I am trying to modify some of the theorem environments defined in ntheorem
. I wanted the definitions in my document to be boxed the way in which ntheorem
does for the framed theorem classes which they define by:
\theoremclass{Theorem}
\theoremstyle{break}
\newframedtheorem{importantTheorem}[Theorem]{Theorem}
and so I modified the above code to the following:
\theoremclass{Theorem}
\theoremstyle{break}
\newframedtheorem{defn}[Theorem]{Definition}
then in my document called up an instance of a definition by:
\begin{defn}[Logical Equivalance] Two propositions are said to be logically equivalent iff ...
\end{defn}
Now, I wish to modify their shaded theorem environment which is coded as follows:
\theoremclass{Theorem}
\theoremstyle{break}
\newshadedtheorem{moreImportantTheorem}[Theorem]{Theorem}
I tried the following:
\theoremclass{Theorem}
\theoremstyle{break}
\newshadedtheorem{prop}[Theorem]{Proposition}
but keep getting the error:
Undefined control sequence: begin{prop}
Can anyone help me with this?
\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}
Thanks!!!
I am now added more "theorem"-like enviornments and am getting even more errors and would ask that some one please help me use mdframed to fix the problem if that is possible. Here is a current and up-to-date 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}
Best Answer
You can fix your MWE by adding the line
which is needed for both of your subsequent theorem-like environments,
prop
anddefn
.Note that this MWE relies upon the
pstricks
package, so needs to be compiled through thelatex->dvips->ps2pdf
unless you want to follow the instructions in How to use PSTricks in pdfLaTeX?For all of your framing needs I would highly recommend the
mdframed
package, which addresses the many short comings of its competitors.Here's a version of the previous MWE using the
mdframed
package; note that this package does not rely upon thepstricks
package (in contrast to the previous method). As such, you can (easily) compile this document withpdflatex
.Of course, the
mdframed
package can be told to usepstricks
ortikz
if you wish, but that is beyond the scope of the question- see the manual for more details.Update, following the question edit.
With the additional theorem-like environments, this MWE works- note that you can't define a theorem-like environment twice using
\newtheorem