\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\_sf\_mastr (TMazP5PXGL6h8YX4QA8uQTAnfKq5eUzgMA1)}
\maketitle
Let $ m2\_subset\_1 : {{\iota}{\unicdbdjdcd}{{\iota}{\unicdbdjdcd}{{\iota}{\unicdbdjdcd}{o}}}}$ be given.
Let $ k1\_numbers : {\iota}$ be given.
Let $ k5\_numbers : {\iota}$ be given.
Let $ k11\_scmfsa\_2 : {{\iota}{\unicdbdjdcd}{\iota}}$ be given.
Let $ k1\_amistd\_1 : {{\iota}{\unicdbdjdcd}{{\iota}{\unicdbdjdcd}{{\iota}{\unicdbdjdcd}{{\iota}{\unicdbdjdcd}{\iota}}}}}$ be given.
Let $ np\_\_3 : {\iota}$ be given.
Let $ k1\_scmfsa\_2 : {\iota}$ be given.
Let $ k1\_tarski : {{\iota}{\unicdbdjdcd}{\iota}}$ be given.
Assume the following.
\begin{equation}
\begin{array}{c}{\unicdcdadad} X0 .  (m2\_subset\_1~X0~k1\_numbers~k5\_numbers){\unicdbdjdcd}({\unicdcdadad} X1 .  \\
(m2\_subset\_1~X1~k1\_numbers~k5\_numbers){\unicdbdjdcd}(k1\_amistd\_1~np\_\_3~k1\_scmfsa\_2~\\
X1~(k11\_scmfsa\_2~X0)=k1\_tarski~X0))\end{array}
\label{hyp:ehddddpfdhdgngggdhbgbdad}
\end{equation}
Assume the following.
\begin{equation}
\begin{array}{c}{\unicdcdadad} X0 .  {\unicdcdadad} X1 .  (X1=k1\_tarski~X0){\unicdbdjded}({\unicdcdadad} X2 .  (X2 \in X1){\unicdbdjded}\\
(X2=X0))\end{array}
\label{hyp:egbdpfehbgchdhlgjg}
\end{equation}
\begin{theorem}\label{thm:ehgdpfdhggpfngbgdhehch}
$$\begin{array}{c}{\unicdcdadad} X0 .  (m2\_subset\_1~X0~k1\_numbers~k5\_numbers){\unicdbdjdcd}({\unicdcdadad} X1 .  \\
(m2\_subset\_1~X1~k1\_numbers~k5\_numbers){\unicdbdjdcd}((k11\_scmfsa\_2~X0=k11\_scmfsa\_2~\\
X1){\unicdbdjdcd}(X0=X1)))\end{array}$$
\end{theorem}
\end{document}
