YES(?,O(n^3))

We are left with following problem, upon which TcT provides the
certificate YES(?,O(n^3)).

Strict Trs:
  { from(X) -> cons(X, n__from(s(X)))
  , from(X) -> n__from(X)
  , sel(s(X), cons(Y, Z)) -> sel(X, activate(Z))
  , sel(0(), cons(X, Y)) -> X
  , activate(X) -> X
  , activate(n__from(X)) -> from(X) }
Obligation:
  innermost runtime complexity
Answer:
  YES(?,O(n^3))

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

 safe(from) = {1}, safe(cons) = {1, 2}, safe(n__from) = {1},
 safe(s) = {1}, safe(sel) = {2}, safe(0) = {}, safe(activate) = {1}

and precedence

 sel > from, sel > activate, activate > from .

Following symbols are considered recursive:

 {from, sel, activate}

The recursion depth is 3.

For your convenience, here are the satisfied ordering constraints:

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

Hurray, we answered YES(?,O(n^3))