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\section*{Homework 9}

\begin{enumerate}

\item It is true (I confirmed it) that

\end{center}

\noindent

holds, or equivalently

\begin{center}

$L(r) = \{\}$ \;\;implies\;\; $\varnothing$ occurs in $r$.

\end{center}

\noindent

You can prove either version by induction on $r$. The best way to

make more formal what is meant by `$\varnothing$ occurs in $r$', you can define

the following function:

\begin{center}

\begin{tabular}{@ {}l@ {\hspace{2mm}}c@ {\hspace{2mm}}l@ {}}

$occurs(\varnothing)$      & $\dn$ & $true$\\

$occurs(\epsilon)$           & $\dn$ &  $f\!alse$\\

$occurs (c)$                    & $\dn$ &  $f\!alse$\\

$occurs (r_1 + r_2)$       & $\dn$ &  $occurs(r_1) \vee occurs(r_2)$\\ 

$occurs (r_1 \cdot r_2)$ & $\dn$ &  $occurs(r_1) \vee occurs(r_2)$\\

$occurs (r^*)$                & $\dn$ & $occurs(r)$ \\

\end{tabular}

\end{center}

\noindent

Now you can prove 

\begin{center}

$L(r) = \{\}$ \;\;implies\;\; $occurs(r)$.

\end{center}

\noindent

The interesting cases are $r_1 + r_2$ and $r^*$.

The other direction is not true, that is if $occurs(r)$ then $L(r) = \{\}$. A counter example

is $\varnothing + a$: although $\varnothing$ occurs in this regular expression, the corresponding

language is not empty. The obvious extension to include the not-regular expression, $\sim r$,

also leads to an incorrect statement. Suppose we add the clause

\begin{center}

\begin{tabular}{@ {}l@ {\hspace{2mm}}c@ {\hspace{2mm}}l@ {}}

$occurs(\sim r)$      & $\dn$ & $occurs(r)$

\end{tabular}

\end{center}  

\noindent

to the definition above, then it will not be true that

\begin{center}

$L(r) = \{\}$ \;\;implies\;\; $occurs(r)$.

\end{center}

\noindent

Assume the alphabet contains just $a$ and $b$, find a counter example to this

property.

\end{enumerate}

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