\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{l5\_jordan1d (TMUZZi1QKfCDRnt9TCvDegUZmb97ZzhMyTf)}
\maketitle
Let $ k2\_xboole\_0 : {{\iota}{\unicdbdjdcd}{{\iota}{\unicdbdjdcd}{\iota}}}$ be given.
Assume the following.
\begin{equation}
\begin{array}{c}{\unicdcdadad} X0 .  {\unicdcdadad} X1 .  {\unicdcdadad} X2 .  (X2=k2\_xboole\_0~X0~X1){\unicdbdjded}({\unicdcdadad} X3 .  \\
(X3 \in X2){\unicdbdjded}((X3 \in X0){\unicdcdcdid}(X3 \in X1)))\end{array}
\label{hyp:egddpfihcgpgpgmgfgpfad}
\end{equation}
\begin{theorem}\label{thm:mgfdpfkgpgchegbgogbdeg}
$$\begin{array}{c}{\unicdcdadad} X0 .  {\unicdcdadad} X1 .  {\unicdcdadad} X2 .  {\unicdcdadad} X3 .  {\unicdcdadad} X4 .  (X0 \in k2\_xboole\_0~\\
(k2\_xboole\_0~(k2\_xboole\_0~X1~X2)~X3)~X4){\unicdbdjded}(\preoh (\preoh X0 \in X1){\unicdcdcdhd}((\preoh X0 \in X2){\unicdcdcdhd}\\
((\preoh X0 \in X3){\unicdcdcdhd}(\preoh X0 \in X4))))\end{array}$$
\end{theorem}
\end{document}
