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Stéphane GALLAND committed cb68492

Use the macro \setminus for differences between two sets.

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chapters/chapter2.tex

 \subsection{Separating the lexical analyzer and the parser}
 \begin{frame}{Lexical Analyzer and Parser}
 	\begin{itemize}
-	\item The lexical Analyzer generally does not control the execution flow of the compiler.
+	\item The lexical analyzer generally does not control the execution flow of the compiler.
 	\vfill
 		\begin{center}
 			\includegraphics[width=.9\linewidth]{lexical_parser_relation}
 		\inlineexample{$A = \left\{a,b,c,\delta\right\}$}
 	\vfill
 	\item A \emph{string} over an alphabet is a finite sequence of symbols drawn from that alphabet. \\
-		\inlineexample{$s \in S=\permut(\powerset(A)) - \{\emptyset\} = \left\{a,b,c,\delta,ab,ac,a\delta,\dots\right\}$}
+		\inlineexample{$s \in S=\permut(\powerset(A)) \setminus \{\emptyset\} = \left\{a,b,c,\delta,ab,ac,a\delta,\dots\right\}$}
 	\item The length of a string $s$, usually written $|s|$, is the number of occurrences of symbols in $s$.
 	\vfill
 	\item A \emph{language} is any countable set of strings over some fixed alphabet. \\
 		\end{tabularx}
 	\end{block}
 	\begin{block}{Concatenation of the languages $L$ and $M$}
-		$LM = \left\{st | s\in L \wedge t \in M\right\}$
+		$LM = \left\{st | s\in L, t \in M\right\}$
 		\begin{tabularx}{\linewidth}{@{}lX@{}}
 		\insertexamplelabel & Let $L = \left\{ a, b ,c \right\}$ and $M = \left\{ d, e \right\}$ \\
 		& then $L \cup M = \left\{ ad, ae, bd, be, cd, ce \right\}$
 		$Dstates$ \affect $\{ T \}$\;
 		$Unmarked$ \affect $\{ T \}$\;
 		\While{$\exists T \in Umarked$}{
-			$Unmarked$ \affect $Unmarked - \{T\}$\;
+			$Unmarked$ \affect $Unmarked \setminus \{T\}$\;
 			\ForEach{input symbol $a$}{
 				$U$ \affect \epsilonclosure(\move($T$,$a$))\;
 				\If{$U \not\in DStates$}{