\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\_ordinal6 (TMcasmipfum4P9kybmPDBUcHirXHXhoJJ1V)}
\maketitle
Let $ v3\_ordinal1 : {{\iota}{\unicdbdjdcd}{o}}$ be given.
Let $ k1\_ordinal6 : {{\iota}{\unicdbdjdcd}{\iota}}$ be given.
Let $ k2\_tarski : {{\iota}{\unicdbdjdcd}{{\iota}{\unicdbdjdcd}{\iota}}}$ be given.
Let $ np\_\_2 : {\iota}$ be given.
Let $ r1\_tarski : {{\iota}{\unicdbdjdcd}{{\iota}{\unicdbdjdcd}{o}}}$ be given.
Let $ k2\_ordinal1 : {{\iota}{\unicdbdjdcd}{\iota}}$ be given.
Let $ k5\_card\_1 : {{\iota}{\unicdbdjdcd}{\iota}}$ be given.
Let $ v1\_finset\_1 : {{\iota}{\unicdbdjdcd}{o}}$ be given.
Let $ k2\_wellord2 : {{\iota}{\unicdbdjdcd}{\iota}}$ be given.
Let $ k1\_wellord2 : {{\iota}{\unicdbdjdcd}{\iota}}$ be given.
Let $ k1\_ordinal1 : {{\iota}{\unicdbdjdcd}{\iota}}$ be given.
Let $ r1\_ordinal1 : {{\iota}{\unicdbdjdcd}{{\iota}{\unicdbdjdcd}{o}}}$ be given.
Let $ v1\_xboole\_0 : {{\iota}{\unicdbdjdcd}{o}}$ be given.
Let $ k2\_xboole\_0 : {{\iota}{\unicdbdjdcd}{{\iota}{\unicdbdjdcd}{\iota}}}$ be given.
Let $ k1\_tarski : {{\iota}{\unicdbdjdcd}{\iota}}$ 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 .  {{\unicdcdadad} X1 .  {{({{X0}{\unicdcdgdad}{X1}})}{\unicdbdjdcd}{({{{k5\_card\_1}~{({{{k2\_tarski}~{X0}}~{X1}})}}={np\_\_2}})}}}}\label{hyp:ehfdhdpfdgbgchegpfcd}
\end{equation}
Assume the following.
\begin{equation}
\begin{array}{c}{\unicdcdadad} X0 .  (v3\_ordinal1~X0){\unicdbdjdcd}({\unicdcdadad} X1 .  (v1\_finset\_1~X1){\unicdbdjdcd}((\\
r1\_tarski~X1~X0){\unicdbdjdcd}(k2\_wellord2~(k1\_wellord2~X1)=k5\_card\_1~X1)))\end{array}
\label{hyp:ehddgdpfdgbgchegpffd}
\end{equation}
Assume the following.
\begin{equation}
\begin{array}{c}{\unicdcdadad} X0 .  {\unicdcdadad} X1 .  {\unicdcdadad} X2 .  (r1\_tarski~(k2\_tarski~X0~X1)~\\
X2){\unicdbdjded}((X0 \in X2){\unicdcdcdhd}(X1 \in X2))\end{array}
\label{hyp:ehddcdpfkhggngjgdhdgpfbd}
\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}
\begin{array}{c}{\unicdcdadad} X0 .  {\unicdcdadad} X1 .  ((v3\_ordinal1~X0){\unicdcdcdhd}(v3\_ordinal1~X1)){\unicdbdjdcd}(\\
r1\_ordinal1~X0~X0)\end{array}
\label{hyp:chfgggmgfgihjgghjgehjhpfchbdpfpgchegjgogbgmgbd}
\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 .  {{\unicdcdadad} X1 .  {{v1\_finset\_1}~{({{{k2\_tarski}~{X0}}~{X1}})}}}}\label{hyp:ggdgcdpfggjgogdhfgehpfbd}
\end{equation}
Assume the following.
\begin{equation}
{{\unicdcdadad} X0 .  {{{k1\_ordinal6}~{X0}}={{k2\_wellord2}~{({{k1\_wellord2}~{({{k2\_ordinal1}~{X0}})}})}}}}\label{hyp:egedpfpgchegjgogbgmggd}
\end{equation}
Assume the following.
\begin{equation}
{{\unicdcdadad} X0 .  {{{k1\_ordinal1}~{X0}}={{{k2\_xboole\_0}~{X0}}~{({{k1\_tarski}~{X0}})}}}}\label{hyp:egbdpfpgchegjgogbgmgbd}
\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:ehbdgdpfpgchegjgogbgmggd}
$$\begin{array}{c}{\unicdcdadad} X0 .  (v3\_ordinal1~X0){\unicdbdjdcd}({\unicdcdadad} X1 .  (v3\_ordinal1~X1){\unicdbdjdcd}((\\
X0{\unicdcdgdad}X1){\unicdbdjdcd}(k1\_ordinal6~(k2\_tarski~X0~X1)=np\_\_2)))\end{array}$$
\end{theorem}
\end{document}
