\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}