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tcb theorem

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EthanDeng
2022-04-18 18:31:16 +08:00
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elegantbook-en.tex
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@@ -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]