We are left with following problem, upon which TcT provides the
certificate YES(?,O(n^1)).

Strict Trs:
  { f(X) -> n__f(X)
  , f(f(a())) -> f(g(n__f(n__a())))
  , a() -> n__a()
  , activate(X) -> X
  , activate(n__f(X)) -> f(activate(X))
  , activate(n__a()) -> a() }
Obligation:
  innermost runtime complexity
Answer:
  YES(?,O(n^1))

Arguments of following rules are not normal-forms:

{ f(f(a())) -> f(g(n__f(n__a()))) }

All above mentioned rules can be savely removed.

We are left with following problem, upon which TcT provides the
certificate YES(?,O(n^1)).

Strict Trs:
  { f(X) -> n__f(X)
  , a() -> n__a()
  , activate(X) -> X
  , activate(n__f(X)) -> f(activate(X))
  , activate(n__a()) -> a() }
Obligation:
  innermost runtime complexity
Answer:
  YES(?,O(n^1))

The input was oriented with the instance of 'Small Polynomial Path
Order (PS,1-bounded)' as induced by the safe mapping

 safe(f) = {1}, safe(a) = {}, safe(n__f) = {1}, safe(n__a) = {},
 safe(activate) = {}

and precedence

 activate > f, activate > a .

Following symbols are considered recursive:

 {activate}

The recursion depth is 1.

For your convenience, here are the satisfied ordering constraints:

                f(; X) > n__f(; X)        
                                          
                   a() > n__a()           
                                          
          activate(X;) > X                
                                          
  activate(n__f(; X);) > f(; activate(X;))
                                          
     activate(n__a();) > a()              
                                          

Hurray, we answered YES(?,O(n^1))