/export/starexec/sandbox/solver/bin/starexec_run_ttt2 /export/starexec/sandbox/benchmark/theBenchmark.xml /export/starexec/sandbox/output/output_files


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YES

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

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