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:
  { first(0(), X) -> nil()
  , first(s(X), cons(Y)) -> cons(Y)
  , from(X) -> cons(X) }
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:
  { first(0(), X) -> nil()
  , first(s(X), cons(Y)) -> cons(Y)
  , from(X) -> cons(X) }

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(first) = {1, 2}, safe(0) = {}, safe(nil) = {}, safe(s) = {1},
   safe(cons) = {1}, safe(from) = {}
  
  and precedence
  
   first ~ from .
  
  Following symbols are considered recursive:
  
   {first}
  
  The recursion depth is 1.
  
  For your convenience, here are the satisfied ordering constraints:
  
               first(; 0(),  X) > nil()    
                                           
    first(; s(; X),  cons(; Y)) > cons(; Y)
                                           
                       from(X;) > cons(; X)
                                           

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:
  { first(0(), X) -> nil()
  , first(s(X), cons(Y)) -> cons(Y)
  , from(X) -> cons(X) }
Obligation:
  innermost runtime complexity
Answer:
  YES(O(1),O(1))

Empty rules are trivially bounded

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