\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{t5\_arytm\_0 (TMagSsgFhH62nG758DMz68A4Js6G8hnuzJm)}
\maketitle
Let $ r1\_xboole\_0 : {{\iota}{\unicdbdjdcd}{{\iota}{\unicdbdjdcd}{o}}}$ be given.
Let $ k2\_arytm\_2 : {\iota}$ be given.
Let $ k2\_zfmisc\_1 : {{\iota}{\unicdbdjdcd}{{\iota}{\unicdbdjdcd}{\iota}}}$ be given.
Let $ k1\_tarski : {{\iota}{\unicdbdjdcd}{\iota}}$ be given.
Let $ k11\_arytm\_3 : {\iota}$ be given.
Let $ r1\_tarski : {{\iota}{\unicdbdjdcd}{{\iota}{\unicdbdjdcd}{o}}}$ be given.
Let $ k4\_tarski : {{\iota}{\unicdbdjdcd}{{\iota}{\unicdbdjdcd}{\iota}}}$ be given.
Assume the following.
\begin{equation}
\begin{array}{c}{\unicdcdadad} X0 .  {\unicdcdadad} X1 .  {\unicdcdadad} X2 .  {\unicdcdadad} X3 .  \preoh (r1\_tarski~X0~(k2\_zfmisc\_1~\\
X1~X2)){\unicdcdcdhd}((X3 \in X0){\unicdcdcdhd}({\unicdcdadad} X4 .  {\unicdcdadad} X5 .  \preoh (X4 \in X1){\unicdcdcdhd}((X5 \in X2){\unicdcdcdhd}\\
(X3=k4\_tarski~X4~X5))))\end{array}
\label{hyp:ehidedpfkhggngjgdhdgpfbd}
\end{equation}
Assume the following.
\begin{equation}
\begin{array}{c}{\unicdcdadad} X0 .  {\unicdcdadad} X1 .  (\preoh (\preoh r1\_xboole\_0~X0~X1){\unicdcdcdhd}({\unicdcdadad} X2 .  \preoh (X2 \in \\
X0){\unicdcdcdhd}(X2 \in X1))){\unicdcdcdhd}(\preoh ({\unicdcdaddd} X2 .  (X2 \in X0){\unicdcdcdhd}(X2 \in X1)){\unicdcdcdhd}(r1\_xboole\_0~\\
X0~X1))\end{array}
\label{hyp:ehddpfihcgpgpgmgfgpfad}
\end{equation}
Assume the following.
\begin{equation}
{{\unicdcdadad} X0 .  {\preoh{{{{{k4\_tarski}~{k11\_arytm\_3}}~{X0}} \in {k2\_arytm\_2}}}}}\label{hyp:ehddpfbgchjhehngpfcd}
\end{equation}
Assume the following.
\begin{equation}
{{\unicdcdadad} X0 .  {{\unicdcdadad} X1 .  {{{r1\_tarski}~{X0}}~{X0}}}}\label{hyp:chfgggmgfgihjgghjgehjhpfchbdpfehbgchdhlgjg}
\end{equation}
Assume the following.
\begin{equation}
\begin{array}{c}{\unicdcdadad} X0 .  {\unicdcdadad} X1 .  (X1=k1\_tarski~X0){\unicdbdjded}({\unicdcdadad} X2 .  (X2 \in X1){\unicdbdjded}\\
(X2=X0))\end{array}
\label{hyp:egbdpfehbgchdhlgjg}
\end{equation}
\begin{theorem}\label{thm:ehfdpfbgchjhehngpfad}
$$\begin{array}{c}r1\_xboole\_0~k2\_arytm\_2~(k2\_zfmisc\_1~(k1\_tarski~k11\_arytm\_3)~\\
k2\_arytm\_2)\end{array}$$
\end{theorem}
\end{document}
