\item \self Consider the propositional formula
$\varphi = ((p \rightarrow q)\wedge(\neg p \rightarrow \neg q)) \rightarrow r$.

\begin{enumerate}
  \item Fill out the truth table for $\varphi$ and its
      subformulas.

\begin{tabular}{|c|c|c||c|c|c|c|c|c|}
\hline
$p$&$q$&$r$&$\;\neg p\;$&$\;\neg q\;$&$(p \rightarrow q)$&$(\neg p \rightarrow \neg q)$&$(p \rightarrow q)\wedge (\neg p \rightarrow \neg q)$&$\quad\varphi\quad$\\
\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 $\varphi$ unsatisfiable?
\item  Is the negation of $\varphi$ valid?
\item Give a formula $\psi$ that is semantically equivalent to
    $\varphi$, but does not use the ``$\rightarrow$'' connective.
\end{enumerate}