如何将一组方程式置于其中央并在其旁边放置一组固定的右标签？

Question

像这样：

一种两列网格布局，其中左列为 80% 宽度且居中对齐，而右列左对齐。

Answer 1

以下是该图片的复制版：

\documentclass{article}
\usepackage{amsmath,array,booktabs}

\newcommand{\hmtag}[2][hm]{[\textsc{#2}]_{\textsc{#1}}}
\newcommand{\entails}[1][hm]{\vdash_{\textsc{#1}}}
\newcommand{\hmcolon}{{\mspace{2mu}:\mspace{2mu}}}

\begin{document}

\begin{figure}
\centering
\fbox{%
  $\begin{array}{>{\displaystyle}cl}
  \addlinespace[2ex]
  \frac{\tau\prec\Gamma(x)}{\Gamma\entails x\hmcolon\tau} & \hmtag{Var}
  \\ \addlinespace[3ex]
  \frac{\Gamma\entails e_1\hmcolon\tau_1\to\tau_2\quad \Gamma\entails e_2\hmcolon\tau_1}
       {\Gamma\entails e_1\ e_2\hmcolon\tau_2} & \hmtag{App}
  \\ \addlinespace[3ex]
  \frac{\Gamma\backslash x\cup\{x\hmcolon\tau_1\} \entails e\hmcolon\tau_2}
       {\Gamma\entails \lambda x\to e_1\hmcolon\tau_1\to\tau_2} & \hmtag{Abs}
  \\ \addlinespace[3ex]
  \frac{\Gamma\entails e_1\hmcolon\tau_1 \quad
        \Gamma\backslash x\cup\{x\colon\textit{generalize}(\Gamma,\tau_1)\}
          \entails e_2\hmcolon\tau_2}
       {\Gamma\entails\textbf{let $x=e_1$ in $e_2\hmcolon\tau_2$}} & \hmtag{Let}
  \\ \addlinespace[2ex]
  \end{array}$%
}

\caption{Hindley-Milner type inference rules}
\end{figure}

\end{document}

在正文中，使用flalign*和\Cen来自https://tex.stackexchange.com/a/209732/4427

\documentclass{article}
\usepackage{amsmath,array,booktabs}

\usepackage{lipsum} % just for the example

\newcommand{\hmtag}[2][hm]{[\textsc{#2}]_{\textsc{#1}}}
\newcommand{\entails}[1][hm]{\vdash_{\textsc{#1}}}
\newcommand{\hmcolon}{{\mspace{2mu}:\mspace{2mu}}}

\makeatletter
\newcommand{\Cen}[2]{%
  \ifmeasuring@
    #2%
  \else
    \makebox[\ifcase\expandafter #1\maxcolumn@widths\fi]{$\displaystyle#2$}%
  \fi
}
\makeatother


\begin{document}

\lipsum*[2]
\begin{flalign*}
&&\Cen{3}{
    \frac{\tau\prec\Gamma(x)}{\Gamma\entails x\hmcolon\tau}
  } &&& \hmtag{Var}
\\
&&\Cen{3}{
    \frac{\Gamma\entails e_1\hmcolon\tau_1\to\tau_2\quad \Gamma\entails e_2\hmcolon\tau_1}
         {\Gamma\entails e_1\ e_2\hmcolon\tau_2}
  } &&& \hmtag{App}
\\
&&\Cen{3}{
  \frac{\Gamma\backslash x\cup\{x\hmcolon\tau_1\} \entails e\hmcolon\tau_2}
       {\Gamma\entails \lambda x\to e_1\hmcolon\tau_1\to\tau_2}
  } &&& \hmtag{Abs}
\\
&&\Cen{3}{
    \frac{\Gamma\entails e_1\hmcolon\tau_1 \quad
          \Gamma\backslash x\cup\{x\colon\textit{generalize}(\Gamma,\tau_1)\}
          \entails e_2\hmcolon\tau_2}
         {\Gamma\entails\textbf{let $x=e_1$ in $e_2\hmcolon\tau_2$}}
  } &&& \hmtag{Let}
\end{flalign*}
\lipsum[3]

\begin{figure}
\centering
\fbox{%
  $\begin{array}{>{\displaystyle}cl}
  \addlinespace[2ex]
  \frac{\tau\prec\Gamma(x)}{\Gamma\entails x\hmcolon\tau} &\hmtag{Var}
  \\ \addlinespace[3ex]
  \frac{\Gamma\entails e_1\hmcolon\tau_1\to\tau_2\quad \Gamma\entails e_2\hmcolon\tau_1}
       {\Gamma\entails e_1\ e_2\hmcolon\tau_2} &\hmtag{App}
  \\ \addlinespace[3ex]
  \frac{\Gamma\backslash x\cup\{x\hmcolon\tau_1\} \entails e\hmcolon\tau_2}
       {\Gamma\entails \lambda x\to e_1\hmcolon\tau_1\to\tau_2} &\hmtag{Abs}
  \\ \addlinespace[3ex]
  \frac{\Gamma\entails e_1\hmcolon\tau_1 \quad
        \Gamma\backslash x\cup\{x\colon\textit{generalize}(\Gamma,\tau_1)\}
          \entails e_2\hmcolon\tau_2}
       {\Gamma\entails\textbf{let $x=e_1$ in $e_2\hmcolon\tau_2$}} &\hmtag{Let}
  \\ \addlinespace[2ex]
  \end{array}$%
}

\caption{Hindley-Milner type inference rules}
\end{figure}

\end{document}

Answer 2

这些标签是右对齐的。使用嵌套的\makebox，（当我有机会时），可以将左对齐的标签移到右侧。可选参数\mytag设置 上方和下方的间隙equation*。

\documentclass{article}
\usepackage{stackengine,amsmath,lipsum}
\stackMath
\newcommand\mytag[3][2pt]{%
  \renewcommand\useanchorwidth{T}
  \renewcommand\stacktype{L}
  \addstackgap[#1]{\stackon[0pt]{\displaystyle#2}{\makebox[\textwidth][r]{#3}}}
}
\begin{document}
\lipsum[3]
\begin{equation*}
\mytag{\min_{x_{ij},\delta_i} \sum_{i=1}^m (P_i^{svr} + P_i^{cool})}
  {$\text{[VAR]}_\text{HM}$}
\end{equation*}
\begin{equation*}
\mytag{E = \frac{1}{n!}mc^2}
  {$\text{[APP]}_\text{HM}$}
\end{equation*}
\begin{equation*}
\mytag{y = \frac{Ax^2 + Bx + C}{x - a}}
  {$\text{[ABS]}_\text{HM}$}
\end{equation*}
\begin{equation*}
\mytag{y = mx + b}
  {$\text{[LET]}_\text{HM}$}
\end{equation*}
\lipsum[2]
\end{document}

Answer 3

\newtagform基于的和\usetagform命令mathtools以及环境，具有实数方程标签的解决方案framed：

\documentclass{article}
\usepackage{mathtools, caption, eqparbox, framed}
\newtagform{hmstring}[\scshape\eqparbox{HM}]{[}{\textsubscript{hm}]}
\usepackage[x11names]{xcolor} \colorlet{framecolor}{OrangeRed3}
        \def\FrameCommand{\fboxrule=2\FrameRule\fboxsep=0\FrameSep \fcolorbox{framecolor}{white}}%

\usepackage[showframe]{geometry} \newcommand{\entails}[1][hm]{\vdash_{\textsc{#1}}}
\newcommand{\hmcolon}{{\mspace{2mu}:\mspace{2mu}}}

\begin{document}

 \begin{framed}
\usetagform{hmstring}
 \begin{gather}
 \frac{\tau\prec\Gamma(x)}{\Gamma\entails x\hmcolon\tau} \tag{Var}\label{var}
  \\[2ex]
   \frac{\Gamma\entails e_1\hmcolon\tau_1\to\tau_2\quad \Gamma\entails e_2\hmcolon\tau_1}
       {\Gamma\entails e_1\ e_2\hmcolon\tau_2} \tag{App}
  \\[2ex]
  \frac{\Gamma\backslash x\cup\{x\hmcolon\tau_1\} \entails e\hmcolon\tau_2}
       {\Gamma\entails \lambda x\to e_1\hmcolon\tau_1\to\tau_2}\tag{Abs}
  \\[2ex] \frac{\Gamma\entails e_1\hmcolon\tau_1 \quad
        \Gamma\backslash x\cup\{x\colon\textit{generalize}(\Gamma,\tau_1)\}
          \entails e_2\hmcolon\tau_2}
       {\Gamma\entails\textbf{let $x=e_1$ in $e_2\hmcolon\tau_2$}} \tag{Let}
\end{gather}
 \end{framed}%
\captionof{figure}{Hindley-Milner type inference rules}

\end{document}