\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{t65\_quatern3 (TMYYNsEKLre52Q9DzGqmCdgKwqBhnHHPTw3)}
\maketitle
Let $ v1\_quaterni : {{\iota}{\unicdbdjdcd}{o}}$ be given.
Let $ k27\_quaterni : {{\iota}{\unicdbdjdcd}{{\iota}{\unicdbdjdcd}{\iota}}}$ be given.
Let $ k26\_quaterni : {{\iota}{\unicdbdjdcd}{{\iota}{\unicdbdjdcd}{\iota}}}$ be given.
Let $ k29\_quaterni : {{\iota}{\unicdbdjdcd}{{\iota}{\unicdbdjdcd}{\iota}}}$ be given.
Let $ m1\_subset\_1 : {{\iota}{\unicdbdjdcd}{{\iota}{\unicdbdjdcd}{o}}}$ be given.
Let $ k1\_quaterni : {\iota}$ be given.
Assume the following.
\begin{equation}
\begin{array}{c}{\unicdcdadad} X0 .  (v1\_quaterni~X0){\unicdbdjdcd}({\unicdcdadad} X1 .  (v1\_quaterni~X1){\unicdbdjdcd}({\unicdcdadad} X2 .  \\
(v1\_quaterni~X2){\unicdbdjdcd}(k27\_quaterni~(k29\_quaterni~X0~X1)~X2=k29\_quaterni~\\
(k27\_quaterni~X0~X2)~(k27\_quaterni~X1~X2))))\end{array}
\label{hyp:ehcdddpfbhfhbgehfgchogcd}
\end{equation}
Assume the following.
\begin{equation}
\begin{array}{c}{\unicdcdadad} X0 .  (v1\_quaterni~X0){\unicdbdjdcd}({\unicdcdadad} X1 .  (v1\_quaterni~X1){\unicdbdjdcd}({\unicdcdadad} X2 .  \\
(v1\_quaterni~X2){\unicdbdjdcd}(k27\_quaterni~(k26\_quaterni~X0~X1)~X2=k26\_quaterni~\\
(k27\_quaterni~X0~X2)~(k27\_quaterni~X1~X2))))\end{array}
\label{hyp:ehbdidpfbhfhbgehfgchogcd}
\end{equation}
Assume the following.
\begin{equation}
\begin{array}{c}{\unicdcdadad} X0 .  {\unicdcdadad} X1 .  ((v1\_quaterni~X0){\unicdcdcdhd}(v1\_quaterni~X1)){\unicdbdjdcd}(\\
m1\_subset\_1~(k29\_quaterni~X0~X1)~k1\_quaterni)\end{array}
\label{hyp:egehpflgcdjdpfbhfhbgehfgchogjg}
\end{equation}
Assume the following.
\begin{equation}
{{\unicdcdadad} X0 .  {{({{{m1\_subset\_1}~{X0}}~{k1\_quaterni}})}{\unicdbdjdcd}{({{v1\_quaterni}~{X0}})}}}\label{hyp:dgdgbdpfbhfhbgehfgchogjg}
\end{equation}
\begin{theorem}\label{thm:ehgdfdpfbhfhbgehfgchogdd}
$$\begin{array}{c}{\unicdcdadad} X0 .  (v1\_quaterni~X0){\unicdbdjdcd}({\unicdcdadad} X1 .  (v1\_quaterni~X1){\unicdbdjdcd}({\unicdcdadad} X2 .  \\
(v1\_quaterni~X2){\unicdbdjdcd}({\unicdcdadad} X3 .  (v1\_quaterni~X3){\unicdbdjdcd}(k27\_quaterni~\\
(k26\_quaterni~(k29\_quaterni~X0~X1)~X2)~X3=k26\_quaterni~(k29\_quaterni~\\
(k27\_quaterni~X0~X3)~(k27\_quaterni~X1~X3))~(k27\_quaterni~X2~X3)))))\end{array}$$
\end{theorem}
\end{document}
