\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{t10\_ordinal3 (TMZaBPQPfJAjZ1j9yVWM9o3eRbN5DfHrVkV)}
\maketitle
Let $ v3\_ordinal1 : {{\iota}{\unicdbdjdcd}{o}}$ be given.
Let $ k2\_ordinal2 : {{\iota}{\unicdbdjdcd}{\iota}}$ be given.
Let $ k1\_tarski : {{\iota}{\unicdbdjdcd}{\iota}}$ be given.
Let $ k2\_ordinal1 : {{\iota}{\unicdbdjdcd}{\iota}}$ be given.
Let $ k1\_setfam\_1 : {{\iota}{\unicdbdjdcd}{\iota}}$ be given.
Assume the following.
\begin{equation}
\begin{array}{c}{\unicdcdadad} X0 .  (v3\_ordinal1~X0){\unicdbdjdcd}(k2\_ordinal1~(k1\_tarski~X0)=k1\_tarski~\\
X0)\end{array}
\label{hyp:ehhdpfpgchegjgogbgmgdd}
\end{equation}
Assume the following.
\begin{equation}
{{\unicdcdadad} X0 .  {{{k1\_setfam\_1}~{({{k1\_tarski}~{X0}})}}={X0}}}\label{hyp:ehbdadpfdhfgehggbgngpfbd}
\end{equation}
Assume the following.
\begin{equation}
{{\unicdcdadad} X0 .  {{{k2\_ordinal2}~{X0}}={{k1\_setfam\_1}~{({{k2\_ordinal1}~{X0}})}}}}\label{hyp:egcdpfpgchegjgogbgmgcd}
\end{equation}
\begin{theorem}\label{thm:ehbdadpfpgchegjgogbgmgdd}
${{\unicdcdadad} X0 .  {{({{v3\_ordinal1}~{X0}})}{\unicdbdjdcd}{({{{k2\_ordinal2}~{({{k1\_tarski}~{X0}})}}={X0}})}}}$.
\end{theorem}
\end{document}
