这也许分为两个部分，1) 我希望使用定理计数器（及其相关的“样式”）而不是内置于 \begin{equation} 中的样式对某些方程进行编号。根据 MWE，方程 (1.1) 将显示为 1.3（无括号和粗体）。2) 是否可以将定理编号“拆分”或围绕一系列方程进行显示，就像 \split 对方程编号所做的那样？我已经破解了 \newtheoremstyle 以提供适当的编号（如 MWE 中显示的 1.2），但此编号超出了对齐方程的范围。我不确定我需要在哪里进行调整。任何想法都将不胜感激。
\documentclass[12pt,leqno]{book}
\usepackage{amsmath, mathtools,amssymb,amscd,amsthm,amstext}
\usepackage{indentfirst}
\usepackage{changepage}
% HACKING \adjustwidth so that it has equal vertical whitespace above/below
\usepackage{etoolbox}
\makeatletter
\apptocmd\adjustwidth{\@inlabelfalse\@newlistfalse}
\makeatother
\setlength\parindent{1.2cm}
% Indentation
\def\changemargin#1#2{\list{}{\rightmargin#2\leftmargin#1}\item[]}
\let\endchangemargin=\endlist
% \swapnumbers puts number ahead of heading - as in 3.4 Definition, rather than Definition 3.4
\swapnumbers
% Redefine theorem style
\newtheoremstyle{mytheoremstyle} % name
{\topsep} % Space above
{\topsep} % Space below
{\itshape} % Body font
{5mm} % Indent amount
{\bfseries} % Theorem head font
{.} % Punctuation after theorem head
{0.5em} % Space after theorem head
{} % Theorem head spec (can be left empty, meaning ‘normal’)
\theoremstyle{mytheoremstyle}
\newtheorem{theorem}{Theorem}[chapter]
\newtheorem{corollary}[theorem]{Corollary}
% Creates a theorem-numbered entry
\newtheoremstyle{dotless}
{}
{}
{}
{}
{\bfseries}
{}
{ }
{}
\theoremstyle{dotless}
\newtheorem{line_eq}[theorem]{}
\begin{document}
\chapter{}
The next result is also an almost immediate consequence of the preceding theorem.
% Corollary 2.5
\begin{adjustwidth}{.6cm}{.6cm}
\begin{corollary}
The additive inverse of an element $a$ of a ring $R$, whose existence is
asserted by Property \textup{\textbf{P}}$_4$, in unique.
\end{corollary}
\noindent\textsc{proof}. To prove this statement, suppose that $a+x= 0$ and that
$a+y=0$. Then $a=x=a+y$, and one of the cancellation laws of addition shows
at once that $x=y$.
\end{adjustwidth}
\vspace{5mm}
\noindent Here's some statements:
\begin{line_eq}
\begin{align*}
&(\textup{i}) & -(-a) &= a, \\
&(\textup{ii}) & -(a+b) &= -a-b, \\
&(\textup{iii}) & -(a-b) &= -a+b, \\
&(\textup{iv}) & (a-b)-c &= a-(b+c). \\
\end{align*}
\end{line_eq}
\noindent
Let us prove the second of these ...
\begin{equation}
a = -(-a)
\end{equation}
\end{document}
答案1
抱歉，MWE 不太简单，但我想展示一个比较全面的例子。我希望有定理、推论、定义和具体的方程式在一章的所有章节中都使用同一个计数器，然后在下一章中重置。我试图自己记录示例以展示代码的重要部分，因为我可能是最后一个解释 Latex 细节的人。我在原始帖子中提出的上述两个问题都在这里得到解决。感谢那些评论者，他们让这项工作如我所设想的那样顺利进行。
\documentclass[12pt, leqno]{book}
\usepackage{mathtools,amsthm}
\usepackage{changepage} % Use the \adjustwidth environment
\usepackage{enumerate}
\usepackage[shortlabels]{enumitem}
%% Set equation numbering to chapter, resetting
%% on next chapter
\numberwithin{equation}{chapter}
% Define theorem style - indentation + italics
\newtheoremstyle{mytheoremstyle}
{\topsep}
{\topsep}
{\itshape}
{5mm}
{\bfseries}
{.}
{0.5em}
{}
\theoremstyle{mytheoremstyle}
\newtheorem{theorem}[equation]{Theorem} %% Theorems and Corollaries are
\newtheorem{corollary}[equation]{Corollary} %% now tied to equation numbering
% Define definition style - indentation only
\newtheoremstyle{mydefinitionstyle}
{\topsep}
{\topsep}
{}
{5mm}
{\bfseries}
{.}
{0.5em}
{}
\theoremstyle{mydefinitionstyle}
\newtheorem{definition}[equation]{Definition} %% Defn tied to equation numbering
%% Define tagform to remove brackets/braces around equation tag
\newtagform{nobrackets}[\textbf]{}{}
%% Apply tagform to document
\usetagform{nobrackets}
%% Inserts spaces, as in "\blank{3cm}"
\newcommand{\blank}[1]{\hspace*{#1}}
\begin{document}
\chapter{Fundamentals}
\section{Basic Concepts}
Start with an equation:
\begin{equation} %% Equation 1.1
1 = 1 + 0
\end{equation}
We next prove the following theorem.
\begin{adjustwidth}{.6cm}{.6cm} %% Theorem 1.2
\begin{theorem}
\textsc{(Cancellation Laws of Addition)}. If $a$, $b$, and $c$ are elements of a ring $R$,
the following are true:
\begin{enumerate}[label=\textup{(}\roman*\textup{)}]
\item If $a+c = b+c$, then $a=b$,
\item If $c+a=c+b$, then $a=b$.
\end{enumerate}
\end{theorem}
\noindent \textsc{proof}. We proceed to prove the first statement of this theorem. Let us therefore assume that
\usetagform{default} %% Reset equation numbering to "default" behavior
\begin{equation}
a + c = b + c. \tag{1} %% Tag should be generated autom. but for now hard-coded
\end{equation}
By \textbf{P}$_4$, there exists an element $t$ of $R$ such that
\begin{equation}
c + t = 0. \tag{2}
\end{equation}
Now if follows from Equation (1) that ...
\end{adjustwidth}
\vspace{5mm}
\noindent This leads to the following corollary.
\usetagform{nobrackets} %% Future equations now will follow "modified" behavior
\begin{adjustwidth}{.6cm}{.6cm} %% Corollary 1.3
\begin{corollary}
Given ring $R$, then for all $a,b,c \in R$, then we have the following:
\end{corollary}
\begin{equation}
\begin{aligned}
&\text{(i)} & \blank{1cm} -(-a) &= a, \blank{6cm} \\
&\text{(ii)} & \blank{1cm} -(a+b) &= -a-b, \\
&\text{(iii)} & \blank{1cm} -(a-b) &= -a+b, \\
&\text{(iv)} & \blank{1cm} (a-b)-c &= a-(b+c). \\
\end{aligned}
\end{equation}
\end{adjustwidth}
\begin{adjustwidth}{.6cm}{.6cm} % Definition 1.5
\begin{definition}
Let $a$ be an element of a ring $R$ with unity $e$. If there exists an element
$s$ of $R$ such that
\begin{equation*}
as = sa = e,
\end{equation*}
then $s$ is called the \textit{multiplicative inverse} of $a$.
\end{definition}
\end{adjustwidth}
\end{document}