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

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