This was one of my exam questions and the answer is apparently no.
Can someone explain why because I don't understand.
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$\begingroup$ @Jut, please do not use the comments here to advertise one of your other posts. That is not an appropriate use of comments. $\endgroup$– D.W. ♦Dec 20, 2022 at 6:14
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1 Answer
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There exist languages which are Non-RE and their complement is also Non-RE, let $L$ be such a language we can say that
$L \cup L^{c} = \Sigma^{*}$
Here both languages are Non-RE but their union is $\Sigma^{*}$ which is regular as well as RE also.
