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{\bf Question}

Show that

\begin{itemize}
\item[i)]
$T-(S_1\cap S_2)=(T-S_1)\cup[(T\cap S_1)-S_2]$

\item[ii)]
$S_1\cup S_2=[(S_1)^C\cap(S_2)^C]^C$

\end{itemize}

\vspace{0.25in}

{\bf Answer}

\begin{itemize}
\item[i)]
$(T-S_1)\cup[(T\cap S_1)-S_2]$

$=(T\cap(S_1)^C)\cup(T\cap S_1\cap(S_2)^C)$

$=[(T\cap(S_1)^C)\cup T]\cap[(T\cap(S_1)^C)\cup
S_1]\cap[T\cap(S_1)^C\cup(S_2)^C]$

$\hspace{3in}$(Distributive law)

$=T\cap[T\cup S_1]\cap[T\cup(S_2)^C]\cap[(S_1)^C\cup(S_2)^C]$

$\hspace{1in}$(using $A\cup(A\cap B)=A$ and $(A-B)\cup B=A\cup B$)

$=T\cap(S_1\cap S_2)^C$

$\hspace{1in}$ (using $A\cap(A\cup B)=A$ twice and De-Morgan's
Law)

$=T-(S_1\cap S_2)$



\item[ii)]
$(S_1\cup S_2)^C=(S_1)^C\cap(S_2)^C \hspace{1.5in}$ by De-Morgan

Therefore $S_1\cup S_2=[(S_1)^C\cap(S_2)^C]^C \hspace{0.5in}$
(note $((S)^C)^C=S$)

\end{itemize}



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