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
  Arctic Interpretation Processor:
   dimension: 1
   interpretation:
    [activate#](x0) = x0 + 4,
    
    [g#](x0) = 1,
    
    [a#] = 0,
    
    [f#](x0) = x0 + 0,
    
    [activate](x0) = 2x0 + 13,
    
    [g](x0) = 8,
    
    [a] = 2,
    
    [n__g](x0) = 0,
    
    [f](x0) = 2x0 + 0,
    
    [n__f](x0) = 1x0,
    
    [n__a] = 0
   orientation:
    f#(n__f(n__a())) = 1 >= 0 = f#(n__a())
    
    f#(n__f(n__a())) = 1 >= 0 = f#(n__g(f(n__a())))
    
    activate#(n__f(X)) = 1X + 4 >= X + 0 = f#(X)
    
    activate#(n__a()) = 4 >= 0 = a#()
    
    activate#(n__g(X)) = 4 >= 1 = g#(X)
    
    f(n__f(n__a())) = 3 >= 2 = f(n__g(f(n__a())))
    
    f(X) = 2X + 0 >= 1X = n__f(X)
    
    a() = 2 >= 0 = n__a()
    
    g(X) = 8 >= 0 = n__g(X)
    
    activate(n__f(X)) = 3X + 13 >= 2X + 0 = f(X)
    
    activate(n__a()) = 13 >= 2 = a()
    
    activate(n__g(X)) = 13 >= 8 = g(X)
    
    activate(X) = 2X + 13 >= X = X
   problem:
    DPs:
     
    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
   Qed