答案1
您可以使用普通数组和 来执行此操作pstricks
:我将第 5 列和第 8 列中的最后元素定义为\rnode
s 并与相关节点连接相连。请注意,以这种方式使用的代码的 pstricks 部分没有 latex 的维度,因此我添加了一个仅包含垂直空间的补充行,以防止与以下文本重叠。
\documentclass{article}
\usepackage{array}
\usepackage{bm}
\usepackage[semibold, condensed]{cabin}
\usepackage[usestackEOL]{stackengine}
\usepackage{pst-node}
\begin{document}
\[
\setlength{\extrarowheight}{3pt}
\begin{array}{ |*{8}{c|}}
\hline
\bm{A} & \bm{B} & \bm{C} & \bm{B\cap C}& \bm{A \cup ∪ (B \cap C)}& \bm{A \cup B}& \bm{A \cup C}& \bm{(A \cup B)\cap(A \cup C)} \\
\hline
0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\
0 & 0 & 1 & 0 & 0 & 0 & 1 & 0 \\
0 & 1 & 0 & 0 & 0 & 1 & 0 & 0 \\
0 & 1 & 1 & 1 & 1 & 1 & 1 & 1 \\
1 & 0 & 0 & 0 & 1 & 1 & 1 & 1 \\
1 & 0 & 1 & 0 & 1 & 1 & 1 & 1 \\
1 & 1 & 0 & 0 & 1 & 1 & 1 & 1 \\
1 & 1 & 1 & 1 & \rnode{A}{1} & 1 & 1 & \rnode{B}{1} \\
\hline
\multicolumn{8}{c}{\vspace{4ex}}%
\ncbar[linewidth=0.6pt, arrows=<->, arrowinset=0.12,angle=-90, nodesep=1.5ex]{A}{B}
\nbput{\sffamily\bfseries\small \Shortunderstack{Since these columns are identical, we conclude \\%
that $\;\bm{A\cup(B\cap C)=(A \cup B) \cap (A \cup C)}$}}
\end{array}
\]
\end{document}
答案2
让我扩展我的评论来回答:
\documentclass{article}
\usepackage{bm}
\usepackage{tikz}
\usetikzlibrary{arrows.meta,
tikzmark}
\usepackage{lipsum}
\begin{document}
\[
\renewcommand\arraystretch{1.2}
\begin{array}{ |*{8}{c|}}
\hline
\bm{A} & \bm{B} & \bm{C} & \bm{B\cap C}& \bm{A\cup (B\cap C)}
& \bm{A\cup B} & \bm{A\cup C} & \bm{(A\cup B)\cap (A \cup C)} \\
\hline
0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\
0 & 0 & 1 & 0 & 0 & 0 & 1 & 0 \\
0 & 1 & 0 & 0 & 0 & 1 & 0 & 0 \\
0 & 1 & 1 & 1 & 1 & 1 & 1 & 1 \\
1 & 0 & 0 & 0 & 1 & 1 & 1 & 1 \\
1 & 0 & 1 & 0 & 1 & 1 & 1 & 1 \\
1 & 1 & 0 & 0 & 1 & 1 & 1 & 1 \\
1 & 1 & 1 & 1 & \tikzmarknode{a}{1}
& 1 & 1 & \tikzmarknode{b}{1} \\
\hline
\end{array}
\vspace{4\baselineskip} % <--- space for arrow
\tikz[>=Straight Barb, overlay,remember picture]%
{
\draw[<->, semithick, shorten <=2ex, shorten >=2ex]
(a) -- ++ (0,-1) -|
node[align=center,font=\bfseries, pos=0.25, below]
{Since these columns are identical, we conclude \\%
that $\;\bm{A\cup(A\cap C)=(A\cup B)\cap (A\cup C)}$}
(b);
}
\]
\lipsum[1]
\end{document}