# Prove or disprove, If A ≤p B and B is NP-hard, then A is in NP-hard

Intuitively if A can reduce to B, and B is NP-Hard, A might be NP Hard but maybe not. If there is a way to solve A that does not involve reducing to B, it might be faster.

How do I formally disprove this statement using a counter example or some well known algorithms, or using some other technique?

Let A be the empty language. It is trivially reducible to, say 3SAT: transform any input into the formula $$a \land \lnot a$$. But only the empty language is reducible to the empty language, thus A is not NP-hard.