/export/starexec/sandbox/solver/bin/starexec_run_ttt2-1.17+nonreach /export/starexec/sandbox/benchmark/theBenchmark.xml /export/starexec/sandbox/output/output_files


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YES

Problem:
 f(x1) -> n(c(n(a(x1))))
 c(f(x1)) -> f(n(a(c(x1))))
 n(a(x1)) -> c(x1)
 c(c(x1)) -> c(x1)
 n(s(x1)) -> f(s(s(x1)))
 n(f(x1)) -> f(n(x1))

Proof:
 String Reversal Processor:
  f(x1) -> a(n(c(n(x1))))
  f(c(x1)) -> c(a(n(f(x1))))
  a(n(x1)) -> c(x1)
  c(c(x1)) -> c(x1)
  s(n(x1)) -> s(s(f(x1)))
  f(n(x1)) -> n(f(x1))
  Matrix Interpretation Processor: dim=3
   
   interpretation:
              [1 1 0]  
    [s](x0) = [0 0 0]x0
              [1 1 0]  ,
    
              [1 0 0]  
    [c](x0) = [0 0 0]x0
              [0 0 0]  ,
    
              [1 0 0]     [0]
    [n](x0) = [0 1 0]x0 + [1]
              [0 0 0]     [0],
    
              [1 0 0]     [0]
    [a](x0) = [0 0 0]x0 + [0]
              [0 0 0]     [1],
    
              [1 0 0]     [0]
    [f](x0) = [0 1 0]x0 + [0]
              [0 0 0]     [1]
   orientation:
            [1 0 0]     [0]    [1 0 0]     [0]                 
    f(x1) = [0 1 0]x1 + [0] >= [0 0 0]x1 + [0] = a(n(c(n(x1))))
            [0 0 0]     [1]    [0 0 0]     [1]                 
    
               [1 0 0]     [0]    [1 0 0]                   
    f(c(x1)) = [0 0 0]x1 + [0] >= [0 0 0]x1 = c(a(n(f(x1))))
               [0 0 0]     [1]    [0 0 0]                   
    
               [1 0 0]     [0]    [1 0 0]          
    a(n(x1)) = [0 0 0]x1 + [0] >= [0 0 0]x1 = c(x1)
               [0 0 0]     [1]    [0 0 0]          
    
               [1 0 0]      [1 0 0]          
    c(c(x1)) = [0 0 0]x1 >= [0 0 0]x1 = c(x1)
               [0 0 0]      [0 0 0]          
    
               [1 1 0]     [1]    [1 1 0]                
    s(n(x1)) = [0 0 0]x1 + [0] >= [0 0 0]x1 = s(s(f(x1)))
               [1 1 0]     [1]    [1 1 0]                
    
               [1 0 0]     [0]    [1 0 0]     [0]           
    f(n(x1)) = [0 1 0]x1 + [1] >= [0 1 0]x1 + [1] = n(f(x1))
               [0 0 0]     [1]    [0 0 0]     [0]           
   problem:
    f(x1) -> a(n(c(n(x1))))
    f(c(x1)) -> c(a(n(f(x1))))
    a(n(x1)) -> c(x1)
    c(c(x1)) -> c(x1)
    f(n(x1)) -> n(f(x1))
   String Reversal Processor:
    f(x1) -> n(c(n(a(x1))))
    c(f(x1)) -> f(n(a(c(x1))))
    n(a(x1)) -> c(x1)
    c(c(x1)) -> c(x1)
    n(f(x1)) -> f(n(x1))
    Matrix Interpretation Processor: dim=1
     
     interpretation:
      [c](x0) = x0,
      
      [n](x0) = x0,
      
      [a](x0) = x0,
      
      [f](x0) = 4x0 + 1
     orientation:
      f(x1) = 4x1 + 1 >= x1 = n(c(n(a(x1))))
      
      c(f(x1)) = 4x1 + 1 >= 4x1 + 1 = f(n(a(c(x1))))
      
      n(a(x1)) = x1 >= x1 = c(x1)
      
      c(c(x1)) = x1 >= x1 = c(x1)
      
      n(f(x1)) = 4x1 + 1 >= 4x1 + 1 = f(n(x1))
     problem:
      c(f(x1)) -> f(n(a(c(x1))))
      n(a(x1)) -> c(x1)
      c(c(x1)) -> c(x1)
      n(f(x1)) -> f(n(x1))
     Matrix Interpretation Processor: dim=2
      
      interpretation:
                 [1 0]  
       [c](x0) = [2 0]x0,
       
                 [1 2]     [0]
       [n](x0) = [2 0]x0 + [1],
       
                 [1 0]  
       [a](x0) = [0 0]x0,
       
                 [2 0]     [2]
       [f](x0) = [0 2]x0 + [1]
      orientation:
                  [2 0]     [2]    [2 0]     [2]                 
       c(f(x1)) = [4 0]x1 + [4] >= [4 0]x1 + [3] = f(n(a(c(x1))))
       
                  [1 0]     [0]    [1 0]          
       n(a(x1)) = [2 0]x1 + [1] >= [2 0]x1 = c(x1)
       
                  [1 0]      [1 0]          
       c(c(x1)) = [2 0]x1 >= [2 0]x1 = c(x1)
       
                  [2 4]     [4]    [2 4]     [2]           
       n(f(x1)) = [4 0]x1 + [5] >= [4 0]x1 + [3] = f(n(x1))
      problem:
       c(f(x1)) -> f(n(a(c(x1))))
       n(a(x1)) -> c(x1)
       c(c(x1)) -> c(x1)
      String Reversal Processor:
       f(c(x1)) -> c(a(n(f(x1))))
       a(n(x1)) -> c(x1)
       c(c(x1)) -> c(x1)
       Matrix Interpretation Processor: dim=1
        
        interpretation:
         [c](x0) = x0 + 1,
         
         [n](x0) = x0 + 8,
         
         [a](x0) = x0,
         
         [f](x0) = 9x0 + 8
        orientation:
         f(c(x1)) = 9x1 + 17 >= 9x1 + 17 = c(a(n(f(x1))))
         
         a(n(x1)) = x1 + 8 >= x1 + 1 = c(x1)
         
         c(c(x1)) = x1 + 2 >= x1 + 1 = c(x1)
        problem:
         f(c(x1)) -> c(a(n(f(x1))))
        Bounds Processor:
         bound: 0
         enrichment: match
         automaton:
          final states: {1}
          transitions:
           f0(2) -> 3*
           f50() -> 2*
           c0(5) -> 1*
           a0(4) -> 5*
           n0(3) -> 4*
           1 -> 3*
         problem:
          
         Qed