\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{t89\_xxreal\_3 (TMNASCtjQQhocWfEhJi8UvCeL5DGfKcUJS5)}
\maketitle
Let $ v1\_xxreal\_0 : {{\iota}{\unicdbdjdcd}{o}}$ be given.
Let $ k4\_xxreal\_3 : {{\iota}{\unicdbdjdcd}{{\iota}{\unicdbdjdcd}{\iota}}}$ be given.
Let $ k1\_xxreal\_0 : {\iota}$ be given.
Let $ np\_\_1 : {\iota}$ be given.
Let $ k2\_xxreal\_0 : {\iota}$ be given.
Let $ v1\_xboole\_0 : {{\iota}{\unicdbdjdcd}{o}}$ be given.
Let $ k1\_xboole\_0 : {\iota}$ be given.
Let $ k2\_xxreal\_3 : {{\iota}{\unicdbdjdcd}{\iota}}$ be given.
Let $ m1\_subset\_1 : {{\iota}{\unicdbdjdcd}{{\iota}{\unicdbdjdcd}{o}}}$ be given.
Let $ k1\_numbers : {\iota}$ be given.
Let $ v2\_xxreal\_0 : {{\iota}{\unicdbdjdcd}{o}}$ be given.
Let $ m2\_subset\_1 : {{\iota}{\unicdbdjdcd}{{\iota}{\unicdbdjdcd}{{\iota}{\unicdbdjdcd}{o}}}}$ be given.
Let $ k5\_numbers : {\iota}$ be given.
Let $ np\_\_0 : {\iota}$ be given.
Let $ k6\_numbers : {\iota}$ be given.
Assume the following.
\begin{equation}
{{\unicdcdadad} X0 .  {{({{v1\_xboole\_0}~{X0}})}{\unicdbdjdcd}{({{X0}={k1\_xboole\_0}})}}}\label{hyp:ehgdpfcgpgpgmgfg}
\end{equation}
Assume the following.
\begin{equation}
{{{k2\_xxreal\_3}~{k2\_xxreal\_0}}={k1\_xxreal\_0}}\label{hyp:ehfdpfihihchfgbgmgpfdd}
\end{equation}
Assume the following.
\begin{equation}
\begin{array}{c}{\unicdcdadad} X0 .  {\unicdcdadad} X1 .  (m1\_subset\_1~X0~X1){\unicdbdjdcd}((v1\_xboole\_0~X1){\unicdcdcdid}\\
(X0 \in X1))\end{array}
\label{hyp:ehcdpfdhfhcgdhfgeh}
\end{equation}
Assume the following.
\begin{equation}
{\preoh{{{k2\_xxreal\_0} \in {k1\_numbers}}}}\label{hyp:ehbdbdpfihihchfgbgmgpfad}
\end{equation}
Assume the following.
\begin{equation}
\begin{array}{c}((v2\_xxreal\_0~np\_\_1){\unicdcdcdhd}(m2\_subset\_1~np\_\_1~k1\_numbers~k5\_numbers)){\unicdcdcdhd}\\
((m1\_subset\_1~np\_\_1~k5\_numbers){\unicdcdcdhd}(m1\_subset\_1~np\_\_1~k1\_numbers))\end{array}
\label{hyp:dhahdgbdpfogfhngfgchbgmgdh}
\end{equation}
Assume the following.
\begin{equation}
{{v1\_xboole\_0}~{np\_\_0}}\label{hyp:dhahdgadpfcgpgpgmgfg}
\end{equation}
Assume the following.
\begin{equation}
{{k6\_numbers}={k1\_xboole\_0}}\label{hyp:chfgegfgggjgogjgehjgpgogpflggdpfogfhngcgfgchdh}
\end{equation}
Assume the following.
\begin{equation}
\begin{array}{c}{\unicdcdadad} X0 .  (v1\_xxreal\_0~X0){\unicdbdjdcd}({\unicdcdadad} X1 .  (v1\_xxreal\_0~X1){\unicdbdjdcd}(\preoh \\
(k4\_xxreal\_3~X0~X1 \in k1\_numbers){\unicdcdcdhd}((\preoh (X0 \in k1\_numbers){\unicdcdcdhd}(X1 \in k1\_numbers)){\unicdcdcdhd}\\
(k4\_xxreal\_3~X0~X1{\unicdcdgdad}k6\_numbers))))\end{array}
\label{hyp:mgjdjdpfihihchfgbgmgpfdd}
\end{equation}
Assume the following.
\begin{equation}
\begin{array}{c}{\unicdcdadad} X0 .  (v1\_xxreal\_0~X0){\unicdbdjdcd}((k2\_xxreal\_3~X0 \in k1\_numbers){\unicdbdjded}(\\
X0 \in k1\_numbers))\end{array}
\label{hyp:mgbdfdpfihihchfgbgmgpfdd}
\end{equation}
Assume the following.
\begin{equation}
{{v1\_xxreal\_0}~{k2\_xxreal\_0}}\label{hyp:ggdgcdpfihihchfgbgmgpfad}
\end{equation}
Assume the following.
\begin{equation}
{{v1\_xxreal\_0}~{k1\_xxreal\_0}}\label{hyp:ggdgbdpfihihchfgbgmgpfad}
\end{equation}
Assume the following.
\begin{equation}
{\preoh{{{v1\_xboole\_0}~{k1\_numbers}}}}\label{hyp:ggdgbdpfogfhngcgfgchdh}
\end{equation}
Assume the following.
\begin{equation}
{{k1\_xxreal\_0}={k1\_numbers}}\label{hyp:egcdpfihihchfgbgmgpfad}
\end{equation}
\begin{theorem}\label{thm:ehidjdpfihihchfgbgmgpfdd}
$$\begin{array}{c}{\unicdcdadad} X0 .  (v1\_xxreal\_0~X0){\unicdbdjdcd}((k4\_xxreal\_3~k1\_xxreal\_0~X0{\unicdcdgdad}np\_\_1){\unicdcdcdhd}\\
(k4\_xxreal\_3~k2\_xxreal\_0~X0{\unicdcdgdad}np\_\_1))\end{array}$$
\end{theorem}
\end{document}
