\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{t53\_scmyciel (TMJt4aJCEYa7WtfRgg13EL6VfxbhEWGzpvJ)}
\maketitle
Let $ v4\_scmyciel : {{\iota}{\unicdbdjdcd}{o}}$ be given.
Let $ v5\_scmyciel : {{\iota}{\unicdbdjdcd}{o}}$ be given.
Let $ k3\_tarski : {{\iota}{\unicdbdjdcd}{\iota}}$ be given.
Let $ k2\_tarski : {{\iota}{\unicdbdjdcd}{{\iota}{\unicdbdjdcd}{\iota}}}$ be given.
Let $ k5\_scmyciel : {{\iota}{\unicdbdjdcd}{\iota}}$ be given.
Assume the following.
\begin{equation}
\begin{array}{c}{\unicdcdadad} X0 .  {\unicdcdadad} X1 .  {\unicdcdadad} X2 .  ((X1 \in X0){\unicdcdcdhd}(X2 \in X0)){\unicdbdjdcd}(k2\_tarski~\\
X1~X2 \in k5\_scmyciel~X0)\end{array}
\label{hyp:ehddedpfdhdgngjhdgjgfgmg}
\end{equation}
Assume the following.
\begin{equation}
\begin{array}{c}{\unicdcdadad} X0 .  (v4\_scmyciel~X0){\unicdbdjdcd}((v5\_scmyciel~X0){\unicdbdjded}(X0=k5\_scmyciel~\\
(k3\_tarski~X0)))\end{array}
\label{hyp:egbdddpfdhdgngjhdgjgfgmg}
\end{equation}
\begin{theorem}\label{thm:ehfdddpfdhdgngjhdgjgfgmg}
$$\begin{array}{c}{\unicdcdadad} X0 .  ((v4\_scmyciel~X0){\unicdcdcdhd}(v5\_scmyciel~X0)){\unicdbdjdcd}({\unicdcdadad} X1 .  \\
{\unicdcdadad} X2 .  ((X1 \in k3\_tarski~X0){\unicdcdcdhd}(X2 \in k3\_tarski~X0)){\unicdbdjdcd}(k2\_tarski~\\
X1~X2 \in X0))\end{array}$$
\end{theorem}
\end{document}
