# Is the set of regular languages closed under set difference?

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?

• Your proof is correct and probably the easiest way to go about it.
– 5xum
Commented Aug 29, 2016 at 12:20
• 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). Commented Sep 1, 2016 at 17:46

## 1 Answer

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)