
diff -r 4758a6155878 -r 1ab41c59e3d3 hw/proof.tex
--- a/hw/proof.tex	Thu Sep 26 10:39:23 2013 +0100
+++ /dev/null	Thu Jan 01 00:00:00 1970 +0000
@@ -1,210 +0,0 @@
-\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}
-
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