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

Proof:
 Matrix Interpretation Processor: dim=2
  
  interpretation:
             [1 0]     [1]
   [b](x0) = [0 0]x0 + [0],
   
             [1 1]     [0]
   [c](x0) = [0 0]x0 + [1],
   
             [1 1]     [0]
   [a](x0) = [0 0]x0 + [1]
  orientation:
                 [1 1]     [2]    [1 1]     [2]              
   b(b(c(x1))) = [0 0]x1 + [0] >= [0 0]x1 + [0] = b(a(a(x1)))
   
                 [1 1]     [2]    [1 0]     [2]              
   a(c(c(x1))) = [0 0]x1 + [1] >= [0 0]x1 + [1] = a(a(b(x1)))
   
                 [1 0]     [2]    [1 0]     [2]              
   a(c(b(x1))) = [0 0]x1 + [1] >= [0 0]x1 + [1] = c(a(b(x1)))
   
                 [1 1]     [2]    [1 0]     [2]              
   a(c(c(x1))) = [0 0]x1 + [1] >= [0 0]x1 + [1] = c(c(b(x1)))
   
                 [1 1]     [2]    [1 1]     [1]              
   a(a(a(x1))) = [0 0]x1 + [1] >= [0 0]x1 + [1] = a(b(a(x1)))
   
                 [1 1]     [1]    [1 1]     [1]              
   a(b(a(x1))) = [0 0]x1 + [1] >= [0 0]x1 + [1] = a(b(c(x1)))
  problem:
   strict:
    b(b(c(x1))) -> b(a(a(x1)))
    a(c(c(x1))) -> a(a(b(x1)))
   weak:
    a(c(b(x1))) -> c(a(b(x1)))
    a(c(c(x1))) -> c(c(b(x1)))
    a(b(a(x1))) -> a(b(c(x1)))
  Arctic Interpretation Processor:
   dimension: 2
   interpretation:
              [0  -&]  
    [b](x0) = [0  -&]x0,
    
              [0 0]  
    [c](x0) = [1 1]x0,
    
              [0  0 ]  
    [a](x0) = [-& 0 ]x0
   orientation:
                  [0 0]      [0 0]                
    b(b(c(x1))) = [0 0]x1 >= [0 0]x1 = b(a(a(x1)))
    
                  [2 2]      [0  -&]                
    a(c(c(x1))) = [2 2]x1 >= [0  -&]x1 = a(a(b(x1)))
    
                  [1  -&]      [0  -&]                
    a(c(b(x1))) = [1  -&]x1 >= [1  -&]x1 = c(a(b(x1)))
    
                  [2 2]      [1  -&]                
    a(c(c(x1))) = [2 2]x1 >= [2  -&]x1 = c(c(b(x1)))
    
                  [0 0]      [0 0]                
    a(b(a(x1))) = [0 0]x1 >= [0 0]x1 = a(b(c(x1)))
   problem:
    strict:
     b(b(c(x1))) -> b(a(a(x1)))
    weak:
     a(c(b(x1))) -> c(a(b(x1)))
     a(c(c(x1))) -> c(c(b(x1)))
     a(b(a(x1))) -> a(b(c(x1)))
   RT Transformation Processor:
    b(b(c(x1))) -> b(a(a(x1)))
    a(c(b(x1))) -> c(a(b(x1)))
    a(c(c(x1))) -> c(c(b(x1)))
    a(b(a(x1))) -> a(b(c(x1)))
    Matrix Interpretation Processor: dim=2
     
     interpretation:
                [2 0]  
      [b](x0) = [0 0]x0,
      
                [1 1]     [0]
      [c](x0) = [1 1]x0 + [2],
      
                [1 1]  
      [a](x0) = [1 1]x0
     orientation:
                    [4 4]      [4 4]                
      b(b(c(x1))) = [0 0]x1 >= [0 0]x1 = b(a(a(x1)))
      
                    [4 0]     [2]    [4 0]     [0]              
      a(c(b(x1))) = [4 0]x1 + [2] >= [4 0]x1 + [2] = c(a(b(x1)))
      
                    [4 4]     [6]    [4 0]     [2]              
      a(c(c(x1))) = [4 4]x1 + [6] >= [4 0]x1 + [4] = c(c(b(x1)))
      
                    [2 2]      [2 2]                
      a(b(a(x1))) = [2 2]x1 >= [2 2]x1 = a(b(c(x1)))
     problem:
      b(b(c(x1))) -> b(a(a(x1)))
      a(b(a(x1))) -> a(b(c(x1)))
     Matrix Interpretation Processor: dim=3
      
      interpretation:
                 [1 1 0]     [0]
       [b](x0) = [0 0 0]x0 + [1]
                 [0 0 0]     [1],
       
                 [1 0 1]  
       [c](x0) = [0 1 0]x0
                 [0 0 0]  ,
       
                 [1 1 0]  
       [a](x0) = [0 0 1]x0
                 [0 0 0]  
      orientation:
                     [1 1 1]     [1]    [1 1 1]     [0]              
       b(b(c(x1))) = [0 0 0]x1 + [1] >= [0 0 0]x1 + [1] = b(a(a(x1)))
                     [0 0 0]     [1]    [0 0 0]     [1]              
       
                     [1 1 1]     [1]    [1 1 1]     [1]              
       a(b(a(x1))) = [0 0 0]x1 + [1] >= [0 0 0]x1 + [1] = a(b(c(x1)))
                     [0 0 0]     [0]    [0 0 0]     [0]              
      problem:
       a(b(a(x1))) -> a(b(c(x1)))
      Matrix Interpretation Processor: dim=3
       
       interpretation:
                  [1 0 0]  
        [b](x0) = [1 0 0]x0
                  [0 1 1]  ,
        
                  [1 0 0]  
        [c](x0) = [0 0 0]x0
                  [0 1 0]  ,
        
                  [1 0 1]     [0]
        [a](x0) = [0 0 0]x0 + [1]
                  [0 1 0]     [0]
       orientation:
                      [1 1 1]     [1]    [1 1 0]     [0]              
        a(b(a(x1))) = [0 0 0]x1 + [1] >= [0 0 0]x1 + [1] = a(b(c(x1)))
                      [1 0 1]     [0]    [1 0 0]     [0]              
       problem:
        
       Qed