\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{t61\_ordinal5 (TMUpNxbAkocJJVTjjdsjLc37GNE1sVLW8Fz)}
\maketitle
Let $ v3\_ordinal1 : {{\iota}{\unicdbdjdcd}{o}}$ be given.
Let $ k6\_numbers : {\iota}$ be given.
Let $ k7\_ordinal3 : {{\iota}{\unicdbdjdcd}{{\iota}{\unicdbdjdcd}{\iota}}}$ be given.
Let $ k12\_ordinal2 : {{\iota}{\unicdbdjdcd}{{\iota}{\unicdbdjdcd}{\iota}}}$ be given.
Let $ k5\_ordinal3 : {{\iota}{\unicdbdjdcd}{{\iota}{\unicdbdjdcd}{\iota}}}$ be given.
Let $ k10\_ordinal2 : {{\iota}{\unicdbdjdcd}{{\iota}{\unicdbdjdcd}{\iota}}}$ be given.
Let $ k1\_xboole\_0 : {\iota}$ be given.
Let $ k6\_ordinal3 : {{\iota}{\unicdbdjdcd}{{\iota}{\unicdbdjdcd}{\iota}}}$ be given.
Let $ k11\_ordinal2 : {{\iota}{\unicdbdjdcd}{{\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}(k5\_ordinal3~\\
(k10\_ordinal2~X0~X1)~X0=X1))\end{array}
\label{hyp:ehfdcdpfpgchegjgogbgmgdd}
\end{equation}
Assume the following.
\begin{equation}
\begin{array}{c}{\unicdcdadad} X0 .  (v3\_ordinal1~X0){\unicdbdjdcd}({\unicdcdadad} X1 .  (v3\_ordinal1~X1){\unicdbdjdcd}(\preoh \\
(X0{\unicdcdgdad}k1\_xboole\_0){\unicdcdcdhd}(k12\_ordinal2~X0~X1=k1\_xboole\_0)))\end{array}
\label{hyp:ehcdcdpfpgchegjgogbgmged}
\end{equation}
Assume the following.
\begin{equation}
{{k6\_numbers}={k1\_xboole\_0}}\label{hyp:chfgegfgggjgogjgehjgpgogpflggdpfogfhngcgfgchdh}
\end{equation}
Assume the following.
\begin{equation}
\begin{array}{c}{\unicdcdadad} X0 .  {\unicdcdadad} X1 .  ((v3\_ordinal1~X0){\unicdcdcdhd}(v3\_ordinal1~X1)){\unicdbdjdcd}(\\
v3\_ordinal1~(k6\_ordinal3~X0~X1))\end{array}
\label{hyp:egehpflggdpfpgchegjgogbgmgdd}
\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}
Assume the following.
\begin{equation}
\begin{array}{c}{\unicdcdadad} X0 .  {\unicdcdadad} X1 .  ((v3\_ordinal1~X0){\unicdcdcdhd}(v3\_ordinal1~X1)){\unicdbdjdcd}(\\
v3\_ordinal1~(k11\_ordinal2~X0~X1))\end{array}
\label{hyp:egehpflgbdbdpfpgchegjgogbgmgcd}
\end{equation}
Assume the following.
\begin{equation}
\begin{array}{c}{\unicdcdadad} X0 .  (v3\_ordinal1~X0){\unicdbdjdcd}({\unicdcdadad} X1 .  (v3\_ordinal1~X1){\unicdbdjdcd}(k7\_ordinal3~\\
X0~X1=k5\_ordinal3~X0~(k11\_ordinal2~(k6\_ordinal3~X0~X1)~X1)))\end{array}
\label{hyp:eghdpfpgchegjgogbgmgdd}
\end{equation}
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}(((X1{\unicdcdgdad}k1\_xboole\_0){\unicdbdjdcd}((X2=k6\_ordinal3~X0~X1){\unicdbdjded}\\
({\unicdcdaddd} X3 .  (v3\_ordinal1~X3){\unicdcdcdhd}((X0=k10\_ordinal2~(k11\_ordinal2~\\
X2~X1)~X3){\unicdcdcdhd}(X3 \in X1))))){\unicdcdcdhd}((X1=k1\_xboole\_0){\unicdbdjdcd}((X2=k6\_ordinal3~\\
X0~X1){\unicdbdjded}(X2=k1\_xboole\_0))))))\end{array}
\label{hyp:eggdpfpgchegjgogbgmgdd}
\end{equation}
\begin{theorem}\label{thm:ehgdbdpfpgchegjgogbgmgfd}
$$\begin{array}{c}{\unicdcdadad} X0 .  (v3\_ordinal1~X0){\unicdbdjdcd}({\unicdcdadad} X1 .  (v3\_ordinal1~X1){\unicdbdjdcd}({\unicdcdadad} X2 .  \\
(v3\_ordinal1~X2){\unicdbdjdcd}((X0{\unicdcdgdad}k6\_numbers){\unicdbdjdcd}(k7\_ordinal3~X1~(k12\_ordinal2~\\
X0~X2) \in k12\_ordinal2~X0~X2))))\end{array}$$
\end{theorem}
\end{document}
