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

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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.

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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
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@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

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