\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{t92\_zfmisc\_1 (TMKXsZteHoQsigcqraYaaw9wZqFS7dBPKsX)}
\maketitle
Let $ k2\_zfmisc\_1 : {{\iota}{\unicdbdjdcd}{{\iota}{\unicdbdjdcd}{\iota}}}$ be given.
Let $ k4\_tarski : {{\iota}{\unicdbdjdcd}{{\iota}{\unicdbdjdcd}{\iota}}}$ be given.
Assume the following.
\begin{equation}
\begin{array}{c}{\unicdcdadad} X0 .  {\unicdcdadad} X1 .  {\unicdcdadad} X2 .  {\unicdcdadad} X3 .  ({\unicdcdadad} X4 .  {\unicdcdadad} X5 .  \\
(k4\_tarski~X4~X5 \in k2\_zfmisc\_1~X0~X1){\unicdbdjded}(k4\_tarski~X4~X5 \in k2\_zfmisc\_1~\\
X2~X3)){\unicdbdjdcd}(k2\_zfmisc\_1~X0~X1=k2\_zfmisc\_1~X2~X3)\end{array}
\label{hyp:ehidjdpfkhggngjgdhdgpfbd}
\end{equation}
Assume the following.
\begin{equation}
{{\unicdcdadad} X0 .  {{\unicdcdadad} X1 .  {{({{\unicdcdadad} X2 .  {{({{X2} \in {X0}})}{\unicdbdjded}{({{X2} \in {X1}})}}})}{\unicdbdjdcd}{({{X0}={X1}})}}}}\label{hyp:ehbdpfehbgchdhlgjg}
\end{equation}
Assume the following.
\begin{equation}
\begin{array}{c}{\unicdcdadad} X0 .  {\unicdcdadad} X1 .  {\unicdcdadad} X2 .  {\unicdcdadad} X3 .  (k4\_tarski~X0~X1 \in \\
k2\_zfmisc\_1~X2~X3){\unicdbdjded}((X0 \in X2){\unicdcdcdhd}(X1 \in X3))\end{array}
\label{hyp:mgedgdpfkhggngjgdhdgpfbd}
\end{equation}
\begin{theorem}\label{thm:ehjdcdpfkhggngjgdhdgpfbd}
${{\unicdcdadad} X0 .  {{\unicdcdadad} X1 .  {{({{{{k2\_zfmisc\_1}~{X0}}~{X0}}={{{k2\_zfmisc\_1}~{X1}}~{X1}}})}{\unicdbdjdcd}{({{X0}={X1}})}}}}$.
\end{theorem}
\end{document}
