互联网上有很多关于如何在 LaTeX 中为定理着色的评论，其中很多都超出了我的理解复杂程度。
目前，我使用了一种相当简单的模式，它允许我为定理、引理等着色，但全部使用相同的颜色。这种模式是否可以轻松扩展，以允许引理、推论、定义、示例等都具有自己的颜色？例如，引理为浅蓝色，定理为浅粉色等。代码如下。
我不太理解为什么这样做有效，但似乎所有着色环境都使用相同的全局值。如果我可以在每个环境中shadecolor局部定义不同的值，那么可能会有希望。1) 我不知道这个概念是否有意义，2) 我不知道语法。shadecolor\newenvironment
\documentclass[a4paper]{article}
\usepackage{xcolor}
\usepackage{amsthm}
\usepackage{framed}
\theoremstyle{plain}% default
\newtheorem{prototheorem}{Theorem}[section]
\colorlet{shadecolor}{orange!15}
\newenvironment{theorem}
{\begin{shaded}\begin{prototheorem}}
{\end{prototheorem}\end{shaded}}
\newtheorem{protolemma}[prototheorem]{Lemma}
\newenvironment{lemma}
{\begin{shaded}\begin{protolemma}}
{\end{protolemma}\end{shaded}}
\newtheorem{protocorollary}[prototheorem]{Corollary}
\newenvironment{corollary}
{\begin{shaded}\begin{protocorollary}}
{\end{protocorollary}\end{shaded}}
\theoremstyle{definition}
\newtheorem{protonotation}{Notation}[section]
\newenvironment{notation}
{\begin{shaded}\begin{protonotation}}
{\end{protonotation}\end{shaded}}
\newtheorem{protoexample}{Example}[section]
\newenvironment{example}
{\begin{shaded}\begin{protoexample}}
{\end{protoexample}\end{shaded}}
\newtheorem{protodefinition}{Definition}[section]
\newenvironment{definition}
{\begin{shaded}\begin{protodefinition}}
{\end{protodefinition}\end{shaded}}
\newcommand{\powerset}[1]{\rho(#1)}
\begin{document}
\section{Introduction}
\begin{theorem}
\label{theorem.188} %% used
If $V \subseteq U$, and if $F$ is a set of binary functions defined on
$U$, then there exists a unique $clos_F(V)$ and it is closed under $F$.
\end{theorem}
\begin{proof}
First we show, by construction, that at least one closed superset of $V$ exists.
Define a sequence $\{\Phi_n\}_{n=0}^{\infty}$ of sets as follows:
\begin{itemize}
\item $\Phi_0 = V$
\item If $i>0$, then $\Phi_{i} = \Phi_{i-1} \cup \bigcup\limits_{f \in F}\{f(x,y) \mid x,y\in \Phi_{i-1}\}$
\end{itemize}
Define the set $\Phi = \bigcup\limits_{i=0}^{\infty}\Phi_i$. We know
that $V = \Phi_0 \subseteq \bigcup\limits_{i=0}^{\infty}\Phi_i$. Next,
let $\alpha \in \Phi$, $\beta \in \Phi$, $f \in F$; take $n \geq 0$
such that $\alpha,\beta \in \Phi_n$. By definition
$f(\alpha,\beta) \in \Phi_{n+1} \subseteq \Phi$. Thus $\Phi$ closed under $F$.
\end{proof}
\begin{example} \label{example.363}
Let $V= \{\{1,2\},\{2,3\}\}$, and $F$ but the set containing the
set-union and set-intersection operations, denoted
$F=\{\cup,\cap\}$. Then $clos_F(V) =
\{\emptyset,\{1,2\},\{2\},\{2,3\},\{1,2,3\}\}$, because if we take $\alpha,
\beta \in \{\emptyset,\{1,2\},\{2\},\{2,3\},\{1,2,3\}\}$ then both $\alpha
\cup \beta$ and $\alpha \cap \beta$ are also therein.
\end{example}
\begin{definition}
\label{def.vhat}
If $V\subseteq U$, and $F$ is the set of three primitive set operations
union, intersection, and relative complement, ($F = \{\cup, \cap, \setminus \}$) then we denote
$clos_F(V)$ simply by $\sigma(V)$ and call it the \emph{Sigma algebra} of $V$. Moreover,
each element of $\sigma(V)$ is called a \emph{Boolean combination} of elements of $V$.
\end{definition}
\section{Conclusion}
\begin{corollary}
\label{corol.467}
If $V\subseteq U$, $\sigma(V)$ exists and is unique.
\end{corollary}
\begin{proof}
Simple application of Theorem~\ref{theorem.188}.
\end{proof}
\begin{example}
Let $V= \{\{1,2\},\{2,3\}\}$ as in Example~\ref{example.363}.\\
$\sigma(V)= \{\emptyset,\{1\},\{2\},\{3\},\{1,2\},\{1,3\},\{2,3\},\{1,2,3\}\}$.
$\sigma(V)= \powerset{\{1,2,3\}}$.
\end{example}
\end{document}
答案1
只需\colorlet{shadecolor}{orange!15}在不同的环境中定义，使用相应的颜色即可。详情如下：
旧代码：
\colorlet{shadecolor}{orange!15} % global definition \newenvironment{lemma} {\begin{shaded}\begin{protolemma}} {\end{protolemma}\end{shaded}}新代码：
\newenvironment{lemma} {\colorlet{shadecolor}{blue!15}\begin{shaded}\begin{protolemma}} % light blue for lemmas only {\end{protolemma}\end{shaded}}
完整示例：
\documentclass[a4paper]{article}
\usepackage{xcolor}
\usepackage{amsthm}
\usepackage{framed}
\theoremstyle{plain}% default
\newtheorem{prototheorem}{Theorem}[section]
\newenvironment{theorem}
{\colorlet{shadecolor}{orange!15}\begin{shaded}\begin{prototheorem}}
{\end{prototheorem}\end{shaded}}
\newtheorem{protolemma}[prototheorem]{Lemma}
\newenvironment{lemma}
{\colorlet{shadecolor}{blue!15}\begin{shaded}\begin{protolemma}}
{\end{protolemma}\end{shaded}}
\newtheorem{protocorollary}[prototheorem]{Corollary}
\newenvironment{corollary}
{\colorlet{shadecolor}{pink!15}\begin{shaded}\begin{protocorollary}}
{\end{protocorollary}\end{shaded}}
\theoremstyle{definition}
\newtheorem{protonotation}{Notation}[section]
\newenvironment{notation}
{\colorlet{shadecolor}{green!15}\begin{shaded}\begin{protonotation}}
{\end{protonotation}\end{shaded}}
\newtheorem{protoexample}{Example}[section]
\newenvironment{example}
{\colorlet{shadecolor}{red!15}\begin{shaded}\begin{protoexample}}
{\end{protoexample}\end{shaded}}
\newtheorem{protodefinition}{Definition}[section]
\newenvironment{definition}
{\colorlet{shadecolor}{black!15}\begin{shaded}\begin{protodefinition}}
{\end{protodefinition}\end{shaded}}
\newcommand{\powerset}[1]{\rho(#1)}
\begin{document}
\section{Introduction}
\begin{theorem}
\label{theorem.188} %% used
If $V \subseteq U$, and if $F$ is a set of binary functions defined on
$U$, then there exists a unique $clos_F(V)$ and it is closed under $F$.
\end{theorem}
\begin{example} \label{example.363}
Let $V= \{\{1,2\},\{2,3\}\}$, and $F$ but the set containing the
set-union and set-intersection operations, denoted
$F=\{\cup,\cap\}$. Then $clos_F(V) =
\{\emptyset,\{1,2\},\{2\},\{2,3\},\{1,2,3\}\}$, because if we take $\alpha,
\beta \in \{\emptyset,\{1,2\},\{2\},\{2,3\},\{1,2,3\}\}$ then both $\alpha
\cup \beta$ and $\alpha \cap \beta$ are also therein.
\end{example}
\begin{lemma}
\label{def.vhat}
If $V\subseteq U$, and $F$ is the set of three primitive set operations
union, intersection, and relative complement, ($F = \{\cup, \cap, \setminus \}$) then we denote
$clos_F(V)$ simply by $\sigma(V)$ and call it the \emph{Sigma algebra} of $V$. Moreover,
each element of $\sigma(V)$ is called a \emph{Boolean combination} of elements of $V$.
\end{lemma}
\section{Conclusion}
\begin{corollary}
\label{corol.467}
If $V\subseteq U$, $\sigma(V)$ exists and is unique.
\end{corollary}
\begin{definition}
Let $V= \{\{1,2\},\{2,3\}\}$ as in Example~\ref{example.363}.\\
$\sigma(V)= \{\emptyset,\{1\},\{2\},\{3\},\{1,2\},\{1,3\},\{2,3\},\{1,2,3\}\}$.
$\sigma(V)= \powerset{\{1,2,3\}}$.
\end{definition}
\end{document}
