\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{t57\_flang\_3 (TMMswacsVc1RR2gzxEvFQj187qVTHuGsb9D)}
\maketitle
Let $ m1\_subset\_1 : {{\iota}{\unicdbdjdcd}{{\iota}{\unicdbdjdcd}{o}}}$ be given.
Let $ k1\_zfmisc\_1 : {{\iota}{\unicdbdjdcd}{\iota}}$ be given.
Let $ k8\_afinsq\_1 : {{\iota}{\unicdbdjdcd}{\iota}}$ be given.
Let $ k2\_flang\_3 : {{\iota}{\unicdbdjdcd}{{\iota}{\unicdbdjdcd}{\iota}}}$ be given.
Let $ k8\_flang\_1 : {{\iota}{\unicdbdjdcd}{{\iota}{\unicdbdjdcd}{\iota}}}$ be given.
Let $ k2\_flang\_1 : {{\iota}{\unicdbdjdcd}{\iota}}$ be given.
Let $ k3\_catalan2 : {{\iota}{\unicdbdjdcd}{\iota}}$ be given.
Let $ k6\_flang\_1 : {{\iota}{\unicdbdjdcd}{{\iota}{\unicdbdjdcd}{{\iota}{\unicdbdjdcd}{\iota}}}}$ be given.
Assume the following.
\begin{equation}
\begin{array}{c}{\unicdcdadad} X0 .  {\unicdcdadad} X1 .  (m1\_subset\_1~X1~(k1\_zfmisc\_1~(k8\_afinsq\_1~\\
X0))){\unicdbdjdcd}((k2\_flang\_1~X0 \in k2\_flang\_3~X0~X1){\unicdbdjded}(k2\_flang\_1~X0 \in X1))\end{array}
\label{hyp:ehfdgdpfggmgbgoghgpfdd}
\end{equation}
Assume the following.
\begin{equation}
\begin{array}{c}{\unicdcdadad} X0 .  {\unicdcdadad} X1 .  (m1\_subset\_1~X1~(k1\_zfmisc\_1~(k3\_catalan2~\\
X0))){\unicdbdjdcd}((k2\_flang\_1~X0 \in X1){\unicdbdjdcd}((k8\_flang\_1~X0~X1=k6\_flang\_1~X0~(\\
k8\_flang\_1~X0~X1)~X1){\unicdcdcdhd}(k8\_flang\_1~X0~X1=k6\_flang\_1~X0~X1~(k8\_flang\_1~\\
X0~X1))))\end{array}
\label{hyp:ehfdedpfggmgbgoghgpfbd}
\end{equation}
Assume the following.
\begin{equation}
\begin{array}{c}{\unicdcdadad} X0 .  {\unicdcdadad} X1 .  (m1\_subset\_1~X1~(k1\_zfmisc\_1~(k8\_afinsq\_1~\\
X0))){\unicdbdjdcd}(k2\_flang\_3~X0~X1=k6\_flang\_1~X0~(k8\_flang\_1~X0~X1)~X1)\end{array}
\label{hyp:ehfdcdpfggmgbgoghgpfdd}
\end{equation}
Assume the following.
\begin{equation}
\begin{array}{c}{\unicdcdadad} X0 .  {\unicdcdadad} X1 .  (m1\_subset\_1~X1~(k1\_zfmisc\_1~(k3\_catalan2~\\
X0))){\unicdbdjdcd}(k2\_flang\_1~X0 \in k8\_flang\_1~X0~X1)\end{array}
\label{hyp:ehedidpfggmgbgoghgpfbd}
\end{equation}
Assume the following.
\begin{equation}
{{\unicdcdadad} X0 .  {{{k3\_catalan2}~{X0}}={{k8\_afinsq\_1}~{X0}}}}\label{hyp:chfgegfgggjgogjgehjgpgogpflgddpfdgbgehbgmgbgogcd}
\end{equation}
\begin{theorem}\label{thm:ehfdhdpfggmgbgoghgpfdd}
$$\begin{array}{c}{\unicdcdadad} X0 .  {\unicdcdadad} X1 .  (m1\_subset\_1~X1~(k1\_zfmisc\_1~(k8\_afinsq\_1~\\
X0))){\unicdbdjdcd}((k2\_flang\_3~X0~X1=k8\_flang\_1~X0~X1){\unicdbdjded}(k2\_flang\_1~X0 \in X1))\end{array}$$
\end{theorem}
\end{document}
