\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}
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\newcommand\unicdcdadjd{\notin}
\newcommand\unicdbdjded{\Leftrightarrow}
\newcommand\unicdcdcdid{\lor}
\newcommand\unicdcdcdhd{\land}
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\newcommand\unicdcdaddd{\exists}
\newcommand\unicdcdidid{\not\subseteq}
\newcommand\preoh[1]{\lnot #1}
\newcommand\mname[1]{{\mathsf{#1}}}
\begin{document}
\title{l78\_card\_2 (TMUh9wKAGRThxUsFPSUVr7XFrcS1DjNkJQo)}
\maketitle
Let $ np\_\_2 : {\iota}$ be given.
Let $ k5\_card\_1 : {{\iota}{\unicdbdjdcd}{\iota}}$ be given.
Let $ k2\_tarski : {{\iota}{\unicdbdjdcd}{{\iota}{\unicdbdjdcd}{\iota}}}$ be given.
Let $ k2\_xboole\_0 : {{\iota}{\unicdbdjdcd}{{\iota}{\unicdbdjdcd}{\iota}}}$ be given.
Let $ k1\_tarski : {{\iota}{\unicdbdjdcd}{\iota}}$ be given.
Let $ v1\_finset\_1 : {{\iota}{\unicdbdjdcd}{o}}$ be given.
Assume the following.
\begin{equation}
{{\unicdcdadad} X0 .  {{\unicdcdadad} X1 .  {{({{X0}{\unicdcdgdad}{X1}})}{\unicdbdjdcd}{({{{k5\_card\_1}~{({{{k2\_tarski}~{X0}}~{X1}})}}={np\_\_2}})}}}}\label{hyp:ehfdhdpfdgbgchegpfcd}
\end{equation}
Assume the following.
\begin{equation}
\begin{array}{c}{\unicdcdadad} X0 .  {\unicdcdadad} X1 .  k2\_tarski~X0~X1=k2\_xboole\_0~(k1\_tarski~\\
X0)~(k1\_tarski~X1)\end{array}
\label{hyp:ehbdpffgogfhngdhfgehbd}
\end{equation}
Assume the following.
\begin{equation}
{{\unicdcdadad} X0 .  {{({{v1\_finset\_1}~{X0}})}{\unicdbdjdcd}{({{{k5\_card\_1}~{({{k5\_card\_1}~{X0}})}}={{k5\_card\_1}~{X0}}})}}}\label{hyp:ahchpgkgfgdgehjgghjgehjhpflgfdpfdgbgchegpfbd}
\end{equation}
Assume the following.
\begin{equation}
{{\unicdcdadad} X0 .  {{\unicdcdadad} X1 .  {{v1\_finset\_1}~{({{{k2\_tarski}~{X0}}~{X1}})}}}}\label{hyp:ggdgcdpfggjgogdhfgehpfbd}
\end{equation}
\begin{theorem}\label{thm:mghdidpfdgbgchegpfcd}
${{np\_\_2}={{k5\_card\_1}~{np\_\_2}}}$.
\end{theorem}
\end{document}
