Suppose A∪B is NP-hard, does it follow that A is NP-hard or B is NP-hard?

Equivalently, is the property "not NP-hard" closed under union?

A related question was already asked:

Are NP-complete sets formed from two other sets only if at least one is NP-hard?

but it seems no consensus was formed.


1 Answer 1


If $P = NP$, then the only not-$NP$-hard sets are $\emptyset$ and $\Sigma^*$, and the statement trivially holds. So let us assume $P \neq NP$.

Then whenever $X$ is $NP$-hard, both $X$ and $X^C$ must contain an infinite c.e. set, namely "the image of the positive SAT-instances under the reduction" and "the image of the negative SAT-instances under the reduction". For if either of these sets were finite, we could get a P-algorithm for SAT.

By diagonalization, we can construct some set $X$ such that neither $X$ nor $X^C$ contain an infinite c.e.-set. Then $SAT \cap X$ and $SAT \cap X^C$ do not contain infinite c.e. sets, hence are both not $NP$-hard. But their union is SAT, and hence $NP$-hard.

To get a decidable counterexample, it should suffice to note that if $X$ is NP-hard, we not only have infinite c.e.~subsets of $X$ and $X^C$, but actually ones which can be computed in exponential time. Diagonalizing against those should doable in a decidable way; but I have not worked out the details.


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