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+\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}
+
+\noindent
+We 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}
+
+\noindent
+With 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}
+
+\noindent
+Let us try out an induction proof. Recall the definition
+
+\begin{center}
+$Der\, c\, A \dn \{ s\;\mid\; c\!::\!s \in A\}$
+\end{center}
+
+\noindent
+whereby $A$ 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}
+
+\noindent
+by induction over the regular expression $r$.
+
+
+\newpage
+\noindent
+{\bf Proof}
+
+\noindent
+According 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}
+
+\noindent
+These 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 see
+that
+
+\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 string
+in $A\,@\,B$ 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 $L(der\,c\,(r^*)) = L(der\,c\,r\cdot r^*)$ and
+further $L(der\,c\,r) \,@\, L(r^*)$. 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 defined
+as $\bigcup_{n \ge 0} L(r)$. 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
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+%%% End: