Is there any instance where a problem $A$ can be reduced to a problem $B$ where $B$ is easier to solve than $A$?
I've been learning about NP-Hardness recently and seems that the answer is no. Whenever we show $ A \leq_p B$ its also said that $B$ must therefore be at least as hard as $A$. If this is always the case then is reducing a problem into another only useful for proving NP-Hardness/NP-Completeness? Is there no way to leverage reductions to find an easier problem to solve than the one you originally start out with?