1 - Regular Expression Overview
2 - Use of regular expressions
3 - Definition of regular expressions
![](https://1.bp.blogspot.com/-JoGvQ01qSyU/Wim8_HaM1eI/AAAAAAAAAU0/_9Tdaq5a4NkniXLyM44exTf9qp7PLhrmwCEwYBhgL/s1600/3-definition-of-regular-expressions.png "\subsection{Definition of rational (regular) expression} Let now $\ \Sigma $ \ a finite alphabet and $\underline{\Sigma }$ a copy of $% \Sigma .$ We can then define a \textbf{regular expression} \textbf{RE} ( called also \textbf{rational expression} ) recursively as follow : \begin{definition} A \ regular expression is a word of $\underline{\Sigma }\cup \{(,),+,\cdot ,\ast \}$ such that the set \textbf{RE} of regular expressions satisfy the following axiomes :\newline $1-$ $\ \emptyset $ and $\underline{a}\in RE$ for all $a\in \Sigma $\newline $2-$ \ If \ $E$ $\ and\ \ F$ are tow regular expressions in $RE$ then $% E+F,E.F$ \ and \ $E^{\ast }$ are also. \end{definition} Now we can define a mono\\"{\i}de morphisme $M:RE\longrightarrow P(\Sigma ^{\ast })$ by posing : \begin{eqnarray*} M(\emptyset ) &=&\{\emptyset \}\text{ }and\text{ }M(a)=\{a\} \\ M(E+F) &=&M(E)\cup M(F) \\ M(E.F) &=&M(E).M(F) \\ M(E^{\ast }) &=&(M(E))^{\ast } \end{eqnarray*} Thus we can deduce : \begin{theorem} A language \ $L$ over $\Sigma $ is rational ( regular ) if and only if there exists a regular \ expression $E$ such that $L=M(E)$ \end{theorem}")
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