\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{t15\_ordinal6 (TMTYRCwAJ9BQoz1gyrnVrVVSTxcTWd6ECPW)}
\maketitle
Let $ v3\_ordinal1 : {{\iota}{\unicdbdjdcd}{o}}$ be given.
Let $ k1\_ordinal6 : {{\iota}{\unicdbdjdcd}{\iota}}$ be given.
Let $ k1\_tarski : {{\iota}{\unicdbdjdcd}{\iota}}$ be given.
Let $ np\_\_1 : {\iota}$ be given.
Let $ r1\_tarski : {{\iota}{\unicdbdjdcd}{{\iota}{\unicdbdjdcd}{o}}}$ be given.
Let $ k2\_ordinal1 : {{\iota}{\unicdbdjdcd}{\iota}}$ be given.
Let $ k1\_ordinal1 : {{\iota}{\unicdbdjdcd}{\iota}}$ be given.
Let $ k2\_wellord2 : {{\iota}{\unicdbdjdcd}{\iota}}$ be given.
Let $ k1\_wellord2 : {{\iota}{\unicdbdjdcd}{\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 .  (r1\_tarski~X1~X0){\unicdbdjdcd}(k2\_ordinal1~\\
X1=X1))\end{array}
\label{hyp:ehgdpfpgchegjgogbgmgdd}
\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 .  {\unicdcdadad} X1 .  (v3\_ordinal1~X1){\unicdbdjdcd}((r1\_tarski~(k1\_tarski~\\
X0)~X1){\unicdbdjdcd}(k2\_wellord2~(k1\_wellord2~(k1\_tarski~X0))=np\_\_1))\end{array}
\label{hyp:ehddhdpfdgbgchegpffd}
\end{equation}
Assume the following.
\begin{equation}
{{\unicdcdadad} X0 .  {{\unicdcdadad} X1 .  {{({{{r1\_tarski}~{({{k1\_tarski}~{X0}})}}~{X1}})}{\unicdbdjded}{({{X0} \in {X1}})}}}}\label{hyp:ehddbdpfkhggngjgdhdgpfbd}
\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}
Assume the following.
\begin{equation}
{{\unicdcdadad} X0 .  {{{k1\_ordinal6}~{X0}}={{k2\_wellord2}~{({{k1\_wellord2}~{({{k2\_ordinal1}~{X0}})}})}}}}\label{hyp:egedpfpgchegjgogbgmggd}
\end{equation}
\begin{theorem}\label{thm:ehbdfdpfpgchegjgogbgmggd}
${{\unicdcdadad} X0 .  {{({{v3\_ordinal1}~{X0}})}{\unicdbdjdcd}{({{{k1\_ordinal6}~{({{k1\_tarski}~{X0}})}}={np\_\_1}})}}}$.
\end{theorem}
\end{document}
