\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{t70\_eqrel\_1 (TMJCx2n8D1RCT6sxQwtC3YGwhrreSFVw4kn)}
\maketitle
Let $ m1\_eqrel\_1 : {{\iota}{\unicdbdjdcd}{{\iota}{\unicdbdjdcd}{o}}}$ be given.
Let $ m1\_subset\_1 : {{\iota}{\unicdbdjdcd}{{\iota}{\unicdbdjdcd}{o}}}$ be given.
Let $ k1\_zfmisc\_1 : {{\iota}{\unicdbdjdcd}{\iota}}$ be given.
Let $ r1\_xboole\_0 : {{\iota}{\unicdbdjdcd}{{\iota}{\unicdbdjdcd}{o}}}$ be given.
Let $ k5\_setfam\_1 : {{\iota}{\unicdbdjdcd}{{\iota}{\unicdbdjdcd}{\iota}}}$ be given.
Let $ k1\_xboole\_0 : {\iota}$ be given.
Assume the following.
\begin{equation}
\begin{array}{c}{\unicdcdadad} X0 .  {\unicdcdadad} X1 .  {\unicdcdadad} X2 .  ((X0 \in X1){\unicdcdcdhd}(m1\_subset\_1~X1~(k1\_zfmisc\_1~\\
X2))){\unicdbdjdcd}(m1\_subset\_1~X0~X2)\end{array}
\label{hyp:ehedpfdhfhcgdhfgeh}
\end{equation}
Assume the following.
\begin{equation}
\begin{array}{c}{\unicdcdadad} X0 .  {\unicdcdadad} X1 .  (\preoh (\preoh r1\_xboole\_0~X0~X1){\unicdcdcdhd}({\unicdcdadad} X2 .  \preoh (X2 \in \\
X0){\unicdcdcdhd}(X2 \in X1))){\unicdcdcdhd}(\preoh ({\unicdcdaddd} X2 .  (X2 \in X0){\unicdcdcdhd}(X2 \in X1)){\unicdcdcdhd}(r1\_xboole\_0~\\
X0~X1))\end{array}
\label{hyp:ehddpfihcgpgpgmgfgpfad}
\end{equation}
Assume the following.
\begin{equation}
\begin{array}{c}{\unicdcdadad} X0 .  {\unicdcdadad} X1 .  (m1\_eqrel\_1~X1~X0){\unicdbdjdcd}(m1\_subset\_1~X1~(k1\_zfmisc\_1~\\
(k1\_zfmisc\_1~X0)))\end{array}
\label{hyp:egehpfngbdpffgbhchfgmgpfbd}
\end{equation}
Assume the following.
\begin{equation}
\begin{array}{c}{\unicdcdadad} X0 .  {\unicdcdadad} X1 .  (m1\_subset\_1~X1~(k1\_zfmisc\_1~(k1\_zfmisc\_1~\\
X0))){\unicdbdjdcd}((m1\_eqrel\_1~X1~X0){\unicdbdjded}((k5\_setfam\_1~X0~X1=X0){\unicdcdcdhd}({\unicdcdadad} X2 .  \\
(m1\_subset\_1~X2~(k1\_zfmisc\_1~X0)){\unicdbdjdcd}((X2 \in X1){\unicdbdjdcd}((X2{\unicdcdgdad}k1\_xboole\_0){\unicdcdcdhd}\\
({\unicdcdadad} X3 .  (m1\_subset\_1~X3~(k1\_zfmisc\_1~X0)){\unicdbdjdcd}(\preoh (X3 \in X1){\unicdcdcdhd}((X2{\unicdcdgdad}\\
X3){\unicdcdcdhd}(\preoh r1\_xboole\_0~X2~X3)))))))))\end{array}
\label{hyp:egedpffgbhchfgmgpfbd}
\end{equation}
\begin{theorem}\label{thm:ehhdadpffgbhchfgmgpfbd}
$$\begin{array}{c}{\unicdcdadad} X0 .  {\unicdcdadad} X1 .  (m1\_eqrel\_1~X1~X0){\unicdbdjdcd}({\unicdcdadad} X2 .  {\unicdcdadad} X3 .  \\
{\unicdcdadad} X4 .  ((X2 \in X3){\unicdcdcdhd}((X3 \in X1){\unicdcdcdhd}((X2 \in X4){\unicdcdcdhd}(X4 \in X1)))){\unicdbdjdcd}(X3=X4))\end{array}$$
\end{theorem}
\end{document}
