YES

We show the termination of the TRS R:

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

-- SCC decomposition.

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

p1: f#(n__f(n__a())) -> f#(n__g(f(n__a())))
p2: f#(n__f(n__a())) -> f#(n__a())
p3: activate#(n__f(X)) -> f#(X)
p4: activate#(n__a()) -> a#()
p5: activate#(n__g(X)) -> g#(X)

and R consists of:

r1: f(n__f(n__a())) -> f(n__g(f(n__a())))
r2: f(X) -> n__f(X)
r3: a() -> n__a()
r4: g(X) -> n__g(X)
r5: activate(n__f(X)) -> f(X)
r6: activate(n__a()) -> a()
r7: activate(n__g(X)) -> g(X)
r8: activate(X) -> X

The estimated dependency graph contains the following SCCs:

  (no SCCs)