\documentclass{article}
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\newtheorem{lemma}{Lemma}
\newtheorem{proposition}{Proposition}
\newtheorem{definition}{Definition}
\newtheorem{example}{Example}
\newtheorem{conjecture}{Conjecture}
\newcommand\unicdcdgdad{\not=}
\newcommand\unicdcdcdbg{\cup}
\newcommand\unicdcdbdgd{\setminus}
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\newcommand\unicdbdjdcd{\Rightarrow}
\newcommand\unicdcdadjd{\notin}
\newcommand\unicdbdjded{\Leftrightarrow}
\newcommand\unicdcdcdid{\lor}
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\newcommand\unicdcdidid{\not\subseteq}
\newcommand\preoh[1]{\lnot #1}
\newcommand\mname[1]{{\mathsf{#1}}}
\begin{document}
\title{t16\_polyeq\_3 (TMGLShgHpWASxk3eJxtgUCs7DR4ZNedEYSc)}
\maketitle
Let $ m1\_subset\_1 : {{\iota}{\unicdbdjdcd}{{\iota}{\unicdbdjdcd}{o}}}$ be given.
Let $ k2\_numbers : {\iota}$ be given.
Let $ k1\_polyeq\_1 : {{\iota}{\unicdbdjdcd}{{\iota}{\unicdbdjdcd}{{\iota}{\unicdbdjdcd}{\iota}}}}$ be given.
Let $ k6\_numbers : {\iota}$ be given.
Let $ k1\_binop\_2 : {{\iota}{\unicdbdjdcd}{\iota}}$ be given.
Let $ k6\_binop\_2 : {{\iota}{\unicdbdjdcd}{{\iota}{\unicdbdjdcd}{\iota}}}$ be given.
Let $ v1\_xcmplx\_0 : {{\iota}{\unicdbdjdcd}{o}}$ be given.
Let $ k4\_xcmplx\_0 : {{\iota}{\unicdbdjdcd}{\iota}}$ be given.
Let $ k7\_xcmplx\_0 : {{\iota}{\unicdbdjdcd}{{\iota}{\unicdbdjdcd}{\iota}}}$ be given.
Assume the following.
\begin{equation}
\begin{array}{c}{\unicdcdadad} X0 .  (v1\_xcmplx\_0~X0){\unicdbdjdcd}({\unicdcdadad} X1 .  (v1\_xcmplx\_0~X1){\unicdbdjdcd}({\unicdcdadad} X2 .  \\
(v1\_xcmplx\_0~X2){\unicdbdjdcd}((k1\_polyeq\_1~X0~X1~X2=k6\_numbers){\unicdbdjdcd}((X0=k6\_numbers){\unicdcdcdid}\\
(X2=k4\_xcmplx\_0~(k7\_xcmplx\_0~X1~X0))))))\end{array}
\label{hyp:ehbdpfahpgmgjhfgbhpfbd}
\end{equation}
Assume the following.
\begin{equation}
\begin{array}{c}{\unicdcdadad} X0 .  {\unicdcdadad} X1 .  ((v1\_xcmplx\_0~X0){\unicdcdcdhd}(v1\_xcmplx\_0~X1)){\unicdbdjdcd}(\\
k6\_binop\_2~X0~X1=k7\_xcmplx\_0~X0~X1)\end{array}
\label{hyp:chfgegfgggjgogjgehjgpgogpflggdpfcgjgogpgahpfcd}
\end{equation}
Assume the following.
\begin{equation}
{{\unicdcdadad} X0 .  {{({{v1\_xcmplx\_0}~{X0}})}{\unicdbdjdcd}{({{{k1\_binop\_2}~{X0}}={{k4\_xcmplx\_0}~{X0}}})}}}\label{hyp:chfgegfgggjgogjgehjgpgogpflgbdpfcgjgogpgahpfcd}
\end{equation}
Assume the following.
\begin{equation}
\begin{array}{c}{\unicdcdadad} X0 .  {\unicdcdadad} X1 .  ((v1\_xcmplx\_0~X0){\unicdcdcdhd}(v1\_xcmplx\_0~X1)){\unicdbdjdcd}(\\
m1\_subset\_1~(k6\_binop\_2~X0~X1)~k2\_numbers)\end{array}
\label{hyp:egehpflggdpfcgjgogpgahpfcd}
\end{equation}
Assume the following.
\begin{equation}
{{\unicdcdadad} X0 .  {{({{{m1\_subset\_1}~{X0}}~{k2\_numbers}})}{\unicdbdjdcd}{({{v1\_xcmplx\_0}~{X0}})}}}\label{hyp:dgdgddpfihdgngahmgihpfad}
\end{equation}
\begin{theorem}\label{thm:ehbdgdpfahpgmgjhfgbhpfdd}
$$\begin{array}{c}{\unicdcdadad} X0 .  (m1\_subset\_1~X0~k2\_numbers){\unicdbdjdcd}({\unicdcdadad} X1 .  (m1\_subset\_1~\\
X1~k2\_numbers){\unicdbdjdcd}({\unicdcdadad} X2 .  (m1\_subset\_1~X2~k2\_numbers){\unicdbdjdcd}((k1\_polyeq\_1~\\
X0~X1~X2=k6\_numbers){\unicdbdjdcd}((X0=k6\_numbers){\unicdcdcdid}(X2=k1\_binop\_2~(k6\_binop\_2~\\
X1~X0))))))\end{array}$$
\end{theorem}
\end{document}
