\documentclass{article}
\newtheorem{axiom}{Axiom}
\newtheorem{theorem}{Theorem}
\newtheorem{lemma}{Lemma}
\newtheorem{proposition}{Proposition}
\newtheorem{definition}{Definition}
\newtheorem{example}{Example}
\newtheorem{conjecture}{Conjecture}
\newcommand\unicdcdgdad{\not=}
\newcommand\unicdcdcdbg{\cup}
\newcommand\unicdcdbdgd{\setminus}
\newcommand\unicdcdcdjd{\cap}
\newcommand\unicdcdadad{\forall}
\newcommand\unicdbdjdcd{\Rightarrow}
\newcommand\unicdcdadjd{\notin}
\newcommand\unicdbdjded{\Leftrightarrow}
\newcommand\unicdcdcdid{\lor}
\newcommand\unicdcdcdhd{\land}
\newcommand\uniddcgcg{\lambda}
\newcommand\unicdcdaddd{\exists}
\newcommand\unicdcdidid{\not\subseteq}
\newcommand\preoh[1]{\lnot #1}
\newcommand\mname[1]{{\mathsf{#1}}}
\begin{document}
\title{t1\_dilworth (TMHPMVdTNDhgJsJf68bsLmeYJvpbtkcDciu)}
\maketitle
Let $ k6\_subset\_1 : {{\iota}{\unicdbdjdcd}{{\iota}{\unicdbdjdcd}{\iota}}}$ be given.
Let $ k2\_xboole\_0 : {{\iota}{\unicdbdjdcd}{{\iota}{\unicdbdjdcd}{\iota}}}$ be given.
Let $ k1\_tarski : {{\iota}{\unicdbdjdcd}{\iota}}$ be given.
Let $ k4\_xboole\_0 : {{\iota}{\unicdbdjdcd}{{\iota}{\unicdbdjdcd}{\iota}}}$ be given.
Assume the following.
\begin{equation}
{{\unicdcdadad} X0 .  {{\unicdcdadad} X1 .  {{({{{{k4\_xboole\_0}~{X0}}~{({{k1\_tarski}~{X1}})}}={X0}})}{\unicdbdjded}{({\preoh{{{X1} \in {X0}}}})}}}}\label{hyp:ehfdhdpfkhggngjgdhdgpfbd}
\end{equation}
Assume the following.
\begin{equation}
\begin{array}{c}{\unicdcdadad} X0 .  {\unicdcdadad} X1 .  {\unicdcdadad} X2 .  k4\_xboole\_0~(k4\_xboole\_0~X0~\\
X1)~X2=k4\_xboole\_0~X0~(k2\_xboole\_0~X1~X2)\end{array}
\label{hyp:ehedbdpfihcgpgpgmgfgpfbd}
\end{equation}
Assume the following.
\begin{equation}
{{\unicdcdadad} X0 .  {{\unicdcdadad} X1 .  {{{{k6\_subset\_1}~{X0}}~{X1}}={{{k4\_xboole\_0}~{X0}}~{X1}}}}}\label{hyp:chfgegfgggjgogjgehjgpgogpflggdpfdhfhcgdhfgehpfbd}
\end{equation}
Assume the following.
\begin{equation}
{{\unicdcdadad} X0 .  {{\unicdcdadad} X1 .  {{{{k2\_xboole\_0}~{X0}}~{X1}}={{{k2\_xboole\_0}~{X1}}~{X0}}}}}\label{hyp:dgpgngngfhehbgehjgghjgehjhpflgcdpfihcgpgpgmgfgpfad}
\end{equation}
\begin{theorem}\label{thm:ehbdpfegjgmghhpgchehig}
$$\begin{array}{c}{\unicdcdadad} X0 .  {\unicdcdadad} X1 .  {\unicdcdadad} X2 .  (\preoh X2 \in X0){\unicdbdjdcd}(k6\_subset\_1~X0~(k2\_xboole\_0~\\
X1~(k1\_tarski~X2))=k6\_subset\_1~X0~X1)\end{array}$$
\end{theorem}
\end{document}
