YES(?,O(n^2))

We are left with following problem, upon which TcT provides the
certificate YES(?,O(n^2)).

Strict Trs:
  { from(X) -> cons(X, n__from(s(X)))
  , from(X) -> n__from(X)
  , cons(X1, X2) -> n__cons(X1, X2)
  , 2ndspos(s(N), cons(X, n__cons(Y, Z))) ->
    rcons(posrecip(activate(Y)), 2ndsneg(N, activate(Z)))
  , 2ndspos(0(), Z) -> rnil()
  , activate(X) -> X
  , activate(n__from(X)) -> from(X)
  , activate(n__cons(X1, X2)) -> cons(X1, X2)
  , 2ndsneg(s(N), cons(X, n__cons(Y, Z))) ->
    rcons(negrecip(activate(Y)), 2ndspos(N, activate(Z)))
  , 2ndsneg(0(), Z) -> rnil()
  , pi(X) -> 2ndspos(X, from(0()))
  , plus(s(X), Y) -> s(plus(X, Y))
  , plus(0(), Y) -> Y
  , times(s(X), Y) -> plus(Y, times(X, Y))
  , times(0(), Y) -> 0()
  , square(X) -> times(X, X) }
Obligation:
  innermost runtime complexity
Answer:
  YES(?,O(n^2))

Arguments of following rules are not normal-forms:

{ 2ndspos(s(N), cons(X, n__cons(Y, Z))) ->
  rcons(posrecip(activate(Y)), 2ndsneg(N, activate(Z)))
, 2ndsneg(s(N), cons(X, n__cons(Y, Z))) ->
  rcons(negrecip(activate(Y)), 2ndspos(N, 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^2)).

Strict Trs:
  { from(X) -> cons(X, n__from(s(X)))
  , from(X) -> n__from(X)
  , cons(X1, X2) -> n__cons(X1, X2)
  , 2ndspos(0(), Z) -> rnil()
  , activate(X) -> X
  , activate(n__from(X)) -> from(X)
  , activate(n__cons(X1, X2)) -> cons(X1, X2)
  , 2ndsneg(0(), Z) -> rnil()
  , pi(X) -> 2ndspos(X, from(0()))
  , plus(s(X), Y) -> s(plus(X, Y))
  , plus(0(), Y) -> Y
  , times(s(X), Y) -> plus(Y, times(X, Y))
  , times(0(), Y) -> 0()
  , square(X) -> times(X, X) }
Obligation:
  innermost runtime complexity
Answer:
  YES(?,O(n^2))

The input was oriented with the instance of 'Small Polynomial Path
Order (PS,2-bounded)' as induced by the safe mapping

 safe(from) = {1}, safe(cons) = {1, 2}, safe(n__from) = {1},
 safe(s) = {1}, safe(2ndspos) = {1, 2}, safe(0) = {},
 safe(rnil) = {}, safe(n__cons) = {1, 2}, safe(rcons) = {1, 2},
 safe(posrecip) = {1}, safe(activate) = {1}, safe(2ndsneg) = {},
 safe(negrecip) = {1}, safe(pi) = {}, safe(plus) = {2},
 safe(times) = {}, safe(square) = {}

and precedence

 from > cons, from > plus, activate > from, activate > cons,
 activate > 2ndspos, activate > plus, 2ndsneg > from,
 2ndsneg > cons, 2ndsneg > plus, 2ndsneg > times, pi > from,
 pi > cons, pi > 2ndspos, pi > plus, times > cons, times > plus,
 square > from, square > cons, square > 2ndspos, square > 2ndsneg,
 square > plus, square > times, cons ~ plus, activate ~ pi .

Following symbols are considered recursive:

 {cons, plus, times}

The recursion depth is 2.

For your convenience, here are the satisfied ordering constraints:

                       from(; X) > cons(; X,  n__from(; s(; X)))
                                                                
                       from(; X) > n__from(; X)                 
                                                                
                 cons(; X1,  X2) > n__cons(; X1,  X2)           
                                                                
              2ndspos(; 0(),  Z) > rnil()                       
                                                                
                   activate(; X) > X                            
                                                                
        activate(; n__from(; X)) > from(; X)                    
                                                                
  activate(; n__cons(; X1,  X2)) > cons(; X1,  X2)              
                                                                
               2ndsneg(0(),  Z;) > rnil()                       
                                                                
                          pi(X;) > 2ndspos(; X,  from(; 0()))   
                                                                
                 plus(s(; X); Y) > s(; plus(X; Y))              
                                                                
                    plus(0(); Y) > Y                            
                                                                
              times(s(; X),  Y;) > plus(Y; times(X,  Y;))       
                                                                
                 times(0(),  Y;) > 0()                          
                                                                
                      square(X;) > times(X,  X;)                
                                                                

Hurray, we answered YES(?,O(n^2))