YES

Problem:
 f(X) -> g(n__h(n__f(X)))
 h(X) -> n__h(X)
 f(X) -> n__f(X)
 activate(n__h(X)) -> h(activate(X))
 activate(n__f(X)) -> f(activate(X))
 activate(X) -> X

Proof:
 DP Processor:
  DPs:
   activate#(n__h(X)) -> activate#(X)
   activate#(n__h(X)) -> h#(activate(X))
   activate#(n__f(X)) -> activate#(X)
   activate#(n__f(X)) -> f#(activate(X))
  TRS:
   f(X) -> g(n__h(n__f(X)))
   h(X) -> n__h(X)
   f(X) -> n__f(X)
   activate(n__h(X)) -> h(activate(X))
   activate(n__f(X)) -> f(activate(X))
   activate(X) -> X
  Arctic Interpretation Processor:
   dimension: 1
   usable rules:
    
   interpretation:
    [activate#](x0) = 1x0,
    
    [h#](x0) = 0,
    
    [f#](x0) = 2,
    
    [activate](x0) = x0 + -16,
    
    [h](x0) = 2x0,
    
    [g](x0) = 0,
    
    [n__h](x0) = 8x0 + 15,
    
    [n__f](x0) = 1x0 + 14,
    
    [f](x0) = 0
   orientation:
    activate#(n__h(X)) = 9X + 16 >= 1X = activate#(X)
    
    activate#(n__h(X)) = 9X + 16 >= 0 = h#(activate(X))
    
    activate#(n__f(X)) = 2X + 15 >= 1X = activate#(X)
    
    activate#(n__f(X)) = 2X + 15 >= 2 = f#(activate(X))
    
    f(X) = 0 >= 0 = g(n__h(n__f(X)))
    
    h(X) = 2X >= 8X + 15 = n__h(X)
    
    f(X) = 0 >= 1X + 14 = n__f(X)
    
    activate(n__h(X)) = 8X + 15 >= 2X + -14 = h(activate(X))
    
    activate(n__f(X)) = 1X + 14 >= 0 = f(activate(X))
    
    activate(X) = X + -16 >= X = X
   problem:
    DPs:
     
    TRS:
     f(X) -> g(n__h(n__f(X)))
     h(X) -> n__h(X)
     f(X) -> n__f(X)
     activate(n__h(X)) -> h(activate(X))
     activate(n__f(X)) -> f(activate(X))
     activate(X) -> X
   Qed