\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}
