\documentclass{article}
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\newtheorem{lemma}{Lemma}
\newtheorem{proposition}{Proposition}
\newtheorem{definition}{Definition}
\newtheorem{example}{Example}
\newtheorem{conjecture}{Conjecture}
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\begin{document}
\title{t18\_square\_1 (TMSS7jr26CRzGw4jfx3C5wGdHUEvd5p698p)}
\maketitle
Let $ k7\_square\_1 : {{\iota}{\unicdbdjdcd}{\iota}}$ be given.
Let $ np\_\_1 : {\iota}$ be given.
Let $ k3\_xcmplx\_0 : {{\iota}{\unicdbdjdcd}{{\iota}{\unicdbdjdcd}{\iota}}}$ be given.
Let $ k2\_xcmplx\_0 : {{\iota}{\unicdbdjdcd}{{\iota}{\unicdbdjdcd}{\iota}}}$ be given.
Let $ k4\_xcmplx\_0 : {{\iota}{\unicdbdjdcd}{\iota}}$ be given.
Let $ np\_\_0 : {\iota}$ be given.
Let $ r1\_xxreal\_0 : {{\iota}{\unicdbdjdcd}{{\iota}{\unicdbdjdcd}{o}}}$ be given.
Let $ m1\_subset\_1 : {{\iota}{\unicdbdjdcd}{{\iota}{\unicdbdjdcd}{o}}}$ be given.
Let $ k1\_numbers : {\iota}$ be given.
Let $ k6\_square\_1 : {{\iota}{\unicdbdjdcd}{\iota}}$ be given.
Let $ c5\_\_xreal\_0 : {\iota}$ be given.
Let $ k1\_arytm\_0 : {{\iota}{\unicdbdjdcd}{{\iota}{\unicdbdjdcd}{\iota}}}$ be given.
Let $ c3\_\_xreal\_0 : {\iota}$ be given.
Let $ k6\_numbers : {\iota}$ be given.
Let $ v1\_xreal\_0 : {{\iota}{\unicdbdjdcd}{o}}$ be given.
Let $ k3\_square\_1 : {{\iota}{\unicdbdjdcd}{\iota}}$ be given.
Let $ v1\_xcmplx\_0 : {{\iota}{\unicdbdjdcd}{o}}$ be given.
Assume the following.
\begin{equation}
{{{{k3\_xcmplx\_0}~{np\_\_1}}~{np\_\_1}}={np\_\_1}}\label{hyp:chbhcffgbgmgnefhmgehpfpflgddpfihdgngahmgihpfadpfpfchbdpfchbdpfchbd}
\end{equation}
Assume the following.
\begin{equation}
{{{{k2\_xcmplx\_0}~{np\_\_1}}~{({{k4\_xcmplx\_0}~{np\_\_1}})}}={np\_\_0}}\label{hyp:chbhcffgbgmgbeegegpfpflgcdpfihdgngahmgihpfadpfpfchbdpfchngbdpfchad}
\end{equation}
Assume the following.
\begin{equation}
{{{r1\_xxreal\_0}~{np\_\_0}}~{np\_\_1}}\label{hyp:chbhmefgdhdhpechfebhfhbgmgpfpfchbdpfihihchfgbgmgpfadpfpfchadpfchbd}
\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}
{{{m1\_subset\_1}~{np\_\_1}}~{k1\_numbers}}\label{hyp:mgcdjdpfbgchjhehngpfad}
\end{equation}
Assume the following.
\begin{equation}
\begin{array}{c}(c5\_\_xreal\_0=k4\_xcmplx\_0~np\_\_1){\unicdcdcdhd}(k1\_arytm\_0~c3\_\_xreal\_0~c5\_\_xreal\_0=\\
k6\_numbers)\end{array}
\label{hyp:mgbdidpfihchfgbgmgpfad}
\end{equation}
Assume the following.
\begin{equation}
{{{{k2\_xcmplx\_0}~{np\_\_1}}~{({{k4\_xcmplx\_0}~{np\_\_1}})}}={k6\_numbers}}\label{hyp:mgbdddpfihchfgbgmgpfad}
\end{equation}
Assume the following.
\begin{equation}
\begin{array}{c}{\unicdcdadad} X0 .  (v1\_xreal\_0~X0){\unicdbdjdcd}((r1\_xxreal\_0~k6\_numbers~X0){\unicdbdjdcd}({\unicdcdadad} X1 .  \\
(v1\_xreal\_0~X1){\unicdbdjdcd}((X1=k6\_square\_1~X0){\unicdbdjded}((r1\_xxreal\_0~k6\_numbers~\\
X1){\unicdcdcdhd}(k3\_square\_1~X1=X0)))))\end{array}
\label{hyp:egcdpfdhbhfhbgchfgpfbd}
\end{equation}
Assume the following.
\begin{equation}
{{\unicdcdadad} X0 .  {{({{v1\_xcmplx\_0}~{X0}})}{\unicdbdjdcd}{({{{k3\_square\_1}~{X0}}={{{k3\_xcmplx\_0}~{X0}}~{X0}}})}}}\label{hyp:egbdpfdhbhfhbgchfgpfbd}
\end{equation}
Assume the following.
\begin{equation}
{{\unicdcdadad} X0 .  {{({{{m1\_subset\_1}~{X0}}~{k1\_numbers}})}{\unicdbdjdcd}{({{v1\_xreal\_0}~{X0}})}}}\label{hyp:dgdgbdpfihchfgbgmgpfad}
\end{equation}
Assume the following.
\begin{equation}
{{\unicdcdadad} X0 .  {{({{{m1\_subset\_1}~{X0}}~{k1\_numbers}})}{\unicdbdjdcd}{({{v1\_xcmplx\_0}~{X0}})}}}\label{hyp:dgdgbdpfihdgngahmgihpfad}
\end{equation}
\begin{theorem}\label{thm:ehbdidpfdhbhfhbgchfgpfbd}
${{{k7\_square\_1}~{np\_\_1}}={np\_\_1}}$.
\end{theorem}
\end{document}
