YES

Problem:
 f(s(X),Y) -> h(s(f(h(Y),X)))

Proof:
 DP Processor:
  DPs:
   f#(s(X),Y) -> f#(h(Y),X)
  TRS:
   f(s(X),Y) -> h(s(f(h(Y),X)))
  Usable Rule Processor:
   DPs:
    f#(s(X),Y) -> f#(h(Y),X)
   TRS:
    
   ADG Processor:
    DPs:
     f#(s(X),Y) -> f#(h(Y),X)
    TRS:
     
    graph:
     
    Qed