使用 enumitem 定义一个新列表，该列表使用来自周围环境的 ref

Question

考虑以下 MWE

\documentclass{article}

\usepackage{amsmath,amsthm}
\usepackage{enumitem}

\theoremstyle{plain}
\newtheorem{theorem}{Theorem}

\begin{document}

\begin{theorem}[Great result]
  \label{th:great}
  Let 1 be the number one.
  Then:
  \begin{enumerate}[label=\alph*), ref={\ref{th:great}.\alph*}]
    \item if $x=1$ and $y=x$, then $y=1$;
          \label{th:x-eq-y}
    \item if $x<1$ and $y=x$, then $y<1$.
  \end{enumerate}
\end{theorem}

\begin{theorem}[Lesser result]
  \label{th:lesser}
  Let 0 be the number zero.
  Then:
  \begin{enumerate}[label=\alph*), ref={\ref{th:lesser}.\alph*}]
    \item if $x=0$ and $y=x$, then $y=0$;
          \label{th:other}
    \item if $x<0$ and $y=x$, then $y<0$.
  \end{enumerate}
\end{theorem}

Wow, that Theorem~\ref{th:great} was great.
Proving Theorem~\ref{th:x-eq-y} was particularly challenging, much more than Theorem~\ref{th:other}.
\end{document}

它按预期工作。

如何使用\newlist/定义新列表\setlist，以便根据周围theorem环境的 ref 适当地设置 ref？

理想情况下，我希望能够只写

\begin{theorem}[Great result]
  \label{th:great}
  Let 1 be the number one.
  Then:
  \begin{thenumerate}
    \item if $x=1$ and $y=x$, then $y=1$;
          \label{th:x-eq-y}
    \item if $x<1$ and $y=x$, then $y<1$.
  \end{thenumerate}
\end{theorem}

并得到\ref{th:x-eq-y}产品“1.a”。

有关的：关于 enumitem 包中自定义 ref 的一个问题

Answer 1

使用\thetheorem：

\documentclass{article}

\usepackage{amsmath,amsthm}
\usepackage{enumitem}

\theoremstyle{plain}
\newtheorem{theorem}{Theorem}

\newenvironment{thenumerate}[1][]
 {\enumerate[label=\alph*\textup{)},ref=\thetheorem.\alph*),#1]}
 {\endenumerate}

\begin{document}

\begin{theorem}[Great result]
\label{th:great}
Let $1$ be the number one. Then:
\begin{thenumerate}
\item\label{th:x-eq-y} if $x=1$ and $y=x$, then $y=1$;
\item if $x<1$ and $y=x$, then $y<1$.
\end{thenumerate}
\end{theorem}

\begin{theorem}[Lesser result]
\label{th:lesser}
Let $0$ be the number zero. Then:
\begin{thenumerate}
\item\label{th:other} if $x=0$ and $y=x$, then $y=0$;
\item if $x<0$ and $y=x$, then $y<0$.
\end{thenumerate}
\end{theorem}

Wow, that Theorem~\ref{th:great} was great. Proving Theorem~\ref{th:x-eq-y} was 
particularly challenging, much more than Theorem~\ref{th:other}.

\end{document}

只要你根据环境对类似定理的环境进行编号theorem，比如

\newtheorem{lemma}[theorem]{Lemma}

继承正确的号码不会有问题。

以下变体也适用于不同环境的不同计数器。如果在同一级别有显式计数器，则会出错\refstepcounter（这通常不会发生）。

\documentclass{article}

\usepackage{amsmath,amsthm}
\usepackage{enumitem}

\theoremstyle{plain}
\newtheorem{theorem}{Theorem}
\newtheorem{lemma}{Lemma}

\makeatletter
\newenvironment{thenumerate}[1][]
 {\edef\thistheorem{\@currentlabel}%
  \enumerate[label=\alph*\textup{)},ref=\thistheorem.\alph*),#1]}
 {\endenumerate}
\makeatother

\begin{document}

\begin{theorem}[Great result]
\label{th:great}
Let $1$ be the number one. Then:
\begin{thenumerate}
\item\label{th:x-eq-y} if $x=1$ and $y=x$, then $y=1$;
\item if $x<1$ and $y=x$, then $y<1$.
\end{thenumerate}
\end{theorem}

\begin{lemma}[Lesser result]
\label{th:lesser}
Let $0$ be the number zero. Then:
\begin{thenumerate}
\item\label{th:other} if $x=0$ and $y=x$, then $y=0$;
\item if $x<0$ and $y=x$, then $y<0$.
\end{thenumerate}
\end{lemma}

Wow, that Theorem~\ref{th:great} was great. Proving Theorem~\ref{th:x-eq-y} was 
particularly challenging, much more than Lemma~\ref{th:other}.

\end{document}

Answer 2

这只是对 egregs 答案的补充。没有理由为此使用额外的环境。只要我们在环境中，就添加额外的配置theorem。缺点：这必须添加到每个类似环境的 thm 中。

我用它来控制教学材料等中的枚举格式是否一致。

\documentclass{article}

\usepackage{amsmath,amsthm}
\usepackage{enumitem,etoolbox}
\SetEnumitemKey{:thmrefs}{
  label=\alph*\textup{)},
  ref=\thetheorem.\alph*)
}
\theoremstyle{plain}
\newtheorem{theorem}{Theorem}
\AtBeginEnvironment{theorem}{
  \setlist*[enumerate]{:thmrefs}
}

\begin{document}

\begin{theorem}[Great result]
\label{th:great}
Let $1$ be the number one. Then:
\begin{enumerate}
\item\label{th:x-eq-y} if $x=1$ and $y=x$, then $y=1$;
\item if $x<1$ and $y=x$, then $y<1$.
\end{enumerate}
\end{theorem}

\begin{theorem}[Lesser result]
\label{th:lesser}
Let $0$ be the number zero. Then:
\begin{enumerate}
\item\label{th:other} if $x=0$ and $y=x$, then $y=0$;
\item if $x<0$ and $y=x$, then $y<0$.
\end{enumerate}
\end{theorem}

Wow, that Theorem~\ref{th:great} was great. Proving Theorem~\ref{th:x-eq-y} was 
particularly challenging, much more than Theorem~\ref{th:other}.

\end{document}