\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{t11\_ordinal6 (TMQcpnxARxufu8MRyNUMeMoK4PdqrSwsbY9)}
\maketitle
Let $ v3\_ordinal1 : {{\iota}{\unicdbdjdcd}{o}}$ be given.
Let $ r1\_ordinal1 : {{\iota}{\unicdbdjdcd}{{\iota}{\unicdbdjdcd}{o}}}$ be given.
Let $ k3\_ordinal5 : {{\iota}{\unicdbdjdcd}{\iota}}$ be given.
Let $ k1\_ordinal1 : {{\iota}{\unicdbdjdcd}{\iota}}$ be given.
Assume the following.
\begin{equation}
{{\unicdcdadad} X0 .  {{\unicdcdadad} X1 .  {{({{X0} \in {{k1\_ordinal1}~{X1}}})}{\unicdbdjded}{({{({{X0} \in {X1}})}{\unicdcdcdid}{({{X0}={X1}})}})}}}}\label{hyp:ehidpfpgchegjgogbgmgbd}
\end{equation}
Assume the following.
\begin{equation}
{{\unicdcdadad} X0 .  {{X0} \in {{k1\_ordinal1}~{X0}}}}\label{hyp:ehgdpfpgchegjgogbgmgbd}
\end{equation}
Assume the following.
\begin{equation}
\begin{array}{c}{\unicdcdadad} X0 .  (v3\_ordinal1~X0){\unicdbdjdcd}({\unicdcdadad} X1 .  (v3\_ordinal1~X1){\unicdbdjdcd}((\\
r1\_ordinal1~X0~X1){\unicdbdjded}(\preoh X1 \in X0)))\end{array}
\label{hyp:ehedpfpgchegjgogbgmggd}
\end{equation}
Assume the following.
\begin{equation}
\begin{array}{c}{\unicdcdadad} X0 .  (v3\_ordinal1~X0){\unicdbdjdcd}({\unicdcdadad} X1 .  (v3\_ordinal1~X1){\unicdbdjdcd}((\\
X0 \in X1){\unicdbdjdcd}(k3\_ordinal5~X0 \in k3\_ordinal5~X1)))\end{array}
\label{hyp:ehededpfpgchegjgogbgmgfd}
\end{equation}
Assume the following.
\begin{equation}
\begin{array}{c}{\unicdcdadad} X0 .  (v3\_ordinal1~X0){\unicdbdjdcd}({\unicdcdadad} X1 .  (v3\_ordinal1~X1){\unicdbdjdcd}((\\
X0 \in k1\_ordinal1~X1){\unicdbdjded}(r1\_ordinal1~X0~X1)))\end{array}
\label{hyp:ehcdcdpfpgchegjgogbgmgbd}
\end{equation}
Assume the following.
\begin{equation}
{{\unicdcdadad} X0 .  {{\unicdcdadad} X1 .  {{({{v3\_ordinal1}~{X1}})}{\unicdbdjdcd}{({{({{X0} \in {X1}})}{\unicdbdjdcd}{({{v3\_ordinal1}~{X0}})}})}}}}\label{hyp:ehbdddpfpgchegjgogbgmgbd}
\end{equation}
Assume the following.
\begin{equation}
{{\unicdcdadad} X0 .  {{({{v3\_ordinal1}~{X0}})}{\unicdbdjdcd}{({{v3\_ordinal1}~{({{k3\_ordinal5}~{X0}})}})}}}\label{hyp:egehpflgddpfpgchegjgogbgmgfd}
\end{equation}
Assume the following.
\begin{equation}
\begin{array}{c}{\unicdcdadad} X0 .  {\unicdcdadad} X1 .  ((v3\_ordinal1~X0){\unicdcdcdhd}(v3\_ordinal1~X1)){\unicdbdjdcd}(\\
(r1\_ordinal1~X0~X1){\unicdcdcdid}(r1\_ordinal1~X1~X0))\end{array}
\label{hyp:dgpgogogfgdgehfgegogfgdhdhpfchbdpfpgchegjgogbgmgbd}
\end{equation}
\begin{theorem}\label{thm:ehbdbdpfpgchegjgogbgmggd}
$$\begin{array}{c}{\unicdcdadad} X0 .  (v3\_ordinal1~X0){\unicdbdjdcd}({\unicdcdadad} X1 .  (v3\_ordinal1~X1){\unicdbdjdcd}((\\
r1\_ordinal1~X0~X1){\unicdbdjded}(r1\_ordinal1~(k3\_ordinal5~X0)~(k3\_ordinal5~\\
X1))))\end{array}$$
\end{theorem}
\end{document}
