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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} & \T & \F & \T & \F \\\hline\textbf{F} &\textbf{F} &\textbf{T} & \T & \T & \T & \F \\\hline\textbf{F} &\textbf{T} &\textbf{F} & \T & \F & \T & \F \\\hline\textbf{F} &\textbf{T} &\textbf{T} & \T & \T & \T & \T \\\hline\textbf{T} &\textbf{F} &\textbf{F} & \F & \T & \T & \F \\\hline\textbf{T} &\textbf{F} &\textbf{T} & \F & \T & \F & \F \\\hline\textbf{T} &\textbf{T} &\textbf{F} & \F & \T & \T & \T \\\hline\textbf{T} &\textbf{T} &\textbf{T} & \F & \T & \F & \F \\\hline\end{tabular}
\item Is $\varphi$ satisfiable? \\\quad Yes.\item Is $\varphi$ valid? \\\quad No.\item Give a formula $\psi$ that semantically entails $\varphi$. \\\quad For any formula $\varphi$ it holds that $\bot \models \varphi$, we can therefore choose $\psi = \bot$. We could also represent $\varphi$ as DNF: $(\neg p \land q \land r) \lor (p \land q \land \neg r)$. This is an equivalent formula and therefore semantically entails $\varphi$.\item Give a formula $\psi$ such that $\varphi$ semantically entails $\psi$. \\\quad For any formula $\varphi$ it holds that $\varphi \models \top$. We could also again choose $\varphi$ in DNF.
\end{enumerate}
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