\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}
\newcommand\uniddcgcg{\lambda}
\newcommand\unicdcdaddd{\exists}
\newcommand\unicdcdidid{\not\subseteq}
\newcommand\preoh[1]{\lnot #1}
\newcommand\mname[1]{{\mathsf{#1}}}
\begin{document}
\title{l34\_mod\_2 (TMJbLqZ4FXvg45tNxVxyM7oPX12EHVYeacZ)}
\maketitle
Let $ m1\_subset\_1 : {{\iota}{\unicdbdjdcd}{{\iota}{\unicdbdjdcd}{o}}}$ be given.
Let $ k1\_enumset1 : {{\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.
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 .  {\unicdcdadad} X1 .  {\unicdcdadad} X2 .  {\unicdcdadad} X3 .  (X3=k1\_enumset1~\\
X0~X1~X2){\unicdbdjded}({\unicdcdadad} X4 .  (X4 \in X3){\unicdbdjded}(\preoh (X4{\unicdcdgdad}X0){\unicdcdcdhd}((X4{\unicdcdgdad}X1){\unicdcdcdhd}(X4{\unicdcdgdad}X2))))\end{array}
\label{hyp:egbdpffgogfhngdhfgehbd}
\end{equation}
\begin{theorem}\label{thm:mgddedpfngpgegpfcd}
$$\begin{array}{c}{\unicdcdaddd} X0 .  (m1\_subset\_1~X0~(k1\_enumset1~k6\_numbers~np\_\_1~np\_\_2)){\unicdcdcdhd}\\
(X0=k6\_numbers)\end{array}$$
\end{theorem}
\end{document}
