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
  Arctic Interpretation Processor:
   dimension: 1
   interpretation:
    [activate#](x0) = 4x0 + -16,
    
    [f#](x0) = x0 + -16,
    
    [activate](x0) = 4x0 + 0,
    
    [g](x0) = -5x0 + 4,
    
    [n__f](x0) = -3x0 + 0,
    
    [f](x0) = x0 + 2,
    
    [a] = 5
   orientation:
    f#(f(a())) = 5 >= 4 = f#(g(n__f(a())))
    
    activate#(n__f(X)) = 1X + 4 >= X + -16 = f#(X)
    
    f(f(a())) = 5 >= 4 = f(g(n__f(a())))
    
    f(X) = X + 2 >= -3X + 0 = n__f(X)
    
    activate(n__f(X)) = 1X + 4 >= X + 2 = f(X)
    
    activate(X) = 4X + 0 >= X = X
   problem:
    DPs:
     
    TRS:
     f(f(a())) -> f(g(n__f(a())))
     f(X) -> n__f(X)
     activate(n__f(X)) -> f(X)
     activate(X) -> X
   Qed