\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{t59\_ordinal5 (TMR7nMnFbYc5j72hvDYNMMAM1AZvSRZuS5R)}
\maketitle
Let $ v3\_ordinal1 : {{\iota}{\unicdbdjdcd}{o}}$ be given.
Let $ k6\_numbers : {\iota}$ be given.
Let $ np\_\_1 : {\iota}$ be given.
Let $ k5\_ordinal5 : {{\iota}{\unicdbdjdcd}{{\iota}{\unicdbdjdcd}{\iota}}}$ be given.
Let $ k6\_ordinal3 : {{\iota}{\unicdbdjdcd}{{\iota}{\unicdbdjdcd}{\iota}}}$ be given.
Let $ k12\_ordinal2 : {{\iota}{\unicdbdjdcd}{{\iota}{\unicdbdjdcd}{\iota}}}$ be given.
Let $ k11\_ordinal2 : {{\iota}{\unicdbdjdcd}{{\iota}{\unicdbdjdcd}{\iota}}}$ be given.
Let $ k7\_ordinal3 : {{\iota}{\unicdbdjdcd}{{\iota}{\unicdbdjdcd}{\iota}}}$ be given.
Let $ k1\_ordinal1 : {{\iota}{\unicdbdjdcd}{\iota}}$ be given.
Assume the following.
\begin{equation}
\begin{array}{c}{\unicdcdadad} X0 .  (v3\_ordinal1~X0){\unicdbdjdcd}({\unicdcdadad} X1 .  (v3\_ordinal1~X1){\unicdbdjdcd}({\unicdcdadad} X2 .  \\
(v3\_ordinal1~X2){\unicdbdjdcd}((X0 \in k11\_ordinal2~X1~X2){\unicdbdjdcd}((k6\_ordinal3~X0~X2 \in \\
X1){\unicdcdcdhd}(k7\_ordinal3~X0~X2 \in X2)))))\end{array}
\label{hyp:ehgdhdpfpgchegjgogbgmgdd}
\end{equation}
Assume the following.
\begin{equation}
\begin{array}{c}{\unicdcdadad} X0 .  (v3\_ordinal1~X0){\unicdbdjdcd}({\unicdcdadad} X1 .  (v3\_ordinal1~X1){\unicdbdjdcd}((\\
np\_\_1 \in X0){\unicdbdjdcd}(X1 \in k12\_ordinal2~X0~(k1\_ordinal1~(k5\_ordinal5~X1~X0)))))\end{array}
\label{hyp:ehfdhdpfpgchegjgogbgmgfd}
\end{equation}
Assume the following.
\begin{equation}
\begin{array}{c}{\unicdcdadad} X0 .  (v3\_ordinal1~X0){\unicdbdjdcd}({\unicdcdadad} X1 .  (v3\_ordinal1~X1){\unicdbdjdcd}(k12\_ordinal2~\\
X0~(k1\_ordinal1~X1)=k11\_ordinal2~X0~(k12\_ordinal2~X0~X1)))\end{array}
\label{hyp:ehededpfpgchegjgogbgmgcd}
\end{equation}
Assume the following.
\begin{equation}
\begin{array}{c}{\unicdcdadad} X0 .  {\unicdcdadad} X1 .  ((v3\_ordinal1~X0){\unicdcdcdhd}(v3\_ordinal1~X1)){\unicdbdjdcd}(\\
v3\_ordinal1~(k12\_ordinal2~X0~X1))\end{array}
\label{hyp:egehpflgbdcdpfpgchegjgogbgmgcd}
\end{equation}
\begin{theorem}\label{thm:ehfdjdpfpgchegjgogbgmgfd}
$$\begin{array}{c}{\unicdcdadad} X0 .  (v3\_ordinal1~X0){\unicdbdjdcd}({\unicdcdadad} X1 .  (v3\_ordinal1~X1){\unicdbdjdcd}({\unicdcdadad} X2 .  \\
(v3\_ordinal1~X2){\unicdbdjdcd}(((k6\_numbers \in X0){\unicdcdcdhd}((np\_\_1 \in X1){\unicdcdcdhd}(X2=k5\_ordinal5~\\
X0~X1))){\unicdbdjdcd}(k6\_ordinal3~X0~(k12\_ordinal2~X1~X2) \in X1))))\end{array}$$
\end{theorem}
\end{document}
