\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}
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\newcommand\unicdcdidid{\not\subseteq}
\newcommand\preoh[1]{\lnot #1}
\newcommand\mname[1]{{\mathsf{#1}}}
\begin{document}
\title{l10\_uniform1 (TMPEPfnBa19P42bPBSAtb672qAiiVqDe8cK)}
\maketitle
Let $ k5\_topmetr : {\iota}$ be given.
Let $ k3\_pcomps\_1 : {{\iota}{\unicdbdjdcd}{\iota}}$ be given.
Let $ k2\_topmetr : {{\iota}{\unicdbdjdcd}{{\iota}{\unicdbdjdcd}{\iota}}}$ be given.
Let $ k6\_numbers : {\iota}$ be given.
Let $ np\_\_1 : {\iota}$ be given.
Let $ k4\_topmetr : {{\iota}{\unicdbdjdcd}{{\iota}{\unicdbdjdcd}{\iota}}}$ be given.
Let $ k17\_borsuk\_1 : {\iota}$ be given.
Assume the following.
\begin{equation}
{{{{k4\_topmetr}~{k6\_numbers}}~{np\_\_1}}={k17\_borsuk\_1}}\label{hyp:ehcdadpfehpgahngfgehch}
\end{equation}
Assume the following.
\begin{equation}
{{k5\_topmetr}={k17\_borsuk\_1}}\label{hyp:chfgegfgggjgogjgehjgpgogpflgfdpfehpgahngfgehch}
\end{equation}
Assume the following.
\begin{equation}
{{{{k4\_topmetr}~{k6\_numbers}}~{np\_\_1}}={{k3\_pcomps\_1}~{({{{k2\_topmetr}~{k6\_numbers}}~{np\_\_1}})}}}\label{hyp:mgjdpffhogjgggpgchngbd}
\end{equation}
\begin{theorem}\label{thm:mgbdadpffhogjgggpgchngbd}
${{k5\_topmetr}={{k3\_pcomps\_1}~{({{{k2\_topmetr}~{k6\_numbers}}~{np\_\_1}})}}}$.
\end{theorem}
\end{document}
