/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(c(x1)))
 c(f(x1)) -> f(c(c(x1)))
 c(c(x1)) -> c(x1)
 n(s(x1)) -> f(s(s(x1)))
 n(f(x1)) -> f(n(x1))

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