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

Strict Trs:
  { from(X) -> cons(X, n__from(n__s(X)))
  , from(X) -> n__from(X)
  , sel(0(), cons(X, Y)) -> X
  , sel(s(X), cons(Y, Z)) -> sel(X, activate(Z))
  , s(X) -> n__s(X)
  , activate(X) -> X
  , activate(n__from(X)) -> from(activate(X))
  , activate(n__s(X)) -> s(activate(X)) }
Obligation:
  innermost runtime complexity
Answer:
  YES(?,O(n^1))

Arguments of following rules are not normal-forms:

{ sel(s(X), cons(Y, Z)) -> sel(X, activate(Z)) }

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:
  { from(X) -> cons(X, n__from(n__s(X)))
  , from(X) -> n__from(X)
  , sel(0(), cons(X, Y)) -> X
  , s(X) -> n__s(X)
  , activate(X) -> X
  , activate(n__from(X)) -> from(activate(X))
  , activate(n__s(X)) -> s(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(from) = {1}, safe(cons) = {1, 2}, safe(n__from) = {1},
 safe(n__s) = {1}, safe(sel) = {}, safe(0) = {}, safe(s) = {1},
 safe(activate) = {}

and precedence

 activate > from, activate > s .

Following symbols are considered recursive:

 {activate}

The recursion depth is 1.

For your convenience, here are the satisfied ordering constraints:

                  from(; X) > cons(; X,  n__from(; n__s(; X)))
                                                              
                  from(; X) > n__from(; X)                    
                                                              
  sel(0(),  cons(; X,  Y);) > X                               
                                                              
                     s(; X) > n__s(; X)                       
                                                              
               activate(X;) > X                               
                                                              
    activate(n__from(; X);) > from(; activate(X;))            
                                                              
       activate(n__s(; X);) > s(; activate(X;))               
                                                              

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