\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{t23\_ordinal5 (TMQxD1RQgP636c7JqaNuwnAh84pfJSdUT5H)}
\maketitle
Let $ v3\_ordinal1 : {{\iota}{\unicdbdjdcd}{o}}$ be given.
Let $ k6\_numbers : {\iota}$ be given.
Let $ r1\_ordinal1 : {{\iota}{\unicdbdjdcd}{{\iota}{\unicdbdjdcd}{o}}}$ be given.
Let $ k1\_ordinal5 : {{\iota}{\unicdbdjdcd}{{\iota}{\unicdbdjdcd}{\iota}}}$ be given.
Let $ k1\_ordinal1 : {{\iota}{\unicdbdjdcd}{\iota}}$ be given.
Let $ np\_\_1 : {\iota}$ be given.
Let $ k1\_xboole\_0 : {\iota}$ be given.
Let $ v1\_xboole\_0 : {{\iota}{\unicdbdjdcd}{o}}$ be given.
Assume the following.
\begin{equation}
\begin{array}{c}{\unicdcdadad} X0 .  (v3\_ordinal1~X0){\unicdbdjdcd}({\unicdcdadad} X1 .  (v3\_ordinal1~X1){\unicdbdjdcd}({\unicdcdadad} X2 .  \\
(v3\_ordinal1~X2){\unicdbdjdcd}(((r1\_ordinal1~X0~X1){\unicdcdcdhd}(k6\_numbers \in X2)){\unicdbdjdcd}(r1\_ordinal1~\\
(k1\_ordinal5~X2~X0)~(k1\_ordinal5~X2~X1)))))\end{array}
\label{hyp:ehcdbdpfpgchegjgogbgmgfd}
\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){\unicdbdjded}(r1\_ordinal1~(k1\_ordinal1~X0)~X1)))\end{array}
\label{hyp:ehcdbdpfpgchegjgogbgmgbd}
\end{equation}
Assume the following.
\begin{equation}
{{\unicdcdadad} X0 .  {{({{v3\_ordinal1}~{X0}})}{\unicdbdjdcd}{({{{{k1\_ordinal5}~{X0}}~{np\_\_1}}={X0}})}}}\label{hyp:ehbdgdpfpgchegjgogbgmgfd}
\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}
{{k6\_numbers}={k1\_xboole\_0}}\label{hyp:chfgegfgggjgogjgehjgpgogpflggdpfogfhngcgfgchdh}
\end{equation}
Assume the following.
\begin{equation}
{{np\_\_1}={{k1\_ordinal1}~{k1\_xboole\_0}}}\label{hyp:mgbdedpfpgchegjgogbgmgdd}
\end{equation}
Assume the following.
\begin{equation}
\begin{array}{c}{\unicdcdadad} X0 .  (v3\_ordinal1~X0){\unicdbdjdcd}((\preoh v1\_xboole\_0~(k1\_ordinal1~X0)){\unicdcdcdhd}\\
(v3\_ordinal1~(k1\_ordinal1~X0)))\end{array}
\label{hyp:ggdgcdpfpgchegjgogbgmgbd}
\end{equation}
\begin{theorem}\label{thm:ehcdddpfpgchegjgogbgmgfd}
$$\begin{array}{c}{\unicdcdadad} X0 .  (v3\_ordinal1~X0){\unicdbdjdcd}({\unicdcdadad} X1 .  (v3\_ordinal1~X1){\unicdbdjdcd}((\\
(k6\_numbers \in X0){\unicdcdcdhd}(k6\_numbers \in X1)){\unicdbdjdcd}(r1\_ordinal1~X0~(k1\_ordinal5~\\
X0~X1))))\end{array}$$
\end{theorem}
\end{document}
