\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{l6\_petri\_2 (TMJL39J15PD9j53CayPCSyPaKdkHL2WHZEN)}
\maketitle
Let $ v1\_xboole\_0 : {{\iota}{\unicdbdjdcd}{o}}$ be given.
Let $ r1\_tarski : {{\iota}{\unicdbdjdcd}{{\iota}{\unicdbdjdcd}{o}}}$ be given.
Let $ k2\_petri\_2 : {{\iota}{\unicdbdjdcd}{{\iota}{\unicdbdjdcd}{\iota}}}$ be given.
Let $ k9\_setfam\_1 : {{\iota}{\unicdbdjdcd}{\iota}}$ be given.
Let $ k4\_partfun1 : {{\iota}{\unicdbdjdcd}{{\iota}{\unicdbdjdcd}{\iota}}}$ be given.
Let $ k1\_funct\_2 : {{\iota}{\unicdbdjdcd}{{\iota}{\unicdbdjdcd}{\iota}}}$ be given.
Let $ k1\_zfmisc\_1 : {{\iota}{\unicdbdjdcd}{\iota}}$ be given.
Let $ m1\_subset\_1 : {{\iota}{\unicdbdjdcd}{{\iota}{\unicdbdjdcd}{o}}}$ be given.
Let $ k9\_funct\_2 : {{\iota}{\unicdbdjdcd}{{\iota}{\unicdbdjdcd}{\iota}}}$ be given.
Assume the following.
\begin{equation}
\begin{array}{c}{\unicdcdadad} X0 .  {\unicdcdadad} X1 .  r1\_tarski~(k1\_funct\_2~X0~X1)~(k4\_partfun1~\\
X0~X1)\end{array}
\label{hyp:ehhdcdpfggfhogdgehpfcd}
\end{equation}
Assume the following.
\begin{equation}
\begin{array}{c}{\unicdcdadad} X0 .  {\unicdcdadad} X1 .  (r1\_tarski~X0~X1){\unicdbdjdcd}(r1\_tarski~(k1\_zfmisc\_1~\\
X0)~(k1\_zfmisc\_1~X1))\end{array}
\label{hyp:ehgdhdpfkhggngjgdhdgpfbd}
\end{equation}
Assume the following.
\begin{equation}
\begin{array}{c}{\unicdcdadad} X0 .  {\unicdcdadad} X1 .  {\unicdcdadad} X2 .  {\unicdcdadad} X3 .  ((r1\_tarski~X0~X1){\unicdcdcdhd}\\
(r1\_tarski~X2~X3)){\unicdbdjdcd}(r1\_tarski~(k4\_partfun1~X0~X2)~(k4\_partfun1~\\
X1~X3))\end{array}
\label{hyp:ehfdadpfahbgchehggfhogbd}
\end{equation}
Assume the following.
\begin{equation}
\begin{array}{c}{\unicdcdadad} X0 .  {\unicdcdadad} X1 .  (m1\_subset\_1~X0~(k1\_zfmisc\_1~X1)){\unicdbdjded}(r1\_tarski~\\
X0~X1)\end{array}
\label{hyp:ehddpfdhfhcgdhfgeh}
\end{equation}
Assume the following.
\begin{equation}
\begin{array}{c}{\unicdcdadad} X0 .  {\unicdcdadad} X1 .  {\unicdcdadad} X2 .  ((r1\_tarski~X0~X1){\unicdcdcdhd}(r1\_tarski~\\
X1~X2)){\unicdbdjdcd}(r1\_tarski~X0~X2)\end{array}
\label{hyp:ehbdpfihcgpgpgmgfgpfbd}
\end{equation}
Assume the following.
\begin{equation}
{{\unicdcdadad} X0 .  {{\unicdcdadad} X1 .  {{{r1\_tarski}~{X0}}~{X0}}}}\label{hyp:chfgggmgfgihjgghjgehjhpfchbdpfehbgchdhlgjg}
\end{equation}
Assume the following.
\begin{equation}
{{\unicdcdadad} X0 .  {{{k9\_setfam\_1}~{X0}}={{k1\_zfmisc\_1}~{X0}}}}\label{hyp:chfgegfgggjgogjgehjgpgogpflgjdpfdhfgehggbgngpfbd}
\end{equation}
Assume the following.
\begin{equation}
\begin{array}{c}{\unicdcdadad} X0 .  {\unicdcdadad} X1 .  (\preoh v1\_xboole\_0~X1){\unicdbdjdcd}(k9\_funct\_2~X0~X1=k1\_funct\_2~\\
X0~X1)\end{array}
\label{hyp:chfgegfgggjgogjgehjgpgogpflgjdpfggfhogdgehpfcd}
\end{equation}
Assume the following.
\begin{equation}
\begin{array}{c}{\unicdcdadad} X0 .  {\unicdcdadad} X1 .  (\preoh v1\_xboole\_0~X0){\unicdbdjdcd}((\preoh v1\_xboole\_0~(k2\_petri\_2~\\
X0~X1)){\unicdcdcdhd}(m1\_subset\_1~(k2\_petri\_2~X0~X1)~(k1\_zfmisc\_1~(k1\_zfmisc\_1~\\
(k9\_funct\_2~X1~X0)))))\end{array}
\label{hyp:egehpflgcdpfahfgehchjgpfcd}
\end{equation}
\begin{theorem}\label{thm:mggdpfahfgehchjgpfcd}
$$\begin{array}{c}{\unicdcdadad} X0 .  (\preoh v1\_xboole\_0~X0){\unicdbdjdcd}({\unicdcdadad} X1 .  {\unicdcdadad} X2 .  (r1\_tarski~\\
X1~X2){\unicdbdjdcd}(r1\_tarski~(k2\_petri\_2~X0~X1)~(k9\_setfam\_1~(k4\_partfun1~\\
X2~X0))))\end{array}$$
\end{theorem}
\end{document}
