答案1
有点晚了,但这里有一个解决方案。这个想法是在环境中的合适位置pstricks
添加空节点( ) ,然后用节点连接将它们连接起来。您可以查看包文档以了解详细信息。\pnodes
tabular
\ncbar
pst-node
\documentclass{article}
\usepackage[utf8]{inputenc}
\usepackage[T1]{fontenc}
\usepackage{fourier}
\usepackage{array}
\usepackage{pstricks-add}
\usepackage{auto-pst-pdf}
\newcommand\lAnd{\mathrel{\&}}
\begin{document}
\begin{postscript}
\psset{angle=180, linewidth=0.6pt}
\renewcommand\arraystretch{1.25}
\newcounter{tabenum}\setcounter{tabenum}{0}
\newcommand{\tabitem}{\refstepcounter{tabenum} \thetabenum.}
\begin{tabular}{r@{\enspace}>{$}l<{$}@{\qquad}l}
\pnode(-0.4em,0.6ex){H1}\tabitem & A\lAnd B & hypothesis: goal $\neg\neg A\lAnd\neg\neg B$ \\
\tabitem & A & $1,\lAnd E$ \\
\tabitem & B & $1, \lAnd E$ \\
\pnode(-0.4em,0.6ex){H2}\tabitem & \neg A & hypothesis: goal $\bot$ \\
\tabitem & \bot & $4, 1, \subset E$ \\
\tabitem & \neg\neg A \pnode(0, 2.25ex){G2} & $4{-}5, \subset I$ \\
\pnode(-0.4em,0.6ex){H3}\tabitem & \neg B & hypothesis: goal $\bot$ \\
\tabitem & \bot & $7, 3, \subset E$ \\
\tabitem & \neg\neg B \pnode(0, 2.25ex){G3} & $7{-}8, \subset I$ \\
\tabitem & \neg\neg A\lAnd\neg\neg B & $6, 9, \lAnd I$ \\
\tabitem & A\lAnd B \supset \neg\neg A\lAnd\neg\neg B \pnode(0, 2.3ex){G1}
\end{tabular}%\end{listliketab}
\ncbar[armA=20pt]{H1}{G1}
\ncbar{H2}{G2}
\ncbar{H3}{G3}
\end{postscript}
\end{document}