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

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