Templates

\documentclass[12pt]{report}
\begin{document}

	\begin{center}
		\section*{Equations And Proof}
	\end{center}
	
	\begin{enumerate}
		\item A table is used with two columns
		\item To increase the line-height use the package \texttt{setspace} and add \texttt{\textbackslash onehalfspacing} after the beginning of your document
	\end{enumerate}
	
	\begin{center}
		Proving $(q \Leftrightarrow (\neg p \lor \neg q)) \Leftrightarrow (\neg p \land q)$
		\begin{tabular}{l p{0.5\linewidth}}
			$q \Leftrightarrow (\neg p \lor \neg q)$ \\
			$(q \land (\neg p \lor \neg q)) \lor (\neg q \land \neg(\neg p \lor \neg q))$  & Equivalence \\
			$(q \land (\neg p \lor \neg q)) \lor (\neg q \land (\neg\neg p \land \neg\neg q))$ & DeMorgan \\
			$ (q \land (\neg p \lor \neg q)) \lor (\neg q \land (p \land q)) $ & Double Negation \\
			$(q \land (\neg p \lor \neg q)) \lor (\bot \land p) $  & Complement \& Associative \\
			$(q \land (\neg p \lor \neg q))$ & $(\bot \land p)$ Bottom Always False \\
			$(q \land \neg p) \lor (q \land \neg q)$ & Distributive \\
			$q \land \neg p$ \\
			\\
			Therefore $(q \Leftrightarrow (\neg p \lor \neg q)) \Leftrightarrow (\neg p \land q)$
		\end{tabular}
    \end{center}
	
\end{document}