\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{t10\_quatern3 (TMdBwQFFkFPnE1tXGMwkYicdC1cnktKJxJG)}
\maketitle
Let $ v1\_quaterni : {{\iota}{\unicdbdjdcd}{o}}$ be given.
Let $ k18\_quaterni : {{\iota}{\unicdbdjdcd}{\iota}}$ be given.
Let $ k6\_numbers : {\iota}$ be given.
Let $ k19\_quaterni : {{\iota}{\unicdbdjdcd}{\iota}}$ be given.
Let $ k20\_quaterni : {{\iota}{\unicdbdjdcd}{\iota}}$ be given.
Let $ k17\_quaterni : {{\iota}{\unicdbdjdcd}{\iota}}$ be given.
Let $ m1\_subset\_1 : {{\iota}{\unicdbdjdcd}{{\iota}{\unicdbdjdcd}{o}}}$ be given.
Let $ k1\_numbers : {\iota}$ be given.
Let $ k6\_quaterni : {{\iota}{\unicdbdjdcd}{{\iota}{\unicdbdjdcd}{{\iota}{\unicdbdjdcd}{{\iota}{\unicdbdjdcd}{\iota}}}}}$ be given.
Let $ k5\_arytm\_0 : {{\iota}{\unicdbdjdcd}{{\iota}{\unicdbdjdcd}{\iota}}}$ be given.
Let $ k5\_funct\_4 : {{\iota}{\unicdbdjdcd}{{\iota}{\unicdbdjdcd}{{\iota}{\unicdbdjdcd}{{\iota}{\unicdbdjdcd}{{\iota}{\unicdbdjdcd}{\iota}}}}}}$ be given.
Let $ np\_\_1 : {\iota}$ be given.
Assume the following.
\begin{equation}
\begin{array}{c}{\unicdcdadad} X0 .  (m1\_subset\_1~X0~k1\_numbers){\unicdbdjdcd}({\unicdcdadad} X1 .  (m1\_subset\_1~\\
X1~k1\_numbers){\unicdbdjdcd}(k6\_quaterni~X0~X1~k6\_numbers~k6\_numbers=k5\_arytm\_0~\\
X0~X1))\end{array}
\label{hyp:ehjdbdpfbhfhbgehfgchogjg}
\end{equation}
Assume the following.
\begin{equation}
\begin{array}{c}{\unicdcdadad} X0 .  (v1\_quaterni~X0){\unicdbdjdcd}(X0=k6\_quaterni~(k17\_quaterni~X0)~\\
(k18\_quaterni~X0)~(k19\_quaterni~X0)~(k20\_quaterni~X0))\end{array}
\label{hyp:ehcdedpfbhfhbgehfgchogjg}
\end{equation}
Assume the following.
\begin{equation}
\begin{array}{c}{\unicdcdadad} X0 .  (v1\_quaterni~X0){\unicdbdjdcd}(m1\_subset\_1~(k20\_quaterni~X0)~k1\_numbers)\end{array}
\label{hyp:egehpflgcdadpfbhfhbgehfgchogjg}
\end{equation}
Assume the following.
\begin{equation}
\begin{array}{c}{\unicdcdadad} X0 .  (v1\_quaterni~X0){\unicdbdjdcd}(m1\_subset\_1~(k17\_quaterni~X0)~k1\_numbers)\end{array}
\label{hyp:egehpflgbdhdpfbhfhbgehfgchogjg}
\end{equation}
Assume the following.
\begin{equation}
\begin{array}{c}{\unicdcdadad} X0 .  (m1\_subset\_1~X0~k1\_numbers){\unicdbdjdcd}({\unicdcdadad} X1 .  (m1\_subset\_1~\\
X1~k1\_numbers){\unicdbdjdcd}(((X1=k6\_numbers){\unicdbdjdcd}(k5\_arytm\_0~X0~X1=X0)){\unicdcdcdhd}((\\
X1{\unicdcdgdad}k6\_numbers){\unicdbdjdcd}(k5\_arytm\_0~X0~X1=k5\_funct\_4~k1\_numbers~k6\_numbers~\\
np\_\_1~X0~X1))))\end{array}
\label{hyp:egfdpfbgchjhehngpfad}
\end{equation}
\begin{theorem}\label{thm:ehbdadpfbhfhbgehfgchogdd}
$$\begin{array}{c}{\unicdcdadad} X0 .  (v1\_quaterni~X0){\unicdbdjdcd}(((k18\_quaterni~X0=k6\_numbers){\unicdcdcdhd}\\
((k19\_quaterni~X0=k6\_numbers){\unicdcdcdhd}(k20\_quaterni~X0=k6\_numbers))){\unicdbdjdcd}\\
(X0=k17\_quaterni~X0))\end{array}$$
\end{theorem}
\end{document}
