YES(?,O(n^1))

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

Strict Trs:
  { f(X1, X2) -> n__f(X1, X2)
  , f(g(X), Y) -> f(X, n__f(n__g(X), activate(Y)))
  , g(X) -> n__g(X)
  , activate(X) -> X
  , activate(n__f(X1, X2)) -> f(activate(X1), X2)
  , activate(n__g(X)) -> g(activate(X)) }
Obligation:
  innermost runtime complexity
Answer:
  YES(?,O(n^1))

Arguments of following rules are not normal-forms:

{ f(g(X), Y) -> f(X, n__f(n__g(X), activate(Y))) }

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(X1, X2) -> n__f(X1, X2)
  , g(X) -> n__g(X)
  , activate(X) -> X
  , activate(n__f(X1, X2)) -> f(activate(X1), X2)
  , activate(n__g(X)) -> g(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, 2}, safe(g) = {1}, safe(n__f) = {1, 2},
 safe(n__g) = {1}, safe(activate) = {}

and precedence

 activate > f, activate > g, f ~ g .

Following symbols are considered recursive:

 {activate}

The recursion depth is 1.

For your convenience, here are the satisfied ordering constraints:

                f(; X1,  X2) > n__f(; X1,  X2)        
                                                      
                      g(; X) > n__g(; X)              
                                                      
                activate(X;) > X                      
                                                      
  activate(n__f(; X1,  X2);) > f(; activate(X1;),  X2)
                                                      
        activate(n__g(; X);) > g(; activate(X;))      
                                                      

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