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

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

The input is overlay and right-linear. Switching to innermost
rewriting.

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

Strict Trs:
  { f(X) -> g(n__h(n__f(X)))
  , f(X) -> n__f(X)
  , h(X) -> n__h(X)
  , activate(X) -> X
  , activate(n__h(X)) -> h(activate(X))
  , activate(n__f(X)) -> f(activate(X)) }
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(g) = {1}, safe(n__h) = {1}, safe(n__f) = {1},
 safe(h) = {1}, safe(activate) = {}

and precedence

 activate > f, activate > h .

Following symbols are considered recursive:

 {activate}

The recursion depth is 1.

For your convenience, here are the satisfied ordering constraints:

                f(; X) > g(; n__h(; n__f(; X)))
                                               
                f(; X) > n__f(; X)             
                                               
                h(; X) > n__h(; X)             
                                               
          activate(X;) > X                     
                                               
  activate(n__h(; X);) > h(; activate(X;))     
                                               
  activate(n__f(; X);) > f(; activate(X;))     
                                               

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