\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}
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\newcommand\unicdcdaddd{\exists}
\newcommand\unicdcdidid{\not\subseteq}
\newcommand\preoh[1]{\lnot #1}
\newcommand\mname[1]{{\mathsf{#1}}}
\begin{document}
\title{t2\_arytm\_2 (TMGdUNKTpZfTf42of8snFfRxbg4yRSxtZUz)}
\maketitle
Let $ r1\_tarski : {{\iota}{\unicdbdjdcd}{{\iota}{\unicdbdjdcd}{o}}}$ be given.
Let $ k4\_ordinal1 : {\iota}$ be given.
Let $ k2\_arytm\_2 : {\iota}$ be given.
Let $ k5\_arytm\_3 : {\iota}$ be given.
Assume the following.
\begin{equation}
\begin{array}{c}{\unicdcdadad} X0 .  {\unicdcdadad} X1 .  {\unicdcdadad} X2 .  ((r1\_tarski~X0~X1){\unicdcdcdhd}(r1\_tarski~\\
X1~X2)){\unicdbdjdcd}(r1\_tarski~X0~X2)\end{array}
\label{hyp:ehbdpfihcgpgpgmgfgpfbd}
\end{equation}
Assume the following.
\begin{equation}
{{{r1\_tarski}~{k5\_arytm\_3}}~{k2\_arytm\_2}}\label{hyp:ehbdpfbgchjhehngpfcd}
\end{equation}
Assume the following.
\begin{equation}
{{{r1\_tarski}~{k4\_ordinal1}}~{k5\_arytm\_3}}\label{hyp:mgedddpfbgchjhehngpfdd}
\end{equation}
\begin{theorem}\label{thm:ehcdpfbgchjhehngpfcd}
${{{r1\_tarski}~{k4\_ordinal1}}~{k2\_arytm\_2}}$.
\end{theorem}
\end{document}
