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\begin{frame}
\frametitle{Disproving Semantic Entailment}

\begin{block}{}
How to disprove $\alpha_1,\ldots,\alpha_n \;\models \; \beta$?
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That is, $\alpha_1,\ldots,\alpha_n \;\nmodels \; \beta$?
\bigskip
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Find a valuation (assignment of truth values to variables)
that
\begin{itemize}
\item makes $\alpha_1,\ldots,\alpha_n$ true, and
\item $\beta$ false.
\end{itemize}
\end{block}
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\begin{exampleblock}{}
Do we have \quad $p \vee q \;\models\; p \to q$ \quad ?
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\begin{center}
\begin{tabular}{|c|c|c|c|}
\hline
\thd $p$ & \thd $q$ & \thd $p \vee q$ & \thd $p \to q$ \\
\hline
$\F$ & $\F$ & $\F$ & $\T$\\
\hline
$\F$ & $\T$ & $\T$ & $\T$\\
\hline
$\T$ & $\F$ & \malert{1}{3}{$\T$} & \malert{1}{3}{$\F$}\\
\hline
$\T$ & $\T$ & $\T$ & $\T$\\
\hline
\end{tabular}
\end{center}
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When $p$ is $\T$ and $q$ is $\F$,
then $p \vee q$ is $\T$ and $p \to q$ is $\F$.
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Conclusion: $p \vee q \;\not\models\; p \to q$\;.
\end{exampleblock}
\end{frame}