我尝试将前两行的“=”符号与其余行对齐，但没有对齐？我在这里做错了什么吗？
\documentclass[conference]{IEEEtran}
\IEEEoverridecommandlockouts
\usepackage{amsmath,amssymb,amsfonts}
\usepackage{xcolor}
\def\BibTeX{{\rm B\kern-.05em{\sc i\kern-.025em b}\kern-.08em
T\kern-.1667em\lower.7ex\hbox{E}\kern-.125emX}}
\usepackage{mathtools}
\usepackage{graphicx}
\usepackage{textcomp}
\usepackage{xcolor}
\usepackage{tikz}
\usetikzlibrary{shapes.multipart,shapes,shadows,arrows,decorations.markings,trees,positioning,decorations.markings,calc,fit,chains,intersections,decorations.pathreplacing}
\usepackage[underline=true]{pgf-umlsd}
\usepackage{flexisym}
\usepackage{siunitx}
\usepackage{float}
\usepackage{supertabular,booktabs}
\usepackage{enumitem,kantlipsum}
\usepackage{tcolorbox}
\usepackage{longtable}
\usepackage{tikz-qtree}
\usetikzlibrary{positioning, shapes.geometric}
\usepackage{arydshln}
\usepackage{upgreek}
\usepackage{algorithm}
\usepackage{algorithmic}
\usepackage{float}
\DeclareMathSymbol{\al} {\mathord}{letters}{"0B}
\DeclareMathSymbol{\be} {\mathord}{letters}{"0C}
\DeclareMathSymbol{\ga} {\mathord}{letters}{"0D}
\DeclareMathSymbol{\de} {\mathord}{letters}{"0E}
\DeclareMathSymbol{\ep}{\mathord}{letters}{"0F}
\DeclareMathSymbol{\ze} {\mathord}{letters}{"10}
\DeclareMathSymbol{\et} {\mathord}{letters}{"11}
\DeclareMathSymbol{\thet} {\mathord}{letters}{"12}
\DeclareMathSymbol{\io} {\mathord}{letters}{"13}
\DeclareMathSymbol{\ka} {\mathord}{letters}{"14}
\DeclareMathSymbol{\la} {\mathord}{letters}{"15}
\DeclareMathSymbol{\m} {\mathord}{letters}{"16}
\DeclareMathSymbol{\n} {\mathord}{letters}{"17}
\DeclareMathSymbol{\x} {\mathord}{letters}{"18}
\DeclareMathSymbol{\p} {\mathord}{letters}{"19}
\DeclareMathSymbol{\rh} {\mathord}{letters}{"1A}
\DeclareMathSymbol{\sig} {\mathord}{letters}{"1B}
\DeclareMathSymbol{\ta} {\mathord}{letters}{"1C}
\DeclareMathSymbol{\up}{\mathord}{letters}{"1D}
\DeclareMathSymbol{\ph} {\mathord}{letters}{"1E}
\DeclareMathSymbol{\ch} {\mathord}{letters}{"1F}
\DeclareMathSymbol{\ps} {\mathord}{letters}{"20}
\DeclareMathSymbol{\om} {\mathord}{letters}{"21}
\DeclareMathSymbol{\va}{\mathord}{letters}{"22}
\DeclareMathSymbol{\va}{\mathord}{letters}{"23}
\DeclareMathSymbol{\va} {\mathord}{letters}{"24}
\DeclareMathSymbol{\va} {\mathord}{letters}{"27}
\begin{document}
\begin{align*}
\sig_{ij} &=
\begin{cases}
A &= g^{r_4}\\
C &= g^{\frac{1}{r_4 + H(m)}}\\
\begin{rcases}
D &= \{K^{a}\}^{r_4}.\{K^{b}\}^{r_4}.\{K^{c}\}^{r_4}. ...\\
&= g^{r_4{(\prod_{\forall k}^{}r_k)}(\Sigma_{\forall k}^{}AASK_k)}\\
E_i &= {D_{k}}^{r_4} = g^{r_4{(\prod_{\forall k}^{}r_k)}(\frac{q_x(0)}{t_i})}\\
\end{rcases} {\scriptstyle \mid AA_k \in GetAA(\mathcal{T})}
\end{cases}
\end{align*}
\end{document}
答案1
一种非常手动的方法，使用array：
\documentclass[]{article}
\usepackage[]{mathtools}
\begin{document}
\begin{equation*}
\sigma_{ij} = \left\{
\begin{array}{@{}r@{{}={}}l@{}l}
A & g^{r_4}\\
C & g^{\frac{1}{r_4 + H(m)}}\\
D & \{K^\alpha\}^{r_4}\cdot\{K^\beta\}^{r_4}\cdot\{K^\gamma\}^{r_4} \ldots\\
& g^{r_4{(\prod_{\forall k}^{}r_k)}(\sum_{\forall k}^{}AASK_k)}
&
\smash
{%
\left.\begin{array}{@{}c@{}}
\vphantom{\{K^\alpha\}^{r_4}\{K^\beta\}^{r_4}\{K^\gamma\}^{r_4}}\\
\vphantom{g^{r_4{(\prod_{\forall k}^{}r_k)}(\sum_{\forall
k}^{}AASK_k)}}\\
\vphantom{{D_{k}}^{r_4}=g^{r_4{(\prod_{\forall
k}^{}r_k)}(\frac{q_x(0)}{t_i})}}\\
\end{array}\right\} \scriptstyle \mid AA_k \in GetAA(\mathcal{T})
}
\\
E_i & {D_{k}}^{r_4}
= g^{r_4{(\prod_{\forall k}^{}r_k)}(\frac{q_x(0)}{t_i})}\\
\end{array}\right.
\end{equation*}
\end{document}
答案2
使用nicematrix：
\documentclass[conference]{IEEEtran}
\IEEEoverridecommandlockouts
\usepackage{nicematrix} % <---
\usetikzlibrary{decorations.pathreplacing,
calligraphy}
\tikzset{
B/.style = {decorate, % style for braces
decoration={calligraphic brace, amplitude=4pt,
raise=1pt, mirror},% for mirroring of brace
thick,
pen colour=black}
}
\usepackage{lipsum} % for dummy text
\begin{document}
\lipsum[1]
\[\setlength\arraycolsep{1pt}
\hspace*{-4em}\begin{NiceArray}{RCL}%
[code-after={\tikz{\draw[B] (1-1.north -| 5-1.west) --
node[left=2mm] {$\sigma_{ij}^{\vphantom{h}} =$} (5-1.south west);
\draw[B] (3-3.east |- 5-3.south) --
node[right=2mm,font=\scriptsize] {$\mid AA_k \in GetAA(\mathcal{T})$}
(3-3.north east) ;
}
}
]
A &=& g^{r_4}\\
C &=& g^{\frac{1}{r_4 + H(m)}}\\
D &=& \{K^\alpha\}^{r_4}\cdot\{K^\beta\}^{r_4}\cdot\{K^\gamma\}^{r_4} \cdot\ldots \\
&=& g^{r_4{(\prod_{\forall k}^{}r_k)}(\sum_{\forall k}^{}AASK_k)} \\
E_i &=& {D_{k}}^{r_4}
= g^{r_4{(\prod_{\forall k}^{}r_k)}(\frac{q_x(0)}{t_i})} \\
\end{NiceArray}
\]
\lipsum[2-7]
\end{document}
经过两次编译后MWE的结果是：
答案3
\[
\sig_{ij} =\left\{\!
\begin{aligned}
A &= g^{r_4}\\
C &= g^{\frac{1}{r_4 + H(m)}}\\
D &= \{K^\al\}^{r_4}.\{K^\be\}^{r_4}.\{K^\ga\}^{r_4}. ...\\
&= g^{r_4{(\prod_{\forall k}^{}r_k)}(\Sigma_{\forall k}^{}AASK_k)}
\qquad\smash{\left.\rule{0pt}{2\normalbaselineskip}\right\} {\scriptstyle \mid AA_k \in GetAA(\mathcal{T})} } \\
E_i &= {D_{k}}^{r_4} = g^{r_4{(\prod_{\forall k}^{}r_k)}(\frac{q_x(0)}{t_i})}
\end{aligned}\right.
\]
答案4
以下解决方案采用 1 个cases环境和 2 个环境。它还通过减少参数 和 的值aligned来确保材料适合两列文档设置的窄列，这两个参数分别控制放置在和周围的空白填充量。IEEEtran\medmuskipthickmuskip\cdot=
\documentclass[conference]{IEEEtran}
\IEEEoverridecommandlockouts
\newcommand\vn[1]{\mathit{#1}} % for "variable name"
\usepackage{mathtools,amssymb,lipsum}
\begin{document}
\[
\medmuskip=1mu % default: 4mu
\thickmuskip=2mu % default: 5mu
\sigma_{ij} =
\begin{cases}\!
\begin{aligned}
A &= g^{r_4}\\
C &= g^{1/(r_4 + H(m))}
\end{aligned}\\
\mkern-8mu\begin{rcases}
\begin{aligned}
D
&=\{K^{\alpha}\}^{r_4}\cdot\{K^{\beta}\}^{r_4}\cdot\{K^{\gamma}\}^{r_4}\cdot \dots \\[0.5ex]
&= g^{r_4{(\prod_{\forall k}r_k)}(\sum_{\forall k} \vn{AASK}_k)}\\[1ex]
E_i
&= {D_{k}}^{r_4}
= g^{r_4{(\prod_{\forall k}r_k)}(q_x(0)/t_i)}
\end{aligned}
\end{rcases} \scriptstyle \bigm\vert \vn{AA}_k \in \vn{GetAA}(\mathcal{T})
\end{cases}
\]
\lipsum % generate several paragraphs of filler text
\end{document}