\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\_jgraph\_2 (TMSkc5uJ4RpNJdWSiSXozJ7ezog7YjuCMPt)}
\maketitle
Let $ k17\_euclid : {{\iota}{\unicdbdjdcd}{\iota}}$ be given.
Let $ k4\_struct\_0 : {{\iota}{\unicdbdjdcd}{\iota}}$ be given.
Let $ k15\_euclid : {{\iota}{\unicdbdjdcd}{\iota}}$ be given.
Let $ np\_\_2 : {\iota}$ be given.
Let $ k6\_numbers : {\iota}$ be given.
Let $ k18\_euclid : {{\iota}{\unicdbdjdcd}{\iota}}$ be given.
Let $ k19\_euclid : {{\iota}{\unicdbdjdcd}{{\iota}{\unicdbdjdcd}{\iota}}}$ be given.
Let $ k1\_xboole\_0 : {\iota}$ be given.
Assume the following.
\begin{equation}
{{{k4\_struct\_0}~{({{k15\_euclid}~{np\_\_2}})}}={{{k19\_euclid}~{k6\_numbers}}~{k6\_numbers}}}\label{hyp:ehfdedpffgfhdgmgjgeg}
\end{equation}
Assume the following.
\begin{equation}
{{k6\_numbers}={k1\_xboole\_0}}\label{hyp:chfgegfgggjgogjgehjgpgogpflggdpfogfhngcgfgchdh}
\end{equation}
Assume the following.
\begin{equation}
{{{k18\_euclid}~{({{{k19\_euclid}~{k6\_numbers}}~{k6\_numbers}})}}={k6\_numbers}}\label{hyp:mgfdpfehpgahchfgbgmgcd}
\end{equation}
Assume the following.
\begin{equation}
{{{k17\_euclid}~{({{{k19\_euclid}~{k6\_numbers}}~{k6\_numbers}})}}={k6\_numbers}}\label{hyp:mgedpfehpgahchfgbgmgcd}
\end{equation}
\begin{theorem}\label{thm:ehddpfkghgchbgahigpfcd}
$$\begin{array}{c}(k17\_euclid~(k4\_struct\_0~(k15\_euclid~np\_\_2))=k6\_numbers){\unicdcdcdhd}(\\
k18\_euclid~(k4\_struct\_0~(k15\_euclid~np\_\_2))=k6\_numbers)\end{array}$$
\end{theorem}
\end{document}
