\begin{frame}{Decidability of Subsets} \begin{goal}{Theorem} It is decidable for regular languages $L_1$ and $L_2$ if $L_1 \subseteq L_2$. \end{goal} \pause \begin{proof} We have \begin{talign} L_1 \subseteq L_2 \quad\iff\quad \mpause[1]{ L_1 \setminus L_2 = \emptyset } \end{talign} \pause\pause The language $L_1 \setminus L_2$ is regular. \pause\medskip Finally, emptyness is decidable. \end{proof} \end{frame}