YES

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

Proof:
 DP Processor:
  DPs:
   f#(f(a())) -> f#(g(n__f(n__a())))
   activate#(n__f(X)) -> activate#(X)
   activate#(n__f(X)) -> f#(activate(X))
   activate#(n__a()) -> a#()
  TRS:
   f(f(a())) -> f(g(n__f(n__a())))
   f(X) -> n__f(X)
   a() -> n__a()
   activate(n__f(X)) -> f(activate(X))
   activate(n__a()) -> a()
   activate(X) -> X
  KBO Processor:
   argument filtering:
    pi(a) = []
    pi(f) = [0]
    pi(n__a) = []
    pi(n__f) = [0]
    pi(g) = []
    pi(activate) = [0]
    pi(f#) = 0
    pi(a#) = []
    pi(activate#) = 0
   weight function:
    w0 = 1
    w(a#) = w(f#) = w(g) = w(n__f) = w(n__a) = w(f) = w(a) = 1
    w(activate#) = w(activate) = 0
   precedence:
    f# ~ activate ~ g > a > n__a ~ f > activate# ~ a# ~ n__f
   problem:
    DPs:
     
    TRS:
     f(f(a())) -> f(g(n__f(n__a())))
     f(X) -> n__f(X)
     a() -> n__a()
     activate(n__f(X)) -> f(activate(X))
     activate(n__a()) -> a()
     activate(X) -> X
   Qed