\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{t20\_arytm\_0 (TMJdyKi5uG7TmnXpNTZHtMwGJwhYDsFyBBs)}
\maketitle
Let $ m1\_subset\_1 : {{\iota}{\unicdbdjdcd}{{\iota}{\unicdbdjdcd}{o}}}$ be given.
Let $ k1\_numbers : {\iota}$ be given.
Let $ k6\_numbers : {\iota}$ be given.
Let $ k2\_arytm\_0 : {{\iota}{\unicdbdjdcd}{{\iota}{\unicdbdjdcd}{\iota}}}$ be given.
Let $ k4\_arytm\_0 : {{\iota}{\unicdbdjdcd}{\iota}}$ be given.
Let $ np\_\_1 : {\iota}$ be given.
Assume the following.
\begin{equation}
\begin{array}{c}{\unicdcdadad} X0 .  (m1\_subset\_1~X0~k1\_numbers){\unicdbdjdcd}({\unicdcdadad} X1 .  (m1\_subset\_1~\\
X1~k1\_numbers){\unicdbdjdcd}((X1=np\_\_1){\unicdbdjdcd}(k2\_arytm\_0~X0~X1=X0)))\end{array}
\label{hyp:ehbdjdpfbgchjhehngpfad}
\end{equation}
Assume the following.
\begin{equation}
\begin{array}{c}{\unicdcdadad} X0 .  (m1\_subset\_1~X0~k1\_numbers){\unicdbdjdcd}({\unicdcdadad} X1 .  (m1\_subset\_1~\\
X1~k1\_numbers){\unicdbdjdcd}({\unicdcdadad} X2 .  (m1\_subset\_1~X2~k1\_numbers){\unicdbdjdcd}(k2\_arytm\_0~\\
X0~(k2\_arytm\_0~X1~X2)=k2\_arytm\_0~(k2\_arytm\_0~X0~X1)~X2)))\end{array}
\label{hyp:ehbdddpfbgchjhehngpfad}
\end{equation}
Assume the following.
\begin{equation}
\begin{array}{c}{\unicdcdadad} X0 .  (m1\_subset\_1~X0~k1\_numbers){\unicdbdjdcd}(m1\_subset\_1~(k4\_arytm\_0~\\
X0)~k1\_numbers)\end{array}
\label{hyp:egehpflgedpfbgchjhehngpfad}
\end{equation}
Assume the following.
\begin{equation}
\begin{array}{c}{\unicdcdadad} X0 .  {\unicdcdadad} X1 .  ((m1\_subset\_1~X0~k1\_numbers){\unicdcdcdhd}(m1\_subset\_1~\\
X1~k1\_numbers)){\unicdbdjdcd}(m1\_subset\_1~(k2\_arytm\_0~X0~X1)~k1\_numbers)\end{array}
\label{hyp:egehpflgcdpfbgchjhehngpfad}
\end{equation}
Assume the following.
\begin{equation}
\begin{array}{c}{\unicdcdadad} X0 .  (m1\_subset\_1~X0~k1\_numbers){\unicdbdjdcd}({\unicdcdadad} X1 .  (m1\_subset\_1~\\
X1~k1\_numbers){\unicdbdjdcd}(((X0{\unicdcdgdad}k6\_numbers){\unicdbdjdcd}((X1=k4\_arytm\_0~X0){\unicdbdjded}(k2\_arytm\_0~\\
X0~X1=np\_\_1))){\unicdcdcdhd}((X0=k6\_numbers){\unicdbdjdcd}((X1=k4\_arytm\_0~X0){\unicdbdjded}(X1=k6\_numbers)))))\end{array}
\label{hyp:egedpfbgchjhehngpfad}
\end{equation}
\begin{theorem}\label{thm:ehcdadpfbgchjhehngpfad}
$$\begin{array}{c}{\unicdcdadad} X0 .  (m1\_subset\_1~X0~k1\_numbers){\unicdbdjdcd}({\unicdcdadad} X1 .  (m1\_subset\_1~\\
X1~k1\_numbers){\unicdbdjdcd}((X1{\unicdcdgdad}k6\_numbers){\unicdbdjdcd}(k2\_arytm\_0~(k2\_arytm\_0~X0~\\
X1)~(k4\_arytm\_0~X1)=X0)))\end{array}$$
\end{theorem}
\end{document}
