\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{t21\_mod\_2 (TMZuFjGshBVtfhdtvrAahCZ5uJf4FMA1Td2)}
\maketitle
Let $ v1\_xboole\_0 : {{\iota}{\unicdbdjdcd}{o}}$ be given.
Let $ v1\_classes2 : {{\iota}{\unicdbdjdcd}{o}}$ be given.
Let $ m1\_subset\_1 : {{\iota}{\unicdbdjdcd}{{\iota}{\unicdbdjdcd}{o}}}$ be given.
Let $ k6\_numbers : {\iota}$ be given.
Let $ np\_\_1 : {\iota}$ be given.
Let $ np\_\_2 : {\iota}$ be given.
Let $ k1\_xboole\_0 : {\iota}$ be given.
Let $ k1\_ordinal1 : {{\iota}{\unicdbdjdcd}{\iota}}$ be given.
Let $ np\_\_0 : {\iota}$ be given.
Assume the following.
\begin{equation}
{{\unicdcdadad} X0 .  {{({{v1\_xboole\_0}~{X0}})}{\unicdbdjdcd}{({{X0}={k1\_xboole\_0}})}}}\label{hyp:ehgdpfcgpgpgmgfg}
\end{equation}
Assume the following.
\begin{equation}
{{\unicdcdadad} X0 .  {{({{({\preoh{{{v1\_xboole\_0}~{X0}}}})}{\unicdcdcdhd}{({{v1\_classes2}~{X0}})}})}{\unicdbdjdcd}{({{k1\_xboole\_0} \in {X0}})}}}\label{hyp:ehfdgdpfdgmgbgdhdhfgdhcd}
\end{equation}
Assume the following.
\begin{equation}
\begin{array}{c}{\unicdcdadad} X0 .  ((\preoh v1\_xboole\_0~X0){\unicdcdcdhd}(v1\_classes2~X0)){\unicdbdjdcd}({\unicdcdadad} X1 .  \\
(m1\_subset\_1~X1~X0){\unicdbdjdcd}(m1\_subset\_1~(k1\_ordinal1~X1)~X0))\end{array}
\label{hyp:ehcdadpfngpgegpfcd}
\end{equation}
Assume the following.
\begin{equation}
{{\unicdcdadad} X0 .  {{\unicdcdadad} X1 .  {{({{X0} \in {X1}})}{\unicdbdjdcd}{({{{m1\_subset\_1}~{X0}}~{X1}})}}}}\label{hyp:ehbdpfdhfhcgdhfgeh}
\end{equation}
Assume the following.
\begin{equation}
{{v1\_xboole\_0}~{np\_\_0}}\label{hyp:dhahdgadpfcgpgpgmgfg}
\end{equation}
Assume the following.
\begin{equation}
{{{k1\_ordinal1}~{np\_\_1}}={np\_\_2}}\label{hyp:chbhdffhdgdgpfpflgbdpfpgchegjgogbgmgbdpfpfchbdpfchcd}
\end{equation}
Assume the following.
\begin{equation}
{{{k1\_ordinal1}~{np\_\_0}}={np\_\_1}}\label{hyp:chbhdffhdgdgpfpflgbdpfpgchegjgogbgmgbdpfpfchadpfchbd}
\end{equation}
Assume the following.
\begin{equation}
{{k6\_numbers}={k1\_xboole\_0}}\label{hyp:chfgegfgggjgogjgehjgpgogpflggdpfogfhngcgfgchdh}
\end{equation}
\begin{theorem}\label{thm:ehcdbdpfngpgegpfcd}
$$\begin{array}{c}{\unicdcdadad} X0 .  ((\preoh v1\_xboole\_0~X0){\unicdcdcdhd}(v1\_classes2~X0)){\unicdbdjdcd}((m1\_subset\_1~\\
k6\_numbers~X0){\unicdcdcdhd}((m1\_subset\_1~np\_\_1~X0){\unicdcdcdhd}(m1\_subset\_1~np\_\_2~X0)))\end{array}$$
\end{theorem}
\end{document}
