YES

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

Proof:
 DP Processor:
  DPs:
   activate#(n__f(X)) -> activate#(X)
   activate#(n__f(X)) -> f#(activate(X))
   activate#(n__g(X)) -> activate#(X)
   activate#(n__g(X)) -> g#(activate(X))
   activate#(n__a()) -> a#()
  TRS:
   f(f(a())) -> c(n__f(n__g(n__f(n__a()))))
   f(X) -> n__f(X)
   g(X) -> n__g(X)
   a() -> n__a()
   activate(n__f(X)) -> f(activate(X))
   activate(n__g(X)) -> g(activate(X))
   activate(n__a()) -> a()
   activate(X) -> X
  Arctic Interpretation Processor:
   dimension: 1
   usable rules:
    
   interpretation:
    [activate#](x0) = 2x0,
    
    [a#] = 0,
    
    [g#](x0) = 0,
    
    [f#](x0) = 0,
    
    [activate](x0) = 0,
    
    [g](x0) = 1x0 + 2,
    
    [c](x0) = x0 + 1,
    
    [n__g](x0) = 4x0 + 1,
    
    [n__f](x0) = 1x0 + 2,
    
    [n__a] = 0,
    
    [f](x0) = x0 + 1,
    
    [a] = 1
   orientation:
    activate#(n__f(X)) = 3X + 4 >= 2X = activate#(X)
    
    activate#(n__f(X)) = 3X + 4 >= 0 = f#(activate(X))
    
    activate#(n__g(X)) = 6X + 3 >= 2X = activate#(X)
    
    activate#(n__g(X)) = 6X + 3 >= 0 = g#(activate(X))
    
    activate#(n__a()) = 2 >= 0 = a#()
    
    f(f(a())) = 1 >= 7 = c(n__f(n__g(n__f(n__a()))))
    
    f(X) = X + 1 >= 1X + 2 = n__f(X)
    
    g(X) = 1X + 2 >= 4X + 1 = n__g(X)
    
    a() = 1 >= 0 = n__a()
    
    activate(n__f(X)) = 0 >= 1 = f(activate(X))
    
    activate(n__g(X)) = 0 >= 2 = g(activate(X))
    
    activate(n__a()) = 0 >= 1 = a()
    
    activate(X) = 0 >= X = X
   problem:
    DPs:
     
    TRS:
     f(f(a())) -> c(n__f(n__g(n__f(n__a()))))
     f(X) -> n__f(X)
     g(X) -> n__g(X)
     a() -> n__a()
     activate(n__f(X)) -> f(activate(X))
     activate(n__g(X)) -> g(activate(X))
     activate(n__a()) -> a()
     activate(X) -> X
   Qed