\documentclass[12pt,titlepage]{article}

\usepackage{amsmath}
\usepackage{mathrsfs}
\usepackage{amsfonts}
\usepackage{amssymb}
\usepackage{amsthm}
\usepackage{mathtools}
\usepackage{graphicx}
\usepackage{color}
\usepackage{ucs}
\usepackage[utf8x]{inputenc}
\usepackage{xparse}
\usepackage{tikz}
\usepackage{hyperref}

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% math-mode versions of \rlap, etc
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\theoremstyle{plain}
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\newtheorem{lemma}{Lemma}
\newtheorem{prop}{Proposition}
\newtheorem{cor}{Corollary}
\newtheorem*{utheorem}{Theorem}
\newtheorem*{ulemma}{Lemma}
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\theoremstyle{definition}
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\theoremstyle{remark}
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\newtheorem{note}{Note}
\newtheorem*{uremark}{Remark}
\newtheorem*{unote}{Note}

%-------------------------------------------------------------------

\begin{document}

%-------------------------------------------------------------------

\section*{Theorems}

[[!include contents]]

Instiki supports something resembling the \href{ftp://ftp.ams.org/pub/tex/doc/amscls/amsthdoc.pdf}{amsthm} Theorem Environments.

The following input

\begin{verbatim}+-- {: .un_defn}
###### Definition
Let $H$ be a subgroup of a group $G$.  A *left coset* of
$H$ in $G$ is a subset of $G$ that is of the form $x H$,
where $x \in G$ and $x H = \{ x h : h \in H \}$.

Similarly a *right coset* of $H$ in $G$ is a subset of $G$
that is of the form $H x$, where $H x = \{ h x : h \in H\}$.
=--\end{verbatim}
produces an \emph{un-numbered} \textbf{Definition}:

\begin{udefn}
Let $H$ be a subgroup of a group $G$. A \emph{left coset} of $H$ in $G$ is a subset of $G$ that is of the form $x H$, where $x \in G$ and $x H = \{ x h : h \in H \}$.

Similarly a \emph{right coset} of $H$ in $G$ is a subset of $G$ that is of the form $H x$, where $H x = \{ h x : h \in H \}$.

\end{udefn}
Similarly, the input

\begin{verbatim} +-- {: .num_lemma #LeftCosetsDisjoint}
 ###### Lemma
 Let $H$ be a subgroup of a group $G$, and let $x$ and $y$ be elements
 of $G$. Suppose that $x H \cap y H$ is non-empty. Then $x H = y H$.
 =--

 +-- {: .proof}
 ###### Proof
 Let $z$ be some element of $x H \cap y H$.  Then $z = x a$ for some
 $a \in H$, and $z = y b$ for some $b \in H$.  If $h$ is any element
 of $H$ then $a h \in H$ and $a^{-1}h \in H$, since $H$ is a subgroup
 of $G$. But $z h = x(a h)$ and $x h = z(a^{-1}h)$ for all $h \in H$.
 Therefore $z H \subset x H$ and $x H \subset z H$, and thus
 $x H = z H$. Similarly $y H = z H$, and thus $x H = y H$, as required.
 =--

 +-- {: .num_lemma #SizeOfLeftCoset}
 ###### Lemma
 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 $H$.
 =--

 +-- {: .num_theorem #Lagrange}
 ###### Theorem
 **(Lagrange's Theorem)**. 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{verbatim}
produces a \emph{numbered} \textbf{Lemma}, followed by a \textbf{Proof}, followed by another numbered \textbf{Lemma} and a numbered \textbf{Theorem}:

\begin{lemma}
\label{LeftCosetsDisjoint}\hypertarget{LeftCosetsDisjoint}{}
Let $H$ be a subgroup of a group $G$, and let $x$ and $y$ be elements of $G$. Suppose that $x H \cap y H$ is non-empty. Then $x H = y H$.

\end{lemma}
\begin{proof}
Let $z$ be some element of $x H \cap y H$. Then $z = x a$ for some $a \in H$, and $z = y b$ for some $b \in H$. If $h$ is any element of $H$ then $a h \in H$ and $a^{-1}h \in H$, since $H$ is a subgroup of $G$. But $z h = x(a h)$ and $x h = z(a^{-1}h)$ for all $h \in H$. Therefore $z H \subset x H$ and $x H \subset z H$, and thus $x H = z H$. Similarly $y H = z H$, and thus $x H = y H$, as required.

\end{proof}
\begin{lemma}
\label{SizeOfLeftCoset}\hypertarget{SizeOfLeftCoset}{}
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{lemma}
\begin{theorem}
\label{Lagrange}\hypertarget{Lagrange}{}
\textbf{(Lagrange's Theorem)}. 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}
These can be linked-to in the obvious fashion.

\begin{verbatim} See Lemma \ref{LeftCosetsDisjoint}.\end{verbatim}
produces

\begin{quote}%
See Lemma \ref{LeftCosetsDisjoint}.


\end{quote}
as you'd expect. Note that each type of numbered Theorem-like environment has its own number counter.

The key is in the CSS classes.

\begin{verbatim} .proof\end{verbatim}
produces a \textbf{Proof} environment.

\begin{verbatim} .num_*\end{verbatim}
where ``{\colorbox[rgb]{1.00,0.93,1.00}{\tt \char42}}'' is one of

\begin{verbatim} theorem     (for Theorem)
 lemma       (for Lemma)
 prop        (for Proposition)
 cor         (for Corollary)
 defn        (for Definition)
 example     (for Example)
 remark      (for Remark)
 note        (for Note)\end{verbatim}
produces a numbered \textbf{Theorem} (\textbf{Lemma}, \ldots{} etc).

\begin{verbatim}.un_*\end{verbatim}
produces the corresponding \emph{un-numbered} version.

The above examples were drawn from \href{http://www.maths.tcd.ie/~dwilkins/LaTeXPrimer/Theorems.html}{here}, though the \LaTeX\xspace  output (as can be seen, by clicking on the {\colorbox[rgb]{1.00,0.93,1.00}{\tt TeX}} link at the bottom of the page) is, instead, {\colorbox[rgb]{1.00,0.93,1.00}{\tt amsthm}}-compatible.

\hypertarget{extensions}{}\subsubsection*{{Extensions}}\label{extensions}

Extending this facility, with new Theorem-like environments is easy. For example, say we wish to add a \textbf{Conjecture} and a \textbf{Problem} environment.

In our \LaTeX\xspace  template, we will need

\begin{verbatim} \theoremstyle{plain}
 \newtheorem{conjecture}{Conjecture}
 \newtheorem*{uconjecture}{Conjecture}
 \theoremstyle{definition}
 \newtheorem{problem}{Problem}
 \newtheorem*{uproblem}{Problem}\end{verbatim}
(Note the naming convention: the un-numbered version has a ``u'' prepended to the name.)

The CSS styles for our wiki get augmented with

and we're now ready to use our newly-minted CSS class-names, {\colorbox[rgb]{1.00,0.93,1.00}{\tt un\char95conjecture}}, {\colorbox[rgb]{1.00,0.93,1.00}{\tt num\char95conjecture}}, {\colorbox[rgb]{1.00,0.93,1.00}{\tt un\char95problem}} and {\colorbox[rgb]{1.00,0.93,1.00}{\tt num\char95problem}}.



\end{document}