\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{t10\_bspace (TMZsrMmaKK1vjEeksQXoUYcjuh2qDH21Az4)}
\maketitle
Let $ k3\_bspace : {{\iota}{\unicdbdjdcd}{{\iota}{\unicdbdjdcd}{\iota}}}$ be given.
Let $ k4\_struct\_0 : {{\iota}{\unicdbdjdcd}{\iota}}$ be given.
Let $ k2\_bspace : {\iota}$ be given.
Let $ m1\_subset\_1 : {{\iota}{\unicdbdjdcd}{{\iota}{\unicdbdjdcd}{o}}}$ be given.
Let $ u1\_struct\_0 : {{\iota}{\unicdbdjdcd}{\iota}}$ be given.
Let $ k5\_struct\_0 : {{\iota}{\unicdbdjdcd}{\iota}}$ be given.
Assume the following.
\begin{equation}
\begin{array}{c}{\unicdcdadad} X0 .  (m1\_subset\_1~X0~(u1\_struct\_0~k2\_bspace)){\unicdbdjdcd}((X0=k4\_struct\_0~\\
k2\_bspace){\unicdbdjded}(X0{\unicdcdgdad}k5\_struct\_0~k2\_bspace))\end{array}
\label{hyp:ehidpfcgdhahbgdgfg}
\end{equation}
Assume the following.
\begin{equation}
\begin{array}{c}{\unicdcdadad} X0 .  {\unicdcdadad} X1 .  m1\_subset\_1~(k3\_bspace~X0~X1)~(u1\_struct\_0~\\
k2\_bspace)\end{array}
\label{hyp:egehpflgddpfcgdhahbgdgfg}
\end{equation}
Assume the following.
\begin{equation}
\begin{array}{c}{\unicdcdadad} X0 .  {\unicdcdadad} X1 .  ((X1 \in X0){\unicdbdjdcd}(k3\_bspace~X0~X1=k5\_struct\_0~\\
k2\_bspace)){\unicdcdcdhd}((\preoh X1 \in X0){\unicdbdjdcd}(k3\_bspace~X0~X1=k4\_struct\_0~k2\_bspace))\end{array}
\label{hyp:egddpfcgdhahbgdgfg}
\end{equation}
\begin{theorem}\label{thm:ehbdadpfcgdhahbgdgfg}
$$\begin{array}{c}{\unicdcdadad} X0 .  {\unicdcdadad} X1 .  (k3\_bspace~X0~X1=k4\_struct\_0~k2\_bspace){\unicdbdjded}\\
(\preoh X1 \in X0)\end{array}$$
\end{theorem}
\end{document}
