\documentclass{article}+ −
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+ −
\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}+ −
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