Let A be NP-complete and B be NP-hard. Can B be polynomial time reducible to A?

Question

Let A be NP-complete and B be NP-hard. Can B be polynomial time reducible to A?

Ans: I know it can't be. Would the strong reason be, because NP-Complete is a subset of NP-Hard?

Answer 1

Let's first look at naming conventions of NP Hard and NP Complete ( wikipedia ):

NP-hard
Class of decision problems which are at least as hard as the hardest problems in NP. Problems that are NP-hard do not have to be elements of NP; indeed, they may not even be decidable.

NP-complete
Class of decision problems which contains the hardest problems in NP. Each NP-complete problem has to be in NP.

NP-hard (B) are at least as hard as the hardest problem of NP .
Hardest problems of NP are NP-complete (A).

From these two statements, we can say that B is at least as hard as A .

In simpler terms, this means that any algorithm for B immediately gives an algorithm for A. But the inverse is not true, knowing how to solve A doesn't tell us anything about how to solve B. This relation is not symmetric.

This is why NP-hard is not reducible to NP-complete.