\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{t32\_card\_fil (TMJzUBw83wQW1vrz5RZbu9T1bMgw7iJEUVt)}
\maketitle
Let $ v3\_ordinal1 : {{\iota}{\unicdbdjdcd}{o}}$ be given.
Let $ v4\_ordinal1 : {{\iota}{\unicdbdjdcd}{o}}$ be given.
Let $ r1\_tarski : {{\iota}{\unicdbdjdcd}{{\iota}{\unicdbdjdcd}{o}}}$ be given.
Let $ k3\_ordinal2 : {{\iota}{\unicdbdjdcd}{\iota}}$ be given.
Let $ k3\_tarski : {{\iota}{\unicdbdjdcd}{\iota}}$ be given.
Let $ k1\_ordinal1 : {{\iota}{\unicdbdjdcd}{\iota}}$ be given.
Let $ k2\_ordinal1 : {{\iota}{\unicdbdjdcd}{\iota}}$ be given.
Assume the following.
\begin{equation}
\begin{array}{c}{\unicdcdadad} X0 .  {\unicdcdadad} X1 .  (r1\_tarski~X0~X1){\unicdbdjdcd}(r1\_tarski~(k3\_tarski~\\
X0)~(k3\_tarski~X1))\end{array}
\label{hyp:ehhdhdpfkhggngjgdhdgpfbd}
\end{equation}
Assume the following.
\begin{equation}
\begin{array}{c}{\unicdcdadad} X0 .  r1\_tarski~(k3\_ordinal2~X0)~(k1\_ordinal1~(k3\_tarski~\\
(k2\_ordinal1~X0)))\end{array}
\label{hyp:ehhdcdpfpgchegjgogbgmgdd}
\end{equation}
Assume the following.
\begin{equation}
\begin{array}{c}{\unicdcdadad} X0 .  (v3\_ordinal1~X0){\unicdbdjdcd}({\unicdcdadad} X1 .  (r1\_tarski~X1~X0){\unicdbdjdcd}(k2\_ordinal1~\\
X1=X1))\end{array}
\label{hyp:ehgdpfpgchegjgogbgmgdd}
\end{equation}
Assume the following.
\begin{equation}
\begin{array}{c}{\unicdcdadad} X0 .  (v3\_ordinal1~X0){\unicdbdjdcd}({\unicdcdadad} X1 .  (r1\_tarski~X1~X0){\unicdbdjdcd}(v3\_ordinal1~\\
(k3\_tarski~X1)))\end{array}
\label{hyp:ehedpfpgchegjgogbgmgdd}
\end{equation}
Assume the following.
\begin{equation}
{{\unicdcdadad} X0 .  {{({{v3\_ordinal1}~{X0}})}{\unicdbdjdcd}{({{{k3\_tarski}~{({{k1\_ordinal1}~{X0}})}}={X0}})}}}\label{hyp:ehcdpfpgchegjgogbgmgcd}
\end{equation}
Assume the following.
\begin{equation}
{{\unicdcdadad} X0 .  {{({{v4\_ordinal1}~{X0}})}{\unicdbdjded}{({{X0}={{k3\_tarski}~{X0}}})}}}\label{hyp:eggdpfpgchegjgogbgmgbd}
\end{equation}
Assume the following.
\begin{equation}
{{\unicdcdadad} X0 .  {{\unicdcdadad} X1 .  {{({{X0}={X1}})}{\unicdbdjded}{({{({{{r1\_tarski}~{X0}}~{X1}})}{\unicdcdcdhd}{({{{r1\_tarski}~{X1}}~{X0}})}})}}}}\label{hyp:egbdadpfihcgpgpgmgfgpfad}
\end{equation}
\begin{theorem}\label{thm:ehddcdpfdgbgchegpfggjgmg}
$$\begin{array}{c}{\unicdcdadad} X0 .  ((v3\_ordinal1~X0){\unicdcdcdhd}(v4\_ordinal1~X0)){\unicdbdjdcd}({\unicdcdadad} X1 .  \\
((r1\_tarski~X1~X0){\unicdcdcdhd}(k3\_ordinal2~X1=X0)){\unicdbdjdcd}(k3\_tarski~X1=X0))\end{array}$$
\end{theorem}
\end{document}
