YES

Problem:
 f(0(),y) -> 0()
 f(s(x),y) -> f(f(x,y),y)

Proof:
 DP Processor:
  DPs:
   f#(s(x),y) -> f#(x,y)
   f#(s(x),y) -> f#(f(x,y),y)
  TRS:
   f(0(),y) -> 0()
   f(s(x),y) -> f(f(x,y),y)
  KBO Processor:
   argument filtering:
    pi(0) = []
    pi(f) = []
    pi(s) = [0]
    pi(f#) = 0
   usable rules:
    f(0(),y) -> 0()
    f(s(x),y) -> f(f(x,y),y)
   weight function:
    w0 = 1
    w(f#) = w(f) = w(0) = 1
    w(s) = 0
   precedence:
    f# ~ s > f > 0
   problem:
    DPs:
     
    TRS:
     f(0(),y) -> 0()
     f(s(x),y) -> f(f(x,y),y)
   Qed