YES(?,O(n^1))

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

Strict Trs:
  { from(X) -> cons(X, n__from(n__s(X)))
  , from(X) -> n__from(X)
  , 2ndspos(0(), Z) -> rnil()
  , 2ndspos(s(N), cons(X, Z)) -> 2ndspos(s(N), cons2(X, activate(Z)))
  , 2ndspos(s(N), cons2(X, cons(Y, Z))) ->
    rcons(posrecip(Y), 2ndsneg(N, activate(Z)))
  , s(X) -> n__s(X)
  , activate(X) -> X
  , activate(n__from(X)) -> from(activate(X))
  , activate(n__s(X)) -> s(activate(X))
  , 2ndsneg(0(), Z) -> rnil()
  , 2ndsneg(s(N), cons(X, Z)) -> 2ndsneg(s(N), cons2(X, activate(Z)))
  , 2ndsneg(s(N), cons2(X, cons(Y, Z))) ->
    rcons(negrecip(Y), 2ndspos(N, activate(Z)))
  , pi(X) -> 2ndspos(X, from(0()))
  , plus(0(), Y) -> Y
  , plus(s(X), Y) -> s(plus(X, Y))
  , times(0(), Y) -> 0()
  , times(s(X), Y) -> plus(Y, times(X, Y))
  , square(X) -> times(X, X) }
Obligation:
  innermost runtime complexity
Answer:
  YES(?,O(n^1))

Arguments of following rules are not normal-forms:

{ 2ndspos(s(N), cons(X, Z)) -> 2ndspos(s(N), cons2(X, activate(Z)))
, 2ndspos(s(N), cons2(X, cons(Y, Z))) ->
  rcons(posrecip(Y), 2ndsneg(N, activate(Z)))
, 2ndsneg(s(N), cons(X, Z)) -> 2ndsneg(s(N), cons2(X, activate(Z)))
, 2ndsneg(s(N), cons2(X, cons(Y, Z))) ->
  rcons(negrecip(Y), 2ndspos(N, activate(Z)))
, plus(s(X), Y) -> s(plus(X, Y))
, times(s(X), Y) -> plus(Y, times(X, 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:
  { from(X) -> cons(X, n__from(n__s(X)))
  , from(X) -> n__from(X)
  , 2ndspos(0(), Z) -> rnil()
  , s(X) -> n__s(X)
  , activate(X) -> X
  , activate(n__from(X)) -> from(activate(X))
  , activate(n__s(X)) -> s(activate(X))
  , 2ndsneg(0(), Z) -> rnil()
  , pi(X) -> 2ndspos(X, from(0()))
  , plus(0(), Y) -> Y
  , times(0(), Y) -> 0()
  , square(X) -> times(X, 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(from) = {1}, safe(cons) = {1, 2}, safe(n__from) = {1},
 safe(n__s) = {1}, safe(2ndspos) = {1, 2}, safe(0) = {},
 safe(rnil) = {}, safe(s) = {1}, safe(cons2) = {1, 2},
 safe(activate) = {}, safe(rcons) = {1, 2}, safe(posrecip) = {1},
 safe(2ndsneg) = {}, safe(negrecip) = {1}, safe(pi) = {1},
 safe(plus) = {}, safe(times) = {1, 2}, safe(square) = {1}

and precedence

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

Following symbols are considered recursive:

 {2ndspos, activate, times}

The recursion depth is 1.

For your convenience, here are the satisfied ordering constraints:

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

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