\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{t26\_ordinal6 (TMcehTw3pAVf72vAhmoLJrXtuvWPWx6puYC)}
\maketitle
Let $ v1\_ordinal6 : {{\iota}{\unicdbdjdcd}{o}}$ be given.
Let $ k1\_ordinal6 : {{\iota}{\unicdbdjdcd}{\iota}}$ be given.
Let $ k2\_xboole\_0 : {{\iota}{\unicdbdjdcd}{{\iota}{\unicdbdjdcd}{\iota}}}$ be given.
Let $ k10\_ordinal2 : {{\iota}{\unicdbdjdcd}{{\iota}{\unicdbdjdcd}{\iota}}}$ be given.
Let $ k2\_ordinal6 : {{\iota}{\unicdbdjdcd}{\iota}}$ be given.
Let $ k1\_ordinal4 : {{\iota}{\unicdbdjdcd}{{\iota}{\unicdbdjdcd}{\iota}}}$ be given.
Let $ k9\_xtuple\_0 : {{\iota}{\unicdbdjdcd}{\iota}}$ be given.
Let $ k10\_xtuple\_0 : {{\iota}{\unicdbdjdcd}{\iota}}$ be given.
Let $ k2\_ordinal1 : {{\iota}{\unicdbdjdcd}{\iota}}$ be given.
Let $ v1\_relat\_1 : {{\iota}{\unicdbdjdcd}{o}}$ be given.
Let $ v1\_funct\_1 : {{\iota}{\unicdbdjdcd}{o}}$ be given.
Let $ v5\_ordinal1 : {{\iota}{\unicdbdjdcd}{o}}$ be given.
Let $ v1\_ordinal2 : {{\iota}{\unicdbdjdcd}{o}}$ be given.
Let $ v3\_ordinal1 : {{\iota}{\unicdbdjdcd}{o}}$ be given.
Let $ k1\_funct\_1 : {{\iota}{\unicdbdjdcd}{{\iota}{\unicdbdjdcd}{\iota}}}$ be given.
Assume the following.
\begin{equation}
\begin{array}{c}{\unicdcdadad} X0 .  (v1\_ordinal6~X0){\unicdbdjdcd}({\unicdcdadad} X1 .  (v1\_ordinal6~X1){\unicdbdjdcd}((\\
{\unicdcdadad} X2 .  {\unicdcdadad} X3 .  ((X2 \in X0){\unicdcdcdhd}(X3 \in X1)){\unicdbdjdcd}(X2 \in X3)){\unicdbdjdcd}(k2\_ordinal6~\\
(k2\_xboole\_0~X0~X1)=k1\_ordinal4~(k2\_ordinal6~X0)~(k2\_ordinal6~\\
X1))))\end{array}
\label{hyp:ehcdfdpfpgchegjgogbgmggd}
\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}
\begin{array}{c}{\unicdcdadad} X0 .  (v1\_relat\_1~(k2\_ordinal6~X0)){\unicdcdcdhd}((v1\_funct\_1~(k2\_ordinal6~\\
X0)){\unicdcdcdhd}((v5\_ordinal1~(k2\_ordinal6~X0)){\unicdcdcdhd}(v1\_ordinal2~(k2\_ordinal6~\\
X0))))\end{array}
\label{hyp:egehpflgcdpfpgchegjgogbgmggd}
\end{equation}
Assume the following.
\begin{equation}
\begin{array}{c}{\unicdcdadad} X0 .  ((v1\_relat\_1~X0){\unicdcdcdhd}((v5\_ordinal1~X0){\unicdcdcdhd}(v1\_funct\_1~X0))){\unicdbdjdcd}\\
({\unicdcdadad} X1 .  ((v1\_relat\_1~X1){\unicdcdcdhd}((v5\_ordinal1~X1){\unicdcdcdhd}(v1\_funct\_1~\\
X1))){\unicdbdjdcd}({\unicdcdadad} X2 .  ((v1\_relat\_1~X2){\unicdcdcdhd}((v5\_ordinal1~X2){\unicdcdcdhd}(v1\_funct\_1~\\
X2))){\unicdbdjdcd}((X2=k1\_ordinal4~X0~X1){\unicdbdjded}((k9\_xtuple\_0~X2=k10\_ordinal2~\\
(k9\_xtuple\_0~X0)~(k9\_xtuple\_0~X1)){\unicdcdcdhd}(({\unicdcdadad} X3 .  (v3\_ordinal1~\\
X3){\unicdbdjdcd}((X3 \in k9\_xtuple\_0~X0){\unicdbdjdcd}(k1\_funct\_1~X2~X3=k1\_funct\_1~X0~X3))){\unicdcdcdhd}\\
({\unicdcdadad} X3 .  (v3\_ordinal1~X3){\unicdbdjdcd}((X3 \in k9\_xtuple\_0~X1){\unicdbdjdcd}(k1\_funct\_1~\\
X2~(k10\_ordinal2~(k9\_xtuple\_0~X0)~X3)=k1\_funct\_1~X1~X3))))))))\end{array}
\label{hyp:egbdpfpgchegjgogbgmged}
\end{equation}
\begin{theorem}\label{thm:ehcdgdpfpgchegjgogbgmggd}
$$\begin{array}{c}{\unicdcdadad} X0 .  (v1\_ordinal6~X0){\unicdbdjdcd}({\unicdcdadad} X1 .  (v1\_ordinal6~X1){\unicdbdjdcd}((\\
{\unicdcdadad} X2 .  {\unicdcdadad} X3 .  ((X2 \in X0){\unicdcdcdhd}(X3 \in X1)){\unicdbdjdcd}(X2 \in X3)){\unicdbdjdcd}(k1\_ordinal6~\\
(k2\_xboole\_0~X0~X1)=k10\_ordinal2~(k1\_ordinal6~X0)~(k1\_ordinal6~\\
X1))))\end{array}$$
\end{theorem}
\end{document}
