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\section*{Complement Sets}

Consider the following picture:

\begin{center}

\begin{tikzpicture}[fill=gray]

% left hand

\scope

\fill (0,0) circle (1);

\endscope

% outline

\draw (0,0) circle (1) (0,1)  node [text=white,below] {$P(s)$}

      (2,0) node [text=black,above] {$\neg P(s)$}

      (-2,-2) rectangle (3,2) node [text=black,above] {$\Sigma^*$};

\end{tikzpicture}

\end{center}

\noindent

where $\Sigma^*$ is in our case the set of all strings (what follows

also holds for any kind of ``domain'', like the set of all integers or

set of all binary trees, etc). Let us assume $P(s)$ is a property that

is about strings, for example $P(s)$ could be ``the string $s$ has

an even length'', or ``the string $s$ starts with the letter

\texttt{a}''. Every such property carves out a subset of strings from

$\Sigma^*$, which in the picture above is depicted as a grey

circle. This subset of strings is often written as a comprehension like

\begin{equation}\label{set}

\{s \in \Sigma^*\;|\; P(s) \}

\end{equation}

\noindent

meaning all the $s$ (out of $\Sigma^*$) for which the property $P(s)$

is true. If $P(s)$ would not be true then the corresponding string $s$

would be outside the grey area where $\neg P(s)$ holds. Notice that

sometimes the property $P(s)$ holds for every string in

$\Sigma^*$. Then the grey area would fill out the whole rectangle and

the set where $\neg P(s)$ holds is empty. Similarly, the property $P(s)$

holds for no string in which case the grey circle is empty.

Now, we are looking for the complement of the set defined in \eqref{set}.

This complement set is often written as

\[

\overline{\{s \in \Sigma^*\;|\; P(s) \}}

\]

\noindent

It is the area of $\Sigma^*$ which isn't grey, that is $\Sigma^*$

minus $\{s \in \Sigma^*\;|\; P(s) \}$, \textbf{or} written differently

it is the set $\{s \in \Sigma^*\;|\; \neg P(s) \}$. That means it the

set of all the strings where $\neg P(s)$ holds. Consequently we have

for any complement set the equation:

\begin{equation}\label{eq}

\overline{\{s \in \Sigma^*\;|\; P(s) \}} = \{s \in \Sigma^*\;|\; \neg P(s) \}

\end{equation}

\section*{Semantic derivative}

Our semantic derivative $Der\;c\;A$ is nothing else than a property that defines

a subset of strings (inside $\Sigma^*$). The corresponding

property $P(s)$ is $c::s \in A$ because we defined $Der\;c\;A$ as

\[

Der\;c\;A \;\dn\; \{s \in \Sigma^*\;|\; c::s \in A\}

\]

\noindent

That means $Der\;c\;A$ is some grey area inside $\Sigma^*$. Obviously

which subset, or grey area, we are carving out from $\Sigma^*$ depends on what we

choose for $c$ and $A$.\bigskip

\noindent

Let us see how this pans out in a concrete example. For this let

$\Sigma^*$ not be the set of all strings, but only the set of strings

upto a length of 3 over the alphabet $\{a, b\}$. That means $\Sigma^*$

(or the rectangle in the picture above) consists of the strings

\[

  \Sigma^* = \left\{\begin{array}{l}

    $\ensuremath{[]}$\\

    $\ensuremath{[a], [b]}$\\

    $\ensuremath{[aa], [ab], [ba], [bb]}$\\

    $\ensuremath{[aaa], [aab], [aba], [abb], [baa], [bab], [bba], [bbb]}$

    \end{array}\right\}

\] 

\noindent

If we set $A$ to $\{[aaa], [abb], [aa], [bb], []\}$, then $Der\;a\;A$ is the subset

\[

Der\;a\;A = \{[aa], [bb], [a]\}

\]  

\noindent

which is given by the definition of

$Der\;a\;A \dn \{s \in \Sigma^*\;|\;a::s\in A\}$. Now lets look at what

the complement of this set looks like:

\begin{equation}\label{Der}

  \overline{Der\;a\;A} = \left\{\begin{array}{l}

    $\ensuremath{[]}$\\

    $\ensuremath{[b]}$\\

    $\ensuremath{[ab], [ba]}$\\

    $\ensuremath{[aaa], [aab], [aba], [abb], [baa], [bab], [bba], [bbb]}$

    \end{array}\right\}

\end{equation}

\noindent

This can be calculated by ``subtracting'' $\{[aa], [bb], [a]\}$ from

$\Sigma^*$. I let you check whether I did this correctly.  According

to the equation in \eqref{eq} this should be equal to

\[

\overline{Der\;a\;A} = \{s \in \Sigma^*\;|\;a::s\not\in A\}

\]  

\noindent Let us test in turn every string in $\Sigma^*$ and see

whether $a::s$ is in $A$ which we set above to

\[\{[aaa], [abb], [aa], [bb], []\}\]

\noindent

This gives rise to the following table where in the first column are

the strings of $\Sigma^*$ and in the second whether $a::s \in A$ holds. The third column

is the negated version of the second.

\begin{center}

\begin{tabular}{r|c|l}

  s & is $a::s \in A$? & $\neg(a::s \in A) \Leftrightarrow  a::s \not\in A$\\

  \hline

  $[]$    & no & yes\\

  $[a]$   & yes& no\\

  $[b]$   & no & yes\\

  $[aa]$  & yes& no\\

  $[ab]$  & no & yes\\

  $[ba]$  & no & yes\\

  $[bb]$  & yes& no\\

  $[aaa]$ & no & yes\\

  $[aab]$ & no & yes\\

  $[aba]$ & no & yes\\

  $[abb]$ & no & yes\\

  $[baa]$ & no & yes\\

  $[bab]$ & no & yes\\

  $[bba]$ & no & yes\\

  $[bbb]$ & no & yes\\

\end{tabular}    

\end{center}

\noindent

Collecting all the yes in the third row gives you the set in \eqref{Der}. So it works out in this example.

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