\documentclass{article}
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\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}
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\newcommand\unicdbdjdcd{\Rightarrow}
\newcommand\unicdcdadjd{\notin}
\newcommand\unicdbdjded{\Leftrightarrow}
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\newcommand\unicdcdaddd{\exists}
\newcommand\unicdcdidid{\not\subseteq}
\newcommand\preoh[1]{\lnot #1}
\newcommand\mname[1]{{\mathsf{#1}}}
\begin{document}
\title{t153\_zf\_lang1 (TMP7QxYtkUMMu8Ks7tvpyy62paXjSQrENHy)}
\maketitle
Let $ m2\_subset\_1 : {{\iota}{\unicdbdjdcd}{{\iota}{\unicdbdjdcd}{{\iota}{\unicdbdjdcd}{o}}}}$ be given.
Let $ k5\_numbers : {\iota}$ be given.
Let $ k1\_zf\_lang : {\iota}$ be given.
Let $ k5\_zf\_lang1 : {{\iota}{\unicdbdjdcd}{{\iota}{\unicdbdjdcd}{{\iota}{\unicdbdjdcd}{\iota}}}}$ be given.
Let $ k4\_zf\_lang : {{\iota}{\unicdbdjdcd}{{\iota}{\unicdbdjdcd}{\iota}}}$ be given.
Assume the following.
\begin{equation}
\begin{array}{c}{\unicdcdadad} X0 .  (m2\_subset\_1~X0~k5\_numbers~k1\_zf\_lang){\unicdbdjdcd}({\unicdcdadad} X1 .  \\
(m2\_subset\_1~X1~k5\_numbers~k1\_zf\_lang){\unicdbdjdcd}({\unicdcdadad} X2 .  (m2\_subset\_1~\\
X2~k5\_numbers~k1\_zf\_lang){\unicdbdjdcd}({\unicdcdadad} X3 .  (m2\_subset\_1~X3~k5\_numbers~\\
k1\_zf\_lang){\unicdbdjdcd}({\unicdcdadad} X4 .  (m2\_subset\_1~X4~k5\_numbers~k1\_zf\_lang){\unicdbdjdcd}\\
({\unicdcdadad} X5 .  (m2\_subset\_1~X5~k5\_numbers~k1\_zf\_lang){\unicdbdjdcd}((k5\_zf\_lang1~\\
(k4\_zf\_lang~X0~X1)~X2~X3=k4\_zf\_lang~X4~X5){\unicdbdjded}(\preoh (\preoh (X0{\unicdcdgdad}X2){\unicdcdcdhd}((X1{\unicdcdgdad}X2){\unicdcdcdhd}\\
((X4=X0){\unicdcdcdhd}(X5=X1)))){\unicdcdcdhd}((\preoh (X0=X2){\unicdcdcdhd}((X1{\unicdcdgdad}X2){\unicdcdcdhd}((X4=X3){\unicdcdcdhd}(X5=X1)))){\unicdcdcdhd}\\
((\preoh (X0{\unicdcdgdad}X2){\unicdcdcdhd}((X1=X2){\unicdcdcdhd}((X4=X0){\unicdcdcdhd}(X5=X3)))){\unicdcdcdhd}(\preoh (X0=X2){\unicdcdcdhd}((X1=\\
X2){\unicdcdcdhd}((X4=X3){\unicdcdcdhd}(X5=X3)))))))))))))\end{array}
\label{hyp:ehbdfdcdpfkhggpfmgbgoghgbd}
\end{equation}
\begin{theorem}\label{thm:ehbdfdddpfkhggpfmgbgoghgbd}
$$\begin{array}{c}{\unicdcdadad} X0 .  (m2\_subset\_1~X0~k5\_numbers~k1\_zf\_lang){\unicdbdjdcd}({\unicdcdadad} X1 .  \\
(m2\_subset\_1~X1~k5\_numbers~k1\_zf\_lang){\unicdbdjdcd}({\unicdcdadad} X2 .  (m2\_subset\_1~\\
X2~k5\_numbers~k1\_zf\_lang){\unicdbdjdcd}({\unicdcdadad} X3 .  (m2\_subset\_1~X3~k5\_numbers~\\
k1\_zf\_lang){\unicdbdjdcd}({\unicdcdaddd} X4 .  (m2\_subset\_1~X4~k5\_numbers~k1\_zf\_lang){\unicdcdcdhd}\\
({\unicdcdaddd} X5 .  (m2\_subset\_1~X5~k5\_numbers~k1\_zf\_lang){\unicdcdcdhd}(k5\_zf\_lang1~\\
(k4\_zf\_lang~X0~X1)~X2~X3=k4\_zf\_lang~X4~X5))))))\end{array}$$
\end{theorem}
\end{document}
