\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{t6\_flang\_2 (TMFmgXaZFZrPtMphWuGjynbHYW8sU1sFGeR)}
\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 $ k5\_afinsq\_1 : {{\iota}{\unicdbdjdcd}{\iota}}$ be given.
Let $ k6\_flang\_1 : {{\iota}{\unicdbdjdcd}{{\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 $ k1\_flang\_1 : {{\iota}{\unicdbdjdcd}{{\iota}{\unicdbdjdcd}{{\iota}{\unicdbdjdcd}{\iota}}}}$ be given.
Assume the following.
\begin{equation}
\begin{array}{c}{\unicdcdadad} X0 .  {\unicdcdadad} X1 .  {\unicdcdadad} X2 .  ((X0 \in X1){\unicdcdcdhd}(m1\_subset\_1~X1~(k1\_zfmisc\_1~\\
X2))){\unicdbdjdcd}(m1\_subset\_1~X0~X2)\end{array}
\label{hyp:ehedpfdhfhcgdhfgeh}
\end{equation}
Assume the following.
\begin{equation}
\begin{array}{c}{\unicdcdadad} X0 .  {\unicdcdadad} X1 .  {\unicdcdadad} X2 .  (m1\_subset\_1~X2~(k3\_catalan2~\\
X1)){\unicdbdjdcd}({\unicdcdadad} X3 .  (m1\_subset\_1~X3~(k3\_catalan2~X1)){\unicdbdjdcd}((k1\_flang\_1~\\
X1~X2~X3=k5\_afinsq\_1~X0){\unicdbdjded}(((X2=k2\_flang\_1~X1){\unicdcdcdhd}(X3=k5\_afinsq\_1~\\
X0)){\unicdcdcdid}((X3=k2\_flang\_1~X1){\unicdcdcdhd}(X2=k5\_afinsq\_1~X0)))))\end{array}
\label{hyp:ehedpfggmgbgoghgpfbd}
\end{equation}
Assume the following.
\begin{equation}
{{\unicdcdadad} X0 .  {{{k3\_catalan2}~{X0}}={{k8\_afinsq\_1}~{X0}}}}\label{hyp:chfgegfgggjgogjgehjgpgogpflgddpfdgbgehbgmgbgogcd}
\end{equation}
Assume the following.
\begin{equation}
\begin{array}{c}{\unicdcdadad} X0 .  {\unicdcdadad} X1 .  {\unicdcdadad} X2 .  ((m1\_subset\_1~X1~(k1\_zfmisc\_1~\\
(k3\_catalan2~X0))){\unicdcdcdhd}(m1\_subset\_1~X2~(k1\_zfmisc\_1~(k3\_catalan2~\\
X0)))){\unicdbdjdcd}(m1\_subset\_1~(k6\_flang\_1~X0~X1~X2)~(k1\_zfmisc\_1~(k3\_catalan2~\\
X0)))\end{array}
\label{hyp:egehpflggdpfggmgbgoghgpfbd}
\end{equation}
Assume the following.
\begin{equation}
{{\unicdcdadad} X0 .  {{{m1\_subset\_1}~{({{k2\_flang\_1}~{X0}})}}~{({{k3\_catalan2}~{X0}})}}}\label{hyp:egehpflgcdpfggmgbgoghgpfbd}
\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}({\unicdcdadad} X2 .  (m1\_subset\_1~X2~(k1\_zfmisc\_1~(k3\_catalan2~\\
X0))){\unicdbdjdcd}({\unicdcdadad} X3 .  (m1\_subset\_1~X3~(k1\_zfmisc\_1~(k3\_catalan2~\\
X0))){\unicdbdjdcd}((X3=k6\_flang\_1~X0~X1~X2){\unicdbdjded}({\unicdcdadad} X4 .  (X4 \in X3){\unicdbdjded}({\unicdcdaddd} X5 .  \\
(m1\_subset\_1~X5~(k3\_catalan2~X0)){\unicdcdcdhd}({\unicdcdaddd} X6 .  (m1\_subset\_1~X6~\\
(k3\_catalan2~X0)){\unicdcdcdhd}((X5 \in X1){\unicdcdcdhd}((X6 \in X2){\unicdcdcdhd}(X4=k1\_flang\_1~X0~X5~X6)))))))))\end{array}
\label{hyp:egbdpfggmgbgoghgpfbd}
\end{equation}
\begin{theorem}\label{thm:ehgdpfggmgbgoghgpfcd}
$$\begin{array}{c}{\unicdcdadad} X0 .  {\unicdcdadad} X1 .  {\unicdcdadad} X2 .  (m1\_subset\_1~X2~(k1\_zfmisc\_1~\\
(k8\_afinsq\_1~X1))){\unicdbdjdcd}({\unicdcdadad} X3 .  (m1\_subset\_1~X3~(k1\_zfmisc\_1~\\
(k8\_afinsq\_1~X1))){\unicdbdjdcd}((k5\_afinsq\_1~X0 \in k6\_flang\_1~X1~X2~X3){\unicdbdjded}(((\\
k2\_flang\_1~X1 \in X2){\unicdcdcdhd}(k5\_afinsq\_1~X0 \in X3)){\unicdcdcdid}((k5\_afinsq\_1~X0 \in X2){\unicdcdcdhd}\\
(k2\_flang\_1~X1 \in X3)))))\end{array}$$
\end{theorem}
\end{document}
