YES

Problem:
 g(c(x,s(y))) -> g(c(s(x),y))
 f(c(s(x),y)) -> f(c(x,s(y)))
 f(f(x)) -> f(d(f(x)))
 f(x) -> x

Proof:
 DP Processor:
  DPs:
   g#(c(x,s(y))) -> g#(c(s(x),y))
   f#(c(s(x),y)) -> f#(c(x,s(y)))
   f#(f(x)) -> f#(d(f(x)))
  TRS:
   g(c(x,s(y))) -> g(c(s(x),y))
   f(c(s(x),y)) -> f(c(x,s(y)))
   f(f(x)) -> f(d(f(x)))
   f(x) -> x
  Usable Rule Processor:
   DPs:
    g#(c(x,s(y))) -> g#(c(s(x),y))
    f#(c(s(x),y)) -> f#(c(x,s(y)))
    f#(f(x)) -> f#(d(f(x)))
   TRS:
    f(c(s(x),y)) -> f(c(x,s(y)))
    f(f(x)) -> f(d(f(x)))
    f(x) -> x
   Matrix Interpretation Processor: dim=2
    
    usable rules:
     
    interpretation:
     [f#](x0) = [1 0]x0,
     
     [g#](x0) = [0 2]x0,
     
               [0]
     [d](x0) = [3],
     
               [2]
     [f](x0) = [0],
     
                   [0 2]     [0 1]     [3]
     [c](x0, x1) = [0 0]x0 + [1 1]x1 + [0],
     
               [1 1]     [0]
     [s](x0) = [0 1]x0 + [1]
    orientation:
     g#(c(x,s(y))) = [2 4]y + [2] >= [2 2]y = g#(c(s(x),y))
     
     f#(c(s(x),y)) = [0 2]x + [0 1]y + [5] >= [0 2]x + [0 1]y + [4] = f#(c(x,s(y)))
     
     f#(f(x)) = [2] >= [0] = f#(d(f(x)))
     
                    [2]    [2]               
     f(c(s(x),y)) = [0] >= [0] = f(c(x,s(y)))
     
               [2]    [2]             
     f(f(x)) = [0] >= [0] = f(d(f(x)))
     
            [2]         
     f(x) = [0] >= x = x
    problem:
     DPs:
      
     TRS:
      f(c(s(x),y)) -> f(c(x,s(y)))
      f(f(x)) -> f(d(f(x)))
      f(x) -> x
    Qed