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:
 DP Processor:
  DPs:
   f#(x1) -> c#(x1)
   f#(x1) -> c#(c(x1))
   f#(x1) -> n#(c(c(x1)))
   c#(f(x1)) -> c#(x1)
   c#(f(x1)) -> c#(c(x1))
   c#(f(x1)) -> f#(c(c(x1)))
   n#(s(x1)) -> f#(s(s(x1)))
   n#(f(x1)) -> n#(x1)
   n#(f(x1)) -> f#(n(x1))
  TRS:
   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))
  Matrix Interpretation Processor: dim=2
   
   interpretation:
    [n#](x0) = [2 1]x0 + [3],
    
    [c#](x0) = [2 0]x0 + [1],
    
    [f#](x0) = [2 0]x0 + [4],
    
              [1]
    [s](x0) = [2],
    
              [2 1]  
    [n](x0) = [0 2]x0,
    
              [1 0]  
    [c](x0) = [0 0]x0,
    
              [2 0]     [2]
    [f](x0) = [0 2]x0 + [0]
   orientation:
    f#(x1) = [2 0]x1 + [4] >= [2 0]x1 + [1] = c#(x1)
    
    f#(x1) = [2 0]x1 + [4] >= [2 0]x1 + [1] = c#(c(x1))
    
    f#(x1) = [2 0]x1 + [4] >= [2 0]x1 + [3] = n#(c(c(x1)))
    
    c#(f(x1)) = [4 0]x1 + [5] >= [2 0]x1 + [1] = c#(x1)
    
    c#(f(x1)) = [4 0]x1 + [5] >= [2 0]x1 + [1] = c#(c(x1))
    
    c#(f(x1)) = [4 0]x1 + [5] >= [2 0]x1 + [4] = f#(c(c(x1)))
    
    n#(s(x1)) = [7] >= [6] = f#(s(s(x1)))
    
    n#(f(x1)) = [4 2]x1 + [7] >= [2 1]x1 + [3] = n#(x1)
    
    n#(f(x1)) = [4 2]x1 + [7] >= [4 2]x1 + [4] = f#(n(x1))
    
            [2 0]     [2]    [2 0]                
    f(x1) = [0 2]x1 + [0] >= [0 0]x1 = n(c(c(x1)))
    
               [2 0]     [2]    [2 0]     [2]              
    c(f(x1)) = [0 0]x1 + [0] >= [0 0]x1 + [0] = f(c(c(x1)))
    
               [1 0]      [1 0]          
    c(c(x1)) = [0 0]x1 >= [0 0]x1 = c(x1)
    
               [4]    [4]              
    n(s(x1)) = [4] >= [4] = f(s(s(x1)))
    
               [4 2]     [4]    [4 2]     [2]           
    n(f(x1)) = [0 4]x1 + [0] >= [0 4]x1 + [0] = f(n(x1))
   problem:
    DPs:
     
    TRS:
     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))
   Qed