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\item \self Consider the propositional formula $\varphi = (\neg p \imp r) \land (r \imp \neg p) \land q$.
\begin{enumerate} \item Fill out the truth table for $\varphi$ (and its subformulas).
\begin{tabular}{|c|c|c||c|c|c|c|} \hline $p$&$q$&$r$&$\;\neg p\;$&$(\neg p \imp r)$&$(r \imp \neg p)$&$\quad\varphi\quad$\\ \hline \hline \textbf{F} &\textbf{F} &\textbf{F} & & & &\\ \hline \textbf{F} &\textbf{F} &\textbf{T} & & & &\\ \hline \textbf{F} &\textbf{T} &\textbf{F} & & & &\\ \hline \textbf{F} &\textbf{T} &\textbf{T} & & & &\\ \hline \textbf{T} &\textbf{F} &\textbf{F} & & & &\\ \hline \textbf{T} &\textbf{F} &\textbf{T} & & & &\\ \hline \textbf{T} &\textbf{T} &\textbf{F} & & & &\\ \hline \textbf{T} &\textbf{T} &\textbf{T} & & & &\\ \hline \end{tabular}
\item Is the negation of $\varphi$ satisfiable? \item Is the negation of $\varphi$ valid? \item Give a formula $\psi$ that semantically entails $\varphi$ (i.e., it should be the case that $\psi \models \varphi$). \item Give a formula $\psi$ such that $\varphi$ semantically entails $\psi$ (i.e., it should be the case that $\varphi \models \psi$).
\end{enumerate}
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