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

Proof:
 Matrix Interpretation Processor: dim=1
  
  interpretation:
   [d](x0) = x0,
   
   [a](x0) = x0 + 8,
   
   [b](x0) = x0 + 8,
   
   [c](x0) = x0 + 8
  orientation:
   b(d(b(x1))) = x1 + 16 >= x1 + 16 = c(d(b(x1)))
   
   b(a(c(x1))) = x1 + 24 >= x1 + 16 = b(c(x1))
   
   a(d(x1)) = x1 + 8 >= x1 + 8 = d(c(x1))
   
   b(b(b(x1))) = x1 + 24 >= x1 + 24 = a(b(c(x1)))
   
   d(c(x1)) = x1 + 8 >= x1 + 8 = b(d(x1))
   
   d(c(x1)) = x1 + 8 >= x1 + 8 = d(b(d(x1)))
   
   d(a(c(x1))) = x1 + 16 >= x1 + 16 = b(b(x1))
  problem:
   b(d(b(x1))) -> c(d(b(x1)))
   a(d(x1)) -> d(c(x1))
   b(b(b(x1))) -> a(b(c(x1)))
   d(c(x1)) -> b(d(x1))
   d(c(x1)) -> d(b(d(x1)))
   d(a(c(x1))) -> b(b(x1))
  Matrix Interpretation Processor: dim=3
   
   interpretation:
              [1 0 0]     [0]
    [d](x0) = [0 0 0]x0 + [1]
              [0 0 1]     [0],
    
              [1 0 1]  
    [a](x0) = [0 1 1]x0
              [0 1 0]  ,
    
              [1 0 0]  
    [b](x0) = [0 0 0]x0
              [0 1 0]  ,
    
              [1 0 0]     [0]
    [c](x0) = [0 0 0]x0 + [0]
              [0 0 0]     [1]
   orientation:
                  [1 0 0]     [0]    [1 0 0]     [0]              
    b(d(b(x1))) = [0 0 0]x1 + [0] >= [0 0 0]x1 + [0] = c(d(b(x1)))
                  [0 0 0]     [1]    [0 0 0]     [1]              
    
               [1 0 1]     [0]    [1 0 0]     [0]           
    a(d(x1)) = [0 0 1]x1 + [1] >= [0 0 0]x1 + [1] = d(c(x1))
               [0 0 0]     [1]    [0 0 0]     [1]           
    
                  [1 0 0]      [1 0 0]                
    b(b(b(x1))) = [0 0 0]x1 >= [0 0 0]x1 = a(b(c(x1)))
                  [0 0 0]      [0 0 0]                
    
               [1 0 0]     [0]    [1 0 0]     [0]           
    d(c(x1)) = [0 0 0]x1 + [1] >= [0 0 0]x1 + [0] = b(d(x1))
               [0 0 0]     [1]    [0 0 0]     [1]           
    
               [1 0 0]     [0]    [1 0 0]     [0]              
    d(c(x1)) = [0 0 0]x1 + [1] >= [0 0 0]x1 + [1] = d(b(d(x1)))
               [0 0 0]     [1]    [0 0 0]     [1]              
    
                  [1 0 0]     [1]    [1 0 0]             
    d(a(c(x1))) = [0 0 0]x1 + [1] >= [0 0 0]x1 = b(b(x1))
                  [0 0 0]     [0]    [0 0 0]             
   problem:
    b(d(b(x1))) -> c(d(b(x1)))
    a(d(x1)) -> d(c(x1))
    b(b(b(x1))) -> a(b(c(x1)))
    d(c(x1)) -> b(d(x1))
    d(c(x1)) -> d(b(d(x1)))
   Matrix Interpretation Processor: dim=3
    
    interpretation:
               [1 0 0]     [0]
     [d](x0) = [0 0 0]x0 + [1]
               [0 0 0]     [0],
     
               [1 1 0]  
     [a](x0) = [0 1 0]x0
               [0 0 0]  ,
     
               [1 0 0]  
     [b](x0) = [0 0 0]x0
               [0 0 0]  ,
     
               [1 0 0]  
     [c](x0) = [0 0 0]x0
               [0 0 0]  
    orientation:
                   [1 0 0]      [1 0 0]                
     b(d(b(x1))) = [0 0 0]x1 >= [0 0 0]x1 = c(d(b(x1)))
                   [0 0 0]      [0 0 0]                
     
                [1 0 0]     [1]    [1 0 0]     [0]           
     a(d(x1)) = [0 0 0]x1 + [1] >= [0 0 0]x1 + [1] = d(c(x1))
                [0 0 0]     [0]    [0 0 0]     [0]           
     
                   [1 0 0]      [1 0 0]                
     b(b(b(x1))) = [0 0 0]x1 >= [0 0 0]x1 = a(b(c(x1)))
                   [0 0 0]      [0 0 0]                
     
                [1 0 0]     [0]    [1 0 0]             
     d(c(x1)) = [0 0 0]x1 + [1] >= [0 0 0]x1 = b(d(x1))
                [0 0 0]     [0]    [0 0 0]             
     
                [1 0 0]     [0]    [1 0 0]     [0]              
     d(c(x1)) = [0 0 0]x1 + [1] >= [0 0 0]x1 + [1] = d(b(d(x1)))
                [0 0 0]     [0]    [0 0 0]     [0]              
    problem:
     b(d(b(x1))) -> c(d(b(x1)))
     b(b(b(x1))) -> a(b(c(x1)))
     d(c(x1)) -> b(d(x1))
     d(c(x1)) -> d(b(d(x1)))
    String Reversal Processor:
     b(d(b(x1))) -> b(d(c(x1)))
     b(b(b(x1))) -> c(b(a(x1)))
     c(d(x1)) -> d(b(x1))
     c(d(x1)) -> d(b(d(x1)))
     Bounds Processor:
      bound: 0
      enrichment: match
      automaton:
       final states: {10,8,5,1}
       transitions:
        b0(4) -> 1*
        b0(11) -> 12*
        b0(2) -> 9*
        b0(6) -> 7*
        c0(7) -> 5*
        c0(2) -> 3*
        d0(12) -> 10*
        d0(2) -> 11*
        d0(3) -> 4*
        d0(9) -> 8*
        f40() -> 2*
        a0(2) -> 6*
        8 -> 3*
        1 -> 9,12
        5 -> 9*
        10 -> 3*
      problem:
       
      Qed