\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\_card\_2 (TMFCFBW9dU6dwf5wnRsFaDFUxv2dse54LNa)}
\maketitle
Let $ v1\_card\_1 : {{\iota}{\unicdbdjdcd}{o}}$ be given.
Let $ k2\_card\_2 : {{\iota}{\unicdbdjdcd}{{\iota}{\unicdbdjdcd}{\iota}}}$ be given.
Let $ k6\_numbers : {\iota}$ be given.
Let $ k2\_zfmisc\_1 : {{\iota}{\unicdbdjdcd}{{\iota}{\unicdbdjdcd}{\iota}}}$ be given.
Let $ k1\_xboole\_0 : {\iota}$ be given.
Let $ k1\_card\_1 : {{\iota}{\unicdbdjdcd}{\iota}}$ be given.
Assume the following.
\begin{equation}
\begin{array}{c}{\unicdcdadad} X0 .  {\unicdcdadad} X1 .  (k2\_zfmisc\_1~X0~X1=k1\_xboole\_0){\unicdbdjded}((X0=\\
k1\_xboole\_0){\unicdcdcdid}(X1=k1\_xboole\_0))\end{array}
\label{hyp:ehjdadpfkhggngjgdhdgpfbd}
\end{equation}
Assume the following.
\begin{equation}
{{{k1\_card\_1}~{k1\_xboole\_0}}={k1\_xboole\_0}}\label{hyp:ehcdhdpfdgbgchegpfbd}
\end{equation}
Assume the following.
\begin{equation}
{{k6\_numbers}={k1\_xboole\_0}}\label{hyp:chfgegfgggjgogjgehjgpgogpflggdpfogfhngcgfgchdh}
\end{equation}
Assume the following.
\begin{equation}
{{\unicdcdadad} X0 .  {{v1\_card\_1}~{({{k1\_card\_1}~{X0}})}}}\label{hyp:egehpflgbdpfdgbgchegpfbd}
\end{equation}
Assume the following.
\begin{equation}
\begin{array}{c}{\unicdcdadad} X0 .  (v1\_card\_1~X0){\unicdbdjdcd}({\unicdcdadad} X1 .  (v1\_card\_1~X1){\unicdbdjdcd}(k2\_card\_2~\\
X0~X1=k1\_card\_1~(k2\_zfmisc\_1~X0~X1)))\end{array}
\label{hyp:egcdpfdgbgchegpfcd}
\end{equation}
Assume the following.
\begin{equation}
\begin{array}{c}{\unicdcdadad} X0 .  {\unicdcdadad} X1 .  ((v1\_card\_1~X0){\unicdcdcdhd}(v1\_card\_1~X1)){\unicdbdjdcd}(k2\_card\_2~\\
X0~X1=k2\_card\_2~X1~X0)\end{array}
\label{hyp:dgpgngngfhehbgehjgghjgehjhpflgcdpfdgbgchegpfcd}
\end{equation}
\begin{theorem}\label{thm:ehcdadpfdgbgchegpfcd}
${{\unicdcdadad} X0 .  {{({{v1\_card\_1}~{X0}})}{\unicdbdjdcd}{({{{{k2\_card\_2}~{X0}}~{k6\_numbers}}={k6\_numbers}})}}}$.
\end{theorem}
\end{document}
