\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}
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\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{t38\_ordinal5 (TMYHY9tANxbCNN5LmReVkY7n4sMYs9cE1bL)}
\maketitle
Let $ v3\_ordinal1 : {{\iota}{\unicdbdjdcd}{o}}$ be given.
Let $ v4\_ordinal5 : {{\iota}{\unicdbdjdcd}{o}}$ be given.
Let $ k12\_ordinal2 : {{\iota}{\unicdbdjdcd}{{\iota}{\unicdbdjdcd}{\iota}}}$ be given.
Let $ k4\_ordinal1 : {\iota}$ be given.
Let $ r1\_ordinal1 : {{\iota}{\unicdbdjdcd}{{\iota}{\unicdbdjdcd}{o}}}$ be given.
Let $ k10\_ordinal2 : {{\iota}{\unicdbdjdcd}{{\iota}{\unicdbdjdcd}{\iota}}}$ be given.
Let $ np\_\_1 : {\iota}$ be given.
Let $ k11\_ordinal2 : {{\iota}{\unicdbdjdcd}{{\iota}{\unicdbdjdcd}{\iota}}}$ be given.
Let $ v2\_xxreal\_0 : {{\iota}{\unicdbdjdcd}{o}}$ be given.
Let $ m2\_subset\_1 : {{\iota}{\unicdbdjdcd}{{\iota}{\unicdbdjdcd}{{\iota}{\unicdbdjdcd}{o}}}}$ be given.
Let $ k1\_numbers : {\iota}$ be given.
Let $ k5\_numbers : {\iota}$ be given.
Let $ m1\_subset\_1 : {{\iota}{\unicdbdjdcd}{{\iota}{\unicdbdjdcd}{o}}}$ be given.
Let $ v1\_xboole\_0 : {{\iota}{\unicdbdjdcd}{o}}$ be given.
Assume the following.
\begin{equation}
\begin{array}{c}{\unicdcdadad} X0 .  (v3\_ordinal1~X0){\unicdbdjdcd}((r1\_ordinal1~k4\_ordinal1~X0){\unicdbdjdcd}(\\
k10\_ordinal2~np\_\_1~X0=X0))\end{array}
\label{hyp:ehhdedpfdgbgchegpfcd}
\end{equation}
Assume the following.
\begin{equation}
\begin{array}{c}{\unicdcdadad} X0 .  (v3\_ordinal1~X0){\unicdbdjdcd}((k12\_ordinal2~X0~np\_\_1=X0){\unicdcdcdhd}(k12\_ordinal2~\\
np\_\_1~X0=np\_\_1))\end{array}
\label{hyp:ehedgdpfpgchegjgogbgmgcd}
\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}(k12\_ordinal2~(k12\_ordinal2~X0~X1)~X2=k12\_ordinal2~\\
X0~(k11\_ordinal2~X2~X1))))\end{array}
\label{hyp:ehddbdpfpgchegjgogbgmged}
\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}(k12\_ordinal2~X0~(k10\_ordinal2~X1~X2)=k11\_ordinal2~\\
(k12\_ordinal2~X0~X2)~(k12\_ordinal2~X0~X1))))\end{array}
\label{hyp:ehddadpfpgchegjgogbgmged}
\end{equation}
Assume the following.
\begin{equation}
\begin{array}{c}((v2\_xxreal\_0~np\_\_1){\unicdcdcdhd}(m2\_subset\_1~np\_\_1~k1\_numbers~k5\_numbers)){\unicdcdcdhd}\\
((m1\_subset\_1~np\_\_1~k5\_numbers){\unicdcdcdhd}(m1\_subset\_1~np\_\_1~k1\_numbers))\end{array}
\label{hyp:dhahdgbdpfogfhngfgchbgmgdh}
\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}
{{k5\_numbers}={k4\_ordinal1}}\label{hyp:chfgegfgggjgogjgehjgpgogpflgfdpfogfhngcgfgchdh}
\end{equation}
Assume the following.
\begin{equation}
{{({\preoh{{{v1\_xboole\_0}~{k4\_ordinal1}}}})}{\unicdcdcdhd}{({{v3\_ordinal1}~{k4\_ordinal1}})}}\label{hyp:ggdggdpfpgchegjgogbgmgbd}
\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 .  (v3\_ordinal1~X0){\unicdbdjdcd}((v4\_ordinal5~X0){\unicdbdjded}(k12\_ordinal2~\\
k4\_ordinal1~X0=X0))\end{array}
\label{hyp:egfdpfpgchegjgogbgmgfd}
\end{equation}
Assume the following.
\begin{equation}
\begin{array}{c}{\unicdcdadad} X0 .  (v3\_ordinal1~X0){\unicdbdjdcd}({\unicdcdadad} X1 .  (m1\_subset\_1~X1~X0){\unicdbdjdcd}\\
(v3\_ordinal1~X1))\end{array}
\label{hyp:dgdgcdpfpgchegjgogbgmgcd}
\end{equation}
\begin{theorem}\label{thm:ehddidpfpgchegjgogbgmgfd}
$$\begin{array}{c}{\unicdcdadad} X0 .  ((v3\_ordinal1~X0){\unicdcdcdhd}(v4\_ordinal5~X0)){\unicdbdjdcd}(k12\_ordinal2~\\
k4\_ordinal1~(k12\_ordinal2~X0~k4\_ordinal1)=k12\_ordinal2~X0~(k12\_ordinal2~\\
X0~k4\_ordinal1))\end{array}$$
\end{theorem}
\end{document}
