This action confuses me for a while. As you know, once we come up with a problem and we want to know how hard it is, we will do the "reduction". Intuitively, if we prove problem B is as hard as problem A, we can know the complexity boundary of B. But if problem B is harder than A, why we call the reduction process -- reduce problem A to problem B, not reduce problem B to A?
There are several different types of reduction, but generally, if we say we reduce problem A to problem B, it means something like: if we had a way to solve B, we could use that to solve A. Some reductions may allow to use B as a black box. Others might require that "yes" instances of A map to "yes" instances of B, and that "no" instances map as well. Or, it may be something simpler, like "hey, I already have a great solution to problem B, and now with just a simple transform, I can use it to solve problem A." In all forms, it implies something related to "we could solve A if we had solutions to B".
What that means has something to do with what is known. If B is an easy problem to solve, then A reducing to B tells us that solving A is no harder than solving B, plus doing the reduction from A to B. In that case, an easy reduction, to an easy problem, proves that A is easy.
On the other hand, especially in the context of hardness results, we would usually reduce a known difficult problem (A here) to B, and that would show that, because A is hard, but transforming an instance of A into B is easy, B must also be hard, because if B were easy, it would give a solution to the known A problem.
So basically, for an easy reduction from A to B, it proves 2 things: A is not harder than B, and B is not easier than A. Yes, those are the same thing, but usually, you only care about one of them, depending on whether you know A to be difficult, or B to be easy.