MAYBE

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:
 Complexity Transformation Processor:
  strict:
   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))
  weak:
   
  Matrix Interpretation Processor:
   dimension: 1
   max_matrix:
    1
    interpretation:
     [s](x0) = x0 + 1,
     
     [n](x0) = x0,
     
     [c](x0) = x0 + 1,
     
     [f](x0) = x0
    orientation:
     f(x1) = x1 >= x1 + 2 = n(c(c(x1)))
     
     c(f(x1)) = x1 + 1 >= x1 + 2 = f(c(c(x1)))
     
     c(c(x1)) = x1 + 2 >= x1 + 1 = c(x1)
     
     n(s(x1)) = x1 + 1 >= x1 + 2 = f(s(s(x1)))
     
     n(f(x1)) = x1 >= x1 = f(n(x1))
    problem:
     strict:
      f(x1) -> n(c(c(x1)))
      c(f(x1)) -> f(c(c(x1)))
      n(s(x1)) -> f(s(s(x1)))
      n(f(x1)) -> f(n(x1))
     weak:
      c(c(x1)) -> c(x1)
    Matrix Interpretation Processor:
     dimension: 1
     max_matrix:
      1
      interpretation:
       [s](x0) = x0,
       
       [n](x0) = x0,
       
       [c](x0) = x0,
       
       [f](x0) = x0 + 1
      orientation:
       f(x1) = x1 + 1 >= x1 = n(c(c(x1)))
       
       c(f(x1)) = x1 + 1 >= x1 + 1 = f(c(c(x1)))
       
       n(s(x1)) = x1 >= x1 + 1 = f(s(s(x1)))
       
       n(f(x1)) = x1 + 1 >= x1 + 1 = f(n(x1))
       
       c(c(x1)) = x1 >= x1 = c(x1)
      problem:
       strict:
        c(f(x1)) -> f(c(c(x1)))
        n(s(x1)) -> f(s(s(x1)))
        n(f(x1)) -> f(n(x1))
       weak:
        f(x1) -> n(c(c(x1)))
        c(c(x1)) -> c(x1)
      Matrix Interpretation Processor:
       dimension: 2
       max_matrix:
        [1 1]
        [0 1]
        interpretation:
                   [1 0]     [0]
         [s](x0) = [0 0]x0 + [1],
         
                   [1 1]  
         [n](x0) = [0 1]x0,
         
                   [1 0]  
         [c](x0) = [0 0]x0,
         
                     
         [f](x0) = x0
        orientation:
                    [1 0]      [1 0]                
         c(f(x1)) = [0 0]x1 >= [0 0]x1 = f(c(c(x1)))
         
                    [1 0]     [1]    [1 0]     [0]              
         n(s(x1)) = [0 0]x1 + [1] >= [0 0]x1 + [1] = f(s(s(x1)))
         
                    [1 1]      [1 1]             
         n(f(x1)) = [0 1]x1 >= [0 1]x1 = f(n(x1))
         
                       [1 0]                
         f(x1) = x1 >= [0 0]x1 = n(c(c(x1)))
         
                    [1 0]      [1 0]          
         c(c(x1)) = [0 0]x1 >= [0 0]x1 = c(x1)
        problem:
         strict:
          c(f(x1)) -> f(c(c(x1)))
          n(f(x1)) -> f(n(x1))
         weak:
          n(s(x1)) -> f(s(s(x1)))
          f(x1) -> n(c(c(x1)))
          c(c(x1)) -> c(x1)
        Open