\documentclass{article}
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\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}
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\newcommand\unicdcdadjd{\notin}
\newcommand\unicdbdjded{\Leftrightarrow}
\newcommand\unicdcdcdid{\lor}
\newcommand\unicdcdcdhd{\land}
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\newcommand\unicdcdaddd{\exists}
\newcommand\unicdcdidid{\not\subseteq}
\newcommand\preoh[1]{\lnot #1}
\newcommand\mname[1]{{\mathsf{#1}}}
\begin{document}
\title{t73\_zf\_lang (TMVufsUJQFof8gmdtEgXUKBPbTUNZpn4fAF)}
\maketitle
Let $ v1\_zf\_lang : {{\iota}{\unicdbdjdcd}{o}}$ be given.
Let $ m2\_finseq\_1 : {{\iota}{\unicdbdjdcd}{{\iota}{\unicdbdjdcd}{o}}}$ be given.
Let $ k5\_numbers : {\iota}$ be given.
Let $ v4\_zf\_lang : {{\iota}{\unicdbdjdcd}{o}}$ be given.
Let $ r3\_zf\_lang : {{\iota}{\unicdbdjdcd}{{\iota}{\unicdbdjdcd}{o}}}$ be given.
Let $ k20\_zf\_lang : {{\iota}{\unicdbdjdcd}{\iota}}$ be given.
Let $ r1\_zf\_lang : {{\iota}{\unicdbdjdcd}{{\iota}{\unicdbdjdcd}{o}}}$ be given.
Let $ k6\_zf\_lang : {{\iota}{\unicdbdjdcd}{\iota}}$ be given.
Let $ k7\_zf\_lang : {{\iota}{\unicdbdjdcd}{{\iota}{\unicdbdjdcd}{\iota}}}$ be given.
Let $ m2\_subset\_1 : {{\iota}{\unicdbdjdcd}{{\iota}{\unicdbdjdcd}{{\iota}{\unicdbdjdcd}{o}}}}$ be given.
Let $ k1\_zf\_lang : {\iota}$ be given.
Let $ k8\_zf\_lang : {{\iota}{\unicdbdjdcd}{{\iota}{\unicdbdjdcd}{\iota}}}$ be given.
Assume the following.
\begin{equation}
\begin{array}{c}{\unicdcdadad} X0 .  ((v1\_zf\_lang~X0){\unicdcdcdhd}(m2\_finseq\_1~X0~k5\_numbers)){\unicdbdjdcd}({\unicdcdadad} X1 .  \\
((v1\_zf\_lang~X1){\unicdcdcdhd}(m2\_finseq\_1~X1~k5\_numbers)){\unicdbdjdcd}((r1\_zf\_lang~\\
X0~X1){\unicdbdjdcd}(r3\_zf\_lang~X0~X1)))\end{array}
\label{hyp:ehgdbdpfkhggpfmgbgoghg}
\end{equation}
Assume the following.
\begin{equation}
\begin{array}{c}{\unicdcdadad} X0 .  ((v1\_zf\_lang~X0){\unicdcdcdhd}(m2\_finseq\_1~X0~k5\_numbers)){\unicdbdjdcd}({\unicdcdadad} X1 .  \\
((v1\_zf\_lang~X1){\unicdcdcdhd}(m2\_finseq\_1~X1~k5\_numbers)){\unicdbdjdcd}((v4\_zf\_lang~\\
X0){\unicdbdjdcd}((r1\_zf\_lang~X1~X0){\unicdbdjded}(X1=k20\_zf\_lang~X0))))\end{array}
\label{hyp:ehfdgdpfkhggpfmgbgoghg}
\end{equation}
Assume the following.
\begin{equation}
\begin{array}{c}{\unicdcdadad} X0 .  ((v1\_zf\_lang~X0){\unicdcdcdhd}(m2\_finseq\_1~X0~k5\_numbers)){\unicdbdjdcd}({\unicdcdadad} X1 .  \\
((v1\_zf\_lang~X1){\unicdcdcdhd}(m2\_finseq\_1~X1~k5\_numbers)){\unicdbdjdcd}((r1\_zf\_lang~\\
X0~X1){\unicdbdjded}(\preoh (X1{\unicdcdgdad}k6\_zf\_lang~X0){\unicdcdcdhd}(({\unicdcdadad} X2 .  ((v1\_zf\_lang~X2){\unicdcdcdhd}(\\
m2\_finseq\_1~X2~k5\_numbers)){\unicdbdjdcd}((X1{\unicdcdgdad}k7\_zf\_lang~X0~X2){\unicdcdcdhd}(X1{\unicdcdgdad}k7\_zf\_lang~\\
X2~X0))){\unicdcdcdhd}({\unicdcdadad} X2 .  (m2\_subset\_1~X2~k5\_numbers~k1\_zf\_lang){\unicdbdjdcd}\\
(X1{\unicdcdgdad}k8\_zf\_lang~X2~X0))))))\end{array}
\label{hyp:egddjdpfkhggpfmgbgoghg}
\end{equation}
Assume the following.
\begin{equation}
\begin{array}{c}{\unicdcdadad} X0 .  ((v1\_zf\_lang~X0){\unicdcdcdhd}(m2\_finseq\_1~X0~k5\_numbers)){\unicdbdjdcd}((\\
v4\_zf\_lang~X0){\unicdbdjded}({\unicdcdaddd} X1 .  ((v1\_zf\_lang~X1){\unicdcdcdhd}(m2\_finseq\_1~X1~\\
k5\_numbers)){\unicdcdcdhd}(X0=k6\_zf\_lang~X1)))\end{array}
\label{hyp:egbdcdpfkhggpfmgbgoghg}
\end{equation}
\begin{theorem}\label{thm:ehhdddpfkhggpfmgbgoghg}
$$\begin{array}{c}{\unicdcdadad} X0 .  ((v1\_zf\_lang~X0){\unicdcdcdhd}(m2\_finseq\_1~X0~k5\_numbers)){\unicdbdjdcd}((\\
v4\_zf\_lang~X0){\unicdbdjdcd}(r3\_zf\_lang~(k20\_zf\_lang~X0)~X0))\end{array}$$
\end{theorem}
\end{document}
