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

Proof:
 Matrix Interpretation Processor: dim=1
  
  interpretation:
   [c](x0) = x0 + 4,
   
   [d](x0) = x0,
   
   [b](x0) = x0 + 3,
   
   [a](x0) = x0 + 2,
   
   [g](x0) = x0 + 2
  orientation:
   a(x1) = x1 + 2 >= x1 + 2 = g(d(x1))
   
   b(b(b(x1))) = x1 + 9 >= x1 + 8 = c(d(c(x1)))
   
   b(b(x1)) = x1 + 6 >= x1 + 6 = a(g(g(x1)))
   
   c(d(x1)) = x1 + 4 >= x1 + 4 = g(g(x1))
   
   g(g(g(x1))) = x1 + 6 >= x1 + 6 = b(b(x1))
  problem:
   a(x1) -> g(d(x1))
   b(b(x1)) -> a(g(g(x1)))
   c(d(x1)) -> g(g(x1))
   g(g(g(x1))) -> b(b(x1))
  String Reversal Processor:
   a(x1) -> d(g(x1))
   b(b(x1)) -> g(g(a(x1)))
   d(c(x1)) -> g(g(x1))
   g(g(g(x1))) -> b(b(x1))
   Matrix Interpretation Processor: dim=3
    
    interpretation:
               [1 1 0]     [0]
     [c](x0) = [1 1 0]x0 + [1]
               [1 0 0]     [1],
     
               [1 0 1]  
     [d](x0) = [0 1 1]x0
               [1 0 1]  ,
     
               [1 0 0]  
     [b](x0) = [0 0 0]x0
               [0 0 0]  ,
     
               [1 0 0]  
     [a](x0) = [1 0 1]x0
               [1 1 0]  ,
     
               [1 0 0]  
     [g](x0) = [0 0 1]x0
               [0 0 0]  
    orientation:
             [1 0 0]      [1 0 0]             
     a(x1) = [1 0 1]x1 >= [0 0 1]x1 = d(g(x1))
             [1 1 0]      [1 0 0]             
     
                [1 0 0]      [1 0 0]                
     b(b(x1)) = [0 0 0]x1 >= [0 0 0]x1 = g(g(a(x1)))
                [0 0 0]      [0 0 0]                
     
                [2 1 0]     [1]    [1 0 0]             
     d(c(x1)) = [2 1 0]x1 + [2] >= [0 0 0]x1 = g(g(x1))
                [2 1 0]     [1]    [0 0 0]             
     
                   [1 0 0]      [1 0 0]             
     g(g(g(x1))) = [0 0 0]x1 >= [0 0 0]x1 = b(b(x1))
                   [0 0 0]      [0 0 0]             
    problem:
     a(x1) -> d(g(x1))
     b(b(x1)) -> g(g(a(x1)))
     g(g(g(x1))) -> b(b(x1))
    String Reversal Processor:
     a(x1) -> g(d(x1))
     b(b(x1)) -> a(g(g(x1)))
     g(g(g(x1))) -> b(b(x1))
     Matrix Interpretation Processor: dim=3
      
      interpretation:
                 [1 0 0]  
       [d](x0) = [0 0 0]x0
                 [0 0 0]  ,
       
                 [1 1 1]     [1]
       [b](x0) = [0 0 1]x0 + [0]
                 [0 0 0]     [1],
       
                 [1 1 0]     [1]
       [a](x0) = [0 1 0]x0 + [0]
                 [0 0 0]     [1],
       
                 [1 1 1]     [0]
       [g](x0) = [0 0 1]x0 + [0]
                 [0 0 0]     [1]
      orientation:
               [1 1 0]     [1]    [1 0 0]     [0]           
       a(x1) = [0 1 0]x1 + [0] >= [0 0 0]x1 + [0] = g(d(x1))
               [0 0 0]     [1]    [0 0 0]     [1]           
       
                  [1 1 2]     [3]    [1 1 2]     [3]              
       b(b(x1)) = [0 0 0]x1 + [1] >= [0 0 0]x1 + [1] = a(g(g(x1)))
                  [0 0 0]     [1]    [0 0 0]     [1]              
       
                     [1 1 2]     [3]    [1 1 2]     [3]           
       g(g(g(x1))) = [0 0 0]x1 + [1] >= [0 0 0]x1 + [1] = b(b(x1))
                     [0 0 0]     [1]    [0 0 0]     [1]           
      problem:
       b(b(x1)) -> a(g(g(x1)))
       g(g(g(x1))) -> b(b(x1))
      String Reversal Processor:
       b(b(x1)) -> g(g(a(x1)))
       g(g(g(x1))) -> b(b(x1))
       KBO Processor:
        weight function:
         w0 = 1
         w(b) = w(g) = 1
         w(a) = 0
        precedence:
         a > b > g
        problem:
         
        Qed