Prove/Disprove: $L_1, L_2 \in RE-R \implies L_1 \cup L_2 \notin R$ Prove/Disprove: $L_1, L_2 \in RE-R \implies L_1 \cup L_2 \notin R$


My first intuition is "Yes", since we may look at $M_1, M_2$ which accepts $L_1, L_2$, respectively. Then, WLOG there's $w$ such that $M_1$ doesn't halt for, and so the machine $M$ which runs $M_1, M_2$ in parallel, may not halt.
Hint: Let $L$ be any language in $\mathsf{RE} - \mathsf{R}$, and consider
$$ L_1 = 0\Sigma^* \cup 1L, \quad L_2 = 1\Sigma^* \cup 0L. $$