Is it true that if $S\in\left(NP\bigcup coNP\right)$ then $\overline{S}\in NP\bigcap coNP$?

I couldn't find any answer to that question.

My attempt at proving it:

If $S\in\left(NP\bigcup coNP\right)$, then, by set theory, $S=A\bigcup B$ for some $A\in NP\wedge B\in coNP$.

Therefore the complement of $S$ is $\overline{S}=\overline{A\bigcup B}=\overline{A}\bigcap\overline{B}\in NP\bigcap coNP$


1 Answer 1


Your attempt isn't correct.

There is a distinction between $S\in (NP\cup co-NP)$, and $S\in (NP\lor co-NP):=\{A\cup B|A\in NP, B\in co-NP\}$

You assumed both are the same, however the first talks about $S$ being in either $NP$ or $co-NP$, while the second talks about $S$ being a union of something from $NP$ and something else from $co-NP$.

About the question in title, we actually don't know. If $NP\neq co-NP$, then there is some $L\in NP\setminus co-NP$ (or the other way around), which means that $L\in (NP\cup co-NP)$ but $\overline{L}\notin (NP\cap co-NP)$ (since $L\notin NP$ then $\overline{L}\notin co-NP$)

But if $NP=co-NP$, then clearly what you stated would be trivially true.

So essentially, what you tried to prove is equivalent to the open problem "Is $NP=co-NP$ or not?".


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