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:
 Arctic Interpretation Processor:
  dimension: 2
  interpretation:
             [0  -&]  
   [s](x0) = [0  0 ]x0,
   
             [0  1 ]  
   [n](x0) = [-& 1 ]x0,
   
             [0  -&]  
   [c](x0) = [-& -&]x0,
   
             [0  0 ]  
   [f](x0) = [-& 0 ]x0
  orientation:
           [0  0 ]      [0  -&]                
   f(x1) = [-& 0 ]x1 >= [-& -&]x1 = n(c(c(x1)))
   
              [0  0 ]      [0  -&]                
   c(f(x1)) = [-& -&]x1 >= [-& -&]x1 = f(c(c(x1)))
   
              [0  -&]      [0  -&]          
   c(c(x1)) = [-& -&]x1 >= [-& -&]x1 = c(x1)
   
              [1 1]      [0 0]                
   n(s(x1)) = [1 1]x1 >= [0 0]x1 = f(s(s(x1)))
   
              [0  1 ]      [0  1 ]             
   n(f(x1)) = [-& 1 ]x1 >= [-& 1 ]x1 = f(n(x1))
  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))
  Arctic Interpretation Processor:
   dimension: 2
   interpretation:
              [0  -&]  
    [n](x0) = [2  -&]x0,
    
              [0  -&]  
    [c](x0) = [2  0 ]x0,
    
              [2 0]  
    [f](x0) = [3 0]x0
   orientation:
            [2 0]      [0  -&]                
    f(x1) = [3 0]x1 >= [2  -&]x1 = n(c(c(x1)))
    
               [2 0]      [2 0]                
    c(f(x1)) = [4 2]x1 >= [3 0]x1 = f(c(c(x1)))
    
               [0  -&]      [0  -&]          
    c(c(x1)) = [2  0 ]x1 >= [2  0 ]x1 = c(x1)
    
               [2 0]      [2  -&]             
    n(f(x1)) = [4 2]x1 >= [3  -&]x1 = f(n(x1))
   problem:
    c(f(x1)) -> f(c(c(x1)))
    c(c(x1)) -> c(x1)
    n(f(x1)) -> f(n(x1))
   KBO Processor:
    weight function:
     w0 = 1
     w(n) = w(f) = 1
     w(c) = 0
    precedence:
     c > n > f
    problem:
     
    Qed