\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}
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\newcommand\unicdcdaddd{\exists}
\newcommand\unicdcdidid{\not\subseteq}
\newcommand\preoh[1]{\lnot #1}
\newcommand\mname[1]{{\mathsf{#1}}}
\begin{document}
\title{t45\_quatern2 (TMVsu7AK4u6qqFT4ADLDMT7tMW8x5ZqDnXs)}
\maketitle
Let $ m1\_subset\_1 : {{\iota}{\unicdbdjdcd}{{\iota}{\unicdbdjdcd}{o}}}$ be given.
Let $ u1\_struct\_0 : {{\iota}{\unicdbdjdcd}{\iota}}$ be given.
Let $ k16\_quatern2 : {\iota}$ be given.
Let $ k3\_quatern2 : {{\iota}{\unicdbdjdcd}{\iota}}$ be given.
Let $ k6\_numbers : {\iota}$ be given.
Let $ k4\_struct\_0 : {{\iota}{\unicdbdjdcd}{\iota}}$ be given.
Let $ k1\_quatern2 : {\iota}$ be given.
Let $ k21\_quaterni : {\iota}$ be given.
Let $ v1\_quaterni : {{\iota}{\unicdbdjdcd}{o}}$ be given.
Assume the following.
\begin{equation}
{{{k4\_struct\_0}~{k16\_quatern2}}={k1\_quatern2}}\label{hyp:ehddjdpfbhfhbgehfgchogcd}
\end{equation}
Assume the following.
\begin{equation}
{{k1\_quatern2}={k21\_quaterni}}\label{hyp:chfgegfgggjgogjgehjgpgogpflgbdpfbhfhbgehfgchogcd}
\end{equation}
Assume the following.
\begin{equation}
\begin{array}{c}{\unicdcdadad} X0 .  (v1\_quaterni~X0){\unicdbdjdcd}((k3\_quatern2~X0=k6\_numbers){\unicdbdjdcd}(\\
X0=k6\_numbers))\end{array}
\label{hyp:mgbdddpfbhfhbgehfgchogcd}
\end{equation}
Assume the following.
\begin{equation}
{{k21\_quaterni}={k6\_numbers}}\label{hyp:egbdhdpfbhfhbgehfgchogjg}
\end{equation}
Assume the following.
\begin{equation}
\begin{array}{c}{\unicdcdadad} X0 .  (m1\_subset\_1~X0~(u1\_struct\_0~k16\_quatern2)){\unicdbdjdcd}(v1\_quaterni~\\
X0)\end{array}
\label{hyp:dgdgcdpfbhfhbgehfgchogcd}
\end{equation}
\begin{theorem}\label{thm:ehedfdpfbhfhbgehfgchogcd}
$$\begin{array}{c}{\unicdcdadad} X0 .  (m1\_subset\_1~X0~(u1\_struct\_0~k16\_quatern2)){\unicdbdjdcd}((k3\_quatern2~\\
X0=k6\_numbers){\unicdbdjdcd}(X0=k4\_struct\_0~k16\_quatern2))\end{array}$$
\end{theorem}
\end{document}
