\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=}
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\newcommand\unicdcdadjd{\notin}
\newcommand\unicdbdjded{\Leftrightarrow}
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\newcommand\unicdcdidid{\not\subseteq}
\newcommand\preoh[1]{\lnot #1}
\newcommand\mname[1]{{\mathsf{#1}}}
\begin{document}
\title{t88\_zf\_lang (TMRpzXWZYZGRD8ne4a63ziW6AhMJNhZi1SE)}
\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 $ v6\_zf\_lang : {{\iota}{\unicdbdjdcd}{o}}$ be given.
Let $ k29\_zf\_lang : {{\iota}{\unicdbdjdcd}{\iota}}$ be given.
Let $ k2\_xboole\_0 : {{\iota}{\unicdbdjdcd}{{\iota}{\unicdbdjdcd}{\iota}}}$ be given.
Let $ k24\_zf\_lang : {{\iota}{\unicdbdjdcd}{\iota}}$ be given.
Let $ k1\_tarski : {{\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.
Let $ k23\_zf\_lang : {{\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 .  \\
((v1\_zf\_lang~X1){\unicdcdcdhd}(m2\_finseq\_1~X1~k5\_numbers)){\unicdbdjdcd}(k29\_zf\_lang~\\
(k8\_zf\_lang~X0~X1)=k2\_xboole\_0~(k29\_zf\_lang~X1)~(k1\_tarski~(k8\_zf\_lang~\\
X0~X1))))\end{array}
\label{hyp:ehidedpfkhggpfmgbgoghg}
\end{equation}
Assume the following.
\begin{equation}
\begin{array}{c}{\unicdcdadad} X0 .  (m2\_subset\_1~X0~k5\_numbers~k1\_zf\_lang){\unicdbdjdcd}({\unicdcdadad} X1 .  \\
((v1\_zf\_lang~X1){\unicdcdcdhd}(m2\_finseq\_1~X1~k5\_numbers)){\unicdbdjdcd}({\unicdcdadad} X2 .  (\\
(v1\_zf\_lang~X2){\unicdcdcdhd}(m2\_finseq\_1~X2~k5\_numbers)){\unicdbdjdcd}((v6\_zf\_lang~X1){\unicdbdjdcd}\\
((\preoh (X0=k23\_zf\_lang~X1){\unicdcdcdhd}({\unicdcdadad} X3 .  ((v1\_zf\_lang~X3){\unicdcdcdhd}(m2\_finseq\_1~\\
X3~k5\_numbers)){\unicdbdjdcd}(k8\_zf\_lang~X0~X3{\unicdcdgdad}X1))){\unicdcdcdhd}((({\unicdcdaddd} X3 .  ((v1\_zf\_lang~\\
X3){\unicdcdcdhd}(m2\_finseq\_1~X3~k5\_numbers)){\unicdcdcdhd}(k8\_zf\_lang~X0~X3=X1)){\unicdbdjdcd}(X0=\\
k23\_zf\_lang~X1)){\unicdcdcdhd}((\preoh (X2=k24\_zf\_lang~X1){\unicdcdcdhd}({\unicdcdadad} X3 .  (m2\_subset\_1~\\
X3~k5\_numbers~k1\_zf\_lang){\unicdbdjdcd}(k8\_zf\_lang~X3~X2{\unicdcdgdad}X1))){\unicdcdcdhd}(({\unicdcdaddd} X3 .  \\
(m2\_subset\_1~X3~k5\_numbers~k1\_zf\_lang){\unicdcdcdhd}(k8\_zf\_lang~X3~X2=X1)){\unicdbdjdcd}\\
(X2=k24\_zf\_lang~X1))))))))\end{array}
\label{hyp:ehedcdpfkhggpfmgbgoghg}
\end{equation}
Assume the following.
\begin{equation}
\begin{array}{c}{\unicdcdadad} X0 .  ((v1\_zf\_lang~X0){\unicdcdcdhd}(m2\_finseq\_1~X0~k5\_numbers)){\unicdbdjdcd}((\\
v6\_zf\_lang~X0){\unicdbdjded}({\unicdcdaddd} X1 .  (m2\_subset\_1~X1~k5\_numbers~k1\_zf\_lang){\unicdcdcdhd}\\
({\unicdcdaddd} X2 .  ((v1\_zf\_lang~X2){\unicdcdcdhd}(m2\_finseq\_1~X2~k5\_numbers)){\unicdcdcdhd}(\\
X0=k8\_zf\_lang~X1~X2))))\end{array}
\label{hyp:egbdedpfkhggpfmgbgoghg}
\end{equation}
\begin{theorem}\label{thm:ehididpfkhggpfmgbgoghg}
$$\begin{array}{c}{\unicdcdadad} X0 .  ((v1\_zf\_lang~X0){\unicdcdcdhd}(m2\_finseq\_1~X0~k5\_numbers)){\unicdbdjdcd}((\\
v6\_zf\_lang~X0){\unicdbdjdcd}(k29\_zf\_lang~X0=k2\_xboole\_0~(k29\_zf\_lang~(k24\_zf\_lang~\\
X0))~(k1\_tarski~X0)))\end{array}$$
\end{theorem}
\end{document}
