YES(?,O(1))

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

Strict Trs:
  { cons(X1, X2) -> n__cons(X1, X2)
  , activate(X) -> X
  , activate(n__cons(X1, X2)) -> cons(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(1))

The input was oriented with the instance of 'Small Polynomial Path
Order (PS,0-bounded)' as induced by the safe mapping

 safe(2nd) = {}, safe(cons) = {1, 2}, safe(n__cons) = {1, 2},
 safe(activate) = {1}, safe(from) = {1}, safe(n__from) = {1},
 safe(s) = {1}

and precedence

 activate > cons, activate > from, from > cons .

Following symbols are considered recursive:

 {}

The recursion depth is 0.

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

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