\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{t48\_quatern3 (TMNY38akcbYwkv5LiR6fkGNmTGjftnmG3hy)}
\maketitle
Let $ k8\_quatern2 : {{\iota}{\unicdbdjdcd}{\iota}}$ be given.
Let $ k2\_quatern2 : {\iota}$ be given.
Let $ k5\_struct\_0 : {{\iota}{\unicdbdjdcd}{\iota}}$ be given.
Let $ k16\_quatern2 : {\iota}$ be given.
Assume the following.
\begin{equation}
{{{k8\_quatern2}~{({{k5\_struct\_0}~{k16\_quatern2}})}}={{k5\_struct\_0}~{k16\_quatern2}}}\label{hyp:ehedhdpfbhfhbgehfgchogcd}
\end{equation}
Assume the following.
\begin{equation}
{{{k5\_struct\_0}~{k16\_quatern2}}={k2\_quatern2}}\label{hyp:ehddhdpfbhfhbgehfgchogcd}
\end{equation}
\begin{theorem}\label{thm:ehedidpfbhfhbgehfgchogdd}
${{{k8\_quatern2}~{k2\_quatern2}}={k2\_quatern2}}$.
\end{theorem}
\end{document}
