\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{t81\_quatern3 (TMPF6zBeSu1k7R729eVHtZ17wKaDN8fCCbb)}
\maketitle
Let $ v1\_quaterni : {{\iota}{\unicdbdjdcd}{o}}$ be given.
Let $ k1\_quatern3 : {{\iota}{\unicdbdjdcd}{\iota}}$ be given.
Let $ k2\_quatern3 : {{\iota}{\unicdbdjdcd}{\iota}}$ be given.
Let $ k28\_quaterni : {{\iota}{\unicdbdjdcd}{\iota}}$ be given.
Let $ k27\_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}(k27\_quaterni~\\
(k28\_quaterni~X0)~(k28\_quaterni~X1)=k27\_quaterni~X0~X1))\end{array}
\label{hyp:ehcdcdpfbhfhbgehfgchogcd}
\end{equation}
Assume the following.
\begin{equation}
\begin{array}{c}{\unicdcdadad} X0 .  (m1\_subset\_1~X0~k1\_quaterni){\unicdbdjdcd}(k2\_quatern3~X0=k1\_quatern3~\\
X0)\end{array}
\label{hyp:chfgegfgggjgogjgehjgpgogpflgcdpfbhfhbgehfgchogdd}
\end{equation}
Assume the following.
\begin{equation}
\begin{array}{c}{\unicdcdadad} X0 .  (v1\_quaterni~X0){\unicdbdjdcd}(m1\_subset\_1~(k28\_quaterni~X0)~k1\_quaterni)\end{array}
\label{hyp:egehpflgcdidpfbhfhbgehfgchogjg}
\end{equation}
Assume the following.
\begin{equation}
{{\unicdcdadad} X0 .  {{({{v1\_quaterni}~{X0}})}{\unicdbdjdcd}{({{{k1\_quatern3}~{X0}}={{{k27\_quaterni}~{X0}}~{X0}}})}}}\label{hyp:egbdpfbhfhbgehfgchogdd}
\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:ehidbdpfbhfhbgehfgchogdd}
$$\begin{array}{c}{\unicdcdadad} X0 .  (v1\_quaterni~X0){\unicdbdjdcd}(k1\_quatern3~X0=k2\_quatern3~(k28\_quaterni~\\
X0))\end{array}$$
\end{theorem}
\end{document}
