\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{t26\_domain\_1 (TMSrKNLG317pRsUdTKyVRKCWXYGRPgAcojk)}
\maketitle
Let $ v1\_xboole\_0 : {{\iota}{\unicdbdjdcd}{o}}$ be given.
Let $ k1\_subset\_1 : {{\iota}{\unicdbdjdcd}{\iota}}$ be given.
Let $ m1\_subset\_1 : {{\iota}{\unicdbdjdcd}{{\iota}{\unicdbdjdcd}{o}}}$ be given.
Let $ k1\_xboole\_0 : {\iota}$ be given.
Assume the following.
\begin{equation}
{{\unicdcdadad} X0 .  {{({{v1\_xboole\_0}~{X0}})}{\unicdbdjdcd}{({{X0}={k1\_xboole\_0}})}}}\label{hyp:mggdpfcgpgpgmgfg}
\end{equation}
Assume the following.
\begin{equation}
{{\unicdcdadad} X0 .  {{{k1\_subset\_1}~{X0}}={k1\_xboole\_0}}}\label{hyp:egcdpfdhfhcgdhfgehpfbd}
\end{equation}
Assume the following.
\begin{equation}
{{\unicdcdadad} X0 .  {{({{v1\_xboole\_0}~{X0}})}{\unicdbdjded}{({{\unicdcdadad} X1 .  {\preoh{{{X1} \in {X0}}}}})}}}\label{hyp:egbdpfihcgpgpgmgfgpfad}
\end{equation}
\begin{theorem}\label{thm:ehcdgdpfegpgngbgjgogpfbd}
$$\begin{array}{c}{\unicdcdadad} X0 .  (\preoh v1\_xboole\_0~X0){\unicdbdjdcd}(k1\_subset\_1~X0=ReplSep~(toset~\\
({\uniddcgcg} X1 : \iota .  m1\_subset\_1~X1~X0))~({\uniddcgcg} X1 : \iota .  \preoh True)~({\uniddcgcg} X1 : \\
\iota .  X1))\end{array}$$
\end{theorem}
\end{document}
