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)
   Arctic Interpretation Processor:
    dimension: 1
    usable rules:
     
    interpretation:
     [activate#](x0) = 5x0 + 8,
     
     [g#](x0) = 0,
     
     [a#] = 0,
     
     [f#](x0) = 5x0 + -16,
     
     [n__g](x0) = 0,
     
     [f](x0) = -8x0 + 0,
     
     [n__f](x0) = 1x0 + 1,
     
     [n__a] = 1
    orientation:
     f#(n__f(n__a())) = 7 >= 6 = f#(n__a())
     
     f#(n__f(n__a())) = 7 >= 5 = f#(n__g(f(n__a())))
     
     activate#(n__f(X)) = 6X + 8 >= 5X + -16 = f#(X)
     
     activate#(n__a()) = 8 >= 0 = a#()
     
     activate#(n__g(X)) = 8 >= 0 = g#(X)
     
     f(X) = -8X + 0 >= 1X + 1 = n__f(X)
    problem:
     DPs:
      
     TRS:
      f(X) -> n__f(X)
    Qed