YES

We show the termination of the TRS R:

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

and R consists of:

r1: f(n__f(n__a())) -> f(n__g(n__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(activate(X))
r8: activate(X) -> X

The estimated dependency graph contains the following SCCs:

  {p5}


-- Reduction pair.

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

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

and R consists of:

r1: f(n__f(n__a())) -> f(n__g(n__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(activate(X))
r8: activate(X) -> X

The set of usable rules consists of

  (no rules)

Take the monotone reduction pair:

  lexicographic combination of reduction pairs:
  
    1. matrix interpretations:
    
      carrier: N^1
      order: standard order
      interpretations:
        activate#_A(x1) = x1
        n__g_A(x1) = x1 + 1
    
    2. lexicographic path order with precedence:
    
      precedence:
      
        n__g > activate#
      
      argument filter:
    
        pi(activate#) = 1
        pi(n__g) = [1]
    

The next rules are strictly ordered:

  p1
  r1, r2, r3, r4, r5, r6, r7, r8

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