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\newtheorem{theorem}{Theorem}[section]
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\begin{document}
\noindent

%========================================================================
\noindent\framebox[\linewidth]{\shortstack[c]{
\Large{\textbf{Lab07-Propositional Logic}}\vspace{1mm}\\
CS101-�������ѧ���ۿκ���ҵ����ʦ�������h��2016 �＾ѧ��}}
%\vspace{1mm}
\begin{center}
\footnotesize{\color{red}$*$ �ɴ�ӡ��ҳֱ����ҳ������/��ֽװ������ע����\quad ������\_\_\_\_\_\_\_\_\_\_\_\_ \quad ѧ�ţ�\_\_\_\_\_\_\_\_\_\_\_ \quad �༶��\_\_\_\_\_\_}

\footnotesize{\color{blue}$*$ ���Ӱ���ҵ��ֱ���ϴ����γ���վ}
\end{center}

\begin{enumerate}
\item ��$P$��ʾ``��ĳ��ٳ���ÿСʱ65Ӣ��"��$Q$ ��ʾ`` ��ӵ�һ�ų��ٷ���"������������⡣
\begin{enumerate}
\item ������������Ż���
\begin{itemize}
\item ��ĳ��ٳ���65Ӣ���û�ӵ����ٷ�����
\item ��ĳ���������ÿСʱ65Ӣ����ӵ�һ�ų��ٷ�����
\end{itemize}
\item ����Ȼ�����������й�ʽ��
\begin{itemize}
\item $\neg P \vee Q$
\item $\neg P \rightarrow \neg Q$
\end{itemize}
\end{enumerate}

\item ������ֵ����ʽ֤��$\neg(P \leftrightarrow Q)=(P \wedge \neg Q) \vee (\neg P \wedge Q)$��

\begin{center}
\begin{tabular}{c|c|c|c|c|c|c|c|c}
\hline
$P$ & $Q$ & $\neg P$ & $\neg Q$ & $P \wedge \neg Q$ & $\neg P \wedge Q$ & $(P \wedge \neg Q) \vee (\neg P \wedge Q)$ & $P \leftrightarrow Q$ & $\neg(P \leftrightarrow Q)$  \\
\hline
$T$ & $T$ &  &  &  &  &  &  &  \\
\hline
$T$ & $F$ &  &  &  &  &  &  &  \\
\hline
$F$ & $T$ &  &  &  &  &  &  &  \\
\hline
$F$ & $F$ &  &  &  &  &  &  &  \\
\hline
\end{tabular}
\end{center}


\item ���õ�ֵ���㷽��֤�����е�ֵ��ʽ��
\begin{enumerate}
\item $P \wedge (((P \vee Q) \wedge \neg P) \rightarrow Q) = P$.
\item $(((P \wedge Q) \vee (P \wedge \neg Q)) \wedge R) \rightarrow P = T$.
\end{enumerate}


\end{enumerate}
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\end{document}
