# Is there any decidable problem that is NOT NP-HARD?

Is there a proof that there exists a decidable problem that is NOT NP-HARD??

-

## 1 Answer

Very simple answer:

Since you need one $x \in A$, and one $x \not \in A$ for a polynomial time reduction, $A = \emptyset$ cannot be a hard language for NP.

-
haha... wonderful.. thnx –  rakesh Dec 3 '11 at 16:41
and under $P=NP$ only $\emptyset$ and $A^{\ast}$ are not NP-hard under polynomial time reductions. –  sdcvvc Dec 3 '11 at 17:27
@sdcvvc, not NP-hard you mean. –  sxd Dec 3 '11 at 17:32
Yes, thanks for the correction. –  sdcvvc Dec 3 '11 at 17:32