YES

Problem:
 f(n__f(n__a())) -> f(n__g(f(n__a())))
 f(X) -> n__f(X)
 a() -> n__a()
 g(X) -> n__g(X)
 activate(n__f(X)) -> f(X)
 activate(n__a()) -> a()
 activate(n__g(X)) -> g(X)
 activate(X) -> X

Proof:
 DP Processor:
  DPs:
   f#(n__f(n__a())) -> f#(n__a())
   f#(n__f(n__a())) -> f#(n__g(f(n__a())))
   activate#(n__f(X)) -> f#(X)
   activate#(n__a()) -> a#()
   activate#(n__g(X)) -> g#(X)
  TRS:
   f(n__f(n__a())) -> f(n__g(f(n__a())))
   f(X) -> n__f(X)
   a() -> n__a()
   g(X) -> n__g(X)
   activate(n__f(X)) -> f(X)
   activate(n__a()) -> a()
   activate(n__g(X)) -> g(X)
   activate(X) -> X
  Usable Rule Processor:
   DPs:
    f#(n__f(n__a())) -> f#(n__a())
    f#(n__f(n__a())) -> f#(n__g(f(n__a())))
    activate#(n__f(X)) -> f#(X)
    activate#(n__a()) -> a#()
    activate#(n__g(X)) -> g#(X)
   TRS:
    f(X) -> n__f(X)
   CDG Processor:
    DPs:
     f#(n__f(n__a())) -> f#(n__a())
     f#(n__f(n__a())) -> f#(n__g(f(n__a())))
     activate#(n__f(X)) -> f#(X)
     activate#(n__a()) -> a#()
     activate#(n__g(X)) -> g#(X)
    TRS:
     f(X) -> n__f(X)
    graph:
     
    Qed