YES

We show the termination of the TRS R:

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

-- SCC decomposition.

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

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

and R consists of:

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

The estimated dependency graph contains the following SCCs:

  {p2, p4}


-- Reduction pair.

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

p1: activate#(n__f(X)) -> activate#(X)
p2: activate#(n__g(X)) -> activate#(X)

and R consists of:

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

The set of usable rules consists of

  (no rules)

Take the monotone reduction pair:

  lexicographic combination of reduction pairs:
  
    1. lexicographic path order with precedence:
    
      precedence:
      
        activate# > n__g > n__f
      
      argument filter:
    
        pi(activate#) = 1
        pi(n__f) = 1
        pi(n__g) = 1
    
    2. lexicographic path order with precedence:
    
      precedence:
      
        activate# > n__g > n__f
      
      argument filter:
    
        pi(activate#) = 1
        pi(n__f) = [1]
        pi(n__g) = [1]
    

The next rules are strictly ordered:

  p1, p2
  r1, r2, r3, r4, r5, r6, r7, r8

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