/export/starexec/sandbox2/solver/bin/starexec_run_ttt2 /export/starexec/sandbox2/benchmark/theBenchmark.xml /export/starexec/sandbox2/output/output_files


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YES

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

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