\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{t27\_euclid\_5 (TMRquPYx1PNdR6LyFjkUDe6jr55YRikVSLi)}
\maketitle
Let $ m1\_subset\_1 : {{\iota}{\unicdbdjdcd}{{\iota}{\unicdbdjdcd}{o}}}$ be given.
Let $ u1\_struct\_0 : {{\iota}{\unicdbdjdcd}{\iota}}$ be given.
Let $ k15\_euclid : {{\iota}{\unicdbdjdcd}{\iota}}$ be given.
Let $ np\_\_3 : {\iota}$ be given.
Let $ k3\_finseq\_4 : {{\iota}{\unicdbdjdcd}{{\iota}{\unicdbdjdcd}{{\iota}{\unicdbdjdcd}{{\iota}{\unicdbdjdcd}{\iota}}}}}$ be given.
Let $ k1\_numbers : {\iota}$ be given.
Let $ k1\_euclid\_5 : {{\iota}{\unicdbdjdcd}{\iota}}$ be given.
Let $ k2\_euclid\_5 : {{\iota}{\unicdbdjdcd}{\iota}}$ be given.
Let $ k3\_euclid\_5 : {{\iota}{\unicdbdjdcd}{\iota}}$ be given.
Let $ k4\_euclid\_5 : {{\iota}{\unicdbdjdcd}{{\iota}{\unicdbdjdcd}{{\iota}{\unicdbdjdcd}{\iota}}}}$ be given.
Let $ v1\_xreal\_0 : {{\iota}{\unicdbdjdcd}{o}}$ be given.
Let $ k11\_finseq\_1 : {{\iota}{\unicdbdjdcd}{{\iota}{\unicdbdjdcd}{{\iota}{\unicdbdjdcd}{\iota}}}}$ be given.
Let $ v1\_xboole\_0 : {{\iota}{\unicdbdjdcd}{o}}$ be given.
Assume the following.
\begin{equation}
\begin{array}{c}{\unicdcdadad} X0 .  (m1\_subset\_1~X0~(u1\_struct\_0~(k15\_euclid~np\_\_3))){\unicdbdjdcd}\\
(X0=k4\_euclid\_5~(k1\_euclid\_5~X0)~(k2\_euclid\_5~X0)~(k3\_euclid\_5~\\
X0))\end{array}
\label{hyp:ehddpffgfhdgmgjgegpffd}
\end{equation}
Assume the following.
\begin{equation}
\begin{array}{c}{\unicdcdadad} X0 .  {\unicdcdadad} X1 .  {\unicdcdadad} X2 .  ((v1\_xreal\_0~X0){\unicdcdcdhd}((v1\_xreal\_0~\\
X1){\unicdcdcdhd}(v1\_xreal\_0~X2))){\unicdbdjdcd}(k4\_euclid\_5~X0~X1~X2=k11\_finseq\_1~X0~X1~\\
X2)\end{array}
\label{hyp:chfgegfgggjgogjgehjgpgogpflgedpffgfhdgmgjgegpffd}
\end{equation}
Assume the following.
\begin{equation}
\begin{array}{c}{\unicdcdadad} X0 .  {\unicdcdadad} X1 .  {\unicdcdadad} X2 .  {\unicdcdadad} X3 .  ((\preoh v1\_xboole\_0~X0){\unicdcdcdhd}\\
((m1\_subset\_1~X1~X0){\unicdcdcdhd}((m1\_subset\_1~X2~X0){\unicdcdcdhd}(m1\_subset\_1~X3~X0)))){\unicdbdjdcd}\\
(k3\_finseq\_4~X0~X1~X2~X3=k11\_finseq\_1~X1~X2~X3)\end{array}
\label{hyp:chfgegfgggjgogjgehjgpgogpflgddpfggjgogdhfgbhpfed}
\end{equation}
Assume the following.
\begin{equation}
{\preoh{{{v1\_xboole\_0}~{k1\_numbers}}}}\label{hyp:ggdgbdpfogfhngcgfgchdh}
\end{equation}
Assume the following.
\begin{equation}
\begin{array}{c}{\unicdcdadad} X0 .  (m1\_subset\_1~X0~(u1\_struct\_0~(k15\_euclid~np\_\_3))){\unicdbdjdcd}\\
(m1\_subset\_1~(k3\_euclid\_5~X0)~k1\_numbers)\end{array}
\label{hyp:egehpflgddpffgfhdgmgjgegpffd}
\end{equation}
Assume the following.
\begin{equation}
\begin{array}{c}{\unicdcdadad} X0 .  (m1\_subset\_1~X0~(u1\_struct\_0~(k15\_euclid~np\_\_3))){\unicdbdjdcd}\\
(m1\_subset\_1~(k2\_euclid\_5~X0)~k1\_numbers)\end{array}
\label{hyp:egehpflgcdpffgfhdgmgjgegpffd}
\end{equation}
Assume the following.
\begin{equation}
\begin{array}{c}{\unicdcdadad} X0 .  (m1\_subset\_1~X0~(u1\_struct\_0~(k15\_euclid~np\_\_3))){\unicdbdjdcd}\\
(m1\_subset\_1~(k1\_euclid\_5~X0)~k1\_numbers)\end{array}
\label{hyp:egehpflgbdpffgfhdgmgjgegpffd}
\end{equation}
Assume the following.
\begin{equation}
{{\unicdcdadad} X0 .  {{({{{m1\_subset\_1}~{X0}}~{k1\_numbers}})}{\unicdbdjdcd}{({{v1\_xreal\_0}~{X0}})}}}\label{hyp:dgdgbdpfchfgbgmgpfbd}
\end{equation}
\begin{theorem}\label{thm:ehcdhdpffgfhdgmgjgegpffd}
$$\begin{array}{c}{\unicdcdadad} X0 .  (m1\_subset\_1~X0~(u1\_struct\_0~(k15\_euclid~np\_\_3))){\unicdbdjdcd}\\
(X0=k3\_finseq\_4~k1\_numbers~(k1\_euclid\_5~X0)~(k2\_euclid\_5~X0)~\\
(k3\_euclid\_5~X0))\end{array}$$
\end{theorem}
\end{document}
