\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{t16\_ordinal5 (TMWuZCY2L2BoXrFYy6DpfKSb6YojW1rLFvq)}
\maketitle
Let $ v3\_ordinal1 : {{\iota}{\unicdbdjdcd}{o}}$ be given.
Let $ k1\_ordinal5 : {{\iota}{\unicdbdjdcd}{{\iota}{\unicdbdjdcd}{\iota}}}$ be given.
Let $ np\_\_1 : {\iota}$ be given.
Let $ k12\_ordinal2 : {{\iota}{\unicdbdjdcd}{{\iota}{\unicdbdjdcd}{\iota}}}$ be given.
Let $ k1\_ordinal1 : {{\iota}{\unicdbdjdcd}{\iota}}$ be given.
Let $ k6\_numbers : {\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}((k12\_ordinal2~X0~np\_\_1=X0){\unicdcdcdhd}(k12\_ordinal2~\\
np\_\_1~X0=np\_\_1))\end{array}
\label{hyp:ehedgdpfpgchegjgogbgmgcd}
\end{equation}
Assume the following.
\begin{equation}
\begin{array}{c}{\unicdcdadad} X0 .  (v3\_ordinal1~X0){\unicdbdjdcd}({\unicdcdadad} X1 .  (v3\_ordinal1~X1){\unicdbdjdcd}(k1\_ordinal5~\\
X0~(k1\_ordinal1~X1)=k12\_ordinal2~X0~(k1\_ordinal5~X0~X1)))\end{array}
\label{hyp:ehbdedpfpgchegjgogbgmgfd}
\end{equation}
Assume the following.
\begin{equation}
{{\unicdcdadad} X0 .  {{({{v3\_ordinal1}~{X0}})}{\unicdbdjdcd}{({{{{k1\_ordinal5}~{X0}}~{k6\_numbers}}={np\_\_1}})}}}\label{hyp:ehbdddpfpgchegjgogbgmgfd}
\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}
{{v1\_xboole\_0}~{k1\_xboole\_0}}\label{hyp:ggdgbdpfihcgpgpgmgfgpfad}
\end{equation}
Assume the following.
\begin{equation}
{{\unicdcdadad} X0 .  {{({{v1\_xboole\_0}~{X0}})}{\unicdbdjdcd}{({{v3\_ordinal1}~{X0}})}}}\label{hyp:dgdgddpfpgchegjgogbgmgbd}
\end{equation}
\begin{theorem}\label{thm:ehbdgdpfpgchegjgogbgmgfd}
${{\unicdcdadad} X0 .  {{({{v3\_ordinal1}~{X0}})}{\unicdbdjdcd}{({{{{k1\_ordinal5}~{X0}}~{np\_\_1}}={X0}})}}}$.
\end{theorem}
\end{document}
