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:
 RT 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
   interpretation:
    [s](x0) = x0 + 12,
    
    [n](x0) = x0 + 1,
    
    [c](x0) = x0,
    
    [f](x0) = x0 + 6
   orientation:
    f(x1) = x1 + 6 >= x1 + 1 = n(c(c(x1)))
    
    c(f(x1)) = x1 + 6 >= x1 + 6 = f(c(c(x1)))
    
    c(c(x1)) = x1 >= x1 = c(x1)
    
    n(s(x1)) = x1 + 13 >= x1 + 30 = f(s(s(x1)))
    
    n(f(x1)) = x1 + 7 >= x1 + 7 = f(n(x1))
   problem:
    strict:
     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:
     f(x1) -> n(c(c(x1)))
   Matrix Interpretation Processor:
    dimension: 1
    interpretation:
     [s](x0) = x0 + 4,
     
     [n](x0) = x0,
     
     [c](x0) = x0 + 1,
     
     [f](x0) = x0 + 2
    orientation:
     c(f(x1)) = x1 + 3 >= x1 + 4 = f(c(c(x1)))
     
     c(c(x1)) = x1 + 2 >= x1 + 1 = c(x1)
     
     n(s(x1)) = x1 + 4 >= x1 + 10 = f(s(s(x1)))
     
     n(f(x1)) = x1 + 2 >= x1 + 2 = f(n(x1))
     
     f(x1) = x1 + 2 >= x1 + 2 = n(c(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:
      c(c(x1)) -> c(x1)
      f(x1) -> n(c(c(x1)))
    Matrix Interpretation Processor:
     dimension: 2
     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]      [1 0]          
      c(c(x1)) = [0 0]x1 >= [0 0]x1 = c(x1)
      
                    [1 0]                
      f(x1) = x1 >= [0 0]x1 = n(c(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)))
       c(c(x1)) -> c(x1)
       f(x1) -> n(c(c(x1)))
     Open