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

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