\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{t18\_membered (TMdgPjJXvFiqUtZMfBsif2vX8ZVeJRfsAP9)}
\maketitle
Let $ v6\_membered : {{\iota}{\unicdbdjdcd}{o}}$ be given.
Let $ v7\_ordinal1 : {{\iota}{\unicdbdjdcd}{o}}$ be given.
Let $ k5\_numbers : {\iota}$ be given.
Let $ k4\_ordinal1 : {\iota}$ be given.
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}
{{k5\_numbers}={k4\_ordinal1}}\label{hyp:chfgegfgggjgogjgehjgpgogpflgfdpfogfhngcgfgchdh}
\end{equation}
Assume the following.
\begin{equation}
{{\unicdcdadad} X0 .  {{({{v6\_membered}~{X0}})}{\unicdbdjded}{({{\unicdcdadad} X1 .  {{({{X1} \in {X0}})}{\unicdbdjdcd}{({{v7\_ordinal1}~{X1}})}}})}}}\label{hyp:eggdpfngfgngcgfgchfgeg}
\end{equation}
Assume the following.
\begin{equation}
{{\unicdcdadad} X0 .  {{({{v7\_ordinal1}~{X0}})}{\unicdbdjded}{({{X0} \in {k4\_ordinal1}})}}}\label{hyp:egbdcdpfpgchegjgogbgmgbd}
\end{equation}
\begin{theorem}\label{thm:ehbdidpfngfgngcgfgchfgeg}
$$\begin{array}{c}{\unicdcdadad} X0 .  (v6\_membered~X0){\unicdbdjdcd}(({\unicdcdadad} X1 .  (v7\_ordinal1~X1){\unicdbdjdcd}(\\
X1 \in X0)){\unicdbdjdcd}(X0=k5\_numbers))\end{array}$$
\end{theorem}
\end{document}
