\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{t3\_ordinal6 (TMMH7NGq5JHGPxxVBnVh68RHqeyEPxeyamz)}
\maketitle
Let $ v1\_ordinal6 : {{\iota}{\unicdbdjdcd}{o}}$ be given.
Let $ r1\_tarski : {{\iota}{\unicdbdjdcd}{{\iota}{\unicdbdjdcd}{o}}}$ be given.
Let $ k3\_ordinal2 : {{\iota}{\unicdbdjdcd}{\iota}}$ be given.
Let $ k2\_ordinal1 : {{\iota}{\unicdbdjdcd}{\iota}}$ be given.
Let $ v3\_ordinal1 : {{\iota}{\unicdbdjdcd}{o}}$ be given.
Let $ r1\_ordinal1 : {{\iota}{\unicdbdjdcd}{{\iota}{\unicdbdjdcd}{o}}}$ be given.
Assume the following.
\begin{equation}
{{\unicdcdadad} X0 .  {{({{v1\_ordinal6}~{X0}})}{\unicdbdjded}{({{{k2\_ordinal1}~{X0}}={X0}})}}}\label{hyp:ehcdpfpgchegjgogbgmggd}
\end{equation}
Assume the following.
\begin{equation}
{{\unicdcdadad} X0 .  {{v3\_ordinal1}~{({{k3\_ordinal2}~{X0}})}}}\label{hyp:egehpflgddpfpgchegjgogbgmgcd}
\end{equation}
Assume the following.
\begin{equation}
\begin{array}{c}{\unicdcdadad} X0 .  {\unicdcdadad} X1 .  (v3\_ordinal1~X1){\unicdbdjdcd}((X1=k3\_ordinal2~X0){\unicdbdjded}\\
((r1\_tarski~(k2\_ordinal1~X0)~X1){\unicdcdcdhd}({\unicdcdadad} X2 .  (v3\_ordinal1~X2){\unicdbdjdcd}\\
((r1\_tarski~(k2\_ordinal1~X0)~X2){\unicdbdjdcd}(r1\_ordinal1~X1~X2)))))\end{array}
\label{hyp:egddpfpgchegjgogbgmgcd}
\end{equation}
\begin{theorem}\label{thm:ehddpfpgchegjgogbgmggd}
${{\unicdcdadad} X0 .  {{({{v1\_ordinal6}~{X0}})}{\unicdbdjdcd}{({{{r1\_tarski}~{X0}}~{({{k3\_ordinal2}~{X0}})}})}}}$.
\end{theorem}
\end{document}
