\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{t51\_rewrite2 (TMJ7TomxAefLuQQ25ja3pjDyySiowvCzXSF)}
\maketitle
Let $ m1\_subset\_1 : {{\iota}{\unicdbdjdcd}{{\iota}{\unicdbdjdcd}{o}}}$ be given.
Let $ k8\_afinsq\_1 : {{\iota}{\unicdbdjdcd}{\iota}}$ be given.
Let $ k8\_rewrite2 : {{\iota}{\unicdbdjdcd}{{\iota}{\unicdbdjdcd}{{\iota}{\unicdbdjdcd}{\iota}}}}$ be given.
Let $ k6\_rewrite2 : {{\iota}{\unicdbdjdcd}{\iota}}$ be given.
Let $ k4\_flang\_1 : {{\iota}{\unicdbdjdcd}{{\iota}{\unicdbdjdcd}{\iota}}}$ be given.
Let $ k1\_partit\_2 : {{\iota}{\unicdbdjdcd}{{\iota}{\unicdbdjdcd}{\iota}}}$ be given.
Let $ k1\_zfmisc\_1 : {{\iota}{\unicdbdjdcd}{\iota}}$ be given.
Let $ k2\_zfmisc\_1 : {{\iota}{\unicdbdjdcd}{{\iota}{\unicdbdjdcd}{\iota}}}$ be given.
Let $ k4\_subset\_1 : {{\iota}{\unicdbdjdcd}{{\iota}{\unicdbdjdcd}{{\iota}{\unicdbdjdcd}{\iota}}}}$ be given.
Let $ k2\_xboole\_0 : {{\iota}{\unicdbdjdcd}{{\iota}{\unicdbdjdcd}{\iota}}}$ be given.
Let $ k1\_xboole\_0 : {\iota}$ be given.
Assume the following.
\begin{equation}
\begin{array}{c}{\unicdcdadad} X0 .  {\unicdcdadad} X1 .  (m1\_subset\_1~X1~(k8\_afinsq\_1~X0)){\unicdbdjdcd}(k8\_rewrite2~\\
X0~(k1\_partit\_2~(k8\_afinsq\_1~X0)~(k8\_afinsq\_1~X0))~X1=k4\_flang\_1~\\
X0~X1)\end{array}
\label{hyp:ehfdadpfchfghhchjgehfgcd}
\end{equation}
Assume the following.
\begin{equation}
\begin{array}{c}{\unicdcdadad} X0 .  {\unicdcdadad} X1 .  (m1\_subset\_1~X1~(k1\_zfmisc\_1~(k2\_zfmisc\_1~\\
(k8\_afinsq\_1~X0)~(k8\_afinsq\_1~X0)))){\unicdbdjdcd}({\unicdcdadad} X2 .  (m1\_subset\_1~\\
X2~(k8\_afinsq\_1~X0)){\unicdbdjdcd}(k8\_rewrite2~X0~X1~X2=k8\_rewrite2~X0~(k4\_subset\_1~\\
(k2\_zfmisc\_1~(k8\_afinsq\_1~X0)~(k8\_afinsq\_1~X0))~X1~(k6\_rewrite2~\\
(k8\_afinsq\_1~X0)))~X2))\end{array}
\label{hyp:ehedjdpfchfghhchjgehfgcd}
\end{equation}
Assume the following.
\begin{equation}
{{\unicdcdadad} X0 .  {{{{k2\_xboole\_0}~{X0}}~{k1\_xboole\_0}}={X0}}}\label{hyp:ehbdpfcgpgpgmgfg}
\end{equation}
Assume the following.
\begin{equation}
\begin{array}{c}{\unicdcdadad} X0 .  {\unicdcdadad} X1 .  {\unicdcdadad} X2 .  ((m1\_subset\_1~X1~(k1\_zfmisc\_1~\\
X0)){\unicdcdcdhd}(m1\_subset\_1~X2~(k1\_zfmisc\_1~X0))){\unicdbdjdcd}(k4\_subset\_1~X0~X1~X2=\\
k2\_xboole\_0~X1~X2)\end{array}
\label{hyp:chfgegfgggjgogjgehjgpgogpflgedpfdhfhcgdhfgehpfbd}
\end{equation}
Assume the following.
\begin{equation}
\begin{array}{c}{\unicdcdadad} X0 .  m1\_subset\_1~(k6\_rewrite2~X0)~(k1\_zfmisc\_1~(k2\_zfmisc\_1~\\
X0~X0))\end{array}
\label{hyp:egehpflggdpfchfghhchjgehfgcd}
\end{equation}
Assume the following.
\begin{equation}
\begin{array}{c}{\unicdcdadad} X0 .  {\unicdcdadad} X1 .  m1\_subset\_1~(k1\_partit\_2~X0~X1)~(k1\_zfmisc\_1~\\
(k2\_zfmisc\_1~X0~X1))\end{array}
\label{hyp:egehpflgbdpfahbgchehjgehpfcd}
\end{equation}
Assume the following.
\begin{equation}
{{\unicdcdadad} X0 .  {{\unicdcdadad} X1 .  {{{{k1\_partit\_2}~{X0}}~{X1}}={k1\_xboole\_0}}}}\label{hyp:egbdpfahbgchehjgehpfcd}
\end{equation}
Assume the following.
\begin{equation}
{{\unicdcdadad} X0 .  {{\unicdcdadad} X1 .  {{{{k2\_xboole\_0}~{X0}}~{X1}}={{{k2\_xboole\_0}~{X1}}~{X0}}}}}\label{hyp:dgpgngngfhehbgehjgghjgehjhpflgcdpfihcgpgpgmgfgpfad}
\end{equation}
\begin{theorem}\label{thm:ehfdbdpfchfghhchjgehfgcd}
$$\begin{array}{c}{\unicdcdadad} X0 .  {\unicdcdadad} X1 .  (m1\_subset\_1~X1~(k8\_afinsq\_1~X0)){\unicdbdjdcd}(k8\_rewrite2~\\
X0~(k6\_rewrite2~(k8\_afinsq\_1~X0))~X1=k4\_flang\_1~X0~X1)\end{array}$$
\end{theorem}
\end{document}
