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