/export/starexec/sandbox2/solver/bin/starexec_run_ttt2-1.17+nonreach /export/starexec/sandbox2/benchmark/theBenchmark.xml /export/starexec/sandbox2/output/output_files


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YES

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

Proof:
 DP Processor:
  DPs:
   a#(b(b(x1))) -> a#(x1)
   a#(b(b(x1))) -> a#(a(x1))
  TRS:
   a(b(a(x1))) -> b(b(b(b(x1))))
   a(b(b(x1))) -> b(b(a(a(x1))))
  Arctic Interpretation Processor:
   dimension: 2
   usable rules:
    a(b(a(x1))) -> b(b(b(b(x1))))
    a(b(b(x1))) -> b(b(a(a(x1))))
   interpretation:
    [a#](x0) = [0 1]x0 + [0],
    
              [-& 1 ]     [2 ]
    [b](x0) = [0  -&]x0 + [-&],
    
              [0  2 ]     [3 ]
    [a](x0) = [-& 0 ]x0 + [-&]
   orientation:
    a#(b(b(x1))) = [1 2]x1 + [3] >= [0 1]x1 + [0] = a#(x1)
    
    a#(b(b(x1))) = [1 2]x1 + [3] >= [0 2]x1 + [3] = a#(a(x1))
    
                  [2 4]     [5]    [2  -&]     [3]                 
    a(b(a(x1))) = [0 2]x1 + [3] >= [-& 2 ]x1 + [3] = b(b(b(b(x1))))
    
                  [1  3 ]     [4]    [1  3 ]     [4]                 
    a(b(b(x1))) = [-& 1 ]x1 + [2] >= [-& 1 ]x1 + [2] = b(b(a(a(x1))))
   problem:
    DPs:
     a#(b(b(x1))) -> a#(a(x1))
    TRS:
     a(b(a(x1))) -> b(b(b(b(x1))))
     a(b(b(x1))) -> b(b(a(a(x1))))
   Restore Modifier:
    DPs:
     a#(b(b(x1))) -> a#(a(x1))
    TRS:
     a(b(a(x1))) -> b(b(b(b(x1))))
     a(b(b(x1))) -> b(b(a(a(x1))))
    Arctic Interpretation Processor:
     dimension: 2
     usable rules:
      a(b(a(x1))) -> b(b(b(b(x1))))
      a(b(b(x1))) -> b(b(a(a(x1))))
     interpretation:
      [a#](x0) = [2 0]x0 + [1],
      
                [-& 2 ]     [0 ]
      [b](x0) = [0  -&]x0 + [-&],
      
                [0  -&]     [-&]
      [a](x0) = [2  0 ]x0 + [0 ]
     orientation:
      a#(b(b(x1))) = [4 2]x1 + [2] >= [2 0]x1 + [1] = a#(a(x1))
      
                    [4 2]     [2]    [4  -&]     [2]                 
      a(b(a(x1))) = [6 4]x1 + [4] >= [-& 4 ]x1 + [2] = b(b(b(b(x1))))
      
                    [2  -&]     [0]    [2  -&]     [0]                 
      a(b(b(x1))) = [4  2 ]x1 + [2] >= [4  2 ]x1 + [2] = b(b(a(a(x1))))
     problem:
      DPs:
       
      TRS:
       a(b(a(x1))) -> b(b(b(b(x1))))
       a(b(b(x1))) -> b(b(a(a(x1))))
     Qed