\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{l2\_complex1 (TMdmeAUgethAtvHx9nhzec2ycaRCJFXdeF4)}
\maketitle
Let $ v1\_xreal\_0 : {{\iota}{\unicdbdjdcd}{o}}$ be given.
Let $ r1\_xxreal\_0 : {{\iota}{\unicdbdjdcd}{{\iota}{\unicdbdjdcd}{o}}}$ be given.
Let $ k6\_numbers : {\iota}$ be given.
Let $ k2\_xcmplx\_0 : {{\iota}{\unicdbdjdcd}{{\iota}{\unicdbdjdcd}{\iota}}}$ be given.
Let $ k3\_square\_1 : {{\iota}{\unicdbdjdcd}{\iota}}$ be given.
Let $ k3\_xcmplx\_0 : {{\iota}{\unicdbdjdcd}{{\iota}{\unicdbdjdcd}{\iota}}}$ be given.
Let $ v1\_xcmplx\_0 : {{\iota}{\unicdbdjdcd}{o}}$ be given.
Assume the following.
\begin{equation}
\begin{array}{c}{\unicdcdadad} X0 .  (v1\_xreal\_0~X0){\unicdbdjdcd}(r1\_xxreal\_0~k6\_numbers~(k3\_xcmplx\_0~\\
X0~X0))\end{array}
\label{hyp:ehgdddpfihchfgbgmgpfbd}
\end{equation}
Assume the following.
\begin{equation}
\begin{array}{c}{\unicdcdadad} X0 .  (v1\_xreal\_0~X0){\unicdbdjdcd}({\unicdcdadad} X1 .  (v1\_xreal\_0~X1){\unicdbdjdcd}(((r1\_xxreal\_0~\\
k6\_numbers~X0){\unicdcdcdhd}(r1\_xxreal\_0~k6\_numbers~X1)){\unicdbdjdcd}(r1\_xxreal\_0~k6\_numbers~\\
(k2\_xcmplx\_0~X0~X1))))\end{array}
\label{hyp:ehddddpfihchfgbgmgpfbd}
\end{equation}
Assume the following.
\begin{equation}
{{\unicdcdadad} X0 .  {{({{v1\_xreal\_0}~{X0}})}{\unicdbdjdcd}{({{v1\_xreal\_0}~{({{k3\_square\_1}~{X0}})}})}}}\label{hyp:ggdgcdpfdhbhfhbgchfgpfbd}
\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 .  {{({{v1\_xreal\_0}~{X0}})}{\unicdbdjdcd}{({{v1\_xcmplx\_0}~{X0}})}}}\label{hyp:dgdgddpfihchfgbgmgpfad}
\end{equation}
\begin{theorem}\label{thm:mgcdpfdgpgngahmgfgihbd}
$$\begin{array}{c}{\unicdcdadad} X0 .  (v1\_xreal\_0~X0){\unicdbdjdcd}({\unicdcdadad} X1 .  (v1\_xreal\_0~X1){\unicdbdjdcd}(r1\_xxreal\_0~\\
k6\_numbers~(k2\_xcmplx\_0~(k3\_square\_1~X0)~(k3\_square\_1~X1))))\end{array}$$
\end{theorem}
\end{document}
