\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=}
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\newcommand\unicdcdadjd{\notin}
\newcommand\unicdbdjded{\Leftrightarrow}
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\newcommand\unicdcdidid{\not\subseteq}
\newcommand\preoh[1]{\lnot #1}
\newcommand\mname[1]{{\mathsf{#1}}}
\begin{document}
\title{t26\_eqrel\_1 (TMXosWKnA4GS4DcLUKhgbww5D2Po59kdrnK)}
\maketitle
Let $ k6\_eqrel\_1 : {{\iota}{\unicdbdjdcd}{{\iota}{\unicdbdjdcd}{{\iota}{\unicdbdjdcd}{{\iota}{\unicdbdjdcd}{\iota}}}}}$ be given.
Let $ k1\_eqrel\_1 : {{\iota}{\unicdbdjdcd}{\iota}}$ be given.
Let $ k9\_relat\_1 : {{\iota}{\unicdbdjdcd}{{\iota}{\unicdbdjdcd}{\iota}}}$ be given.
Let $ k2\_zfmisc\_1 : {{\iota}{\unicdbdjdcd}{{\iota}{\unicdbdjdcd}{\iota}}}$ be given.
Let $ m1\_subset\_1 : {{\iota}{\unicdbdjdcd}{{\iota}{\unicdbdjdcd}{o}}}$ be given.
Let $ k1\_zfmisc\_1 : {{\iota}{\unicdbdjdcd}{\iota}}$ be given.
Assume the following.
\begin{equation}
\begin{array}{c}{\unicdcdadad} X0 .  {\unicdcdadad} X1 .  {\unicdcdadad} X2 .  (X0 \in X1){\unicdbdjdcd}(k9\_relat\_1~(k2\_zfmisc\_1~\\
X1~X2)~X0=X2)\end{array}
\label{hyp:ehbdgdidpfchfgmgbgehpfbd}
\end{equation}
Assume the following.
\begin{equation}
\begin{array}{c}{\unicdcdadad} X0 .  {\unicdcdadad} X1 .  {\unicdcdadad} X2 .  {\unicdcdadad} X3 .  (m1\_subset\_1~X2~(\\
k1\_zfmisc\_1~(k2\_zfmisc\_1~X0~X1))){\unicdbdjdcd}(k6\_eqrel\_1~X0~X1~X2~X3=k9\_relat\_1~\\
X2~X3)\end{array}
\label{hyp:chfgegfgggjgogjgehjgpgogpflggdpffgbhchfgmgpfbd}
\end{equation}
Assume the following.
\begin{equation}
\begin{array}{c}{\unicdcdadad} X0 .  m1\_subset\_1~(k1\_eqrel\_1~X0)~(k1\_zfmisc\_1~(k2\_zfmisc\_1~\\
X0~X0))\end{array}
\label{hyp:egehpflgbdpffgbhchfgmgpfbd}
\end{equation}
Assume the following.
\begin{equation}
{{\unicdcdadad} X0 .  {{{k1\_eqrel\_1}~{X0}}={{{k2\_zfmisc\_1}~{X0}}~{X0}}}}\label{hyp:egbdpffgbhchfgmgpfbd}
\end{equation}
\begin{theorem}\label{thm:ehcdgdpffgbhchfgmgpfbd}
$$\begin{array}{c}{\unicdcdadad} X0 .  {\unicdcdadad} X1 .  (X1 \in X0){\unicdbdjdcd}(k6\_eqrel\_1~X0~X0~(k1\_eqrel\_1~X0)~\\
X1=X0)\end{array}$$
\end{theorem}
\end{document}
