\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{t134\_zf\_lang1 (TMM2tsuDa5xefeNwvQvAhcwnBUeWKnPw5K4)}
\maketitle
Let $ r1\_tarski : {{\iota}{\unicdbdjdcd}{{\iota}{\unicdbdjdcd}{o}}}$ be given.
Let $ k9\_zf\_lang : {\iota}$ be given.
Let $ k1\_zfmisc\_1 : {{\iota}{\unicdbdjdcd}{\iota}}$ be given.
Let $ k2\_zfmisc\_1 : {{\iota}{\unicdbdjdcd}{{\iota}{\unicdbdjdcd}{\iota}}}$ be given.
Let $ k5\_numbers : {\iota}$ be given.
Let $ v1\_zf\_lang : {{\iota}{\unicdbdjdcd}{o}}$ be given.
Let $ m2\_finseq\_1 : {{\iota}{\unicdbdjdcd}{{\iota}{\unicdbdjdcd}{o}}}$ be given.
Let $ m1\_subset\_1 : {{\iota}{\unicdbdjdcd}{{\iota}{\unicdbdjdcd}{o}}}$ be given.
Let $ v1\_xboole\_0 : {{\iota}{\unicdbdjdcd}{o}}$ be given.
Let $ v1\_funct\_1 : {{\iota}{\unicdbdjdcd}{o}}$ be given.
Let $ v1\_finseq\_1 : {{\iota}{\unicdbdjdcd}{o}}$ be given.
Assume the following.
\begin{equation}
\begin{array}{c}{\unicdcdadad} X0 .  ((v1\_zf\_lang~X0){\unicdcdcdhd}(m2\_finseq\_1~X0~k5\_numbers)){\unicdbdjded}(X0 \in \\
k9\_zf\_lang)\end{array}
\label{hyp:ehedpfkhggpfmgbgoghg}
\end{equation}
Assume the following.
\begin{equation}
\begin{array}{c}{\unicdcdadad} X0 .  {\unicdcdadad} X1 .  (m1\_subset\_1~X0~X1){\unicdbdjdcd}((v1\_xboole\_0~X1){\unicdcdcdid}\\
(X0 \in X1))\end{array}
\label{hyp:ehcdpfdhfhcgdhfgeh}
\end{equation}
Assume the following.
\begin{equation}
{{\unicdcdadad} X0 .  {\preoh{{{v1\_xboole\_0}~{({{k1\_zfmisc\_1}~{X0}})}}}}}\label{hyp:ggdgbdpfdhfhcgdhfgehpfbd}
\end{equation}
Assume the following.
\begin{equation}
\begin{array}{c}{\unicdcdadad} X0 .  {\unicdcdadad} X1 .  (m2\_finseq\_1~X1~X0){\unicdbdjdcd}((v1\_funct\_1~X1){\unicdcdcdhd}(\\
(v1\_finseq\_1~X1){\unicdcdcdhd}(m1\_subset\_1~X1~(k1\_zfmisc\_1~(k2\_zfmisc\_1~k5\_numbers~\\
X0)))))\end{array}
\label{hyp:egehpfngcdpfggjgogdhfgbhpfbd}
\end{equation}
Assume the following.
\begin{equation}
\begin{array}{c}{\unicdcdadad} X0 .  {\unicdcdadad} X1 .  (r1\_tarski~X0~X1){\unicdbdjded}({\unicdcdadad} X2 .  (X2 \in X0){\unicdbdjdcd}\\
(X2 \in X1))\end{array}
\label{hyp:egddpfehbgchdhlgjg}
\end{equation}
\begin{theorem}\label{thm:ehbdddedpfkhggpfmgbgoghgbd}
${{{r1\_tarski}~{k9\_zf\_lang}}~{({{k1\_zfmisc\_1}~{({{{k2\_zfmisc\_1}~{k5\_numbers}}~{k5\_numbers}})}})}}$.
\end{theorem}
\end{document}
