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?
Best 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)