\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{t2\_exchsort (TMdMQzkyoS4DAQJhfYNe4v2J6MTt1EH6sWW)}
\maketitle
Let $ v3\_ordinal1 : {{\iota}{\unicdbdjdcd}{o}}$ be given.
Assume the following.
\begin{equation}
\begin{array}{c}{\unicdcdadad} X0 .  (v3\_ordinal1~X0){\unicdbdjdcd}({\unicdcdadad} X1 .  (v3\_ordinal1~X1){\unicdbdjdcd}(\preoh \\
(\preoh X0 \in X1){\unicdcdcdhd}((X0{\unicdcdgdad}X1){\unicdcdcdhd}(\preoh X1 \in X0))))\end{array}
\label{hyp:ehbdedpfpgchegjgogbgmgbd}
\end{equation}
\begin{theorem}\label{thm:ehcdpffgihdgigdhpgcheh}
$$\begin{array}{c}{\unicdcdadad} X0 .  (v3\_ordinal1~X0){\unicdbdjdcd}({\unicdcdadad} X1 .  (v3\_ordinal1~X1){\unicdbdjdcd}({\unicdcdadad} X2 .  \\
(v3\_ordinal1~X2){\unicdbdjdcd}({\unicdcdadad} X3 .  (v3\_ordinal1~X3){\unicdbdjdcd}(\preoh (X0 \in X1){\unicdcdcdhd}((\\
X2 \in X3){\unicdcdcdhd}((\preoh (X2{\unicdcdgdad}X0){\unicdcdcdhd}((X2{\unicdcdgdad}X1){\unicdcdcdhd}((X3{\unicdcdgdad}X0){\unicdcdcdhd}(X3{\unicdcdgdad}X1)))){\unicdcdcdhd}((\preoh (X2 \in X0){\unicdcdcdhd}\\
(X3=X0)){\unicdcdcdhd}((\preoh (X2 \in X0){\unicdcdcdhd}(X3=X1)){\unicdcdcdhd}((\preoh (X2=X0){\unicdcdcdhd}(X3 \in X1)){\unicdcdcdhd}((\preoh (X2=\\
X0){\unicdcdcdhd}(X3=X1)){\unicdcdcdhd}((\preoh (X2=X0){\unicdcdcdhd}(X1 \in X3)){\unicdcdcdhd}((\preoh (X0 \in X2){\unicdcdcdhd}(X3=X1)){\unicdcdcdhd}(\preoh \\
(X2=X1){\unicdcdcdhd}(X1 \in X3))))))))))))))\end{array}$$
\end{theorem}
\end{document}
