\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=}
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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{l15\_zf\_fund1 (TMLPdZfbZHgNSjBEDyedURnEZBNPRq5BNv3)}
\maketitle
Let $ m1\_subset\_1 : {{\iota}{\unicdbdjdcd}{{\iota}{\unicdbdjdcd}{o}}}$ be given.
Let $ k4\_ordinal1 : {\iota}$ be given.
Let $ k3\_zf\_fund1 : {{\iota}{\unicdbdjdcd}{\iota}}$ be given.
Let $ k2\_zf\_lang : {{\iota}{\unicdbdjdcd}{\iota}}$ be given.
Let $ k5\_card\_1 : {{\iota}{\unicdbdjdcd}{\iota}}$ be given.
Let $ v7\_ordinal1 : {{\iota}{\unicdbdjdcd}{o}}$ be given.
Let $ k2\_finseq\_1 : {{\iota}{\unicdbdjdcd}{\iota}}$ be given.
Let $ k5\_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 $ r1\_xxreal\_0 : {{\iota}{\unicdbdjdcd}{{\iota}{\unicdbdjdcd}{o}}}$ be given.
Let $ np\_\_5 : {\iota}$ be given.
Assume the following.
\begin{equation}
{{\unicdcdadad} X0 .  {{({{v7\_ordinal1}~{X0}})}{\unicdbdjdcd}{({{{k5\_card\_1}~{({{k2\_finseq\_1}~{X0}})}}={X0}})}}}\label{hyp:ehfdhdpfggjgogdhfgbhpfbd}
\end{equation}
Assume the following.
\begin{equation}
\begin{array}{c}{\unicdcdadad} X0 .  (v7\_ordinal1~X0){\unicdbdjdcd}(k5\_card\_1~(k2\_finseq\_1~X0)=k5\_card\_1~\\
X0)\end{array}
\label{hyp:ehfdfdpfggjgogdhfgbhpfbd}
\end{equation}
Assume the following.
\begin{equation}
{{k5\_numbers}={k4\_ordinal1}}\label{hyp:chfgegfgggjgogjgehjgpgogpflgfdpfogfhngcgfgchdh}
\end{equation}
Assume the following.
\begin{equation}
\begin{array}{c}{\unicdcdadad} X0 .  (m1\_subset\_1~X0~k5\_numbers){\unicdbdjdcd}(m2\_subset\_1~(k2\_zf\_lang~\\
X0)~k5\_numbers~k1\_zf\_lang)\end{array}
\label{hyp:egehpflgcdpfkhggpfmgbgoghg}
\end{equation}
Assume the following.
\begin{equation}
\begin{array}{c}{\unicdcdadad} X0 .  (m2\_subset\_1~X0~k5\_numbers~k1\_zf\_lang){\unicdbdjdcd}({\unicdcdadad} X1 .  \\
(m1\_subset\_1~X1~k5\_numbers){\unicdbdjdcd}((X1=k3\_zf\_fund1~X0){\unicdbdjded}(k2\_zf\_lang~\\
X1=X0)))\end{array}
\label{hyp:egddpfkhggpfggfhogegbd}
\end{equation}
Assume the following.
\begin{equation}
\begin{array}{c}k1\_zf\_lang=ReplSep~(toset~({\uniddcgcg} X0 : \iota .  m1\_subset\_1~X0~k5\_numbers))~\\
({\uniddcgcg} X0 : \iota .  r1\_xxreal\_0~np\_\_5~X0)~({\uniddcgcg} X0 : \iota .  X0)\end{array}
\label{hyp:egbdpfkhggpfmgbgoghg}
\end{equation}
Assume the following.
\begin{equation}
{{\unicdcdadad} X0 .  {{({{{m1\_subset\_1}~{X0}}~{k4\_ordinal1}})}{\unicdbdjdcd}{({{v7\_ordinal1}~{X0}})}}}\label{hyp:dgdgidpfpgchegjgogbgmgbd}
\end{equation}
\begin{theorem}\label{thm:mgbdfdpfkhggpfggfhogegbd}
$$\begin{array}{c}{\unicdcdadad} X0 .  (m1\_subset\_1~X0~k4\_ordinal1){\unicdbdjdcd}(X0=k3\_zf\_fund1~(k2\_zf\_lang~\\
(k5\_card\_1~X0)))\end{array}$$
\end{theorem}
\end{document}
