\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}
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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{l66\_arytm\_2 (TMYUmK9zfS8cjtKGCH5pb6KSQGoe4mWQDfz)}
\maketitle
Let $ k12\_arytm\_3 : {\iota}$ be given.
Let $ k5\_arytm\_3 : {\iota}$ be given.
Let $ np\_\_1 : {\iota}$ be given.
Let $ v1\_xboole\_0 : {{\iota}{\unicdbdjdcd}{o}}$ be given.
Let $ v3\_ordinal1 : {{\iota}{\unicdbdjdcd}{o}}$ be given.
Let $ m1\_subset\_1 : {{\iota}{\unicdbdjdcd}{{\iota}{\unicdbdjdcd}{o}}}$ be given.
Assume the following.
\begin{equation}
{{k12\_arytm\_3}={np\_\_1}}\label{hyp:mgfdbdpfbgchjhehngpfcd}
\end{equation}
Assume the following.
\begin{equation}
{\preoh{{{v1\_xboole\_0}~{k5\_arytm\_3}}}}\label{hyp:ggdgddpfbgchjhehngpfdd}
\end{equation}
Assume the following.
\begin{equation}
\begin{array}{c}(\preoh v1\_xboole\_0~k12\_arytm\_3){\unicdcdcdhd}((v3\_ordinal1~k12\_arytm\_3){\unicdcdcdhd}(m1\_subset\_1~\\
k12\_arytm\_3~k5\_arytm\_3))\end{array}
\label{hyp:egehpflgbdcdpfbgchjhehngpfdd}
\end{equation}
Assume the following.
\begin{equation}
\begin{array}{c}{\unicdcdadad} X0 .  {\unicdcdadad} X1 .  ((\preoh v1\_xboole\_0~X0){\unicdbdjdcd}((m1\_subset\_1~X1~X0){\unicdbdjded}\\
(X1 \in X0))){\unicdcdcdhd}((v1\_xboole\_0~X0){\unicdbdjdcd}((m1\_subset\_1~X1~X0){\unicdbdjded}(v1\_xboole\_0~\\
X1)))\end{array}
\label{hyp:egbdpfdhfhcgdhfgehpfbd}
\end{equation}
\begin{theorem}\label{thm:mggdgdpfbgchjhehngpfcd}
${{k12\_arytm\_3} \in {k5\_arytm\_3}}$.
\end{theorem}
\end{document}
