# Horizontal space within proof

I am typesetting a few proofs in LaTeX that have multiple cases, and sometimes I wish to indent each case to make my proof clearer (with about the same amount of horizontal space as a \qquad).

I have looked through many posts on TeX Stack Exchange, and I have tried many of the horizontal spacing commands at “What commands are there for horizontal spacing?” as well as \indent, but to no avail.

Here is a simple example:

\documentclass{article}
\usepackage{amsmath}
\usepackage{amssymb}
\usepackage{amsthm}

\theoremstyle{definition}
\newtheorem{theorem}{Theorem}
\newtheorem{case}{Case}

\begin{document}

\begin{theorem}
$\forall (n \in \mathbb{Z})\; 2 \vert (n^2 + n)$
\end{theorem}

\begin{proof}
\begin{case}
$2 \vert n$ \\
$2 \vert n \iff \exists (k \in \mathbb{Z}) \text{ s.t. } n = 2k$
$\begin{array}{r@{\;}l@{\;}l@{\qquad}l} \implies & n^2 &= (2k)^2 \\ &&= 4k^2 \\ \implies & n^2 + n &= 4k^2 + 2k \\ &&=2(2k^2+k) \\ \multicolumn{3}{@{}c}{\text{Let s = 2k^2 + k. k \in \mathbb{Z} \implies s \in \mathbb{Z}}} \\ \implies & \multicolumn{2}{@{}l}{\exists (s \in \mathbb{Z}) \text{ s.t. } n^2 + n = 2s} \\ \iff & 2 \vert (n^2 + n) \end{array}$
\end{case}

\begin{case}
$2 \hspace{-2.5pt}\not\vert\, n$ \\
$2 \vert n \iff \exists (k \in \mathbb{Z}) \text{ s.t. } n = 2k + 1$
$\begin{array}{r@{\;}l@{\;}l@{\qquad}l} \implies & n^2 &= (2k + 1)^2 \\ &&= 4k^2 + 4k + 1 \\ \implies & n^2 + n &= 4k^2 + 4k + 1 + 2k + 1 \\ &&= 4k^2 + 6k + 2 \\ &&= 2(2k^2 + 3k + 1) \\ \multicolumn{3}{@{}c}{\text{Let s = 2k^2 + 3k + 1. k \in \mathbb{Z} \implies s \in \mathbb{Z}}} \\ \implies & \multicolumn{2}{@{}l}{\exists (s \in \mathbb{Z}) \text{ s.t. } n^2 + n = 2s} \\ \iff & 2 \vert (n^2 + n) \end{array}$
\vspace*{-2.25\baselineskip} \\
\qedhere
\end{case}
\end{proof}

\end{document}


I would do that with an enumerate-like environment, which is easy to customise with enumitem. So I defined a mycases list, with a convenient left margin:

\documentclass{article}
\usepackage{amsmath}
\usepackage{amssymb, bm}
\usepackage{amsthm}
\usepackage{calc}
\newlength{\casewd}
\setlength{\casewd}{\widthof{\bfseries Case 0.}}%
\theoremstyle{definition}
\newtheorem{theorem}{Theorem}
\usepackage{enumitem}
\newlist{mycases}{enumerate}{1}
\setlist[mycases,1]{label = Case \arabic*: , wide=0pt, leftmargin=\dimexpr\casewd + \labelsep, font=\bfseries, topsep=2pt, itemsep=0pt}%

\begin{document}

\begin{theorem}
$\forall (n \in \mathbb{Z})\; 2 \mid(n^2 + n)$
\end{theorem}

\begin{proof}\mbox{}

\begin{mycases}
\item $\bm{2 \mid n.}$ \\
${}\mkern2mu 2 \mid n \iff \exists (k \in \mathbb{Z}) \text{ s.t. } n = 2k$
$\begin{array}{r@{\;}l@{\;}l@{\qquad}l} \implies & n^2 &= (2k)^2 \\ &&= 4k^2 \\ \implies & n^2 + n &= 4k^2 + 2k \\ &&=2(2k^2+k) \\ \multicolumn{3}{@{}c}{\text{Let s = 2k^2 + k. k \in \mathbb{Z} \implies s \in \mathbb{Z}}} \\ \implies & \multicolumn{2}{@{}l}{\exists (s \in \mathbb{Z}) \text{ s.t. } n^2 + n = 2s} \\ \iff & 2 \mid (n^2 + n) \end{array}$
\item $\bm{2\nmid n.}$ \\
${}\mkern2mu 2 \nmid n \iff \exists (k \in \mathbb{Z}) \text{ s.t. } n = 2k + 1$
$\begin{array}{r@{\;}l@{\;}l@{\qquad}l} \implies & n^2 &= (2k + 1)^2 \\ &&= 4k^2 + 4k + 1 \\ \implies & n^2 + n &= 4k^2 + 4k + 1 + 2k + 1 \\ &&= 4k^2 + 6k + 2 \\ &&= 2(2k^2 + 3k + 1) \\ \multicolumn{3}{@{}c}{\text{Let s = 2k^2 + 3k + 1. k \in \mathbb{Z} \implies s \in \mathbb{Z}}} \\ \implies & \multicolumn{2}{@{}l}{\exists (s \in \mathbb{Z}) \text{ s.t. } n^2 + n = 2s} \\ \iff & 2 \mid (n^2 + n) \end{array}$
\vspace*{-2.25\baselineskip} \\
\qedhere
\end{mycases}
\end{proof}

\end{document} 