YES

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

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