\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{t26\_zf\_fund1 (TMdRnEbtVKU1ekZD7Y2S2VFCvyD4gYxATwW)}
\maketitle
Let $ m2\_subset\_1 : {{\iota}{\unicdbdjdcd}{{\iota}{\unicdbdjdcd}{{\iota}{\unicdbdjdcd}{o}}}}$ be given.
Let $ k5\_numbers : {\iota}$ be given.
Let $ k1\_zf\_lang : {\iota}$ be given.
Let $ k4\_zf\_fund1 : {{\iota}{\unicdbdjdcd}{\iota}}$ be given.
Let $ k6\_domain\_1 : {{\iota}{\unicdbdjdcd}{{\iota}{\unicdbdjdcd}{\iota}}}$ be given.
Let $ k3\_zf\_fund1 : {{\iota}{\unicdbdjdcd}{\iota}}$ be given.
Let $ k7\_domain\_1 : {{\iota}{\unicdbdjdcd}{{\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}(k4\_zf\_fund1~(k7\_domain\_1~\\
k1\_zf\_lang~X0~X1)=k7\_domain\_1~k5\_numbers~(k3\_zf\_fund1~X0)~(k3\_zf\_fund1~\\
X1)))\end{array}
\label{hyp:mgcdadpfkhggpfggfhogegbd}
\end{equation}
Assume the following.
\begin{equation}
\begin{array}{c}{\unicdcdadad} X0 .  (m2\_subset\_1~X0~k5\_numbers~k1\_zf\_lang){\unicdbdjdcd}(k4\_zf\_fund1~\\
(k6\_domain\_1~k1\_zf\_lang~X0)=k6\_domain\_1~k5\_numbers~(k3\_zf\_fund1~\\
X0))\end{array}
\label{hyp:mgbdjdpfkhggpfggfhogegbd}
\end{equation}
\begin{theorem}\label{thm:ehcdgdpfkhggpfggfhogegbd}
$$\begin{array}{c}{\unicdcdadad} X0 .  (m2\_subset\_1~X0~k5\_numbers~k1\_zf\_lang){\unicdbdjdcd}({\unicdcdadad} X1 .  \\
(m2\_subset\_1~X1~k5\_numbers~k1\_zf\_lang){\unicdbdjdcd}((k4\_zf\_fund1~(k6\_domain\_1~\\
k1\_zf\_lang~X0)=k6\_domain\_1~k5\_numbers~(k3\_zf\_fund1~X0)){\unicdcdcdhd}(k4\_zf\_fund1~\\
(k7\_domain\_1~k1\_zf\_lang~X0~X1)=k7\_domain\_1~k5\_numbers~(k3\_zf\_fund1~\\
X0)~(k3\_zf\_fund1~X1))))\end{array}$$
\end{theorem}
\end{document}
