\documentclass{article}\usepackage{charter}\usepackage{hyperref}\usepackage{amssymb}\usepackage{amsmath}\newcommand{\dn}{\stackrel{\mbox{\scriptsize def}}{=}}% for definitions\begin{document}\section*{Proof}Recall the definitions for regular expressions and the language associated with a regular expression:\begin{center}\begin{tabular}{c}\begin{tabular}[t]{rcl} $r$ & $::=$ & $\varnothing$ \\ & $\mid$ & $\epsilon$ \\ & $\mid$ & $c$ \\ & $\mid$ & $r_1 \cdot r_2$ \\ & $\mid$ & $r_1 + r_2$ \\ & $\mid$ & $r^*$ \\ \end{tabular}\hspace{10mm}\begin{tabular}[t]{r@{\hspace{1mm}}c@{\hspace{1mm}}l}$L(\varnothing)$ & $\dn$ & $\varnothing$ \\$L(\epsilon)$ & $\dn$ & $\{\texttt{""}\}$ \\$L(c)$ & $\dn$ & $\{\texttt{"}c\texttt{"}\}$ \\$L(r_1 \cdot r_2)$ & $\dn$ & $L(r_1) \,@\, L(r_2)$ \\$L(r_1 + r_2)$ & $\dn$ & $L(r_1) \cup L(r_2)$ \\ $L(r^*)$ & $\dn$ & $\bigcup_{n\ge 0} L(r)^n$ \\ \end{tabular}\end{tabular}\end{center}\noindentWe also defined the notion of a derivative of a regular expression (the derivative with respect to a character):\begin{center}\begin{tabular}{lcl} $der\, c\, (\varnothing)$ & $\dn$ & $\varnothing$ \\ $der\, c\, (\epsilon)$ & $\dn$ & $\varnothing$ \\ $der\, c\, (d)$ & $\dn$ & if $c = d$ then $\epsilon$ else $\varnothing$\\ $der\, c\, (r_1 + r_2)$ & $\dn$ & $(der\, c\, r_1) + (der\, c\, r_2)$ \\ $der\, c\, (r_1 \cdot r_2)$ & $\dn$ & if $nullable(r_1)$\\ & & then $((der\, c\, r_1) \cdot r_2) + (der\, c\, r_2)$\\ & & else $(der\, c\, r_1) \cdot r_2$\\ $der\, c\, (r^*)$ & $\dn$ & $(der\, c\, r) \cdot (r^*)$\\ \end{tabular}\end{center}\noindentWith our definition of regular expressions comes an induction principle. Given a property $P$ over regular expressions. We can establish that $\forall r.\; P(r)$ holds, provided we can show the following:\begin{enumerate}\item $P(\varnothing)$, $P(\epsilon)$ and $P(c)$ all hold,\item $P(r_1 + r_2)$ holds under the induction hypotheses that $P(r_1)$ and $P(r_2)$ hold,\item $P(r_1 \cdot r_2)$ holds under the induction hypotheses that $P(r_1)$ and $P(r_2)$ hold, and\item $P(r^*)$ holds under the induction hypothesis that $P(r)$ holds.\end{enumerate}\noindentLet us try out an induction proof. Recall the definition\begin{center}$Der\, c\, A \dn \{ s\;\mid\; c\!::\!s \in A\}$\end{center}\noindentwhereby $A$ is a set of strings. We like to prove\begin{center}\begin{tabular}{l}$P(r) \dn $ \hspace{4mm} $L(der\,c\,r) = Der\,c\,(L(r))$\end{tabular}\end{center}\noindentby induction over the regular expression $r$.\newpage\noindent{\bf Proof}\noindentAccording to 1.~above we need to prove $P(\varnothing)$, $P(\epsilon)$ and $P(d)$. Lets do this in turn.\begin{itemize}\item First Case: $P(\varnothing)$ is $L(der\,c\,\varnothing) = Der\,c\,(L(\varnothing))$ (a). We have $der\,c\,\varnothing = \varnothing$ and $L(\varnothing) = \varnothing$. We also have $Der\,c\,\varnothing = \varnothing$. Hence we have $\varnothing = \varnothing$ in (a). \item Second Case: $P(\epsilon)$ is $L(der\,c\,\epsilon) = Der\,c\,(L(\epsilon))$ (b). We have $der\,c\,\epsilon = \varnothing$,$L(\varnothing) = \varnothing$ and $L(\epsilon) = \{\texttt{""}\}$. We also have $Der\,c\,\{\texttt{""}\} = \varnothing$. Hence we have $\varnothing = \varnothing$ in (b). \item Third Case: $P(d)$ is $L(der\,c\,d) = Der\,c\,(L(d))$ (c). We need to treat the cases $d = c$ and $d \not= c$. $d = c$: We have $der\,c\,c = \epsilon$ and $L(\epsilon) = \{\texttt{""}\}$. We also have $L(c) = \{\texttt{"}c\texttt{"}\}$ and $Der\,c\,\{\texttt{"}c\texttt{"}\} = \{\texttt{""}\}$. Hence we have $\{\texttt{""}\} = \{\texttt{""}\}$ in (c). $d \not=c$: We have $der\,c\,d = \varnothing$.We also have $Der\,c\,\{\texttt{"}d\texttt{"}\} = \varnothing$. Hence we have $\varnothing = \varnothing$ in (c). \end{itemize}\noindentThese were the easy base cases. Now come the inductive cases.\begin{itemize}\item Fourth Case: $P(r_1 + r_2)$ is $L(der\,c\,(r_1 + r_2)) = Der\,c\,(L(r_1 + r_2))$ (d). This is what we have to show.We can assume already:\begin{center}\begin{tabular}{ll}$P(r_1)$: & $L(der\,c\,r_1) = Der\,c\,(L(r_1))$ (I)\\$P(r_2)$: & $L(der\,c\,r_2) = Der\,c\,(L(r_2))$ (II)\end{tabular}\end{center}We have that $der\,c\,(r_1 + r_2) = (der\,c\,r_1) + (der\,c\,r_2)$ and also $L((der\,c\,r_1) + (der\,c\,r_2)) = L(der\,c\,r_1) \cup L(der\,c\,r_2)$.By (I) and (II) we know that the left-hand side is $Der\,c\,(L(r_1)) \cup Der\,c\,(L(r_2))$. You need to ponder a bit, but you should seethat \begin{center}$Der\,c(A \cup B) = (Der\,c\,A) \cup (Der\,c\,B)$\end{center}holds for every set of strings $A$ and $B$. That means the right-hand side of (d) is also $Der\,c\,(L(r_1)) \cup Der\,c\,(L(r_2))$,because $L(r_1 + r_2) = L(r_1) \cup L(r_2)$. And we are done with the fourth case.\item Fifth Case: $P(r_1 \cdot r_2)$ is $L(der\,c\,(r_1 \cdot r_2)) = Der\,c\,(L(r_1 \cdot r_2))$ (e). We can assume already:\begin{center}\begin{tabular}{ll}$P(r_1)$: & $L(der\,c\,r_1) = Der\,c\,(L(r_1))$ (I)\\$P(r_2)$: & $L(der\,c\,r_2) = Der\,c\,(L(r_2))$ (II)\end{tabular}\end{center}Let us first consider the case where $nullable(r_1)$ holds. Then \[der\,c\,(r_1 \cdot r_2) = ((der\,c\,r_1) \cdot r_2) + (der\,c\,r_2).\]The corresponding language of the right-hand side is \[(L(der\,c\,r_1) \,@\, L(r_2)) \cup L(der\,c\,r_2).\]By the induction hypotheses (I) and (II), this is equal to\[(Der\,c\,(L(r_1)) \,@\, L(r_2)) \cup (Der\,c\,(L(r_2)).\;\;(**)\]We also know that $L(r_1 \cdot r_2) = L(r_1) \,@\,L(r_2)$. We have to know what$Der\,c\,(L(r_1) \,@\,L(r_2))$ is.Let us analyse what$Der\,c\,(A \,@\, B)$ is for arbitrary sets of strings $A$ and $B$. If $A$ does \emph{not}contain the empty string, then every string in $A\,@\,B$ is of the form $s_1 \,@\, s_2$ where$s_1 \in A$ and $s_2 \in B$. So if $s_1$ starts with $c$ then we just have to remove it. Consequently,$Der\,c\,(A \,@\, B) = (Der\,c\,(A)) \,@\, B$. This case does not apply here though, because we already proved that if $r_1$ is nullable, then $L(r_1)$ contains the empty string. In this case, every stringin $A\,@\,B$ is either of the form $s_1 \,@\, s_2$, with $s_1 \in A$ and $s_2 \in B$, or$s_3$ with $s_3 \in B$. This means $Der\,c\,(A \,@\, B) = ((Der\,c\,(A)) \,@\, B) \cup Der\,c\,B$.But this proves that (**) is $Der\,c\,(L(r_1) \,@\, L(r_2))$.Similarly in the case where $r_1$ is \emph{not} nullable.\item Sixth Case: $P(r^*)$ is $L(der\,c\,(r^*)) = Der\,c\,L(r^*)$. We can assume already:\begin{center}\begin{tabular}{ll}$P(r)$: & $L(der\,c\,r) = Der\,c\,(L(r))$ (I)\end{tabular}\end{center}We have $der\,c\,(r^*) = der\,c\,r\cdot r^*$. Which means $L(der\,c\,(r^*)) = L(der\,c\,r\cdot r^*)$ andfurther $L(der\,c\,r) \,@\, L(r^*)$. By induction hypothesis (I) we know that is equal to $(Der\,c\,L(r)) \,@\, L(r^*)$. (*)\end{itemize}Let us now analyse $Der\,c\,L(r^*)$, which is equal to $Der\,c\,((L(r))^*)$. Now $(L(r))^*$ is definedas $\bigcup_{n \ge 0} L(r)$. We can write this as $L(r)^0 \cup \bigcup_{n \ge 1} L(r)$, where we just separated the first union and then let the ``big-union'' start from $1$. Form this we can already infer\begin{center}$Der\,c\,(L(r^*)) = Der\,c\,(L(r)^0 \cup \bigcup_{n \ge 1} L(r)) = (Der\,c\,L(r)^0) \cup Der\,c\,(\bigcup_{n \ge 1} L(r))$\end{center}The first union ``disappears'' since $Der\,c\,(L(r)^0) = \varnothing$.\end{document}%%% Local Variables: %%% mode: latex%%% TeX-master: t%%% End: