\documentclass{article}
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\newtheorem{lemma}{Lemma}
\newtheorem{proposition}{Proposition}
\newtheorem{definition}{Definition}
\newtheorem{example}{Example}
\newtheorem{conjecture}{Conjecture}
\newcommand\unicdcdgdad{\not=}
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\newcommand\unicdcdadjd{\notin}
\newcommand\unicdbdjded{\Leftrightarrow}
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\newcommand\unicdcdidid{\not\subseteq}
\newcommand\preoh[1]{\lnot #1}
\newcommand\mname[1]{{\mathsf{#1}}}
\begin{document}
\title{t87\_quatern3 (TMcDWh1nCPmJ9FjafR7R3zcG6bVvvCrJx1t)}
\maketitle
Let $ k2\_quatern3 : {{\iota}{\unicdbdjdcd}{\iota}}$ be given.
Let $ k28\_quaterni : {{\iota}{\unicdbdjdcd}{\iota}}$ be given.
Let $ k2\_quatern2 : {\iota}$ be given.
Let $ np\_\_1 : {\iota}$ be given.
Let $ v1\_quaterni : {{\iota}{\unicdbdjdcd}{o}}$ be given.
Let $ k1\_quatern3 : {{\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}(k1\_quatern3~X0=k2\_quatern3~(k28\_quaterni~\\
X0))\end{array}
\label{hyp:ehidbdpfbhfhbgehfgchogdd}
\end{equation}
Assume the following.
\begin{equation}
{{{k2\_quatern3}~{k2\_quatern2}}={np\_\_1}}\label{hyp:ehidadpfbhfhbgehfgchogdd}
\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}
{{{m1\_subset\_1}~{k2\_quatern2}}~{k1\_quaterni}}\label{hyp:egehpflgcdpfbhfhbgehfgchogcd}
\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:ehidhdpfbhfhbgehfgchogdd}
${{{k2\_quatern3}~{({{k28\_quaterni}~{k2\_quatern2}})}}={np\_\_1}}$.
\end{theorem}
\end{document}
