% \begin{frame}{$L^R$ is Regular}
%   \begin{block}{Construction 3}
%     Reverse a regular expression (defined by recursion):
%     \begin{talign}
%       \begin{array}{lcl}
%         {\it rev}(\emptyset) &=& \alert{\emptyset} \\
%         {\it rev}(\lambda) &=& \alert{\lambda} \\
%         {\it rev}(a) &=& \alert{a} \hspace*{3cm} (a \in \Sigma) \\
%         {\it rev}(r_1 + r_2) &=& \alert{{\it rev}(r_1) + {\it rev}(r_2)} \\
%         {\it rev}(r_1 \cdot r_2) &=& \alert{{\it rev}(r_2) \cdot {\it rev}(r_1)} \\
%         {\it rev}(r^*) &=& \alert{{\it rev}(r)^*}
%       \end{array}
%     \end{talign}
%     Using induction, it can be shown that
%     \begin{talign}
%       \alert{L({\it rev}(r)) ~=~ L(r)^R}
%     \end{talign}
%   \end{block}
% \end{frame}

\themex{Decidable Properties of Regular Languages}