YES

We show the termination of the TRS R:

  f(f(a())) -> f(g(n__f(n__a())))
  f(X) -> n__f(X)
  a() -> n__a()
  activate(n__f(X)) -> f(activate(X))
  activate(n__a()) -> a()
  activate(X) -> X

-- SCC decomposition.

Consider the dependency pair problem (P, R), where P consists of

p1: f#(f(a())) -> f#(g(n__f(n__a())))
p2: activate#(n__f(X)) -> f#(activate(X))
p3: activate#(n__f(X)) -> activate#(X)
p4: activate#(n__a()) -> a#()

and R consists of:

r1: f(f(a())) -> f(g(n__f(n__a())))
r2: f(X) -> n__f(X)
r3: a() -> n__a()
r4: activate(n__f(X)) -> f(activate(X))
r5: activate(n__a()) -> a()
r6: activate(X) -> X

The estimated dependency graph contains the following SCCs:

  {p3}


-- Reduction pair.

Consider the dependency pair problem (P, R), where P consists of

p1: activate#(n__f(X)) -> activate#(X)

and R consists of:

r1: f(f(a())) -> f(g(n__f(n__a())))
r2: f(X) -> n__f(X)
r3: a() -> n__a()
r4: activate(n__f(X)) -> f(activate(X))
r5: activate(n__a()) -> a()
r6: activate(X) -> X

The set of usable rules consists of

  (no rules)

Take the reduction pair:

  lexicographic combination of reduction pairs:
  
    1. matrix interpretations:
    
      carrier: N^2
      order: standard order
      interpretations:
        activate#_A(x1) = ((1,1),(1,1)) x1
        n__f_A(x1) = ((1,1),(1,1)) x1 + (1,1)
    
    2. matrix interpretations:
    
      carrier: N^2
      order: standard order
      interpretations:
        activate#_A(x1) = ((0,0),(1,0)) x1
        n__f_A(x1) = ((1,1),(0,0)) x1 + (1,1)
    

The next rules are strictly ordered:

  p1

We remove them from the problem.  Then no dependency pair remains.