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
  Usable Rule Processor:
   DPs:
    f#(f(a())) -> f#(g(n__f(a())))
    activate#(n__f(X)) -> f#(X)
   TRS:
    
   Arctic Interpretation Processor:
    dimension: 1
    usable rules:
     
    interpretation:
     [activate#](x0) = 5x0 + -16,
     
     [f#](x0) = x0 + -16,
     
     [g](x0) = x0 + -16,
     
     [n__f](x0) = 1x0 + -8,
     
     [f](x0) = x0 + 4,
     
     [a] = 0
    orientation:
     f#(f(a())) = 4 >= 1 = f#(g(n__f(a())))
     
     activate#(n__f(X)) = 6X + -3 >= X + -16 = f#(X)
    problem:
     DPs:
      
     TRS:
      
    Qed