\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{t21\_square\_1 (TMN8sJcsHHNfq4ZSjyPR1QwwdeetuTMZC6v)}
\maketitle
Let $ r1\_xxreal\_0 : {{\iota}{\unicdbdjdcd}{{\iota}{\unicdbdjdcd}{o}}}$ be given.
Let $ np\_\_2 : {\iota}$ be given.
Let $ k7\_square\_1 : {{\iota}{\unicdbdjdcd}{\iota}}$ be given.
Let $ v1\_xboole\_0 : {{\iota}{\unicdbdjdcd}{o}}$ be given.
Let $ k1\_xboole\_0 : {\iota}$ be given.
Let $ np\_\_4 : {\iota}$ be given.
Let $ v2\_xxreal\_0 : {{\iota}{\unicdbdjdcd}{o}}$ be given.
Let $ m2\_subset\_1 : {{\iota}{\unicdbdjdcd}{{\iota}{\unicdbdjdcd}{{\iota}{\unicdbdjdcd}{o}}}}$ be given.
Let $ k1\_numbers : {\iota}$ be given.
Let $ k5\_numbers : {\iota}$ be given.
Let $ m1\_subset\_1 : {{\iota}{\unicdbdjdcd}{{\iota}{\unicdbdjdcd}{o}}}$ be given.
Let $ np\_\_0 : {\iota}$ be given.
Let $ k6\_square\_1 : {{\iota}{\unicdbdjdcd}{\iota}}$ be given.
Let $ k6\_numbers : {\iota}$ be given.
Let $ v1\_xreal\_0 : {{\iota}{\unicdbdjdcd}{o}}$ be given.
Assume the following.
\begin{equation}
{{\unicdcdadad} X0 .  {{({{v1\_xboole\_0}~{X0}})}{\unicdbdjdcd}{({{X0}={k1\_xboole\_0}})}}}\label{hyp:ehgdpfcgpgpgmgfg}
\end{equation}
Assume the following.
\begin{equation}
{{{k7\_square\_1}~{np\_\_4}}={np\_\_2}}\label{hyp:ehcdadpfdhbhfhbgchfgpfbd}
\end{equation}
Assume the following.
\begin{equation}
\begin{array}{c}((v2\_xxreal\_0~np\_\_4){\unicdcdcdhd}(m2\_subset\_1~np\_\_4~k1\_numbers~k5\_numbers)){\unicdcdcdhd}\\
((m1\_subset\_1~np\_\_4~k5\_numbers){\unicdcdcdhd}(m1\_subset\_1~np\_\_4~k1\_numbers))\end{array}
\label{hyp:dhahdgedpfogfhngfgchbgmgdh}
\end{equation}
Assume the following.
\begin{equation}
\begin{array}{c}((v2\_xxreal\_0~np\_\_2){\unicdcdcdhd}(m2\_subset\_1~np\_\_2~k1\_numbers~k5\_numbers)){\unicdcdcdhd}\\
((m1\_subset\_1~np\_\_2~k5\_numbers){\unicdcdcdhd}(m1\_subset\_1~np\_\_2~k1\_numbers))\end{array}
\label{hyp:dhahdgcdpfogfhngfgchbgmgdh}
\end{equation}
Assume the following.
\begin{equation}
{{v1\_xboole\_0}~{np\_\_0}}\label{hyp:dhahdgadpfcgpgpgmgfg}
\end{equation}
Assume the following.
\begin{equation}
{\preoh{{{{r1\_xxreal\_0}~{np\_\_4}}~{np\_\_2}}}}\label{hyp:chbhmefgdhdhpechfebhfhbgmgpfpfchbdpfihihchfgbgmgpfadpfpfchedpfchcd}
\end{equation}
Assume the following.
\begin{equation}
{{{r1\_xxreal\_0}~{np\_\_0}}~{np\_\_2}}\label{hyp:chbhmefgdhdhpechfebhfhbgmgpfpfchbdpfihihchfgbgmgpfadpfpfchadpfchcd}
\end{equation}
Assume the following.
\begin{equation}
\begin{array}{c}{\unicdcdadad} X0 .  (m1\_subset\_1~X0~k1\_numbers){\unicdbdjdcd}(k7\_square\_1~X0=k6\_square\_1~\\
X0)\end{array}
\label{hyp:chfgegfgggjgogjgehjgpgogpflghdpfdhbhfhbgchfgpfbd}
\end{equation}
Assume the following.
\begin{equation}
{{k6\_numbers}={k1\_xboole\_0}}\label{hyp:chfgegfgggjgogjgehjgpgogpflggdpfogfhngcgfgchdh}
\end{equation}
Assume the following.
\begin{equation}
\begin{array}{c}{\unicdcdadad} X0 .  (v1\_xreal\_0~X0){\unicdbdjdcd}({\unicdcdadad} X1 .  (v1\_xreal\_0~X1){\unicdbdjdcd}(\preoh (r1\_xxreal\_0~\\
k6\_numbers~X0){\unicdcdcdhd}((\preoh r1\_xxreal\_0~X1~X0){\unicdcdcdhd}(r1\_xxreal\_0~(k6\_square\_1~\\
X1)~(k6\_square\_1~X0)))))\end{array}
\label{hyp:mgcdddpfdhbhfhbgchfgpfbd}
\end{equation}
Assume the following.
\begin{equation}
{{\unicdcdadad} X0 .  {{({{{m1\_subset\_1}~{X0}}~{k1\_numbers}})}{\unicdbdjdcd}{({{v1\_xreal\_0}~{X0}})}}}\label{hyp:dgdgbdpfchfgbgmgpfbd}
\end{equation}
\begin{theorem}\label{thm:ehcdbdpfdhbhfhbgchfgpfbd}
${\preoh{{{{r1\_xxreal\_0}~{np\_\_2}}~{({{k7\_square\_1}~{np\_\_2}})}}}}$.
\end{theorem}
\end{document}
