YES

Problem:
 f(x,a()) -> x
 f(x,g(y)) -> f(g(x),y)

Proof:
 DP Processor:
  DPs:
   f#(x,g(y)) -> f#(g(x),y)
  TRS:
   f(x,a()) -> x
   f(x,g(y)) -> f(g(x),y)
  Usable Rule Processor:
   DPs:
    f#(x,g(y)) -> f#(g(x),y)
   TRS:
    
   CDG Processor:
    DPs:
     f#(x,g(y)) -> f#(g(x),y)
    TRS:
     
    graph:
     
    Qed