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\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}
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