YES(?,O(n^1))

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

Strict Trs:
  { first(X1, X2) -> n__first(X1, X2)
  , first(0(), X) -> nil()
  , first(s(X), cons(Y, Z)) -> cons(Y, n__first(X, activate(Z)))
  , activate(X) -> X
  , activate(n__first(X1, X2)) -> first(X1, X2)
  , activate(n__from(X)) -> from(X)
  , from(X) -> cons(X, n__from(s(X)))
  , from(X) -> n__from(X) }
Obligation:
  innermost runtime complexity
Answer:
  YES(?,O(n^1))

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

 safe(first) = {1}, safe(0) = {}, safe(nil) = {}, safe(s) = {1},
 safe(cons) = {1, 2}, safe(n__first) = {1, 2}, safe(activate) = {},
 safe(from) = {1}, safe(n__from) = {1}

and precedence

 first ~ activate, first ~ from, activate ~ from .

Following symbols are considered recursive:

 {first, activate, from}

The recursion depth is 1.

For your convenience, here are the satisfied ordering constraints:

                   first(X2; X1) > n__first(; X1,  X2)                     
                                                                           
                   first(X; 0()) > nil()                                   
                                                                           
    first(cons(; Y,  Z); s(; X)) > cons(; Y,  n__first(; X,  activate(Z;)))
                                                                           
                    activate(X;) > X                                       
                                                                           
  activate(n__first(; X1,  X2);) > first(X2; X1)                           
                                                                           
         activate(n__from(; X);) > from(; X)                               
                                                                           
                       from(; X) > cons(; X,  n__from(; s(; X)))           
                                                                           
                       from(; X) > n__from(; X)                            
                                                                           

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