\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}
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\newcommand\unicdbdjdcd{\Rightarrow}
\newcommand\unicdcdadjd{\notin}
\newcommand\unicdbdjded{\Leftrightarrow}
\newcommand\unicdcdcdid{\lor}
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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\_0 (TMdWUjM1UYwi3vwebiQZGhjz6JR6RQ7Gevn)}
\maketitle
Let $ m1\_subset\_1 : {{\iota}{\unicdbdjdcd}{{\iota}{\unicdbdjdcd}{o}}}$ be given.
Let $ k2\_arytm\_2 : {\iota}$ be given.
Let $ k11\_arytm\_3 : {\iota}$ be given.
Let $ k4\_tarski : {{\iota}{\unicdbdjdcd}{{\iota}{\unicdbdjdcd}{\iota}}}$ be given.
Let $ k1\_numbers : {\iota}$ be given.
Let $ k2\_zfmisc\_1 : {{\iota}{\unicdbdjdcd}{{\iota}{\unicdbdjdcd}{\iota}}}$ be given.
Let $ v1\_xboole\_0 : {{\iota}{\unicdbdjdcd}{o}}$ be given.
Let $ k6\_subset\_1 : {{\iota}{\unicdbdjdcd}{{\iota}{\unicdbdjdcd}{\iota}}}$ be given.
Let $ k4\_xboole\_0 : {{\iota}{\unicdbdjdcd}{{\iota}{\unicdbdjdcd}{\iota}}}$ be given.
Let $ k1\_xboole\_0 : {\iota}$ be given.
Let $ k2\_xboole\_0 : {{\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 \in \\
k2\_zfmisc\_1~X2~X3){\unicdbdjded}((X0 \in X2){\unicdcdcdhd}(X1 \in X3))\end{array}
\label{hyp:ehidhdpfkhggngjgdhdgpfbd}
\end{equation}
Assume the following.
\begin{equation}
\begin{array}{c}{\unicdcdadad} X0 .  {\unicdcdadad} X1 .  (m1\_subset\_1~X0~X1){\unicdbdjdcd}((v1\_xboole\_0~X1){\unicdcdcdid}\\
(X0 \in X1))\end{array}
\label{hyp:ehcdpfdhfhcgdhfgeh}
\end{equation}
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 .  {{{{k6\_subset\_1}~{X0}}~{X1}}={{{k4\_xboole\_0}~{X0}}~{X1}}}}}\label{hyp:chfgegfgggjgogjgehjgpgogpflggdpfdhfhcgdhfgehpfbd}
\end{equation}
Assume the following.
\begin{equation}
{{k11\_arytm\_3}={k1\_xboole\_0}}\label{hyp:chfgegfgggjgogjgehjgpgogpflgbdbdpfbgchjhehngpfdd}
\end{equation}
Assume the following.
\begin{equation}
{\preoh{{{v1\_xboole\_0}~{k2\_arytm\_2}}}}\label{hyp:ggdgcdpfbgchjhehngpfcd}
\end{equation}
Assume the following.
\begin{equation}
\begin{array}{c}{\unicdcdadad} X0 .  {\unicdcdadad} X1 .  {\unicdcdadad} X2 .  (X2=k4\_xboole\_0~X0~X1){\unicdbdjded}({\unicdcdadad} X3 .  \\
(X3 \in X2){\unicdbdjded}((X3 \in X0){\unicdcdcdhd}(\preoh X3 \in X1)))\end{array}
\label{hyp:egfdpfihcgpgpgmgfgpfad}
\end{equation}
Assume the following.
\begin{equation}
\begin{array}{c}{\unicdcdadad} X0 .  {\unicdcdadad} X1 .  {\unicdcdadad} X2 .  (X2=k2\_xboole\_0~X0~X1){\unicdbdjded}({\unicdcdadad} X3 .  \\
(X3 \in X2){\unicdbdjded}((X3 \in X0){\unicdcdcdid}(X3 \in X1)))\end{array}
\label{hyp:egddpfihcgpgpgmgfgpfad}
\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}
Assume the following.
\begin{equation}
\begin{array}{c}k1\_numbers=k6\_subset\_1~(k2\_xboole\_0~k2\_arytm\_2~(k2\_zfmisc\_1~\\
(k1\_tarski~k1\_xboole\_0)~k2\_arytm\_2))~(k1\_tarski~(k4\_tarski~k1\_xboole\_0~\\
k1\_xboole\_0))\end{array}
\label{hyp:egbdpfogfhngcgfgchdh}
\end{equation}
\begin{theorem}\label{thm:ehcdpfbgchjhehngpfad}
$$\begin{array}{c}{\unicdcdadad} X0 .  (m1\_subset\_1~X0~k2\_arytm\_2){\unicdbdjdcd}((X0{\unicdcdgdad}k11\_arytm\_3){\unicdbdjdcd}(\\
k4\_tarski~k11\_arytm\_3~X0 \in k1\_numbers))\end{array}$$
\end{theorem}
\end{document}
