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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.. thnx2011-12-03
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    and under $P=NP$ only $\emptyset$ and $A^{\ast}$ are not NP-hard under polynomial time reductions.2011-12-03
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    @sdcvvc, not NP-hard you mean.2011-12-03
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    Yes, thanks for the correction.2011-12-03