\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{t1\_necklace (TMYHNeMX3dWbqDxNBNCWaJ7toeHzRgyHF2a)}
\maketitle
Let $ np\_\_4 : {\iota}$ be given.
Let $ k2\_enumset1 : {{\iota}{\unicdbdjdcd}{{\iota}{\unicdbdjdcd}{{\iota}{\unicdbdjdcd}{{\iota}{\unicdbdjdcd}{\iota}}}}}$ be given.
Let $ k6\_numbers : {\iota}$ be given.
Let $ np\_\_1 : {\iota}$ be given.
Let $ np\_\_2 : {\iota}$ be given.
Let $ np\_\_3 : {\iota}$ be given.
Let $ k1\_xboole\_0 : {\iota}$ be given.
Assume the following.
\begin{equation}
{{np\_\_4}={{{{{k2\_enumset1}~{k1\_xboole\_0}}~{np\_\_1}}~{np\_\_2}}~{np\_\_3}}}\label{hyp:ehfdcdpfdgbgchegpfbd}
\end{equation}
Assume the following.
\begin{equation}
{{k6\_numbers}={k1\_xboole\_0}}\label{hyp:chfgegfgggjgogjgehjgpgogpflggdpfogfhngcgfgchdh}
\end{equation}
\begin{theorem}\label{thm:ehbdpfogfgdglgmgbgdgfg}
${{np\_\_4}={{{{{k2\_enumset1}~{k6\_numbers}}~{np\_\_1}}~{np\_\_2}}~{np\_\_3}}}$.
\end{theorem}
\end{document}
