\documentclass[12pt,dvips,landscape]{amsart}
% use latex; dvips; ps2pdf to process
\usepackage{pstricks-add}
\usepackage{geometry}
\topmargin-.75in\textheight7.5in
\oddsidemargin-.4in\textwidth10.5in
\usepackage{type1cm}
\newdimen\fontpt
\def\mynewfont{ \font\nextfont = cmr10 at \fontpt \relax \nextfont}
\newcommand{\decreasingdigits}[3]{{\fontpt=#1pt\mynewfont
#2.\kern-0.8em \fontpt=#1pt \mynewfont\dimin#3\end}}
\newcommand\B{\rule[-.4ex]{0pt}{0pt}}
\def\dimin#1{\ifx#1\end \let\next=\relax
\else\divide\fontpt by \magstephalf \multiply\fontpt by 1000 
\mynewfont\negthinspace#1\B\negthinspace \let\next=\dimin\fi \next}
\thispagestyle{empty}
\parskip17pt\parindent0pt
\newrgbcolor{orange}{1 .5 0}
\newrgbcolor{purple}{.5 0 .5}
\begin{document}
\begin{center}
\Huge\sf \textbf{The axioms of Zermelo-Fraenkel set theory
with choice {\orange ZFC}}\\[5pt]
\green In principle all of mathematics can be derived from these
axioms\end{center}

\bigskip

\LARGE\sf
\begin{center}
\begin{tabular}{ll}
\red Extensionality& 
$\forall X\,\forall Y\,[\,X=Y\quad\Leftrightarrow\quad\forall z(z\in X\ \Leftrightarrow\ z\in Y)\,]$\\[10pt]

\red Pairing& $\forall x\,\forall y\,\exists Z\,\forall z\,[\,z\in Z\quad\Leftrightarrow\quad z=x\text{ or }z=y\,]$\\[10pt]

\red Union& 
$\forall X\,\exists Y\,\forall y\,[\,y\in Y\quad\Leftrightarrow\quad\exists Z(Z\in X\text{ and }y\in Z)\,]$\\[10pt]

\red Empty set& 
$\exists X\,\forall y\,[\,y\notin X\,]$ \qquad \blue(this set $X$ is denoted by $\emptyset$)\\[10pt]

\red Infinity& 
$\exists X\,[\,\emptyset\in X\text{ and }\forall x(x\in X\Rightarrow x\cup\{ x\}\in X)\,]$\\[10pt]

\red Power set& 
$\forall X\,\exists Y\,\forall Z\,[\,Z\in Y\quad\Leftrightarrow\quad
\forall z(z\in Z\ \Rightarrow\ z\in X)\,]$\\[10pt]

\red Replacement& $\forall x\in X\,\exists!y\,P(x,y)\quad\Rightarrow
\quad [\,\exists Y\,\forall y\,(y\in Y\ \Leftrightarrow\ \exists x\in X\,(P(x,y)))\,]$\\[10pt]

\red Regularity& $\forall X\,[\,X\ne\emptyset\quad\Rightarrow\quad\exists Y\in X\,(X\cap Y=\emptyset)\,]$\\[10pt]

\begin{tabular}{l}
\red \!Axiom of choice\\
\ \end{tabular}& \begin{tabular}{l}
\!$\forall X\,[\,\emptyset\notin X\text{ and }
\forall Y,Z\in X(Y\ne Z\ \Rightarrow Y\cap Z=\emptyset)\\
\hskip3.8in
\Rightarrow\quad\exists Y\,\forall Z\in X\,\exists!z\in Z\,(z\in Y)\,]$
\end{tabular}
\end{tabular}

\blue
$\forall=$ for all\qquad $\exists!=$ there exists a unique\qquad
$P$ is any formula that does not contain $Y$

$z\in X\cup Y\ \Leftrightarrow\ z\in X\text{ or }z\in Y$
\qquad $z\in X\cap Y\ \Leftrightarrow\ z\in X\text{ and }z\in Y$
\end{center}

\bigskip

\begin{center}\tiny 
Set theory axioms --- 
Math Poster 2007 --- math.chapman.edu
\end{center}
\end{document}