Making enumeration item look better

Question

if the following two conditions hold:
\begin{enumerate}
\item $ (a, b_1) \in R \wedge (a, b_3) \in R \Rightarrow (a, b_2) \in R $
for all $ a \in S_i, b_1, b_2, b_3 \in S_j $ with $ b_1 < b_2 < b_3 $.
\item $ (a_1, b_2) \in R \wedge (a_2, b_1) \in R \Rightarrow (a_1, b_1) \in R \wedge (a_2, b_2) \in R $
for all $ a_1, a_2 \in S_i, b_1, b_2 \in S_j $ with $ a_1 < a_2 $ and $ b_1 < b_2 $.
\end{enumerate}

How can one make this look not so stuffed?

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score 1 · Answer 1 · answered Mar 09 '16 at 18:21

I propose to separate the ‘pure maths’ and the ‘mixed text-math’ on two different lines, the mixed part being right-aligned while beginning at the same place (see image):

\documentclass[a4paper,11pt]{book}
\usepackage[T1]{fontenc}
\usepackage[utf8]{inputenc}
\usepackage{lmodern}
\usepackage{mathtools}
\usepackage{enumitem}
\usepackage{eqparbox}

\begin{document}

if the following two conditions hold:
\begin{enumerate}
  \item $ (a, b_1) \in R \wedge (a, b_3) \in R \Rightarrow (a, b_2) \in R $\\[-1.5\baselineskip]
        \begin{flushright}\eqmakebox[B][l]{%æ
            for all $ a \in S_i$ and all $b_1, b_2, b_3 \in S_j $ with $ b_1 < b_2 < b_3 $.}\end{flushright}
          \item $ (a_1, b_2) \in R \wedge (a_2, b_1) \in R \Rightarrow (a_1, b_1) \in R \wedge (a_2, b_2) \in R $ \\[-1.5\baselineskip]
          \begin{flushright}\eqmakebox[B] [l]{for all $a_1, a_2 \in S_i $ and all $ b_1, b_2 \in S_j $ with $ a_1 < a_2 $ and $ b_1 < b_2 $.}\end{flushright}
\end{enumerate}

\end{document}

score 0 · Accepted Answer · edited Apr 13 '17 at 12:35

You're requesting something that's very subjective. It doesn't look stuffed to me, as long as you're using the appropriate terminology and/or definitions. Perhaps using words instead of symbols will allow for things looking less stuffed:

\documentclass{article}

\usepackage{amsmath}

\begin{document}

\noindent
If the following two conditions hold:
\begin{enumerate}
  \item $ (a, b_1) \in R \wedge (a, b_3) \in R \Rightarrow (a, b_2) \in R $
    for all $ a \in S_i, b_1, b_2, b_3 \in S_j $ with $ b_1 < b_2 < b_3 $.

  \item $ (a_1, b_2) \in R \wedge (a_2, b_1) \in R \Rightarrow (a_1, b_1) \in R \wedge (a_2, b_2) \in R $
    for all $ a_1, a_2 \in S_i, b_1, b_2 \in S_j $ with $ a_1 < a_2 $ and $ b_1 < b_2 $.
\end{enumerate}

\noindent
If the following two conditions hold:
\begin{enumerate}
  \item $\bigl( (a, b_1) \in R \bigr) \wedge \bigl( (a, b_3) \in R \bigr) \Rightarrow (a, b_2) \in R$
    for all $a \in S_i$ and $b_1, b_2, b_3 \in S_j$ with $b_1 < b_2 < b_3$.

  \item $\bigl( (a_1, b_2) \in R \bigr) \wedge \bigl( (a_2, b_1) \in R \bigr) \Rightarrow \bigl( (a_1, b_1) \in R \bigr) \wedge \bigl( (a_2, b_2) \in R \bigr)$
  for all $ a_1, a_2 \in S_i$ and $b_1, b_2 \in S_j$ with $a_1 < a_2$ and $b_1 < b_2$.
\end{enumerate}

\noindent
If the following two conditions hold:
\begin{enumerate}
  \item $\bigl( (a, b_1) \in R \bigr)$ and $\bigl( (a, b_3) \in R \bigr)$ implies that $(a, b_2) \in R$
    for all $a \in S_i$ and $b_1, b_2, b_3 \in S_j$ with $b_1 < b_2 < b_3$.

  \item $\bigl( (a_1, b_2) \in R \bigr)$ and $\bigl( (a_2, b_1) \in R \bigr)$ implies that $\bigl( (a_1, b_1) \in R \bigr)$ and $\bigl( (a_2, b_2) \in R \bigr)$
  for all $ a_1, a_2 \in S_i$ and $b_1, b_2 \in S_j$ with $a_1 < a_2$ and $b_1 < b_2$.
\end{enumerate}

\end{document}

I'd advise against adjusting the inter-word spacing just for the sake of this list.

Making enumeration item look better

2 Answers2