我正在尝试修改 中定义的一些定理环境。我希望我的文档中的定义能够ntheorem像ntheorem框架定理类他们的定义是：
\theoremclass{Theorem}
\theoremstyle{break}
\newframedtheorem{importantTheorem}[Theorem]{Theorem}
因此我将上面的代码修改如下：
\theoremclass{Theorem}
\theoremstyle{break}
\newframedtheorem{defn}[Theorem]{Definition}
然后在我的文档中调用一个定义实例：
\begin{defn}[Logical Equivalance] Two propositions are said to be logically equivalent iff ...
\end{defn}
现在，我想修改他们的阴影定理环境，其编码如下：
\theoremclass{Theorem}
\theoremstyle{break}
\newshadedtheorem{moreImportantTheorem}[Theorem]{Theorem}
我尝试了以下操作：
\theoremclass{Theorem}
\theoremstyle{break}
\newshadedtheorem{prop}[Theorem]{Proposition}
但仍然出现错误：
Undefined control sequence: begin{prop}
谁能帮我这个？
\documentclass[10pt,a4paper]{article}
\usepackage[left=2.50cm,right=2.50cm,top=2.50cm,bottom=2.75cm]{geometry}
\usepackage{amsmath,amssymb,amscd,amstext,amsbsy,array,color,epsfig}
\usepackage{fancyhdr,framed,latexsym,multicol,pstricks,slashed,xcolor}
\usepackage[amsmath,framed,thmmarks]{ntheorem}
\begin{document}
\theoremstyle{marginbreak}
\theoremheaderfont{\bfseries\scshape}
\theorembodyfont{\slshape}
\theoremsymbol{\ensuremath{\star}}
\theoremseparator{:}
\newtheorem{axm}{Axiom}[section]
\theoremstyle{marginbreak}
\theoremheaderfont{\bfseries\scshape}
\theorembodyfont{\slshape}
\theoremsymbol{\ensuremath{\diamondsuit}}
\theoremseparator{:}
\newtheorem{Theorem}{Theorem}[section]
\theoremclass{Theorem}
\theoremstyle{break}
\newshadedtheorem{prop}[Theorem]{Proposition}
\theoremstyle{changebreak}
\theoremsymbol{\ensuremath{\heartsuit}}
\theoremindent0.5cm
\theoremnumbering{greek}
\newtheorem{lem}{Lemma}[section]
\theoremindent0cm
\theoremsymbol{\ensuremath{\spadesuit}}
\theoremnumbering{arabic}
\newtheorem{cor}[Theorem]{Corollary}
\theoremstyle{plain}
\theorembodyfont{\upshape}
\theoremsymbol{\ensuremath{\bullet}}
\theoremseparator{}
\newtheorem{exm}{Example}
\theoremclass{Theorem}
\theoremstyle{plain}
\theoremsymbol{\ensuremath{\clubsuit}}
\newframedtheorem{defn}[Theorem]{Definition}
\theoremheaderfont{\sc}
\theorembodyfont{\upshape}
\theoremstyle{nonumberplain}
\theoremseparator{.}
\theoremsymbol{\rule{1ex}{1ex}}
\newtheorem{proof}{Proof}
\theoremstyle{plain}
\theorembodyfont{\upshape}
\theoremsymbol{\ensuremath{\ast}}
\theoremseparator{.}
\newtheorem{rem}{Remark}
\theoremstyle{plain}
\theorembodyfont{\upshape}
\newtheorem{exc}{Exercise}[section]
\begin{defn}[Logical Equivalance] Two propositions are said to be logically equivalent iff ...
\end{defn}
\begin{prop}
Let $P$ and $Q$ be propositions. Then ...
\end{prop}
\end{document}
谢谢！！！
我现在添加了更多类似“定理”的环境，并且出现了更多错误，如果可能的话，请有人帮助我使用 mdframed 来解决问题。这是当前最新的 mwe：
\documentclass[a4paper,12pt,twoside]{book}
\usepackage[left=2.50cm,right=2.50cm,top=2.50cm,bottom=2.75cm]{geometry}
\usepackage{amsmath,amssymb,amscd,amsbsy,array,color,epsfig}
\usepackage{fancyhdr,framed,latexsym,multicol,pstricks,slashed,xcolor}
\usepackage{picture}
\usepackage{indentfirst}
\usepackage{enumitem}
\usepackage{tikz}
\usepackage{subfig}
\usetikzlibrary{calc,positioning,shapes.geometric}
\setenumerate[1]{label=(\alph*)}
\usepackage[amsmath,framed,thmmarks]{ntheorem}
\newtheorem{Theorem}{Thm}
\theoremclass{Theorem}
\theoremstyle{break}
\shadecolor{blue}
\newshadedtheorem{them}[Theorem]{Theorem}
\theoremclass{Theorem}
\theoremstyle{break}
\shadecolor{gray}
\newshadedtheorem{prop}{Proposition}[section]
\theoremclass{Theorem}
\theoremstyle{plain}
\newframedtheorem{lema}[Theorem]{Lemma}
\theoremclass{Theorem}
\theoremstyle{plain}
\newframedtheorem{coro}[Theorem]{Corollary}
\theoremstyle{plain}
\theoremsymbol{\ensuremath{\blacktriangle}}
\theoremseparator{.}
\theoremprework{\bigskip\hrule}
\theorempostwork{\hrule\bigskip}
\newtheorem{defn}{Definition}
\theoremstyle{plain}
\theorembodyfont{\upshape}
\theoremsymbol{\ensuremath{\bullet}}
\theoremseparator{}
\newtheorem{exam}{Example}
\theoremstyle{plain}
\theorembodyfont{\upshape}
\theoremsymbol{\ensuremath{\bullet}}
\newtheorem{exer}{Exercise}[section]
\theoremheaderfont{\sc}
\theorembodyfont{\color{blue}\bfseries\boldmath}
\theoremstyle{nonumberplain}
\theoremseparator{.}
\theoremsymbol{\rule{1ex}{1ex}}
\newtheorem{proof}{Proof}
\theoremstyle{plain}
\theorembodyfont{\upshape}
\theoremsymbol{\ensuremath{\bigstar}}
\theoremseparator{.}
\newtheorem{remk}{Remark}
\def \all {\forall}
\def \ex {\exists}
\def \imp {\Rightarrow}
\def \limp {\Leftarrow}
\def \iff {\Longleftrightarrow}
\def \contra {\rightarrow\negmedspace\leftarrow}
\def \es {\emptyset}
\def \st {\backepsilon}
\def \bn{\mathbb N}
\def \bz{\mathbb Z}
\def \bq{\mathbb Q}
\def \br{\mathbb R}
\def \bc{\mathbb C}
\def \bp{\mathbb P}
\def \bt{\mathbb T}
\begin{defn}[Statement/Proposition]
Declarative sentences or strings of symbols in mathematics which can be said to have \textit{exactly} one \textit{truth value}, that is, are either true (denoted T), or false (denoted F), are known as \textbf{statements} or \textbf{propositions}.
\end{defn}
\begin{exam}
Hence, the truth value of the negation of a proposition is \textit{merely} the opposite of the truth value of said proposition. Hence, the truth value of the proposition '$7$ is divisible by $2$' is the proposition 'It is not the case that $7$ is not divisible by $2$' or '$7$ is not divisible by $2$' (both of which are true).
\end{exam}
\begin{prop}
Let $P$ and $Q$ be propositions. Then:
\begin{enumerate}
\item $P \imp Q \equiv (\neg Q) \imp (\neg P).$
\item $P \imp Q \not \equiv Q \imp P.$
\end{enumerate}
\end{prop}
\begin{prop}
Let $P,Q,$ and $R$ be propositions. Then:
\begin{enumerate}
\item $P \imp Q \equiv (\neg P) \vee (Q).$
\item $P \iff Q \equiv (P \imp Q) \wedge (Q \imp P).$
\item $\neg(P \imp Q) \equiv (P) \wedge (\neg Q).$
\item $\neg(P \wedge Q) \equiv (P) \imp (\neg Q) \equiv (Q) \imp (\neg P).$
\item $P \imp (Q \imp R) \equiv (P \wedge Q) \imp R.$
\item $P \imp (Q \vee R) \equiv (P \imp Q) \wedge (P \imp R).$
\item $(P \vee Q) \imp R \equiv (P \imp R) \wedge (Q \imp R).$
\end{enumerate}
\end{prop}
\begin{proof}
The proof for the above proposition is left to the reader. All of the above statements may be proved using truth tables.
\end{proof}
\begin{axm}[Field Axioms of $\br$]
On the set $\br$ of real numbers, there are two binary operations, denoted by $\pmb{+}$ and $\pmb{\cdot}$ and called \textbf{addition} and \textbf{multiplication} respectively. These operations satisfy the following properties:
\begin{itemize}
\item[$A_0$] $x,y \in \br \imp x+y \in \br \q \all \, x,y \in \br$. [additive closure]
\item[$A_1$] $x+y=y+x \q \all \, x,y \in \br$. [additive commutativity]
\item[$A_2$] $(x+y)+z=x+(y+z) \q \all \, x,y,z \in \br$. [additive associativity]
\item[$A_3$] There is a unique $0 \in \br \text{ such that } 0+x=x=x+0 \q \all \, x \in \br$. [existence of an additive identity]
\item[$A_4$] There is a unique $-x \in \br \text{ such that } x+(-x)=0=(-x)+x \q \all \, x \in \br$. [existence of an additive inverse]
\item[$M_0$] $x,y \in \br \imp x \cdot y \in \br \q \all \, x,y \in \br$. [multiplicative closure]
\item[$M_1$] $x \cdot y=y \cdot x \q \all \, x,y \in \br$. [multiplicative commutativity]
\item[$M_2$] $(x \cdot y) \cdot z=x \cdot (y \cdot z) \q \all \, x,y,z \in \br$. [multiplicative associativity]
\item[$M_3$] There is a unique $1 \in \br \text{ such that } 1 \cdot x=x=x \cdot 1 \q \all \, x \in \br$. [existence of multiplicative identity]
\item[$M_4$] There is a unique $\nicefrac{1}{x} \in \br \text{ such that } x \cdot (\nicefrac{1}{x})=1=(\nicefrac{1}{x}) \cdot x \q \all x \in \br$. [existence of multiplicative inverse]
\item[$AM_1$] $x \cdot (y+z) = (x \cdot y) + (x \cdot z)$ and $(y + z) \cdot x = (y \cdot x) + (z \cdot x)$. [distributivity]
\end{itemize}
\end{axm}
\begin{rem}
The reader should be familiar with all of the aforementioned field properties. We note that all of the `familiar' properties of algebra (those learned in middle school and high school, for example) may be deduced from this list. We now establish the basic fact that both the additive identity, $0$, and the multiplicative identity are in fact unique; and that multiplication by $0$ always results in $0$.
\end{rem}
\end{document}
答案1
您可以通过添加行来修复您的 MWE
\newtheorem{Theorem}{Thm}
这对于您后续的两个类定理环境prop和都是必需的defn。
% arara: latex
% arara: dvips
% arara: ps2pdf
% !arara: indent: {overwrite: yes}
\documentclass[10pt,a4paper]{article}
\usepackage[left=2.50cm,right=2.50cm,top=2.50cm,bottom=2.75cm]{geometry}
\usepackage{amsmath}
\usepackage{pstricks}
\usepackage{framed}
\usepackage[amsmath,framed,thmmarks]{ntheorem}
\newtheorem{Theorem}{Thm}
\theoremclass{Theorem}
\theoremstyle{break}
\newshadedtheorem{prop}[Theorem]{Proposition}
\theoremclass{Theorem}
\theoremstyle{plain}
\theoremsymbol{\ensuremath{\clubsuit}}
\newframedtheorem{defn}[Theorem]{Definition}
\begin{document}
\begin{defn}[Logical Equivalance]
Two propositions are said to be logically equivalent iff ...
\end{defn}
\begin{prop}
Let $P$ and $Q$ be propositions. Then ...
\end{prop}
\end{document}
请注意，此 MWE 依赖于pstricks包，因此需要通过编译，latex->dvips->ps2pdf除非你想按照如何在 pdfLaTeX 中使用 PSTricks？
对于你所有的框架需求，我强烈推荐mdframed该方案解决了其竞争对手的诸多缺点。
这是之前 MWE 的一个版本，使用mdframed包；请注意，这个包没有依靠包上pstricks（与之前的方法相反）。这样，您可以（轻松地）使用 编译此文档pdflatex。
% arara: pdflatex
% !arara: indent: {overwrite: yes}
\documentclass[10pt,a4paper]{article}
\usepackage[left=2.50cm,right=2.50cm,top=2.50cm,bottom=2.75cm]{geometry}
\usepackage{amsmath}
\usepackage[amsmath,framed,thmmarks]{ntheorem}
\usepackage[ntheorem,xcolor]{mdframed}
\newtheorem{Theorem}{Thm}
\theoremclass{Theorem}
\theoremstyle{break}
\newmdtheoremenv[
outerlinewidth = 2 ,%
roundcorner = 10 pt ,%
leftmargin = 40 ,%
rightmargin = 40 ,%
backgroundcolor=yellow!40,%
outerlinecolor=blue!70!black,%
innertopmargin=\topskip,%
splittopskip = \topskip ,%
ntheorem = true ,%
]{prop}[Theorem]{Proposition}
\theoremstyle{plain}
\theoremsymbol{\ensuremath{\clubsuit}}
%\newframedtheorem{defn}[Theorem]{Definition}
\newmdtheoremenv{defn}[Theorem]{Definition}
\begin{document}
\begin{defn}[Logical Equivalance]
Two propositions are said to be logically equivalent iff ...
\end{defn}
\begin{prop}
Let $P$ and $Q$ be propositions. Then ...
\end{prop}
\end{document}
当然，mdframed可以告诉包使用pstricks或者tikz如果您愿意，但这超出了问题的范围 - 请参阅手册了解更多详细信息。
根据问题编辑进行更新。
有了额外的类定理环境，这个 MWE 就可以工作了。请注意，你不能使用以下方法两次定义类定理环境\newtheorem
% arara: latex
% arara: dvips
% arara: ps2pdf
% !arara: indent: {overwrite: yes}
\documentclass[10pt,a4paper]{article}
\usepackage[left=2.50cm,right=2.50cm,top=2.50cm,bottom=2.75cm]{geometry}
\usepackage{amsmath}
\usepackage{pstricks}
\usepackage{framed}
\usepackage[amsmath,framed,thmmarks]{ntheorem}
\theoremstyle{marginbreak}
\theoremheaderfont{\bfseries\scshape}
\theorembodyfont{\slshape}
\theoremsymbol{\ensuremath{\diamondsuit}}
\theoremseparator{:}
\newtheorem{Theorem}{Theorem}[section]
\theoremclass{Theorem}
\theoremstyle{break}
\newshadedtheorem{prop}[Theorem]{Proposition}
\theoremclass{Theorem}
\theoremstyle{plain}
\theoremsymbol{\ensuremath{\clubsuit}}
\newframedtheorem{defn}[Theorem]{Definition}
\theoremstyle{marginbreak}
\theoremheaderfont{\bfseries\scshape}
\theorembodyfont{\slshape}
\theoremsymbol{\ensuremath{\star}}
\theoremseparator{:}
\newtheorem{axm}{Axiom}[section]
\theoremstyle{changebreak}
\theoremsymbol{\ensuremath{\heartsuit}}
\theoremindent0.5cm
\theoremnumbering{greek}
\newtheorem{lem}{Lemma}[section]
\theoremindent0cm
\theoremsymbol{\ensuremath{\spadesuit}}
\theoremnumbering{arabic}
\newtheorem{cor}[Theorem]{Corollary}
\theoremstyle{plain}
\theorembodyfont{\upshape}
\theoremsymbol{\ensuremath{\bullet}}
\theoremseparator{}
\newtheorem{exm}{Example}
\theoremheaderfont{\sc}
\theorembodyfont{\upshape}
\theoremstyle{nonumberplain}
\theoremseparator{.}
\theoremsymbol{\rule{1ex}{1ex}}
\newtheorem{proof}{Proof}
\theoremstyle{plain}
\theorembodyfont{\upshape}
\theoremsymbol{\ensuremath{\ast}}
\theoremseparator{.}
\newtheorem{rem}{Remark}
\theoremstyle{plain}
\theorembodyfont{\upshape}
\newtheorem{exc}{Exercise}[section]
\begin{document}
\begin{defn}[Logical Equivalance]
Two propositions are said to be logically equivalent iff ...
\end{defn}
\begin{prop}
Let $P$ and $Q$ be propositions. Then ...
\end{prop}
\end{document}