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
  Arctic Interpretation Processor:
   dimension: 1
   interpretation:
    [activate#](x0) = x0 + 1,
    
    [f#](x0) = -16x0 + 0,
    
    [activate](x0) = 4x0 + 1,
    
    [c](x0) = -7x0 + 1,
    
    [n__f](x0) = x0 + -8,
    
    [g](x0) = x0 + -16,
    
    [f](x0) = 1x0 + 0,
    
    [a] = 1
   orientation:
    activate#(n__f(X)) = X + 1 >= -16X + 0 = f#(X)
    
    f(f(a())) = 3 >= 1 = c(n__f(g(f(a()))))
    
    f(X) = 1X + 0 >= X + -8 = n__f(X)
    
    activate(n__f(X)) = 4X + 1 >= 1X + 0 = f(X)
    
    activate(X) = 4X + 1 >= X = X
   problem:
    DPs:
     
    TRS:
     f(f(a())) -> c(n__f(g(f(a()))))
     f(X) -> n__f(X)
     activate(n__f(X)) -> f(X)
     activate(X) -> X
   Qed