\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{t18\_zf\_lang (TMWbfLhKgQ84b4ECpW2djFZHspQmcvzU7Z6)}
\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 $ v2\_zf\_lang : {{\iota}{\unicdbdjdcd}{o}}$ be given.
Let $ k1\_funct\_1 : {{\iota}{\unicdbdjdcd}{{\iota}{\unicdbdjdcd}{\iota}}}$ be given.
Let $ np\_\_1 : {\iota}$ be given.
Let $ k6\_numbers : {\iota}$ be given.
Let $ m2\_subset\_1 : {{\iota}{\unicdbdjdcd}{{\iota}{\unicdbdjdcd}{{\iota}{\unicdbdjdcd}{o}}}}$ be given.
Let $ k1\_zf\_lang : {\iota}$ be given.
Let $ k4\_zf\_lang : {{\iota}{\unicdbdjdcd}{{\iota}{\unicdbdjdcd}{\iota}}}$ be given.
Let $ k5\_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}((k1\_funct\_1~(k4\_zf\_lang~\\
X0~X1)~np\_\_1=k6\_numbers){\unicdcdcdhd}(k1\_funct\_1~(k5\_zf\_lang~X0~X1)~np\_\_1=\\
np\_\_1)))\end{array}
\label{hyp:ehbdfdpfkhggpfmgbgoghg}
\end{equation}
Assume the following.
\begin{equation}
\begin{array}{c}{\unicdcdadad} X0 .  ((v1\_zf\_lang~X0){\unicdcdcdhd}(m2\_finseq\_1~X0~k5\_numbers)){\unicdbdjdcd}((\\
v2\_zf\_lang~X0){\unicdbdjded}({\unicdcdaddd} X1 .  (m2\_subset\_1~X1~k5\_numbers~k1\_zf\_lang){\unicdcdcdhd}\\
({\unicdcdaddd} X2 .  (m2\_subset\_1~X2~k5\_numbers~k1\_zf\_lang){\unicdcdcdhd}(X0=k4\_zf\_lang~\\
X1~X2))))\end{array}
\label{hyp:egbdadpfkhggpfmgbgoghg}
\end{equation}
\begin{theorem}\label{thm:ehbdidpfkhggpfmgbgoghg}
$$\begin{array}{c}{\unicdcdadad} X0 .  ((v1\_zf\_lang~X0){\unicdcdcdhd}(m2\_finseq\_1~X0~k5\_numbers)){\unicdbdjdcd}((\\
v2\_zf\_lang~X0){\unicdbdjdcd}(k1\_funct\_1~X0~np\_\_1=k6\_numbers))\end{array}$$
\end{theorem}
\end{document}
