\documentclass{article}
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\newtheorem{proposition}{Proposition}
\newtheorem{definition}{Definition}
\newtheorem{example}{Example}
\newtheorem{conjecture}{Conjecture}
\newcommand\unicdcdgdad{\not=}
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\begin{document}
\title{t59\_euclid\_8 (TMFy7XSdMmA85fAkT2y3XQ6EL7k3Rf8yhCX)}
\maketitle
Let $ m1\_subset\_1 : {{\iota}{\unicdbdjdcd}{{\iota}{\unicdbdjdcd}{o}}}$ be given.
Let $ k1\_numbers : {\iota}$ be given.
Let $ v1\_xreal\_0 : {{\iota}{\unicdbdjdcd}{o}}$ be given.
Let $ k9\_euclid : {{\iota}{\unicdbdjdcd}{{\iota}{\unicdbdjdcd}{{\iota}{\unicdbdjdcd}{\iota}}}}$ be given.
Let $ np\_\_3 : {\iota}$ be given.
Let $ k1\_euclid\_8 : {{\iota}{\unicdbdjdcd}{{\iota}{\unicdbdjdcd}{{\iota}{\unicdbdjdcd}{\iota}}}}$ be given.
Let $ k11\_binop\_2 : {{\iota}{\unicdbdjdcd}{{\iota}{\unicdbdjdcd}{\iota}}}$ be given.
Assume the following.
\begin{equation}
{{\unicdcdadad} X0 .  {{\unicdcdadad} X1 .  {{({{X0} \in {X1}})}{\unicdbdjdcd}{({{{m1\_subset\_1}~{X0}}~{X1}})}}}}\label{hyp:ehbdpfdhfhcgdhfgeh}
\end{equation}
Assume the following.
\begin{equation}
\begin{array}{c}{\unicdcdadad} X0 .  (m1\_subset\_1~X0~k1\_numbers){\unicdbdjdcd}({\unicdcdadad} X1 .  (m1\_subset\_1~\\
X1~k1\_numbers){\unicdbdjdcd}({\unicdcdadad} X2 .  (m1\_subset\_1~X2~k1\_numbers){\unicdbdjdcd}({\unicdcdadad} X3 .  \\
(m1\_subset\_1~X3~k1\_numbers){\unicdbdjdcd}(k9\_euclid~np\_\_3~(k1\_euclid\_8~X1~\\
X2~X3)~X0=k1\_euclid\_8~(k11\_binop\_2~X0~X1)~(k11\_binop\_2~X0~X2)~(k11\_binop\_2~\\
X0~X3)))))\end{array}
\label{hyp:mgidpffgfhdgmgjgegpfid}
\end{equation}
Assume the following.
\begin{equation}
{{\unicdcdadad} X0 .  {{({{v1\_xreal\_0}~{X0}})}{\unicdbdjded}{({{X0} \in {k1\_numbers}})}}}\label{hyp:egbdpfihchfgbgmgpfad}
\end{equation}
\begin{theorem}\label{thm:ehfdjdpffgfhdgmgjgegpfid}
$$\begin{array}{c}{\unicdcdadad} X0 .  (m1\_subset\_1~X0~k1\_numbers){\unicdbdjdcd}({\unicdcdadad} X1 .  (m1\_subset\_1~\\
X1~k1\_numbers){\unicdbdjdcd}({\unicdcdadad} X2 .  (m1\_subset\_1~X2~k1\_numbers){\unicdbdjdcd}({\unicdcdadad} X3 .  \\
(v1\_xreal\_0~X3){\unicdbdjdcd}(k9\_euclid~np\_\_3~(k1\_euclid\_8~X0~X1~X2)~X3=k1\_euclid\_8~\\
(k11\_binop\_2~X3~X0)~(k11\_binop\_2~X3~X1)~(k11\_binop\_2~X3~X2)))))\end{array}$$
\end{theorem}
\end{document}
