\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{t3\_toprealb (TMQkRm6ChBZoco2QgVsCqyVdy6XG2b2Bsdm)}
\maketitle
Let $ v1\_xreal\_0 : {{\iota}{\unicdbdjdcd}{o}}$ be given.
Let $ k1\_toprealb : {{\iota}{\unicdbdjdcd}{{\iota}{\unicdbdjdcd}{\iota}}}$ be given.
Let $ m1\_subset\_1 : {{\iota}{\unicdbdjdcd}{{\iota}{\unicdbdjdcd}{o}}}$ be given.
Let $ k4\_numbers : {\iota}$ be given.
Let $ k2\_rcomp\_1 : {{\iota}{\unicdbdjdcd}{{\iota}{\unicdbdjdcd}{\iota}}}$ be given.
Let $ k2\_xcmplx\_0 : {{\iota}{\unicdbdjdcd}{{\iota}{\unicdbdjdcd}{\iota}}}$ be given.
Assume the following.
\begin{equation}
\begin{array}{c}{\unicdcdadad} X0 .  (v1\_xreal\_0~X0){\unicdbdjdcd}({\unicdcdadad} X1 .  (v1\_xreal\_0~X1){\unicdbdjdcd}(k1\_toprealb~\\
X0~X1=ReplSep~(toset~({\uniddcgcg} X2 : \iota .  m1\_subset\_1~X2~k4\_numbers))~\\
({\uniddcgcg} X2 : \iota .  True)~({\uniddcgcg} X2 : \iota .  k2\_rcomp\_1~(k2\_xcmplx\_0~X0~X2)~\\
(k2\_xcmplx\_0~X1~X2))))\end{array}
\label{hyp:egbdpfehpgahchfgbgmgcg}
\end{equation}
\begin{theorem}\label{thm:ehddpfehpgahchfgbgmgcg}
$$\begin{array}{c}{\unicdcdadad} X0 .  (v1\_xreal\_0~X0){\unicdbdjdcd}({\unicdcdadad} X1 .  (v1\_xreal\_0~X1){\unicdbdjdcd}({\unicdcdadad} X2 .  \\
(X2 \in k1\_toprealb~X0~X1){\unicdbdjded}({\unicdcdaddd} X3 .  (m1\_subset\_1~X3~k4\_numbers){\unicdcdcdhd}\\
(X2=k2\_rcomp\_1~(k2\_xcmplx\_0~X0~X3)~(k2\_xcmplx\_0~X1~X3)))))\end{array}$$
\end{theorem}
\end{document}
