a(a(x1))))
    problem:
     DPs:
      b#(a(a(x1))) -> b#(b(x1))
     TRS:
      b(a(a(x1))) -> a(a(b(b(x1))))
      b(a(b(x1))) -> a(a(a(a(x1))))
    Restore Modifier:
     DPs:
      b#(a(a(x1))) -> b#(b(x1))
     TRS:
      b(a(a(x1))) -> a(a(b(b(x1))))
      b(a(b(x1))) -> a(a(a(a(x1))))
     Arctic Interpretation Processor:
      dimension: 2
      usable rules:
       b(a(a(x1))) -> a(a(b(b(x1))))
       b(a(b(x1))) -> a(a(a(a(x1))))
      interpretation:
       [b#](x0) = [-& 0 ]x0 + [0],
       
                 [-& 0 ]     [0]
       [a](x0) = [1  -&]x0 + [1],
       
                 [0  3 ]     [1 ]
       [b](x0) = [-& 0 ]x0 + [-&]
      orientation:
       b#(a(a(x1))) = [-& 1 ]x1 + [1] >= [-& 0 ]x1 + [0] = b#(b(x1))
       
                     [1  4 ]     [4]    [1  4 ]     [2]                 
       b(a(a(x1))) = [-& 1 ]x1 + [1] >= [-& 1 ]x1 + [1] = a(a(b(b(x1))))
       
                     [4 7]     [5]    [2  -&]     [2]                 
       b(a(b(x1))) = [1 4]x1 + [2] >= [-& 2 ]x1 + [2] = a(a(a(a(x1))))
      problem:
       DPs:
        
       TRS:
        b(a(a(x1))) -> a(a(b(b(x1))))
        b(a(b(x1))) -> a(a(a(a(x1))))
      Qed