\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{l1\_ami\_4 (TMLxNCZT5CLSAfyvsi2X1TaEwSh54BgREAc)}
\maketitle
Let $ k10\_ami\_3 : {{\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}
{{{k10\_ami\_3}~{k6\_numbers}}{\unicdcdgdad}{{k10\_ami\_3}~{np\_\_1}}}\label{hyp:mghdpfggjgcgpfggfhdhdg}
\end{equation}
Assume the following.
\begin{equation}
{{{k10\_ami\_3}~{np\_\_1}}{\unicdcdgdad}{{k10\_ami\_3}~{np\_\_2}}}\label{hyp:mgbdadpfggjgcgpfggfhdhdg}
\end{equation}
\begin{theorem}\label{thm:mgbdpfbgngjgpfed}
$$\begin{array}{c}(k10\_ami\_3~k6\_numbers{\unicdcdgdad}k10\_ami\_3~np\_\_1){\unicdcdcdhd}(k10\_ami\_3~np\_\_1{\unicdcdgdad}k10\_ami\_3~\\
np\_\_2)\end{array}$$
\end{theorem}
\end{document}
