From b1f7d7b30dec81697276ff3132ac97fe82cb6767 Mon Sep 17 00:00:00 2001 From: EthanDeng Date: Mon, 18 Apr 2022 18:31:16 +0800 Subject: [PATCH] tcb theorem --- elegantbook-cn.tex | 8 ++------ elegantbook-en.tex | 18 ++++++++++-------- 2 files changed, 12 insertions(+), 14 deletions(-) diff --git a/elegantbook-cn.tex b/elegantbook-cn.tex index 728e69f..21709fe 100644 --- a/elegantbook-cn.tex +++ b/elegantbook-cn.tex @@ -1,4 +1,4 @@ -\documentclass[lang=cn,10pt]{elegantbook} +\documentclass[lang=cn,10pt,founder]{elegantbook} \title{ElegantBook：优美的 \LaTeX{} 书籍模板} \subtitle{Elegant\LaTeX{} 经典之作} @@ -180,7 +180,7 @@ tlmgr update --self --all --reinstall-forcibly-removed \end{table} 如果需要自定义颜色的话请选择 \lstinline{nocolor} 选项或者使用 \lstinline{color=none}，然后在导言区定义 structurecolor、main、second、third 颜色，具体方法如下： -\begin{lstlisting}[tabsize=4]] +\begin{lstlisting}[tabsize=4] \definecolor{structurecolor}{RGB}{0,0,0} \definecolor{main}{RGB}{70,70,70} \definecolor{second}{RGB}{115,45,2} @@ -607,10 +607,6 @@ LaTeX Error: \item \lstinline{nofont}：后台会调用 \lstinline{ctex} 宏包并且使用 \lstinline{fontset=none} 选项，不设定中文字体，用户可以自行设置中文字体，具体见后文。 \end{enumerate} -\begin{remark} - 使用 \lstinline{founder} 选项或者 \lstinline{nofont} 时，必须使用 \hologo{XeLaTeX} 进行编译。 -\end{remark} - \subsection{方正字体选项} 由于使用 \lstinline{ctex} 宏包默认调用系统已有的字体，部分系统字体缺失严重，因此，用户希望能够使用其它字体，我们推荐使用方正字体。方正的{\songti 方正书宋}、{\heiti 方正黑体}、{\kaishu 方正楷体}、{\fangsong 方正仿宋}四款字体均可免费试用，且可用于商业用途。用户可以自行从\href{http://www.foundertype.com/}{方正字体官网}下载此四款字体，在下载的时候请\textbf{务必}注意选择 GBK 字符集，也可以使用 \href{https://www.latexstudio.net/}{\LaTeX{} 工作室}提供的\href{https://pan.baidu.com/s/1BgbQM7LoinY7m8yeP25Y7Q}{方正字体，提取码为：njy9} 进行安装。安装时，{\kaishu Win 10 用户请右键选择为全部用户安装，否则会找不到字体。} diff --git a/elegantbook-en.tex b/elegantbook-en.tex index 00e7f67..6defec0 100644 --- a/elegantbook-en.tex +++ b/elegantbook-en.tex @@ -49,7 +49,7 @@ For portable version, simply download lastest ElegantBook-master from GitHub or \section{Update Templates} You can use cmd/terminal to update the tlmgr itself and all the packages, the commands are: \begin{lstlisting} -tlmgr update --self +tlmgr update --self tlmgr update --all \end{lstlisting} @@ -58,14 +58,12 @@ To learn more, please refer to \href{https://tex.stackexchange.com/questions/554 \section{Other Release Versions} If you are a \TeX{} Live 2018/2019/2020/2021 user and would like to update, the official solution is to uninstall the previous version. If you want to save the bother of uninstallation and re-installation, please copy \lstinline{elegantbook.cls} to the installation directory of \TeX{} Live 2022 (default: \lstinline|C:\texlive\2022\texmf-dist\tex\latex\elegantbook| ), run \lstinline{texhash} in cmd. - - \chapter{ElegantBook Settings} - This template is based on the Standard \LaTeX{} book class, so the options of book class work as well (Note that the option of papersize has no effect due to \lstinline{device} option). The default encoding is UTF-8 while \TeX{} Live is recommended. The test environment is Win10/Ubuntu 20.04/macOS + \TeX{} Live 2022/ Mac\TeX{} 2022, either \hologo{pdfLaTeX} or \hologo{XeLaTeX} works fine. \section{Languages} We defined one option named \lstinline{lang} which has two basic values, \lstinline{lang=en} (default) , \lstinline{lang=cn}. Different values will alter the captions of figure/table, abstract name, refname, etc. You can use this option as + \begin{lstlisting} \documentclass[en]{elegantbook} \documentclass[lang=en]{elegantbook} @@ -78,7 +76,7 @@ Besides the two basic language translation, our user provide more options, here \item French translation \lstinline{lang=fr}, provided by \href{https://github.com/abfek66}{abfek66} , please refer to \href{https://github.com/ElegantLaTeX/ElegantBook/issues/85}{Italian translation}; \item Dutch Translation \lstinline{lang=nl}, provided by \href{https://github.com/inktvis75}{inktvis75} , please refer to \href{https://github.com/ElegantLaTeX/ElegantBook/issues/108}{Dutch Translation}; \item Hungarian translation \lstinline{lang=hu}, provided by \href{https://github.com/palkotamas}{palkotamas}, please refer to \href{https://github.com/ElegantLaTeX/ElegantBook/issues/111}{Hungarian translation}; - \item Deutsch translation \lstinline{lang=de}, provided by Lisa, please refer to\href{https://github.com/ElegantLaTeX/ElegantBook/issues/113}{Deutsch translation}; + \item Deutsch translation \lstinline{lang=de}, provided by Lisa, please refer to \href{https://github.com/ElegantLaTeX/ElegantBook/issues/113}{Deutsch translation}; \item Spanish translation \lstinline{lang=es}, provided by Gustavo A. Corradi, please refer to \href{https://github.com/ElegantLaTeX/ElegantBook/issues/133}{Spanish translation}; \item Mongolian translation \lstinline{lang=mn}, provided by \href{https://github.com/Altantsooj}{Altantsooj}, please refer to \href{https://github.com/ElegantLaTeX/ElegantBook/issues/137}{Mongolian translation}; \item Japanese translation \lstinline{lang=jp}, provided by \href{https://github.com/inusturbo}{inusturbo}, please refer to \href{https://github.com/ElegantLaTeX/ElegantBook/issues/172}{Japanese translation}. @@ -532,19 +530,23 @@ Note that a subgroup~$H$ of a group $G$ is itself a left coset of $H$ in $G$. \lipsum[2] \begin{theorem}[Lagrange's Theorem] \label{thm:lg} -Let $G$ be a finite group, and let $H$ be a subgroup of $G$. Then the order of $H$ divides the order of $G$. +Let $G$ be a finite group, and let $H$ be a subgroup of $G$. Then the order of $H$ divides the order of $G$. \end{theorem} \ref{thm:lg} \lipsum[3] - + +\begin{theorem}{theorem name}{label text} + The content of theorem. +\end{theorem} + \begin{proposition}[Size of Left Coset] Let $H$ be a finite subgroup of a group $G$. Then each left coset of $H$ in $G$ has the same number of elements as $H$. \end{proposition} \begin{proof} -Let $z$ be some element of $xH \cap yH$. Then $z = xa$ for some $a \in H$, and $z = yb$ for some $b \in H$. If $h$ is any element of $H$ then $ah \in H$ and $a^{-1}h \in H$, since $H$ is a subgroup of $G$. But $zh = x(ah)$ and $xh = z(a^{-1}h)$ for all $h \in H$. Therefore $zH \subset xH$ and $xH \subset zH$, and thus $xH = zH$. Similarly $yH = zH$, and thus $xH = yH$, as required. + Let $z$ be some element of $xH \cap yH$. Then $z = xa$ for some $a \in H$, and $z = yb$ for some $b \in H$. If $h$ is any element of $H$ then $ah \in H$ and $a^{-1}h \in H$, since $H$ is a subgroup of $G$. But $zh = x(ah)$ and $xh = z(a^{-1}h)$ for all $h \in H$. Therefore $zH \subset xH$ and $xH \subset zH$, and thus $xH = zH$. Similarly $yH = zH$, and thus $xH = yH$, as required. \end{proof} \begin{figure}[htbp]