\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}
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\newcommand\unicdbdjdcd{\Rightarrow}
\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{t11\_catalg\_1 (TMJm5LiGKKZ6Ku2XMwm6Wf9Vw4xhN3c2ESQ)}
\maketitle
Let $ k5\_catalg\_1 : {{\iota}{\unicdbdjdcd}{\iota}}$ be given.
Let $ k4\_tarski : {{\iota}{\unicdbdjdcd}{{\iota}{\unicdbdjdcd}{\iota}}}$ be given.
Let $ k9\_finseq\_1 : {{\iota}{\unicdbdjdcd}{\iota}}$ be given.
Let $ np\_\_1 : {\iota}$ be given.
Let $ k2\_tarski : {{\iota}{\unicdbdjdcd}{{\iota}{\unicdbdjdcd}{\iota}}}$ be given.
Let $ k1\_tarski : {{\iota}{\unicdbdjdcd}{\iota}}$ be given.
Assume the following.
\begin{equation}
\begin{array}{c}{\unicdcdadad} X0 .  {\unicdcdadad} X1 .  {\unicdcdadad} X2 .  {\unicdcdadad} X3 .  (k4\_tarski~X0~X1=\\
k4\_tarski~X2~X3){\unicdbdjdcd}((X0=X2){\unicdcdcdhd}(X1=X3))\end{array}
\label{hyp:ehbdpfihehfhahmgfgpfad}
\end{equation}
Assume the following.
\begin{equation}
{{\unicdcdadad} X0 .  {{\unicdcdadad} X1 .  {{({{{k9\_finseq\_1}~{X0}}={{k9\_finseq\_1}~{X1}}})}{\unicdbdjdcd}{({{X0}={X1}})}}}}\label{hyp:mgfdpfdgbgehbgmghgpfbd}
\end{equation}
Assume the following.
\begin{equation}
{{\unicdcdadad} X0 .  {{{k5\_catalg\_1}~{X0}}={{{k4\_tarski}~{np\_\_1}}~{({{k9\_finseq\_1}~{X0}})}}}}\label{hyp:egidpfdgbgehbgmghgpfbd}
\end{equation}
Assume the following.
\begin{equation}
\begin{array}{c}{\unicdcdadad} X0 .  {\unicdcdadad} X1 .  k4\_tarski~X0~X1=k2\_tarski~(k2\_tarski~X0~\\
X1)~(k1\_tarski~X0)\end{array}
\label{hyp:egfdpfehbgchdhlgjg}
\end{equation}
\begin{theorem}\label{thm:ehbdbdpfdgbgehbgmghgpfbd}
${{\unicdcdadad} X0 .  {{\unicdcdadad} X1 .  {{({{{k5\_catalg\_1}~{X0}}={{k5\_catalg\_1}~{X1}}})}{\unicdbdjdcd}{({{X0}={X1}})}}}}$.
\end{theorem}
\end{document}
