\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{t45\_eqrel\_1 (TMWQMn145xy2j5NCriLKNWgTkhwrRvJq5iW)}
\maketitle
Let $ m1\_eqrel\_1 : {{\iota}{\unicdbdjdcd}{{\iota}{\unicdbdjdcd}{o}}}$ be given.
Let $ k1\_xboole\_0 : {\iota}$ be given.
Assume the following.
\begin{equation}
{{\unicdcdadad} X0 .  {{({{{m1\_eqrel\_1}~{X0}}~{k1\_xboole\_0}})}{\unicdbdjdcd}{({{X0}={k1\_xboole\_0}})}}}\label{hyp:ehddcdpffgbhchfgmgpfbd}
\end{equation}
Assume the following.
\begin{equation}
{{\unicdcdadad} X0 .  {{\unicdcdaddd} X1 .  {{{m1\_eqrel\_1}~{X1}}~{X0}}}}\label{hyp:fgihjgdhehfgogdgfgpfngbdpffgbhchfgmgpfbd}
\end{equation}
\begin{theorem}\label{thm:ehedfdpffgbhchfgmgpfbd}
${{{m1\_eqrel\_1}~{k1\_xboole\_0}}~{k1\_xboole\_0}}$.
\end{theorem}
\end{document}
