\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{l2\_quatern3 (TMVrWahFpk6RRhKrm2paYWbgLtUoVGaKS3w)}
\maketitle
Let $ v1\_quaterni : {{\iota}{\unicdbdjdcd}{o}}$ be given.
Let $ m1\_subset\_1 : {{\iota}{\unicdbdjdcd}{{\iota}{\unicdbdjdcd}{o}}}$ be given.
Let $ k1\_numbers : {\iota}$ be given.
Let $ k17\_quaterni : {{\iota}{\unicdbdjdcd}{\iota}}$ 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.
Assume the following.
\begin{equation}
\begin{array}{c}{\unicdcdadad} X0 .  (v1\_quaterni~X0){\unicdbdjdcd}({\unicdcdadad} X1 .  (m1\_subset\_1~X1~k1\_numbers){\unicdbdjdcd}\\
((X0=X1){\unicdbdjdcd}((k17\_quaterni~X0=X1){\unicdcdcdhd}((k18\_quaterni~X0=k6\_numbers){\unicdcdcdhd}\\
((k19\_quaterni~X0=k6\_numbers){\unicdcdcdhd}(k20\_quaterni~X0=k6\_numbers))))))\end{array}
\label{hyp:mggdjdpfbhfhbgehfgchogjg}
\end{equation}
\begin{theorem}\label{thm:mgcdpfbhfhbgehfgchogdd}
$$\begin{array}{c}{\unicdcdadad} X0 .  (v1\_quaterni~X0){\unicdbdjdcd}((m1\_subset\_1~X0~k1\_numbers){\unicdbdjdcd}((\\
X0=k17\_quaterni~X0){\unicdcdcdhd}((k18\_quaterni~X0=k6\_numbers){\unicdcdcdhd}((k19\_quaterni~\\
X0=k6\_numbers){\unicdcdcdhd}(k20\_quaterni~X0=k6\_numbers)))))\end{array}$$
\end{theorem}
\end{document}
