(*<*)
theory Slides1
imports "LaTeXsugar"
begin
notation (latex output)
set ("_") and
Cons ("_::/_" [66,65] 65)
(*>*)
text_raw {*
\renewcommand{\slidecaption}{Cambridge, 9 November 2010}
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\mode<presentation>{
\begin{frame}
\frametitle{%
\begin{tabular}{@ {}c@ {}}
\LARGE A Formalisation of the\\[-3mm]
\LARGE Myhill-Nerode Theorem\\[-3mm]
\LARGE based on Regular Expressions\\[-3mm]
\large \onslide<2>{\alert{or, Regular Languages Done Right}}\\
\end{tabular}}
\begin{center}
Christian Urban
\end{center}
\begin{center}
joint work with Chunhan Wu and Xingyuan Zhang from the PLA
University of Science and Technology in Nanjing
\end{center}
\end{frame}}
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*}
text_raw {*
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\mode<presentation>{
\begin{frame}[c]
\frametitle{In Textbooks\ldots}
\begin{itemize}
\item A \alert{regular language} is one where there is DFA that
recognises it.\pause
\item Pumping lemma, closure properties of regular languages (closed
under ``negation'') etc are all described and proved in terms of DFAs.\pause
\item Similarly the Myhill-Nerode theorem, which gives necessary and sufficient
conditions for a language being regular (also describes a minimal DFA for a language).
\end{itemize}
\end{frame}}
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*}
text_raw {*
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\mode<presentation>{
\begin{frame}[t]
\frametitle{Really Bad News!}
This is bad news for formalisations in theorem provers. DFAs might
be represented as
\begin{itemize}
\item graphs
\item matrices
\item partial functions
\end{itemize}
All constructions are difficult to reason about.\bigskip\bigskip
\pause
\small
\only<2>{
Constable et al needed (on and off) 18 months for a 3-person team
to formalise automata theory in Nuprl including Myhill-Nerode. There is
only very little other formalised work on regular languages I know of
in Coq, Isabelle and HOL.}
\only<3>{typical textbook reasoning goes like: ``\ldots if \smath{M} and \smath{N} are any two
automata with no inaccessible states \ldots''
}
\end{frame}}
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*}
text_raw {*
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\mode<presentation>{
\begin{frame}[t]
\frametitle{Regular Expressions}
\ldots are a simple datatype:
\only<1>{
\begin{center}\color{blue}
\begin{tabular}{rcl}
rexp & $::=$ & NULL\\
& $\mid$ & EMPTY\\
& $\mid$ & CHR c\\
& $\mid$ & ALT rexp rexp\\
& $\mid$ & SEQ rexp rexp\\
& $\mid$ & STAR rexp
\end{tabular}
\end{center}}
\only<2->{
\begin{center}
\begin{tabular}{rcl}
\smath{r} & \smath{::=} & \smath{0} \\
& \smath{\mid} & \smath{[]}\\
& \smath{\mid} & \smath{c}\\
& \smath{\mid} & \smath{r_1 + r_2}\\
& \smath{\mid} & \smath{r_1 \cdot r_2}\\
& \smath{\mid} & \smath{r^\star}
\end{tabular}
\end{center}}
\only<3->{Induction and recursion principles come for free.}
\end{frame}}
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*}
text_raw {*
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\mode<presentation>{
\begin{frame}[c]
\frametitle{Semantics of Rexps}
\begin{center}
\begin{tabular}{rcl}
\smath{\mathbb{L}(0)} & \smath{=} & \smath{\varnothing}\\
\smath{\mathbb{L}([])} & \smath{=} & \smath{\{[]\}}\\
\smath{\mathbb{L}(c)} & \smath{=} & \smath{\{[c]\}}\\
\smath{\mathbb{L}(r_1 + r_2)} & \smath{=} & \smath{\mathbb{L}(r_1) \cup \mathbb{L}(r_2)}\\
\smath{\mathbb{L}(r_1 \cdot r_2)} & \smath{=} & \smath{\mathbb{L}(r_1)\; ;\; \mathbb{L} (r_2)}\\
\smath{\mathbb{L}(r^\star)} & \smath{=} & \smath{\mathbb{L}(r)^\star}
\end{tabular}
\end{center}
\small
\begin{center}
\begin{tabular}{rcl}
\smath{L_1 ; L_2} & \smath{\dn} & \smath{\{ s_1 @ s_2 \mid s_1 \in L_1 \wedge s_2 \in L_2\}}\bigskip\\
\multicolumn{3}{c}{
\smath{\infer{[] \in L^\star}{}} \hspace{10mm}
\smath{\infer{s_1 @ s_2 \in L^\star}{s_1 \in L & s_2 \in L^\star}}
}
\end{tabular}
\end{center}
\end{frame}}
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*}
text_raw {*
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\mode<presentation>{
\begin{frame}[c]
\frametitle{\LARGE Regular Expression Matching}
\begin{itemize}
\item Harper in JFP'99: ``Functional Pearl: Proof- Directed Debugging''\medskip
\item Yi in JFP'06: ``Educational Pearl: `Proof-Directed Debugging' revisited
for a first-order version''\medskip
\item Owens et al in JFP'09: ``Regular-expression derivatives re-examined''\bigskip\pause
\begin{quote}\small
``Unfortunately, regular expression derivatives have been lost in the
sands of time, and few computer scientists are aware of them.''
\end{quote}
\end{itemize}
\end{frame}}
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*}
text_raw {*
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\mode<presentation>{
\begin{frame}[c]
\begin{center}
\huge\bf Demo
\end{center}
\end{frame}}
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*}
text_raw {*
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\mode<presentation>{
\begin{frame}[c]
\frametitle{\LARGE The Myhill-Nerode Theorem}
\begin{itemize}
\item will help with closure properties of regular languages and
with the pumping lemma.\medskip
\item provides necessary and suf\!ficient conditions for a language being
regular\bigskip\pause
\begin{center}
\smath{x \approx_{L} y \,\dn\, \forall z.\; x @ z \in L \Leftrightarrow y @ z \in L}
\end{center}
\end{itemize}
\end{frame}}
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*}
text_raw {*
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\mode<presentation>{
\begin{frame}[c]
\frametitle{\LARGE The Myhill-Nerode Theorem}
\mbox{}\\[5cm]
\begin{itemize}
\item \smath{\text{finite}\, (U\!N\!IV /\!/ \approx_L) \;\Leftrightarrow\; L\; \text{is regular}}
\end{itemize}
\end{frame}}
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*}
text_raw {*
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\mode<presentation>{
\begin{frame}[c]
\frametitle{\LARGE Equivalence Classes}
\begin{itemize}
\item \smath{L = []}
\begin{center}
\smath{\Big\{\{[]\},\; U\!N\!IV - \{[]\}\Big\}}
\end{center}\bigskip\bigskip
\item \smath{L = [c]}
\begin{center}
\smath{\Big\{\{[]\},\; \{[c]\},\; U\!N\!IV - \{[], [c]\}\Big\}}
\end{center}\bigskip\bigskip
\item \smath{L = \varnothing}
\begin{center}
\smath{\Big\{U\!N\!IV\Big\}}
\end{center}
\end{itemize}
\end{frame}}
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*}
text_raw {*
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\mode<presentation>{
\begin{frame}[c]
\frametitle{\LARGE Regular Languages}
\begin{itemize}
\item \smath{L} is regular \smath{\dn} if there is an automaton \smath{M}
such that \smath{\mathbb{L}(M) = L}\\[1.5cm]
\item Myhill-Nerode:
\begin{center}
\begin{tabular}{l}
finite $\Rightarrow$ regular\\
\;\;\;\smath{\text{finite}\,(U\!N\!IV /\!/ \approx_L) \Rightarrow \exists r. L = \mathbb{L}(r)}\\[3mm]
regular $\Rightarrow$ finite\\
\;\;\;\smath{\text{finite}\, (U\!N\!IV /\!/ \approx_{\mathbb{L}(r)})}
\end{tabular}
\end{center}
\end{itemize}
\end{frame}}
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*}
text_raw {*
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\mode<presentation>{
\begin{frame}[c]
\frametitle{\LARGE Final States}
\mbox{}\\[3cm]
\begin{itemize}
\item \smath{\text{final}_L\,X \dn}\\
\smath{\hspace{6mm}X \in (U\!N\!IV /\!/\approx_L) \;\wedge\; \forall s \in X.\; s \in L}
\smallskip
\item we can prove: \smath{L = \bigcup \{X.\;\text{final}_L\,X\}}
\end{itemize}
\end{frame}}
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*}
text_raw {*
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\mode<presentation>{
\begin{frame}[c]
\frametitle{\LARGE Transitions between\\[-3mm] Equivalence Classes}
\smath{L = \{[c]\}}
\begin{tabular}{@ {\hspace{-7mm}}cc}
\begin{tabular}{c}
\begin{tikzpicture}[shorten >=1pt,node distance=2cm,auto, ultra thick]
\tikzstyle{state}=[circle,thick,draw=blue!75,fill=blue!20,minimum size=0mm]
%\draw[help lines] (0,0) grid (3,2);
\node[state,initial] (q_0) {$R_1$};
\node[state,accepting] (q_1) [above right of=q_0] {$R_2$};
\node[state] (q_2) [below right of=q_0] {$R_3$};
\path[->] (q_0) edge node {c} (q_1)
edge node [swap] {$\Sigma-{c}$} (q_2)
(q_2) edge [loop below] node {$\Sigma$} ();
\end{tikzpicture}
\end{tabular}
&
\begin{tabular}[t]{ll}
\\[-20mm]
\multicolumn{2}{l}{\smath{U\!N\!IV /\!/\approx_L} produces}\\[4mm]
\smath{R_1}: & \smath{\{[]\}}\\
\smath{R_2}: & \smath{\{[c]\}}\\
\smath{R_3}: & \smath{U\!N\!IV - \{[], [c]\}}\\[6mm]
\multicolumn{2}{l}{\onslide<2->{\smath{X \stackrel{c}{\longrightarrow} Y \dn X ; [c] \subseteq Y}}}
\end{tabular}
\end{tabular}
\end{frame}}
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*}
text_raw {*
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\mode<presentation>{
\begin{frame}[c]
\frametitle{\LARGE Systems of Equations}
Inspired by a method of Brzozowski\;'64, we can build an equational system
characterising the equivalence classes:
\begin{center}
\begin{tabular}{@ {\hspace{-20mm}}c}
\\[-13mm]
\begin{tikzpicture}[shorten >=1pt,node distance=2cm,auto, ultra thick]
\tikzstyle{state}=[circle,thick,draw=blue!75,fill=blue!20,minimum size=0mm]
%\draw[help lines] (0,0) grid (3,2);
\node[state,initial] (p_0) {$R_1$};
\node[state,accepting] (p_1) [right of=q_0] {$R_2$};
\path[->] (p_0) edge [bend left] node {a} (p_1)
edge [loop above] node {b} ()
(p_1) edge [loop above] node {a} ()
edge [bend left] node {b} (p_0);
\end{tikzpicture}\\
\\[-13mm]
\end{tabular}
\end{center}
\begin{center}
\begin{tabular}{@ {\hspace{-6mm}}ll@ {\hspace{1mm}}c@ {\hspace{1mm}}l}
& \smath{R_1} & \smath{\equiv} & \smath{R_1;b + R_2;b \onslide<2->{\alert<2>{+ \lambda;[]}}}\\
& \smath{R_2} & \smath{\equiv} & \smath{R_1;a + R_2;a}\medskip\\
\onslide<3->{we can prove}
& \onslide<3->{\smath{R_1}} & \onslide<3->{\smath{=}}
& \onslide<3->{\smath{R_1; \mathbb{L}(b) \,\cup\, R_2;\mathbb{L}(b) \,\cup\, \{[]\};\{[]\}}}\\
& \onslide<3->{\smath{R_2}} & \onslide<3->{\smath{=}}
& \onslide<3->{\smath{R_1; \mathbb{L}(a) \,\cup\, R_2;\mathbb{L}(a)}}\\
\end{tabular}
\end{center}
\end{frame}}
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*}
text_raw {*
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\mode<presentation>{
\begin{frame}<1>[t]
\small
\begin{center}
\begin{tabular}{l@ {\hspace{1mm}}c@ {\hspace{1mm}}ll}
\onslide<1->{\smath{R_1}} & \onslide<1->{\smath{=}}
& \onslide<1->{\smath{R_1; b + R_2; b + \lambda;[]}}\\
\onslide<1->{\smath{R_2}} & \onslide<1->{\smath{=}}
& \onslide<1->{\smath{R_1; a + R_2; a}}\\
& & & \onslide<2->{by Arden}\\
\onslide<2->{\smath{R_1}} & \onslide<2->{\smath{=}}
& \onslide<2->{\smath{R_1; b + R_2; b + \lambda;[]}}\\
\onslide<2->{\smath{R_2}} & \onslide<2->{\smath{=}}
& \only<2>{\smath{R_1; a + R_2; a}}%
\only<3->{\smath{R_1; a\cdot a^\star}}\\
& & & \onslide<3->{by Arden}\\
\onslide<3->{\smath{R_1}} & \onslide<3->{\smath{=}}
& \onslide<3->{\smath{R_2; b \cdot b^\star+ \lambda;b^\star}}\\
\onslide<3->{\smath{R_2}} & \onslide<3->{\smath{=}}
& \onslide<3->{\smath{R_1; a\cdot a^\star}}\\
& & & \onslide<4->{by substitution}\\
\onslide<4->{\smath{R_1}} & \onslide<4->{\smath{=}}
& \onslide<4->{\smath{R_1; a\cdot a^\star \cdot b \cdot b^\star+ \lambda;b^\star}}\\
\onslide<4->{\smath{R_2}} & \onslide<4->{\smath{=}}
& \onslide<4->{\smath{R_1; a\cdot a^\star}}\\
& & & \onslide<5->{by Arden}\\
\onslide<5->{\smath{R_1}} & \onslide<5->{\smath{=}}
& \onslide<5->{\smath{\lambda;b^\star\cdot (a\cdot a^\star \cdot b \cdot b^\star)^\star}}\\
\onslide<5->{\smath{R_2}} & \onslide<5->{\smath{=}}
& \onslide<5->{\smath{R_1; a\cdot a^\star}}\\
& & & \onslide<6->{by substitution}\\
\onslide<6->{\smath{R_1}} & \onslide<6->{\smath{=}}
& \onslide<6->{\smath{\lambda;b^\star\cdot (a\cdot a^\star \cdot b \cdot b^\star)^\star}}\\
\onslide<6->{\smath{R_2}} & \onslide<6->{\smath{=}}
& \onslide<6->{\smath{\lambda; b^\star\cdot (a\cdot a^\star \cdot b \cdot b^\star)^\star
\cdot a\cdot a^\star}}\\
\end{tabular}
\end{center}
\end{frame}}
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*}
text_raw {*
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\mode<presentation>{
\begin{frame}[c]
\frametitle{\LARGE A Variant of Arden's Lemma}
{\bf Arden's Lemma:}
If \smath{[] \not\in A} then
\begin{center}
\smath{X = X; A + \text{something}}
\end{center}
has the (unique) solution
\begin{center}
\smath{X = \text{something} ; A^\star}
\end{center}
\end{frame}}
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*}
text_raw {*
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\mode<presentation>{
\begin{frame}<1->[t]
\small
\begin{center}
\begin{tabular}{l@ {\hspace{1mm}}c@ {\hspace{1mm}}ll}
\onslide<1->{\smath{R_1}} & \onslide<1->{\smath{=}}
& \onslide<1->{\smath{R_1; b + R_2; b + \lambda;[]}}\\
\onslide<1->{\smath{R_2}} & \onslide<1->{\smath{=}}
& \onslide<1->{\smath{R_1; a + R_2; a}}\\
& & & \onslide<2->{by Arden}\\
\onslide<2->{\smath{R_1}} & \onslide<2->{\smath{=}}
& \onslide<2->{\smath{R_1; b + R_2; b + \lambda;[]}}\\
\onslide<2->{\smath{R_2}} & \onslide<2->{\smath{=}}
& \only<2>{\smath{R_1; a + R_2; a}}%
\only<3->{\smath{R_1; a\cdot a^\star}}\\
& & & \onslide<3->{by Arden}\\
\onslide<3->{\smath{R_1}} & \onslide<3->{\smath{=}}
& \onslide<3->{\smath{R_2; b \cdot b^\star+ \lambda;b^\star}}\\
\onslide<3->{\smath{R_2}} & \onslide<3->{\smath{=}}
& \onslide<3->{\smath{R_1; a\cdot a^\star}}\\
& & & \onslide<4->{by substitution}\\
\onslide<4->{\smath{R_1}} & \onslide<4->{\smath{=}}
& \onslide<4->{\smath{R_1; a\cdot a^\star \cdot b \cdot b^\star+ \lambda;b^\star}}\\
\onslide<4->{\smath{R_2}} & \onslide<4->{\smath{=}}
& \onslide<4->{\smath{R_1; a\cdot a^\star}}\\
& & & \onslide<5->{by Arden}\\
\onslide<5->{\smath{R_1}} & \onslide<5->{\smath{=}}
& \onslide<5->{\smath{\lambda;b^\star\cdot (a\cdot a^\star \cdot b \cdot b^\star)^\star}}\\
\onslide<5->{\smath{R_2}} & \onslide<5->{\smath{=}}
& \onslide<5->{\smath{R_1; a\cdot a^\star}}\\
& & & \onslide<6->{by substitution}\\
\onslide<6->{\smath{R_1}} & \onslide<6->{\smath{=}}
& \onslide<6->{\smath{\lambda;b^\star\cdot (a\cdot a^\star \cdot b \cdot b^\star)^\star}}\\
\onslide<6->{\smath{R_2}} & \onslide<6->{\smath{=}}
& \onslide<6->{\smath{\lambda; b^\star\cdot (a\cdot a^\star \cdot b \cdot b^\star)^\star
\cdot a\cdot a^\star}}\\
& & & \onslide<7->{\alert{solved form}}\\
\end{tabular}
\end{center}
\only<8->{
\begin{textblock}{6}(2.5,4)
\begin{block}{}
\begin{minipage}{8cm}\raggedright
\begin{tikzpicture}[shorten >=1pt,node distance=2cm,auto, ultra thick, inner sep=1mm]
\tikzstyle{state}=[circle,thick,draw=blue!75,fill=blue!20,minimum size=0mm]
%\draw[help lines] (0,0) grid (3,2);
\node[state,initial] (p_0) {$R_1$};
\node[state,accepting] (p_1) [right of=q_0] {$R_2$};
\path[->] (p_0) edge [bend left] node {a} (p_1)
edge [loop above] node {b} ()
(p_1) edge [loop above] node {a} ()
edge [bend left] node {b} (p_0);
\end{tikzpicture}
\end{minipage}
\end{block}
\end{textblock}}
\end{frame}}
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*}
text_raw {*
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\mode<presentation>{
\begin{frame}[c]
\frametitle{\LARGE The Equ's Solving Algorithm}
\begin{itemize}
\item The algorithm must terminate: Arden makes one equation smaller;
substitution deletes one variable from the right-hand sides.\bigskip
\item This is still a bit hairy to formalise because of our set-representation
for equations:
\begin{center}
\begin{tabular}{ll}
\smath{\big\{ (X, \{(Y_1, r_1), (Y_2, r_2), \ldots\}),}\\
\mbox{}\hspace{5mm}\smath{\ldots}\\
& \smath{\big\}}
\end{tabular}
\end{center}\pause
\small
They are generated from \smath{U\!N\!IV /\!/ \approx_L}
\end{itemize}
\end{frame}}
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*}
text_raw {*
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\mode<presentation>{
\begin{frame}[c]
\frametitle{\LARGE Other Direction}
One has to prove
\begin{center}
\smath{\text{finite} (U\!N\!IV /\!/ \approx_{\mathbb{L}(r)})}
\end{center}
by induction on \smath{r}. Not trivial, but after a bit
of thinking (by Chunhan), one can prove that if
\begin{center}
\smath{\text{finite} (U\!N\!IV /\!/ \approx_{\mathbb{L}(r_1)})}\hspace{5mm}
\smath{\text{finite} (U\!N\!IV /\!/ \approx_{\mathbb{L}(r_2)})}
\end{center}
then
\begin{center}
\smath{\text{finite} (U\!N\!IV /\!/ \approx_{\mathbb{L}(r_1) \,\cup\, \mathbb{L}(r_2)})}
\end{center}
\end{frame}}
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*}
text_raw {*
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\mode<presentation>{
\begin{frame}[c]
\frametitle{\LARGE What Have We Achieved?}
\begin{itemize}
\item \smath{\text{finite}\, (U\!N\!IV /\!/ \approx_L) \;\Leftrightarrow\; L\; \text{is regular}}
\bigskip\pause
\item regular languages are closed under `inversion'
\begin{center}
\smath{U\!N\!IV /\!/ \approx_L \;\;=\;\; U\!N\!IV /\!/ \approx_{-L}}
\end{center}\pause\bigskip
\item regular expressions are not good if you look for a minimal
one of a language (DFA have this notion)\pause\bigskip
\item if you want to do regular expression matching (see Scott's paper)
\end{itemize}
\only<2>{
\begin{textblock}{10}(4,14)
\small
\smath{x \approx_{L} y \,\dn\, \forall z.\; x @ z \in L \Leftrightarrow y @ z \in L}
\end{textblock}
}
\end{frame}}
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*}
text_raw {*
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\mode<presentation>{
\begin{frame}[c]
\frametitle{\LARGE Conclusion}
\begin{itemize}
\item on balance regular expression are superior to DFAs\bigskip
\item I cannot think of a reason to not teach regular languages
to students this way\bigskip
\item I have never ever seen a proof of Myhill-Nerode based on
regular expressions\bigskip
\item no application, but a lot of fun\bigskip
\item great source of examples
\end{itemize}
\end{frame}}
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*}
(*<*)
end
(*>*)