\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\_classes2 (TMaFcpasHQSBT59CGdcxUp3yk3FS9pSm3dz)}
\maketitle
Let $ v1\_xboole\_0 : {{\iota}{\unicdbdjdcd}{o}}$ be given.
Let $ v1\_classes2 : {{\iota}{\unicdbdjdcd}{o}}$ be given.
Let $ k1\_classes1 : {{\iota}{\unicdbdjdcd}{\iota}}$ be given.
Let $ v1\_ordinal1 : {{\iota}{\unicdbdjdcd}{o}}$ be given.
Let $ v2\_classes1 : {{\iota}{\unicdbdjdcd}{o}}$ be given.
Assume the following.
\begin{equation}
{{\unicdcdadad} X0 .  {{({{v1\_ordinal1}~{X0}})}{\unicdbdjdcd}{({{v1\_classes2}~{({{k1\_classes1}~{X0}})}})}}}\label{hyp:ehedhdpfdgmgbgdhdhfgdhcd}
\end{equation}
Assume the following.
\begin{equation}
{{\unicdcdadad} X0 .  {\preoh{{{v1\_xboole\_0}~{({{k1\_classes1}~{X0}})}}}}}\label{hyp:ggdgbdpfdgmgbgdhdhfgdhbd}
\end{equation}
Assume the following.
\begin{equation}
{{\unicdcdadad} X0 .  {{({{v1\_classes2}~{X0}})}{\unicdbdjdcd}{({{({{v1\_ordinal1}~{X0}})}{\unicdcdcdhd}{({{v2\_classes1}~{X0}})}})}}}\label{hyp:dgdgcdpfdgmgbgdhdhfgdhcd}
\end{equation}
\begin{theorem}\label{thm:ehedidpfdgmgbgdhdhfgdhcd}
$$\begin{array}{c}{\unicdcdadad} X0 .  ((\preoh v1\_xboole\_0~X0){\unicdcdcdhd}(v1\_classes2~X0)){\unicdbdjdcd}((\preoh v1\_xboole\_0~\\
(k1\_classes1~X0)){\unicdcdcdhd}(v1\_classes2~(k1\_classes1~X0)))\end{array}$$
\end{theorem}
\end{document}
