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)
  EDG 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)
   graph:
    f#(s(x),y) -> f#(x,y) -> f#(s(x),y) -> f#(x,y)
    f#(s(x),y) -> f#(x,y) -> f#(s(x),y) -> f#(f(x,y),y)
   SCC Processor:
    #sccs: 1
    #rules: 1
    #arcs: 2/4
    DPs:
     f#(s(x),y) -> f#(x,y)
    TRS:
     f(0(),y) -> 0()
     f(s(x),y) -> f(f(x,y),y)
    KBO Processor:
     argument filtering:
      pi(0) = []
      pi(f) = 0
      pi(s) = [0]
      pi(f#) = 0
     weight function:
      w0 = 1
      w(f#) = w(s) = w(f) = w(0) = 1
     precedence:
      f# ~ s ~ f ~ 0
     problem:
      DPs:
       
      TRS:
       f(0(),y) -> 0()
       f(s(x),y) -> f(f(x,y),y)
     Qed