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Let $\Sigma$ be an alphabet and $L_1,L_2 \subseteq \Sigma^*$ two regular languages.

I know that $REG$ is closed under intersections of regular languages and under complementation of a regular language. My reasoning looks like this:

$L_1 \setminus L_2 = L_1 \cap \overline L_2$

Hence, we have that $REG$ is closed under set differentiation.

Have I overlooked something?

Ștefan
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    Your proof is correct and probably the easiest way to go about it. – 5xum Aug 29 '16 at 12:20
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    Note that your argument is not specific to regular languages: if a class of subsets of a set is closed under intersection and complementation, then it is also closed under set difference (and under union). – J.-E. Pin Sep 01 '16 at 17:46

1 Answers1

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Your proof is correct and probably the easiest way to go about it.

Note that your argument is not specific to regular languages: if a class of subsets of a set is closed under intersection and complementation, then it is also closed under set difference (and under union)

wece
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