YES

We show the termination of the TRS R:

  f(X,X) -> f(a(),n__b())
  b() -> a()
  b() -> n__b()
  activate(n__b()) -> b()
  activate(X) -> X

-- SCC decomposition.

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

p1: f#(X,X) -> f#(a(),n__b())
p2: activate#(n__b()) -> b#()

and R consists of:

r1: f(X,X) -> f(a(),n__b())
r2: b() -> a()
r3: b() -> n__b()
r4: activate(n__b()) -> b()
r5: activate(X) -> X

The estimated dependency graph contains the following SCCs:

  (no SCCs)