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)
  , cons(X1, X2) -> n__cons(X1, X2)
  , 2ndspos(0(), Z) -> rnil()
  , 2ndspos(s(N), cons(X, n__cons(Y, Z))) ->
    rcons(posrecip(activate(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))
  , activate(n__cons(X1, X2)) -> cons(activate(X1), X2)
  , 2ndsneg(0(), Z) -> rnil()
  , 2ndsneg(s(N), cons(X, n__cons(Y, Z))) ->
    rcons(negrecip(activate(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, 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)))
, 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)
  , cons(X1, X2) -> n__cons(X1, X2)
  , 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))
  , activate(n__cons(X1, X2)) -> cons(activate(X1), X2)
  , 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(n__cons) = {1, 2},
 safe(rcons) = {1, 2}, safe(posrecip) = {1}, safe(activate) = {},
 safe(2ndsneg) = {}, safe(negrecip) = {1}, safe(pi) = {1},
 safe(plus) = {}, safe(times) = {1, 2}, safe(square) = {1}

and precedence

 from > cons, 2ndspos > cons, s > cons, activate > from,
 activate > cons, activate > s, 2ndsneg > cons, pi > from,
 pi > cons, pi > 2ndspos, pi > s, pi > 2ndsneg, pi > times,
 plus > from, plus > cons, plus > 2ndspos, plus > s, plus > 2ndsneg,
 plus > times, times > cons, square > from, square > cons,
 square > 2ndspos, square > s, square > 2ndsneg, square > times,
 from ~ s, 2ndspos ~ 2ndsneg, 2ndspos ~ times, activate ~ pi,
 activate ~ plus, activate ~ square, 2ndsneg ~ times, pi ~ plus,
 pi ~ square, plus ~ 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)                    
                                                                  
                cons(; X1,  X2) > n__cons(; X1,  X2)              
                                                                  
             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;))               
                                                                  
  activate(n__cons(; X1,  X2);) > cons(; activate(X1;),  X2)      
                                                                  
              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))