\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{t23\_zf\_lang (TMLjbJDk1kRYmocphh7nqjyPfkBG8XEsJqZ)}
\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 $ v3\_zf\_lang : {{\iota}{\unicdbdjdcd}{o}}$ be given.
Let $ v4\_zf\_lang : {{\iota}{\unicdbdjdcd}{o}}$ be given.
Let $ np\_\_2 : {\iota}$ be given.
Let $ v5\_zf\_lang : {{\iota}{\unicdbdjdcd}{o}}$ be given.
Let $ np\_\_3 : {\iota}$ be given.
Let $ v6\_zf\_lang : {{\iota}{\unicdbdjdcd}{o}}$ be given.
Let $ np\_\_4 : {\iota}$ be given.
Assume the following.
\begin{equation}
\begin{array}{c}{\unicdcdadad} X0 .  ((v1\_zf\_lang~X0){\unicdcdcdhd}(m2\_finseq\_1~X0~k5\_numbers)){\unicdbdjdcd}(\preoh \\
(\preoh v2\_zf\_lang~X0){\unicdcdcdhd}((\preoh v3\_zf\_lang~X0){\unicdcdcdhd}((\preoh v4\_zf\_lang~X0){\unicdcdcdhd}((\preoh v5\_zf\_lang~\\
X0){\unicdcdcdhd}(\preoh v6\_zf\_lang~X0)))))\end{array}
\label{hyp:ehjdpfkhggpfmgbgoghg}
\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){\unicdbdjdcd}(k1\_funct\_1~X0~np\_\_1=np\_\_4))\end{array}
\label{hyp:ehcdcdpfkhggpfmgbgoghg}
\end{equation}
Assume the following.
\begin{equation}
\begin{array}{c}{\unicdcdadad} X0 .  ((v1\_zf\_lang~X0){\unicdcdcdhd}(m2\_finseq\_1~X0~k5\_numbers)){\unicdbdjdcd}((\\
v5\_zf\_lang~X0){\unicdbdjdcd}(k1\_funct\_1~X0~np\_\_1=np\_\_3))\end{array}
\label{hyp:ehcdbdpfkhggpfmgbgoghg}
\end{equation}
Assume the following.
\begin{equation}
\begin{array}{c}{\unicdcdadad} X0 .  ((v1\_zf\_lang~X0){\unicdcdcdhd}(m2\_finseq\_1~X0~k5\_numbers)){\unicdbdjdcd}((\\
v4\_zf\_lang~X0){\unicdbdjdcd}(k1\_funct\_1~X0~np\_\_1=np\_\_2))\end{array}
\label{hyp:ehcdadpfkhggpfmgbgoghg}
\end{equation}
Assume the following.
\begin{equation}
\begin{array}{c}{\unicdcdadad} X0 .  ((v1\_zf\_lang~X0){\unicdcdcdhd}(m2\_finseq\_1~X0~k5\_numbers)){\unicdbdjdcd}((\\
v3\_zf\_lang~X0){\unicdbdjdcd}(k1\_funct\_1~X0~np\_\_1=np\_\_1))\end{array}
\label{hyp:ehbdjdpfkhggpfmgbgoghg}
\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){\unicdbdjdcd}(k1\_funct\_1~X0~np\_\_1=k6\_numbers))\end{array}
\label{hyp:ehbdidpfkhggpfmgbgoghg}
\end{equation}
\begin{theorem}\label{thm:ehcdddpfkhggpfmgbgoghg}
$$\begin{array}{c}{\unicdcdadad} X0 .  ((v1\_zf\_lang~X0){\unicdcdcdhd}(m2\_finseq\_1~X0~k5\_numbers)){\unicdbdjdcd}(\preoh \\
(\preoh (v2\_zf\_lang~X0){\unicdcdcdhd}(k1\_funct\_1~X0~np\_\_1=k6\_numbers)){\unicdcdcdhd}((\preoh (v3\_zf\_lang~\\
X0){\unicdcdcdhd}(k1\_funct\_1~X0~np\_\_1=np\_\_1)){\unicdcdcdhd}((\preoh (v4\_zf\_lang~X0){\unicdcdcdhd}(k1\_funct\_1~\\
X0~np\_\_1=np\_\_2)){\unicdcdcdhd}((\preoh (v5\_zf\_lang~X0){\unicdcdcdhd}(k1\_funct\_1~X0~np\_\_1=np\_\_3)){\unicdcdcdhd}\\
(\preoh (v6\_zf\_lang~X0){\unicdcdcdhd}(k1\_funct\_1~X0~np\_\_1=np\_\_4))))))\end{array}$$
\end{theorem}
\end{document}
