\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}
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\newcommand\unicdcdaddd{\exists}
\newcommand\unicdcdidid{\not\subseteq}
\newcommand\preoh[1]{\lnot #1}
\newcommand\mname[1]{{\mathsf{#1}}}
\begin{document}
\title{t20\_ordinal6 (TMU8gbyAHeJxhJQPoeJMzNj6nBZwuTUTb6Z)}
\maketitle
Let $ k1\_card\_1 : {{\iota}{\unicdbdjdcd}{\iota}}$ be given.
Let $ k9\_xtuple\_0 : {{\iota}{\unicdbdjdcd}{\iota}}$ be given.
Let $ k2\_ordinal6 : {{\iota}{\unicdbdjdcd}{\iota}}$ be given.
Let $ k2\_ordinal1 : {{\iota}{\unicdbdjdcd}{\iota}}$ be given.
Let $ v3\_ordinal1 : {{\iota}{\unicdbdjdcd}{o}}$ be given.
Let $ r1\_tarski : {{\iota}{\unicdbdjdcd}{{\iota}{\unicdbdjdcd}{o}}}$ be given.
Let $ k2\_wellord2 : {{\iota}{\unicdbdjdcd}{\iota}}$ be given.
Let $ k1\_wellord2 : {{\iota}{\unicdbdjdcd}{\iota}}$ be given.
Let $ k1\_ordinal6 : {{\iota}{\unicdbdjdcd}{\iota}}$ be given.
Let $ k10\_xtuple\_0 : {{\iota}{\unicdbdjdcd}{\iota}}$ be given.
Let $ v1\_ordinal6 : {{\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}(k1\_card\_1~\\
X1=k1\_card\_1~(k2\_wellord2~(k1\_wellord2~X1))))\end{array}
\label{hyp:ehhdpfdgbgchegpffd}
\end{equation}
Assume the following.
\begin{equation}
\begin{array}{c}{\unicdcdadad} X0 .  (k9\_xtuple\_0~(k2\_ordinal6~X0)=k1\_ordinal6~X0){\unicdcdcdhd}(k10\_xtuple\_0~\\
(k2\_ordinal6~X0)=k2\_ordinal1~X0)\end{array}
\label{hyp:ehbdidpfpgchegjgogbgmggd}
\end{equation}
Assume the following.
\begin{equation}
{{\unicdcdadad} X0 .  {{v1\_ordinal6}~{({{k2\_ordinal1}~{X0}})}}}\label{hyp:ggdgbdpfpgchegjgogbgmggd}
\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}
\begin{array}{c}{\unicdcdadad} X0 .  (v1\_ordinal6~X0){\unicdbdjded}({\unicdcdaddd} X1 .  (v3\_ordinal1~X1){\unicdcdcdhd}(r1\_tarski~\\
X0~X1))\end{array}
\label{hyp:egbdpfpgchegjgogbgmggd}
\end{equation}
\begin{theorem}\label{thm:ehcdadpfpgchegjgogbgmggd}
$$\begin{array}{c}{\unicdcdadad} X0 .  k1\_card\_1~(k9\_xtuple\_0~(k2\_ordinal6~X0))=k1\_card\_1~\\
(k2\_ordinal1~X0)\end{array}$$
\end{theorem}
\end{document}
