\documentclass{article}
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\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}
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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{l2\_arithm (TMGeZeRwBRgUFcMRZN2U8MKRCxxDXNtSUGm)}
\maketitle
Let $ k4\_xcmplx\_0 : {{\iota}{\unicdbdjdcd}{\iota}}$ be given.
Let $ k6\_numbers : {\iota}$ be given.
Let $ v1\_xcmplx\_0 : {{\iota}{\unicdbdjdcd}{o}}$ be given.
Let $ k2\_xcmplx\_0 : {{\iota}{\unicdbdjdcd}{{\iota}{\unicdbdjdcd}{\iota}}}$ be given.
Let $ m1\_subset\_1 : {{\iota}{\unicdbdjdcd}{{\iota}{\unicdbdjdcd}{o}}}$ be given.
Let $ c2\_\_arytm\_0 : {\iota}$ be given.
Let $ k1\_numbers : {\iota}$ be given.
Assume the following.
\begin{equation}
{{\unicdcdadad} X0 .  {{({{v1\_xcmplx\_0}~{X0}})}{\unicdbdjdcd}{({{{{k2\_xcmplx\_0}~{X0}}~{k6\_numbers}}={X0}})}}}\label{hyp:ehbdpfbgchjgehigng}
\end{equation}
Assume the following.
\begin{equation}
{{{m1\_subset\_1}~{c2\_\_arytm\_0}}~{k1\_numbers}}\label{hyp:egehpfdgcdpfpfbgchjhehngpfad}
\end{equation}
Assume the following.
\begin{equation}
{{c2\_\_arytm\_0}={k6\_numbers}}\label{hyp:egfgpfdgcdpfpfbgchjhehngpfad}
\end{equation}
Assume the following.
\begin{equation}
\begin{array}{c}{\unicdcdadad} X0 .  (v1\_xcmplx\_0~X0){\unicdbdjdcd}({\unicdcdadad} X1 .  (v1\_xcmplx\_0~X1){\unicdbdjdcd}((\\
X1=k4\_xcmplx\_0~X0){\unicdbdjded}(k2\_xcmplx\_0~X0~X1=k6\_numbers)))\end{array}
\label{hyp:eggdpfihdgngahmgihpfad}
\end{equation}
Assume the following.
\begin{equation}
{{\unicdcdadad} X0 .  {{({{{m1\_subset\_1}~{X0}}~{k1\_numbers}})}{\unicdbdjdcd}{({{v1\_xcmplx\_0}~{X0}})}}}\label{hyp:dgdgbdpfihdgngahmgihpfad}
\end{equation}
\begin{theorem}\label{thm:mgcdpfbgchjgehigng}
${{{k4\_xcmplx\_0}~{k6\_numbers}}={k6\_numbers}}$.
\end{theorem}
\end{document}