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\item \self Consider the Boolean functions $\varphi_1$ and $\varphi_2$ over variables $p$,
$q$, and $r$. Their truth table is given below.
\begin{tabular}{|c|c|c|c|c||c|c|}
\hline
$p$&$q$&$r$&$\varphi_1$&$\varphi_2$&$\psi$&$\gamma$\\
\hline
\hline
\textbf{F} &\textbf{F} &\textbf{F} &\textbf{F} &\textbf{F} & &\\
\hline
\textbf{F} &\textbf{F} &\textbf{T} &\textbf{F} &\textbf{F} & &\\
\hline
\textbf{F} &\textbf{T} &\textbf{F} &\textbf{T} &\textbf{T} & &\\
\hline
\textbf{F} &\textbf{T} &\textbf{T} &\textbf{F} &\textbf{F} & &\\
\hline
\textbf{T} &\textbf{F} &\textbf{F} &\textbf{T} &\textbf{F} & &\\
\hline
\textbf{T} &\textbf{F} &\textbf{T} &\textbf{F} &\textbf{T} & &\\
\hline
\textbf{T} &\textbf{T} &\textbf{F} &\textbf{F} &\textbf{F} & &\\
\hline
\textbf{T} &\textbf{T} &\textbf{T} &\textbf{F} &\textbf{F} & &\\
\hline
\end{tabular}
\begin{enumerate}
\item Fill the column for $\psi$ such that $\varphi_1$ entails
$\psi$ (i.e., $\varphi_1 \models \psi$), but $\varphi_2$ does
\emph{not} entail $\psi$ (i.e., $\varphi_2 \nvDash \psi$).
\item Fill the column for $\gamma$ such that $\varphi_1$ implies
$\gamma$ (i.e., $\varphi_1 \rightarrow \gamma$) as well as $\varphi_2$ implies $\gamma$ (i.e., $\varphi_2 \rightarrow \gamma$).
\end{enumerate}