自定义枚举并对齐多个方程

Question

我想要制作以下布局。

我希望能够选择像 (a) 或罗马字母 I 这样的编号。但我不想自己自动编号。我知道我可以用它做到这一点，enumitem但真正的问题是在每个中插入数学\item并保留一个全球的跨项目、=文本进行对齐for all P.. [Identity]。我尝试过\intertext{\item~\phantom}内部对齐align*，但alignat*没有成功。我留下了文本以方便您使用。

\begin{enumerate}[label=(\alph*)]
\begin{alignat*}{3}
    \item 
        P+\mathcal{O}
        &=\mathcal{O}+P=P \quad
        &&\text{para todo $P\in E$.} \quad
        &&\text{[Identidad]}
    \item 
        P+(-P)
        &=\mathcal{O} \quad
        &&\text{para todo $P\in E$.} \quad
        &&\text{[Inverso]}
    \item
        \item 
        (P+Q)+R
        &=P+(Q+R) \quad
        &&\text{para todo $P,Q,R\in E$.} \quad
        &&\text{[Asociatividad]}
    \item 
        P+Q
        &=Q+P \quad
        &&\text{para todo $P,Q\in E$.} \quad
        &&\text{[Conmutatividad]}
\end{alignat*}
\end{enumerate}

上面的代码当然会产生错误LaTeX Error: Command \item invalid in math mode.。非常感谢您的帮助。

Answer 1

我不会对齐等号，因为它们彼此无关。您可以将其用于eqparbox任务。

\documentclass{article}
\usepackage{amsmath,amsthm}

\usepackage{enumitem,eqparbox}

\newcounter{itemalign}
\newcommand{\itemalign}[3]{%
  \item
  \eqmakebox[\theitemalign-A][c]{#1}\quad
  \eqmakebox[\theitemalign-B][l]{#2}\quad
  \eqmakebox[\theitemalign-C][l]{#3}
}
\newenvironment{enumalign}
 {\stepcounter{itemalign}\begin{enumerate}[font=\upshape,label=(\alph*),leftmargin=*]}
 {\end{enumerate}}

\newtheorem{theorem}{Theorem}[section]

\newcommand{\EO}{\mathcal{O}}

\begin{document}

\section{Title}

\begin{theorem}
Let $E$ be an elliptic curve. Then the addition law on $E$ has
the following properties:
\begin{enumalign}
\itemalign{$P+\EO=\EO+P=P$}{for all $P\in E$}{\upshape[Identity]}

\itemalign{$P+(-P)=\EO$}{for all $P\in E$}{\upshape[Inverse]}

\itemalign{$(P+Q)+R=P+(Q+R)$}{for all $P,Q,R\in E$}{\upshape[Associativity]}

\itemalign{$P+Q=Q+P$}{for all $P,Q\in E$}{\upshape[Commutativity]}
\end{enumalign}
\end{theorem}

\end{document}

只需稍加改动，\itemalign您便可以将最终的标签推到右边距，但左对齐。

\newcommand{\itemalign}[3]{%
  \item
  \eqmakebox[\theitemalign-A][c]{#1}\quad
  \eqmakebox[\theitemalign-B][l]{#2}\hfill
  \eqmakebox[\theitemalign-C][l]{#3}\ignorespaces
}

如果你坚持要对齐，

\documentclass{article}
\usepackage{amsmath,amsthm}

\usepackage{enumitem,eqparbox}

\newcounter{itemalign}
\newcommand{\itemalign}[4]{%
  \item
  \eqmakebox[\theitemalign-A][r]{#1}
  \eqmakebox[\theitemalign-B][l]{#2}\quad
  \eqmakebox[\theitemalign-C][l]{#3}\quad
  \eqmakebox[\theitemalign-D][l]{#4}
}
\newenvironment{enumalign}
 {\stepcounter{itemalign}\begin{enumerate}[font=\upshape,label=(\alph*),leftmargin=*]}
 {\end{enumerate}}

\newtheorem{theorem}{Theorem}[section]

\newcommand{\EO}{\mathcal{O}}

\begin{document}

\section{Title}

\begin{theorem}
Let $E$ be an elliptic curve. Then the addition law on $E$ has
the following properties:
\begin{enumalign}
\itemalign{$P+\EO$}{${}=\EO+P=P$}{for all $P\in E$}{\upshape[Identity]}

\itemalign{$P+(-P)$}{${}=\EO$}{for all $P\in E$}{\upshape[Inverse]}

\itemalign{$(P+Q)+R$}{${}=P+(Q+R)$}{for all $P,Q,R\in E$}{\upshape[Associativity]}

\itemalign{$P+Q$}{${}=Q+P$}{for all $P,Q\in E$}{\upshape[Commutativity]}
\end{enumalign}
\end{theorem}

\end{document}

Answer 2

我几乎确信还有其他更优雅的解决方案。但是，这是一个可能的解决方案：

\documentclass[11pt,a4paper]{report}
\usepackage{amsthm,amsmath,amssymb,amsfonts}
\usepackage{lipsum}

\newcounter{fooeq}
\renewcommand{\thefooeq}{\alph{fooeq}}
\newcommand{\myfooeq}{\refstepcounter{fooeq}\textup{(\thefooeq)}}

\newtheorem{theorem}{Theorem}
\begin{document}
\begin{theorem}
\lipsum[2]
\par\noindent
$
\begin{aligned}
\myfooeq && P+\mathcal{O} &= \mathcal{O}+P=P && \text{para todo $P\in E$.} && \textup{[Identidad]}\\
\myfooeq && P+(-P)        &= \mathcal{O}     && \text{para todo $P\in E$.} && \textup{[Inverso]}
\end{aligned}
$
\end{theorem}

\lipsum[2]
\end{document}