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\begin{frame}{Decidability of Subsets}
  \begin{block}{Theorem}
    It is decidable for regular languages $L_1$ and $L_2$ if $L_1 \subseteq L_2$.
  \end{block}
  \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}