(*<*)
theory Slides1
imports "~~/src/HOL/Library/LaTeXsugar"
begin
notation (latex output)
set ("_") and
Cons ("_::/_" [66,65] 65)
(*>*)
text_raw {*
%\renewcommand{\slidecaption}{Cambridge, 9 November 2010}
\renewcommand{\slidecaption}{Nijmegen, 25 August 2011}
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\mode<presentation>{
\begin{frame}
\frametitle{%
\begin{tabular}{@ {}c@ {}}
\Large A Formalisation of the\\[-4mm]
\Large Myhill-Nerode Theorem based on\\[-4mm]
\Large Regular Expressions\\[-4mm]
\Large (Proof Pearl)\\[0mm]
\end{tabular}}
\begin{center}
\begin{tabular}{c@ {\hspace{15mm}}c}
\includegraphics[scale=0.034]{chunhan.jpg} &
\includegraphics[scale=0.034]{xingyuan.jpg}\\[-5mm]
\end{tabular}
\end{center}
\begin{center}
\small joint work with Chunhan Wu and Xingyuan Zhang from the PLA
University of Science and Technology in Nanjing
\end{center}
\begin{center}
\small Christian Urban\\
TU Munich
\end{center}
\end{frame}}
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*}
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\mode<presentation>{
\begin{frame}[c]
\frametitle{In Most Textbooks\ldots}
\begin{itemize}
\item A \alert{regular language} is one where there is a DFA that
recognises it.\bigskip\pause
\end{itemize}
I can think of three reasons why this is a good definition:\medskip
\begin{itemize}
\item string matching via DFAs (yacc)
\item pumping lemma
\item closure properties of regular languages (closed under complement)
\end{itemize}
\end{frame}}
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*}
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\mode<presentation>{
\begin{frame}[c]
\frametitle{\normalsize Formal language theory\ldots\hfill\mbox{}}
\begin{center}
\huge\bf\textcolor{gray}{in Nuprl}
\end{center}
\begin{itemize}
\item Constable, Jackson, Naumov, Uribe\medskip
\item \alert{18 months} for automata theory, Hopcroft \& Ullman chapters 1--11 (including Myhill-Nerode)
\end{itemize}
\end{frame}}
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*}
text_raw {*
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\mode<presentation>{
\begin{frame}[c]
\frametitle{\normalsize Formal language theory\ldots\hfill\mbox{}}
\begin{center}
\huge\bf\textcolor{gray}{in Coq}
\end{center}
\begin{itemize}
\item Filli\^atre, Briais, Braibant and others
\item multi-year effort; a number of results in automata theory, e.g.\medskip
\begin{itemize}
\item Kleene's thm.~by Filli\^atre (\alert{``rather big''})
\item automata theory by Briais (5400 loc)
\item Braibant ATBR library, including Myhill-Nerode ($>\!\!\!>$2000 loc)
\item Mirkin's partial derivative automaton construction (10600 loc)
\end{itemize}
\end{itemize}
\end{frame}}
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*}
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\mode<presentation>{
\begin{frame}[t]
\frametitle{\normalsize Formal language theory\ldots\hfill\mbox{}}
\mbox{}\\[-10mm]\mbox{}
\begin{center}
\huge\bf\textcolor{gray}{in HOL}
\end{center}
\begin{itemize}
\item automata @{text "\<Rightarrow>"} graphs, matrices, functions
\item<2-> combining automata/graphs
\onslide<2->{
\begin{center}
\begin{tabular}{ccc}
\begin{tikzpicture}[scale=1]
%\draw[step=2mm] (-1,-1) grid (1,1);
\draw[rounded corners=1mm, very thick] (-1.0,-0.3) rectangle (-0.2,0.3);
\draw[rounded corners=1mm, very thick] ( 0.2,-0.3) rectangle ( 1.0,0.3);
\node (A) at (-1.0,0.0) [circle, very thick, draw, fill=white, inner sep=0.4mm] {};
\node (B) at ( 0.2,0.0) [circle, very thick, draw, fill=white, inner sep=0.4mm] {};
\node (C) at (-0.2, 0.13) [circle, very thick, draw, fill=white, inner sep=0.4mm] {};
\node (D) at (-0.2,-0.13) [circle, very thick, draw, fill=white, inner sep=0.4mm] {};
\node (E) at (1.0, 0.2) [circle, very thick, draw, fill=white, inner sep=0.4mm] {};
\node (F) at (1.0,-0.0) [circle, very thick, draw, fill=white, inner sep=0.4mm] {};
\node (G) at (1.0,-0.2) [circle, very thick, draw, fill=white, inner sep=0.4mm] {};
\draw (-0.6,0.0) node {\small$A_1$};
\draw ( 0.6,0.0) node {\small$A_2$};
\end{tikzpicture}}
&
\onslide<3->{\raisebox{1.1mm}{\bf\Large$\;\Rightarrow\,$}}
&
\onslide<3->{\begin{tikzpicture}[scale=1]
%\draw[step=2mm] (-1,-1) grid (1,1);
\draw[rounded corners=1mm, very thick] (-1.0,-0.3) rectangle (-0.2,0.3);
\draw[rounded corners=1mm, very thick] ( 0.2,-0.3) rectangle ( 1.0,0.3);
\node (A) at (-1.0,0.0) [circle, very thick, draw, fill=white, inner sep=0.4mm] {};
\node (B) at ( 0.2,0.0) [circle, very thick, draw, fill=white, inner sep=0.4mm] {};
\node (C) at (-0.2, 0.13) [circle, very thick, draw, fill=white, inner sep=0.4mm] {};
\node (D) at (-0.2,-0.13) [circle, very thick, draw, fill=white, inner sep=0.4mm] {};
\node (E) at (1.0, 0.2) [circle, very thick, draw, fill=white, inner sep=0.4mm] {};
\node (F) at (1.0,-0.0) [circle, very thick, draw, fill=white, inner sep=0.4mm] {};
\node (G) at (1.0,-0.2) [circle, very thick, draw, fill=white, inner sep=0.4mm] {};
\draw (C) to [red, very thick, bend left=45] (B);
\draw (D) to [red, very thick, bend right=45] (B);
\draw (-0.6,0.0) node {\small$A_1$};
\draw ( 0.6,0.0) node {\small$A_2$};
\end{tikzpicture}}
\end{tabular}
\end{center}\medskip
\only<4-5>{
\begin{tabular}{@ {}l@ {}}
disjoint union:\\[2mm]
\smath{A_1\uplus A_2 \dn \{(1, x)\,|\, x \in A_1\} \,\cup\, \{(2, y)\,|\, y \in A_2\}}
\end{tabular}}
\end{itemize}
\only<5>{
\begin{textblock}{13.9}(0.7,7.7)
\begin{block}{}
\medskip
\begin{minipage}{14cm}\raggedright
Problems with definition for regularity (Slind):\bigskip\\
\smath{\;\text{is\_regular}(A) \dn \exists M.\;\text{is\_dfa}(M) \wedge {\cal L} (M) = A}\bigskip
\end{minipage}
\end{block}
\end{textblock}}
\medskip
\only<6->{A solution:\;\;\smath{\text{nat}} @{text "\<Rightarrow>"} state nodes}
\end{frame}}
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*}
text_raw {*
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\mode<presentation>{
\begin{frame}[t]
\frametitle{\normalsize Formal language theory\ldots\hfill\mbox{}}
\mbox{}\\[-10mm]\mbox{}
\begin{center}
\huge\bf\textcolor{gray}{in HOL}
\end{center}
\begin{itemize}
\item Kozen's proof of Myhill-Nerode:\\
\hspace{5cm}\alert{inaccessible states}
\end{itemize}\bigskip\bigskip
\begin{center}
\smath{\;\text{is\_regular}(A) \dn \exists M.\;\text{is\_dfa}(M) \wedge {\cal L} (M) = A}
\end{center}
\end{frame}}
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*}
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\mode<presentation>{
\begin{frame}[t]
\frametitle{Regular Expressions}
\mbox{}\\[20mm]\mbox{}
\begin{textblock}{13.9}(0.7,2.2)
\begin{block}{}
\begin{minipage}{13.4cm}\raggedright
{\bf Definition:}\smallskip\\
A language \smath{A} is \alert{regular}, provided there exists a\\
regular expression that matches all strings of \smath{A}.
\end{minipage}
\end{block}
\end{textblock}\pause
{\large\bf\alert{\ldots{}and forget about automata}}\bigskip\bigskip\pause
What we might lose?\pause
\begin{itemize}\renewcommand{\ULthickness}{2pt}
\item pumping lemma\pause
\item closure under complementation\pause
\item \only<6>{regular expression matching}%
\only<7>{\textcolor{red}{\sout{\textcolor{black}{regular expression matching}}}}
\end{itemize}
\end{frame}}
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*}
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\mode<presentation>{
\begin{frame}[t]
\frametitle{Regular Expressions}
\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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*}
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\mode<presentation>{
\begin{frame}[c]
\frametitle{\LARGE The Myhill-Nerode Theorem}
\begin{itemize}
\item provides necessary and suf\!ficient conditions for a language
being regular (pumping lemma only necessary)\medskip
\item will help with closure properties of regular languages\bigskip\pause
\item key is the equivalence relation:\smallskip
\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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*}
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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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*}
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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 {$\Sigma$} ()
(q_1) edge node {$\Sigma$} (q_2);
\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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*}
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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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*}
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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<4->{by Arden}\\
\onslide<4->{\smath{R_1}} & \onslide<4->{\smath{=}}
& \onslide<4->{\smath{R_2; 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 substitution}\\
\onslide<5->{\smath{R_1}} & \onslide<5->{\smath{=}}
& \onslide<5->{\smath{R_1; a\cdot a^\star \cdot b \cdot b^\star+ \lambda;b^\star}}\\
\onslide<5->{\smath{R_2}} & \onslide<5->{\smath{=}}
& \onslide<5->{\smath{R_1; a\cdot a^\star}}\\
& & & \onslide<6->{by Arden}\\
\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{R_1; a\cdot a^\star}}\\
& & & \onslide<7->{by substitution}\\
\onslide<7->{\smath{R_1}} & \onslide<7->{\smath{=}}
& \onslide<7->{\smath{\lambda;b^\star\cdot (a\cdot a^\star \cdot b \cdot b^\star)^\star}}\\
\onslide<7->{\smath{R_2}} & \onslide<7->{\smath{=}}
& \onslide<7->{\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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*}
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\mode<presentation>{
\begin{frame}[c]
\frametitle{\LARGE A Variant of Arden's Lemma}
{\bf Arden's Lemma:}\smallskip
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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*}
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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<4->{by Arden}\\
\onslide<4->{\smath{R_1}} & \onslide<4->{\smath{=}}
& \onslide<4->{\smath{R_2; 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 substitution}\\
\onslide<5->{\smath{R_1}} & \onslide<5->{\smath{=}}
& \onslide<5->{\smath{R_1; a\cdot a^\star \cdot b \cdot b^\star+ \lambda;b^\star}}\\
\onslide<5->{\smath{R_2}} & \onslide<5->{\smath{=}}
& \onslide<5->{\smath{R_1; a\cdot a^\star}}\\
& & & \onslide<6->{by Arden}\\
\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{R_1; a\cdot a^\star}}\\
& & & \onslide<7->{by substitution}\\
\onslide<7->{\smath{R_1}} & \onslide<7->{\smath{=}}
& \onslide<7->{\smath{\lambda;b^\star\cdot (a\cdot a^\star \cdot b \cdot b^\star)^\star}}\\
\onslide<7->{\smath{R_2}} & \onslide<7->{\smath{=}}
& \onslide<7->{\smath{\lambda; b^\star\cdot (a\cdot a^\star \cdot b \cdot b^\star)^\star
\cdot a\cdot a^\star}}\\
\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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*}
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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 We need to maintain the invariant that Arden is applicable
(if \smath{[] \not\in A} then \ldots):\medskip
\begin{center}\small
\begin{tabular}{l@ {\hspace{1mm}}c@ {\hspace{1mm}}ll}
\smath{R_1} & \smath{=} & \smath{R_1; b + R_2; b + \lambda;[]}\\
\smath{R_2} & \smath{=} & \smath{R_1; a + R_2; a}\\
& & & by Arden\\
\smath{R_1} & \smath{=} & \smath{R_1; b + R_2; b + \lambda;[]}\\
\smath{R_2} & \smath{=} & \smath{R_1; a\cdot a^\star}\\
\end{tabular}
\end{center}
\end{itemize}
\end{frame}}
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*}
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\mode<presentation>{
\begin{frame}[c]
\frametitle{\LARGE The Equ's Solving Algorithm}
\begin{itemize}
\item The algorithm 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}\bigskip\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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*}
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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 complementation; this is easy
\begin{center}
\smath{U\!N\!IV /\!/ \approx_L \;\;=\;\; U\!N\!IV /\!/ \approx_{-L}}
\end{center}\pause\bigskip
\item if you want to do regular expression matching (see Scott's paper)\pause\bigskip
\item I cannot yet give definite numbers
\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 Examples}
\begin{itemize}
\item \smath{L \equiv \Sigma^\star 0 \Sigma} is regular
\begin{quote}\small
\begin{tabular}{lcl}
\smath{A_1} & \smath{=} & \smath{\Sigma^\star 00}\\
\smath{A_2} & \smath{=} & \smath{\Sigma^\star 01}\\
\smath{A_3} & \smath{=} & \smath{\Sigma^\star 10 \cup \{0\}}\\
\smath{A_4} & \smath{=} & \smath{\Sigma^\star 11 \cup \{1\} \cup \{[]\}}\\
\end{tabular}
\end{quote}
\item \smath{L \equiv \{ 0^n 1^n \,|\, n \ge 0\}} is not regular
\begin{quote}\small
\begin{tabular}{lcl}
\smath{B_0} & \smath{=} & \smath{\{0^n 1^n \,|\, n \ge 0\}}\\
\smath{B_1} & \smath{=} & \smath{\{0^n 1^{(n-1)} \,|\, n \ge 1\}}\\
\smath{B_2} & \smath{=} & \smath{\{0^n 1^{(n-2)} \,|\, n \ge 2\}}\\
\smath{B_3} & \smath{=} & \smath{\{0^n 1^{(n-3)} \,|\, n \ge 3\}}\\
& \smath{\vdots} &\\
\end{tabular}
\end{quote}
\end{itemize}
\end{frame}}
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*}
text_raw {*
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\mode<presentation>{
\begin{frame}[c]
\frametitle{\LARGE What We Have Not Achieved}
\begin{itemize}
\item regular expressions are not good if you look for a minimal
one for a language (DFAs have this notion)\pause\bigskip
\item Is there anything to be said about context free languages:\medskip
\begin{quote}
A context free language is where every string can be recognised by
a pushdown automaton.
\end{quote}
\end{itemize}
\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, in my opinion\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 lots of fun\bigskip
\item great source of examples
\end{itemize}
\end{frame}}
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*}
(*<*)
end
(*>*)