YES

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

Proof:
 DP Processor:
  DPs:
   f#(f(a())) -> f#(g(n__f(a())))
   activate#(n__f(X)) -> f#(X)
  TRS:
   f(f(a())) -> f(g(n__f(a())))
   f(X) -> n__f(X)
   activate(n__f(X)) -> f(X)
   activate(X) -> X
  TDG Processor:
   DPs:
    f#(f(a())) -> f#(g(n__f(a())))
    activate#(n__f(X)) -> f#(X)
   TRS:
    f(f(a())) -> f(g(n__f(a())))
    f(X) -> n__f(X)
    activate(n__f(X)) -> f(X)
    activate(X) -> X
   graph:
    activate#(n__f(X)) -> f#(X) -> f#(f(a())) -> f#(g(n__f(a())))
    f#(f(a())) -> f#(g(n__f(a()))) -> f#(f(a())) -> f#(g(n__f(a())))
   SCC Processor:
    #sccs: 1
    #rules: 1
    #arcs: 2/4
    DPs:
     f#(f(a())) -> f#(g(n__f(a())))
    TRS:
     f(f(a())) -> f(g(n__f(a())))
     f(X) -> n__f(X)
     activate(n__f(X)) -> f(X)
     activate(X) -> X
    EDG Processor:
     DPs:
      f#(f(a())) -> f#(g(n__f(a())))
     TRS:
      f(f(a())) -> f(g(n__f(a())))
      f(X) -> n__f(X)
      activate(n__f(X)) -> f(X)
      activate(X) -> X
     graph:
      
     Qed