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

Strict Trs:
  { f(X) -> n__f(X)
  , f(f(X)) -> c(n__f(g(n__f(X))))
  , c(X) -> d(activate(X))
  , d(X) -> n__d(X)
  , activate(X) -> X
  , activate(n__f(X)) -> f(X)
  , activate(n__d(X)) -> d(X)
  , h(X) -> c(n__d(X)) }
Obligation:
  innermost runtime complexity
Answer:
  YES(?,O(1))

Arguments of following rules are not normal-forms:

{ f(f(X)) -> c(n__f(g(n__f(X)))) }

All above mentioned rules can be savely removed.

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

Strict Trs:
  { f(X) -> n__f(X)
  , c(X) -> d(activate(X))
  , d(X) -> n__d(X)
  , activate(X) -> X
  , activate(n__f(X)) -> f(X)
  , activate(n__d(X)) -> d(X)
  , h(X) -> c(n__d(X)) }
Obligation:
  innermost runtime complexity
Answer:
  YES(?,O(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(c) = {1}, safe(n__f) = {1}, safe(d) = {1},
 safe(activate) = {1}, safe(h) = {}, safe(n__d) = {1}

and precedence

 c > d, c > activate, activate > f, activate > d, h > c .

Following symbols are considered recursive:

 {}

The recursion depth is 0.

For your convenience, here are the satisfied ordering constraints:

                 f(; X) > n__f(; X)         
                                            
                 c(; X) > d(; activate(; X))
                                            
                 d(; X) > n__d(; X)         
                                            
          activate(; X) > X                 
                                            
  activate(; n__f(; X)) > f(; X)            
                                            
  activate(; n__d(; X)) > d(; X)            
                                            
                  h(X;) > c(; n__d(; X))    
                                            

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