Slides/Slides.thy
changeset 24 f72c82bf59e5
parent 21 6a0538d8ccd5
child 203 5d724fe0e096
--- /dev/null	Thu Jan 01 00:00:00 1970 +0000
+++ b/Slides/Slides.thy	Thu Nov 25 18:54:45 2010 +0000
@@ -0,0 +1,790 @@
+(*<*)
+theory Slides
+imports "LaTeXsugar"
+begin
+
+notation (latex output)
+  set ("_") and
+  Cons  ("_::/_" [66,65] 65) 
+
+(*>*)
+
+
+text_raw {*
+  %\renewcommand{\slidecaption}{Cambridge, 9 November 2010}
+  \renewcommand{\slidecaption}{Munich, 17 November 2010}
+  %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
+  \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}}
+  %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%     
+
+*}
+
+text_raw {*
+  %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
+  \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}}
+  %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%     
+
+*}
+
+
+text_raw {*
+  %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
+  \mode<presentation>{
+  \begin{frame}[t]
+  \frametitle{Really Bad News!}
+
+  DFAs are bad news for formalisations in theorem provers. They might
+  be represented as:
+
+  \begin{itemize}
+  \item graphs
+  \item matrices
+  \item partial functions
+  \end{itemize}
+
+  All constructions are messy to reason about.\bigskip\bigskip 
+  \pause
+
+  \small
+  \only<2>{Alexander and Tobias: ``\ldots automata theory \ldots does not come for free \ldots''} 
+  \only<3>{
+  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<4>{typical textbook reasoning goes like: ``\ldots if \smath{M} and \smath{N} are any two
+  automata with no inaccessible states \ldots''
+  }
+  
+  \end{frame}}
+  %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%     
+
+*}
+
+text_raw {*
+  %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
+  \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}}
+  %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%     
+
+*}
+
+
+text_raw {*
+  %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
+  \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}}
+  %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%     
+
+*}
+
+text_raw {*
+  %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
+  \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}}
+  %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%     
+
+*}
+
+text_raw {*
+  %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
+  \mode<presentation>{
+  \begin{frame}[c]
+
+  \begin{center}
+  \huge\bf Demo
+  \end{center}
+  
+
+  \end{frame}}
+  %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%     
+
+*}
+
+text_raw {*
+  %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
+  \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}}
+  %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%     
+
+*}
+
+text_raw {*
+  %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
+  \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}}
+  %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%     
+
+*}
+
+
+text_raw {*
+  %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
+  \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}}
+  %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%     
+
+*}
+
+
+text_raw {*
+  %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
+  \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}}
+  %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%     
+
+*}
+
+
+text_raw {*
+  %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
+  \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}}
+  %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%     
+*}
+
+
+text_raw {*
+  %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
+  \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}}
+  %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%     
+*}
+
+
+text_raw {*
+  %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
+  \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}}
+  %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%     
+*}
+
+text_raw {*
+  %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
+  \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}}
+  %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%     
+*}
+
+text_raw {*
+  %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
+  \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}}
+  %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%     
+*}
+
+
+text_raw {*
+  %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
+  \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}}
+  %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%     
+*}
+
+text_raw {*
+  %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
+  \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}}
+  %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%     
+*}
+
+text_raw {*
+  %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
+  \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}}
+  %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%     
+*}
+
+
+
+text_raw {*
+  %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
+  \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}}
+  %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%     
+*}
+
+text_raw {*
+  %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
+  \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}}
+  %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%     
+*}
+
+text_raw {*
+  %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
+  \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}}
+  %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%     
+*}
+
+
+text_raw {*
+  %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
+  \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}}
+  %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%     
+*}
+
+
+text_raw {*
+  %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
+  \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}}
+  %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%     
+*}
+
+(*<*)
+end
+(*>*)
\ No newline at end of file