Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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

share|cite|improve this question

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.

share|cite|improve this answer
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

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.