\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{l26\_complex1 (TMdRpEaN6YWxXqsQr9tocNp3unWb8Nw2N7x)}
\maketitle
Let $ v1\_xcmplx\_0 : {{\iota}{\unicdbdjdcd}{o}}$ be given.
Let $ k2\_xcmplx\_0 : {{\iota}{\unicdbdjdcd}{{\iota}{\unicdbdjdcd}{\iota}}}$ be given.
Let $ k5\_arytm\_0 : {{\iota}{\unicdbdjdcd}{{\iota}{\unicdbdjdcd}{\iota}}}$ be given.
Let $ k7\_real\_1 : {{\iota}{\unicdbdjdcd}{{\iota}{\unicdbdjdcd}{\iota}}}$ be given.
Let $ k3\_complex1 : {{\iota}{\unicdbdjdcd}{\iota}}$ be given.
Let $ k4\_complex1 : {{\iota}{\unicdbdjdcd}{\iota}}$ be given.
Assume the following.
\begin{equation}
\begin{array}{c}{\unicdcdadad} X0 .  (v1\_xcmplx\_0~X0){\unicdbdjdcd}(k5\_arytm\_0~(k3\_complex1~X0)~(k4\_complex1~\\
X0)=X0)\end{array}
\label{hyp:mghdpfdgpgngahmgfgihbd}
\end{equation}
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}((X2=k2\_xcmplx\_0~X0~X1){\unicdbdjdcd}(k4\_complex1~X2=k7\_real\_1~\\
(k4\_complex1~X0)~(k4\_complex1~X1)))))\end{array}
\label{hyp:mgcdddpfdgpgngahmgfgihbd}
\end{equation}
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}((X2=k2\_xcmplx\_0~X0~X1){\unicdbdjdcd}(k3\_complex1~X2=k7\_real\_1~\\
(k3\_complex1~X0)~(k3\_complex1~X1)))))\end{array}
\label{hyp:mgcdcdpfdgpgngahmgfgihbd}
\end{equation}
Assume the following.
\begin{equation}
\begin{array}{c}{\unicdcdadad} X0 .  {\unicdcdadad} X1 .  ((v1\_xcmplx\_0~X0){\unicdcdcdhd}(v1\_xcmplx\_0~X1)){\unicdbdjdcd}(\\
v1\_xcmplx\_0~(k2\_xcmplx\_0~X0~X1))\end{array}
\label{hyp:ggdgcdpfihdgngahmgihpfad}
\end{equation}
\begin{theorem}\label{thm:mgcdgdpfdgpgngahmgfgihbd}
$$\begin{array}{c}{\unicdcdadad} X0 .  (v1\_xcmplx\_0~X0){\unicdbdjdcd}({\unicdcdadad} X1 .  (v1\_xcmplx\_0~X1){\unicdbdjdcd}(k2\_xcmplx\_0~\\
X0~X1=k5\_arytm\_0~(k7\_real\_1~(k3\_complex1~X0)~(k3\_complex1~X1))~\\
(k7\_real\_1~(k4\_complex1~X0)~(k4\_complex1~X1))))\end{array}$$
\end{theorem}
\end{document}
