YES

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

Proof:
 DP Processor:
  DPs:
   activate#(n__f(X)) -> f#(X)
  TRS:
   f(f(a())) -> c(n__f(g(f(a()))))
   f(X) -> n__f(X)
   activate(n__f(X)) -> f(X)
   activate(X) -> X
  Usable Rule Processor:
   DPs:
    activate#(n__f(X)) -> f#(X)
   TRS:
    
   Arctic Interpretation Processor:
    dimension: 1
    usable rules:
     
    interpretation:
     [activate#](x0) = -13x0 + 0,
     
     [f#](x0) = x0,
     
     [n__f](x0) = 15x0 + 13
    orientation:
     activate#(n__f(X)) = 2X + 0 >= X = f#(X)
    problem:
     DPs:
      
     TRS:
      
    Qed