\documentclass{article}
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\newtheorem{theorem}{Theorem}
\newtheorem{lemma}{Lemma}
\newtheorem{proposition}{Proposition}
\newtheorem{definition}{Definition}
\newtheorem{example}{Example}
\newtheorem{conjecture}{Conjecture}
\newcommand\unicdcdgdad{\not=}
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\newcommand\unicdcdcdhd{\land}
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\newcommand\unicdcdidid{\not\subseteq}
\newcommand\preoh[1]{\lnot #1}
\newcommand\mname[1]{{\mathsf{#1}}}
\begin{document}
\title{t2\_polyeq\_5 (TMMMzGzwV8B2T33ZdBSN82TN35xKa7sV18J)}
\maketitle
Let $ v1\_xcmplx\_0 : {{\iota}{\unicdbdjdcd}{o}}$ be given.
Let $ k3\_xcmplx\_0 : {{\iota}{\unicdbdjdcd}{{\iota}{\unicdbdjdcd}{\iota}}}$ be given.
Let $ k1\_newton : {{\iota}{\unicdbdjdcd}{{\iota}{\unicdbdjdcd}{\iota}}}$ be given.
Let $ np\_\_3 : {\iota}$ be given.
Let $ m1\_subset\_1 : {{\iota}{\unicdbdjdcd}{{\iota}{\unicdbdjdcd}{o}}}$ be given.
Let $ k2\_numbers : {\iota}$ be given.
Let $ k5\_binop\_2 : {{\iota}{\unicdbdjdcd}{{\iota}{\unicdbdjdcd}{\iota}}}$ be given.
Let $ k1\_polyeq\_3 : {{\iota}{\unicdbdjdcd}{\iota}}$ be given.
Let $ k3\_polyeq\_3 : {{\iota}{\unicdbdjdcd}{\iota}}$ be given.
Let $ k3\_square\_1 : {{\iota}{\unicdbdjdcd}{\iota}}$ be given.
Assume the following.
\begin{equation}
\begin{array}{c}{\unicdcdadad} X0 .  (m1\_subset\_1~X0~k2\_numbers){\unicdbdjdcd}((k1\_newton~X0~np\_\_3=\\
k5\_binop\_2~(k5\_binop\_2~X0~X0)~X0){\unicdcdcdhd}((k1\_newton~X0~np\_\_3=k5\_binop\_2~\\
(k1\_polyeq\_3~X0)~X0){\unicdcdcdhd}(k1\_newton~X0~np\_\_3=k3\_polyeq\_3~X0)))\end{array}
\label{hyp:ehcdhdpfahpgmgjhfgbhpfdd}
\end{equation}
Assume the following.
\begin{equation}
{{\unicdcdadad} X0 .  {{\unicdcdadad} X1 .  {{({{X0} \in {X1}})}{\unicdbdjdcd}{({{{m1\_subset\_1}~{X0}}~{X1}})}}}}\label{hyp:ehbdpfdhfhcgdhfgeh}
\end{equation}
Assume the following.
\begin{equation}
\begin{array}{c}{\unicdcdadad} X0 .  {\unicdcdadad} X1 .  ((v1\_xcmplx\_0~X0){\unicdcdcdhd}(v1\_xcmplx\_0~X1)){\unicdbdjdcd}(\\
k5\_binop\_2~X0~X1=k3\_xcmplx\_0~X0~X1)\end{array}
\label{hyp:chfgegfgggjgogjgehjgpgogpflgfdpfcgjgogpgahpfcd}
\end{equation}
Assume the following.
\begin{equation}
{{\unicdcdadad} X0 .  {{({{v1\_xcmplx\_0}~{X0}})}{\unicdbdjdcd}{({{v1\_xcmplx\_0}~{({{k3\_square\_1}~{X0}})}})}}}\label{hyp:ggdgbdpfdhbhfhbgchfgpfbd}
\end{equation}
Assume the following.
\begin{equation}
{{\unicdcdadad} X0 .  {{({{v1\_xcmplx\_0}~{X0}})}{\unicdbdjded}{({{X0} \in {k2\_numbers}})}}}\label{hyp:egcdpfihdgngahmgihpfad}
\end{equation}
Assume the following.
\begin{equation}
\begin{array}{c}{\unicdcdadad} X0 .  (v1\_xcmplx\_0~X0){\unicdbdjdcd}(k3\_polyeq\_3~X0=k5\_binop\_2~(k3\_square\_1~\\
X0)~X0)\end{array}
\label{hyp:egcdpfahpgmgjhfgbhpfdd}
\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}
\begin{theorem}\label{thm:ehcdpfahpgmgjhfgbhpffd}
$$\begin{array}{c}{\unicdcdadad} X0 .  (v1\_xcmplx\_0~X0){\unicdbdjdcd}(k3\_xcmplx\_0~(k3\_xcmplx\_0~X0~X0)~\\
X0=k1\_newton~X0~np\_\_3)\end{array}$$
\end{theorem}
\end{document}
