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Let's assume that $L_1 = a^nb^{2n}$ and $L_2 = a^na^{2n}$, knowing that $L_1$ is not regular, and $L_2$ is. We also know that regular languages are closed under intersection and union, and complement. What can we say about $L_3 = L_1^\complement$ and $L_4 = L_2^ \complement$?

It seems correct to say $L_4$ is regular, because $L_2$ is, but since $L_1$ is not regular, can we say about $L_3$ the same?

Also, let's assume we have two irregular languages $L_5$ and $L_6$, can we say about their intersection and union as being irregular as well?

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    $\begingroup$ Are you allowed to refer to the closure property that regular languages are closed under complement? $\endgroup$
    – kviiri
    Dec 21, 2022 at 11:58
  • $\begingroup$ Yes, I am allowed $\endgroup$
    – Papa
    Dec 21, 2022 at 12:01
  • $\begingroup$ Please ask only one question per post. $\endgroup$
    – D.W.
    Dec 21, 2022 at 23:52

1 Answer 1

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Since $L_3$ is the complement of $L_1$, $L_1$ is likewise the complement of $L_3$. If $L_3$ was regular, what would that imply regarding the regularity of $L_1$?

For $L_5$ and $L_6$, consider for example the case where the two languages are each others' complements. Can you determine what languages result from the union and intersection of a language and its complement? Are these languages regular?

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  • $\begingroup$ For the last question, the resulting is $L \cap L^\complement = \epsilon$ which is regular. But you re missing something, we only know this: if $L$ is regular, and $L_1$ is regular, then their intersection or union is regular, we can have regular intersection irregular give regular or non regular, and we can have the intersection of two non regular languages give a regular or a non regular. My question is simple, and you repeated my questions $\endgroup$
    – Papa
    Dec 21, 2022 at 12:35
  • $\begingroup$ I am trying to get my facts checked with from someone an expert, that s why I came here $\endgroup$
    – Papa
    Dec 21, 2022 at 12:39

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