YES(?,O(n^1))

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

Strict Trs:
  { 2nd(cons(X, n__cons(Y, Z))) -> activate(Y)
  , cons(X1, X2) -> n__cons(X1, X2)
  , activate(X) -> X
  , activate(n__cons(X1, X2)) -> cons(activate(X1), X2)
  , activate(n__from(X)) -> from(activate(X))
  , activate(n__s(X)) -> s(activate(X))
  , from(X) -> cons(X, n__from(n__s(X)))
  , from(X) -> n__from(X)
  , s(X) -> n__s(X) }
Obligation:
  innermost runtime complexity
Answer:
  YES(?,O(n^1))

Arguments of following rules are not normal-forms:

{ 2nd(cons(X, n__cons(Y, Z))) -> 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:
  { cons(X1, X2) -> n__cons(X1, X2)
  , activate(X) -> X
  , activate(n__cons(X1, X2)) -> cons(activate(X1), X2)
  , activate(n__from(X)) -> from(activate(X))
  , activate(n__s(X)) -> s(activate(X))
  , from(X) -> cons(X, n__from(n__s(X)))
  , from(X) -> n__from(X)
  , s(X) -> n__s(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(2nd) = {}, safe(cons) = {1, 2}, safe(n__cons) = {1, 2},
 safe(activate) = {}, safe(from) = {1}, safe(n__from) = {1},
 safe(n__s) = {1}, safe(s) = {1}

and precedence

 activate > cons, activate > from, activate > s, from > cons,
 s > cons, from ~ s .

Following symbols are considered recursive:

 {activate}

The recursion depth is 1.

For your convenience, here are the satisfied ordering constraints:

                cons(; X1,  X2) > n__cons(; X1,  X2)              
                                                                  
                   activate(X;) > X                               
                                                                  
  activate(n__cons(; X1,  X2);) > cons(; activate(X1;),  X2)      
                                                                  
        activate(n__from(; X);) > from(; activate(X;))            
                                                                  
           activate(n__s(; X);) > s(; activate(X;))               
                                                                  
                      from(; X) > cons(; X,  n__from(; n__s(; X)))
                                                                  
                      from(; X) > n__from(; X)                    
                                                                  
                         s(; X) > n__s(; X)                       
                                                                  

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