YES(O(1),O(n^1))

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

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(1),O(n^1))

We use the processor 'Small Polynomial Path Order (PS,1-bounded)'
to orient following rules strictly.

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) }

The induced complexity on above rules (modulo remaining rules) is
YES(?,O(n^1)) . These rules are moved into the corresponding weak
component(s).

Sub-proof:
----------
  The input was oriented with the instance of 'Small Polynomial Path
  Order (PS,1-bounded)' 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, add > dbl, first > sqr, terms ~ first .
  
  Following symbols are considered recursive:
  
   {terms, first}
  
  The recursion depth is 1.
  
  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)                
                                                     

We return to the main proof.

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

Weak 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(1),O(1))

Empty rules are trivially bounded

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