\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{t2\_pythtrip (TMKgfY6CihwgL6bYkm89hmvehCrmUmYB4gP)}
\maketitle
Let $ v1\_int\_1 : {{\iota}{\unicdbdjdcd}{o}}$ be given.
Let $ v1\_abian : {{\iota}{\unicdbdjdcd}{o}}$ be given.
Let $ k1\_pepin : {{\iota}{\unicdbdjdcd}{\iota}}$ be given.
Let $ k3\_square\_1 : {{\iota}{\unicdbdjdcd}{\iota}}$ be given.
Assume the following.
\begin{equation}
{{\unicdcdadad} X0 .  {{({{v1\_int\_1}~{X0}})}{\unicdbdjdcd}{({{{k1\_pepin}~{X0}}={{k3\_square\_1}~{X0}}})}}}\label{hyp:chfgegfgggjgogjgehjgpgogpflgbdpfahfgahjgog}
\end{equation}
Assume the following.
\begin{equation}
\begin{array}{c}{\unicdcdadad} X0 .  ((v1\_int\_1~X0){\unicdcdcdhd}(\preoh v1\_abian~X0)){\unicdbdjdcd}(\preoh v1\_abian~(k3\_square\_1~\\
X0))\end{array}
\label{hyp:ggdgfdpfahjhehigehchjgah}
\end{equation}
Assume the following.
\begin{equation}
{{\unicdcdadad} X0 .  {{({{({{v1\_int\_1}~{X0}})}{\unicdcdcdhd}{({{v1\_abian}~{X0}})}})}{\unicdbdjdcd}{({{v1\_abian}~{({{k3\_square\_1}~{X0}})}})}}}\label{hyp:ggdgedpfahjhehigehchjgah}
\end{equation}
\begin{theorem}\label{thm:ehcdpfahjhehigehchjgah}
${{\unicdcdadad} X0 .  {{({{v1\_int\_1}~{X0}})}{\unicdbdjdcd}{({{({{v1\_abian}~{X0}})}{\unicdbdjded}{({{v1\_abian}~{({{k1\_pepin}~{X0}})}})}})}}}$.
\end{theorem}
\end{document}
