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

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