YES(?,O(n^2))

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

Strict Trs:
  { terms(N) -> cons(recip(sqr(N)))
  , sqr(0()) -> 0()
  , sqr(s()) -> s()
  , dbl(0()) -> 0()
  , dbl(s()) -> s()
  , add(0(), X) -> X
  , add(s(), Y) -> s()
  , first(0(), X) -> nil()
  , first(s(), cons(Y)) -> cons(Y) }
Obligation:
  innermost runtime complexity
Answer:
  YES(?,O(n^2))

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

 safe(terms) = {1}, safe(cons) = {1}, safe(recip) = {1},
 safe(sqr) = {1}, safe(0) = {}, safe(s) = {}, safe(dbl) = {},
 safe(add) = {}, safe(first) = {1}, safe(nil) = {}

and precedence

 terms > sqr, terms > add, dbl > sqr, dbl > add, add > sqr,
 first > sqr, first > add, terms ~ dbl, terms ~ first, dbl ~ first .

Following symbols are considered recursive:

 {terms, sqr}

The recursion depth is 2.

For your convenience, here are the satisfied ordering constraints:

             terms(; N) > cons(; recip(; sqr(; N)))
                                                   
             sqr(; 0()) > 0()                      
                                                   
             sqr(; s()) > s()                      
                                                   
              dbl(0();) > 0()                      
                                                   
              dbl(s();) > s()                      
                                                   
          add(0(),  X;) > X                        
                                                   
          add(s(),  Y;) > s()                      
                                                   
          first(X; 0()) > nil()                    
                                                   
  first(cons(; Y); s()) > cons(; Y)                
                                                   

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