\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{t5\_arytm\_1 (TMbNN33BCfz96cxp8VXf6YK5AMwAfoi21gr)}
\maketitle
Let $ m1\_subset\_1 : {{\iota}{\unicdbdjdcd}{{\iota}{\unicdbdjdcd}{o}}}$ be given.
Let $ k2\_arytm\_2 : {\iota}$ be given.
Let $ r1\_arytm\_2 : {{\iota}{\unicdbdjdcd}{{\iota}{\unicdbdjdcd}{o}}}$ be given.
Let $ k11\_arytm\_3 : {\iota}$ be given.
Let $ k7\_arytm\_2 : {{\iota}{\unicdbdjdcd}{{\iota}{\unicdbdjdcd}{\iota}}}$ be given.
Assume the following.
\begin{equation}
\begin{array}{c}{\unicdcdadad} X0 .  (m1\_subset\_1~X0~k2\_arytm\_2){\unicdbdjdcd}({\unicdcdadad} X1 .  (m1\_subset\_1~\\
X1~k2\_arytm\_2){\unicdbdjdcd}(\preoh (r1\_arytm\_2~X0~X1){\unicdcdcdhd}({\unicdcdadad} X2 .  (m1\_subset\_1~\\
X2~k2\_arytm\_2){\unicdbdjdcd}(k7\_arytm\_2~X0~X2{\unicdcdgdad}X1))))\end{array}
\label{hyp:ehjdpfbgchjhehngpfcd}
\end{equation}
Assume the following.
\begin{equation}
\begin{array}{c}{\unicdcdadad} X0 .  (m1\_subset\_1~X0~k2\_arytm\_2){\unicdbdjdcd}({\unicdcdadad} X1 .  (m1\_subset\_1~\\
X1~k2\_arytm\_2){\unicdbdjdcd}((k7\_arytm\_2~X0~X1=k11\_arytm\_3){\unicdbdjdcd}(X0=k11\_arytm\_3)))\end{array}
\label{hyp:ehfdpfbgchjhehngpfcd}
\end{equation}
\begin{theorem}\label{thm:ehfdpfbgchjhehngpfbd}
$$\begin{array}{c}{\unicdcdadad} X0 .  (m1\_subset\_1~X0~k2\_arytm\_2){\unicdbdjdcd}({\unicdcdadad} X1 .  (m1\_subset\_1~\\
X1~k2\_arytm\_2){\unicdbdjdcd}(((r1\_arytm\_2~X0~X1){\unicdcdcdhd}(X1=k11\_arytm\_3)){\unicdbdjdcd}(X0=\\
k11\_arytm\_3)))\end{array}$$
\end{theorem}
\end{document}
