\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}
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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\_arytm\_3 (TMXEvNerqru7cYmPAhbDj7Vy5fe8nL5ALKH)}
\maketitle
Let $ np\_\_1 : {\iota}$ be given.
Let $ k4\_ordinal1 : {\iota}$ be given.
Let $ k1\_ordinal1 : {{\iota}{\unicdbdjdcd}{\iota}}$ be given.
Let $ k1\_xboole\_0 : {\iota}$ be given.
Let $ v3\_ordinal1 : {{\iota}{\unicdbdjdcd}{o}}$ be given.
Let $ v7\_ordinal1 : {{\iota}{\unicdbdjdcd}{o}}$ be given.
Assume the following.
\begin{equation}
{{np\_\_1}={{k1\_ordinal1}~{k1\_xboole\_0}}}\label{hyp:mgedhdpfpgchegjgogbgmgcd}
\end{equation}
Assume the following.
\begin{equation}
{{k1\_xboole\_0} \in {k4\_ordinal1}}\label{hyp:mgbdpfbgchjhehngpfdd}
\end{equation}
Assume the following.
\begin{equation}
\begin{array}{c}{\unicdcdadad} X0 .  ((v3\_ordinal1~X0){\unicdcdcdhd}(v7\_ordinal1~X0)){\unicdbdjdcd}(v7\_ordinal1~\\
(k1\_ordinal1~X0))\end{array}
\label{hyp:ggdghdpfpgchegjgogbgmgbd}
\end{equation}
Assume the following.
\begin{equation}
{{\unicdcdadad} X0 .  {{({{v7\_ordinal1}~{X0}})}{\unicdbdjded}{({{X0} \in {k4\_ordinal1}})}}}\label{hyp:egbdcdpfpgchegjgogbgmgbd}
\end{equation}
Assume the following.
\begin{equation}
{{\unicdcdadad} X0 .  {{({{v7\_ordinal1}~{X0}})}{\unicdbdjdcd}{({{v3\_ordinal1}~{X0}})}}}\label{hyp:dgdggdpfpgchegjgogbgmgbd}
\end{equation}
\begin{theorem}\label{thm:mgcdpfbgchjhehngpfdd}
${{np\_\_1} \in {k4\_ordinal1}}$.
\end{theorem}
\end{document}
