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)
    Subterm Criterion Processor:
     simple projection:
      pi(f#) = 0
     problem:
      DPs:
       
      TRS:
       f(0(),y) -> 0()
       f(s(x),y) -> f(f(x,y),y)
     Qed