# Proof by reduction clarification

If I execute a proof by reduction using a given adversary A as a subroutine to A'. How do I know that using adversary A which can solve a given hard subproblem, is the most efficient subroutine to use to solve the larger problem that A' solves?

in reduction you have two problems A and B, A$\leq$B means that if you can solve problem B then you can solve problem A , now in most reductions ,the level of efficiently for the solver for A that you construct is the same as B unless you add more steps. so this means you cant know if its the most efficient , it may be and it may not.
You don't, unless you specify that $A$ is the most efficient one.