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)
  Subterm Criterion Processor:
   simple projection:
    pi(f#) = 1
   problem:
    DPs:
     
    TRS:
     f(x,a()) -> x
     f(x,g(y)) -> f(g(x),y)
   Qed