\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{t20\_classes2 (TMR1L2Npxnx256Q8g3V7UGzsViByGmSB6AR)}
\maketitle
Let $ v2\_classes1 : {{\iota}{\unicdbdjdcd}{o}}$ be given.
Let $ k1\_xboole\_0 : {\iota}$ be given.
Let $ k1\_card\_1 : {{\iota}{\unicdbdjdcd}{\iota}}$ be given.
Let $ k6\_numbers : {\iota}$ be given.
Let $ v4\_ordinal1 : {{\iota}{\unicdbdjdcd}{o}}$ be given.
Assume the following.
\begin{equation}
\begin{array}{c}{\unicdcdadad} X0 .  {\unicdcdadad} X1 .  ((k1\_card\_1~X0=k6\_numbers){\unicdcdcdhd}(k1\_card\_1~\\
X1=k6\_numbers)){\unicdbdjdcd}(X0=X1)\end{array}
\label{hyp:ehgdedpfdgbgchegpfcd}
\end{equation}
Assume the following.
\begin{equation}
{{\unicdcdadad} X0 .  {{({{v2\_classes1}~{X0}})}{\unicdbdjdcd}{({{v4\_ordinal1}~{({{k1\_card\_1}~{X0}})}})}}}\label{hyp:ehbdjdpfdgmgbgdhdhfgdhcd}
\end{equation}
Assume the following.
\begin{equation}
{{k6\_numbers}={k1\_xboole\_0}}\label{hyp:chfgegfgggjgogjgehjgpgogpflggdpfogfhngcgfgchdh}
\end{equation}
Assume the following.
\begin{equation}
{{\unicdcdadad} X0 .  {{{k1\_card\_1}~{({{k1\_card\_1}~{X0}})}}={{k1\_card\_1}~{X0}}}}\label{hyp:ahchpgkgfgdgehjgghjgehjhpflgbdpfdgbgchegpfbd}
\end{equation}
\begin{theorem}\label{thm:ehcdadpfdgmgbgdhdhfgdhcd}
$$\begin{array}{c}{\unicdcdadad} X0 .  (v2\_classes1~X0){\unicdbdjdcd}((X0=k1\_xboole\_0){\unicdcdcdid}((k1\_card\_1~\\
X0{\unicdcdgdad}k6\_numbers){\unicdcdcdhd}((k1\_card\_1~X0{\unicdcdgdad}k1\_xboole\_0){\unicdcdcdhd}(v4\_ordinal1~(\\
k1\_card\_1~X0)))))\end{array}$$
\end{theorem}
\end{document}
