\documentclass[12pt]{amsart}

\usepackage{amsmath, amssymb, amsthm, enumerate}

\newtheorem{theorem}{Theorem}

\begin{document}

\begin{theorem}
Let $m, n \in \mathbb{Z}$. 
If either $m$ or $n$ is even, then $mn$ is even.
\end{theorem}

\begin{proof}
Suppose $m, n \in \mathbb{Z}$ and $m$ or $n$ is even.

Case 1: $m$ is even.
Then by definition there is $a \in \mathbb{Z}$ such that $m = 2a$.
It follows that $mn = (2a)n = 2(an)$.
Moreover, $an \in \mathbb{Z}$.
Hence $mn$ is even.

Case 2: $n$ is even. \dots

In either case $mn$ is even, so we have proved the theorem.

\end{proof}

\end{document}