\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}
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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{t35\_zf\_lang (TMHU2tw7WNMdVPZir8tsUTfSXUWKvq2K8Cm)}
\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 $ v7\_zf\_lang : {{\iota}{\unicdbdjdcd}{o}}$ be given.
Let $ k18\_zf\_lang : {{\iota}{\unicdbdjdcd}{\iota}}$ be given.
Let $ k1\_funct\_1 : {{\iota}{\unicdbdjdcd}{{\iota}{\unicdbdjdcd}{\iota}}}$ be given.
Let $ np\_\_2 : {\iota}$ be given.
Let $ k19\_zf\_lang : {{\iota}{\unicdbdjdcd}{\iota}}$ be given.
Let $ np\_\_3 : {\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}((\\
v7\_zf\_lang~X0){\unicdbdjdcd}(k19\_zf\_lang~X0=k1\_funct\_1~X0~np\_\_3))\end{array}
\label{hyp:egcdjdpfkhggpfmgbgoghg}
\end{equation}
Assume the following.
\begin{equation}
\begin{array}{c}{\unicdcdadad} X0 .  ((v1\_zf\_lang~X0){\unicdcdcdhd}(m2\_finseq\_1~X0~k5\_numbers)){\unicdbdjdcd}((\\
v7\_zf\_lang~X0){\unicdbdjdcd}(k18\_zf\_lang~X0=k1\_funct\_1~X0~np\_\_2))\end{array}
\label{hyp:egcdidpfkhggpfmgbgoghg}
\end{equation}
\begin{theorem}\label{thm:ehddfdpfkhggpfmgbgoghg}
$$\begin{array}{c}{\unicdcdadad} X0 .  ((v1\_zf\_lang~X0){\unicdcdcdhd}(m2\_finseq\_1~X0~k5\_numbers)){\unicdbdjdcd}((\\
v7\_zf\_lang~X0){\unicdbdjdcd}((k18\_zf\_lang~X0=k1\_funct\_1~X0~np\_\_2){\unicdcdcdhd}(k19\_zf\_lang~\\
X0=k1\_funct\_1~X0~np\_\_3)))\end{array}$$
\end{theorem}
\end{document}
