\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{t9\_zf\_lang1 (TMb68v7Q8CtMuSbZPmPcaRuBqAAnuwMBsSu)}
\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 $ m2\_subset\_1 : {{\iota}{\unicdbdjdcd}{{\iota}{\unicdbdjdcd}{{\iota}{\unicdbdjdcd}{o}}}}$ be given.
Let $ k1\_zf\_lang : {\iota}$ be given.
Let $ k23\_zf\_lang : {{\iota}{\unicdbdjdcd}{\iota}}$ be given.
Let $ k13\_zf\_lang : {{\iota}{\unicdbdjdcd}{{\iota}{\unicdbdjdcd}{\iota}}}$ be given.
Let $ k24\_zf\_lang : {{\iota}{\unicdbdjdcd}{\iota}}$ be given.
Let $ v8\_zf\_lang : {{\iota}{\unicdbdjdcd}{o}}$ be given.
Let $ k10\_zf\_lang : {{\iota}{\unicdbdjdcd}{{\iota}{\unicdbdjdcd}{\iota}}}$ be given.
Let $ v9\_zf\_lang : {{\iota}{\unicdbdjdcd}{o}}$ be given.
Let $ k11\_zf\_lang : {{\iota}{\unicdbdjdcd}{{\iota}{\unicdbdjdcd}{\iota}}}$ be given.
Let $ v10\_zf\_lang : {{\iota}{\unicdbdjdcd}{o}}$ be given.
Let $ k12\_zf\_lang : {{\iota}{\unicdbdjdcd}{{\iota}{\unicdbdjdcd}{\iota}}}$ be given.
Let $ v11\_zf\_lang : {{\iota}{\unicdbdjdcd}{o}}$ be given.
Let $ v1\_xboole\_0 : {{\iota}{\unicdbdjdcd}{o}}$ be given.
Let $ m1\_subset\_1 : {{\iota}{\unicdbdjdcd}{{\iota}{\unicdbdjdcd}{o}}}$ be given.
Let $ k1\_zfmisc\_1 : {{\iota}{\unicdbdjdcd}{\iota}}$ be given.
Let $ m1\_finseq\_1 : {{\iota}{\unicdbdjdcd}{{\iota}{\unicdbdjdcd}{o}}}$ be given.
Let $ k4\_ordinal1 : {\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}
\begin{array}{c}{\unicdcdadad} X0 .  ((v1\_zf\_lang~X0){\unicdcdcdhd}(m2\_finseq\_1~X0~k5\_numbers)){\unicdbdjdcd}((\\
\preoh (v8\_zf\_lang~X0){\unicdcdcdhd}({\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}(X0{\unicdcdgdad}k10\_zf\_lang~X1~X2)))){\unicdcdcdhd}((({\unicdcdaddd} X1 .  ((v1\_zf\_lang~\\
X1){\unicdcdcdhd}(m2\_finseq\_1~X1~k5\_numbers)){\unicdcdcdhd}({\unicdcdaddd} X2 .  ((v1\_zf\_lang~X2){\unicdcdcdhd}\\
(m2\_finseq\_1~X2~k5\_numbers)){\unicdcdcdhd}(X0=k10\_zf\_lang~X1~X2))){\unicdbdjdcd}(v8\_zf\_lang~\\
X0)){\unicdcdcdhd}((\preoh (v9\_zf\_lang~X0){\unicdcdcdhd}({\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}(X0{\unicdcdgdad}k11\_zf\_lang~X1~X2)))){\unicdcdcdhd}((({\unicdcdaddd} X1 .  ((v1\_zf\_lang~\\
X1){\unicdcdcdhd}(m2\_finseq\_1~X1~k5\_numbers)){\unicdcdcdhd}({\unicdcdaddd} X2 .  ((v1\_zf\_lang~X2){\unicdcdcdhd}\\
(m2\_finseq\_1~X2~k5\_numbers)){\unicdcdcdhd}(X0=k11\_zf\_lang~X1~X2))){\unicdbdjdcd}(v9\_zf\_lang~\\
X0)){\unicdcdcdhd}((\preoh (v10\_zf\_lang~X0){\unicdcdcdhd}({\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}(X0{\unicdcdgdad}k12\_zf\_lang~X1~X2)))){\unicdcdcdhd}((({\unicdcdaddd} X1 .  ((v1\_zf\_lang~\\
X1){\unicdcdcdhd}(m2\_finseq\_1~X1~k5\_numbers)){\unicdcdcdhd}({\unicdcdaddd} X2 .  ((v1\_zf\_lang~X2){\unicdcdcdhd}\\
(m2\_finseq\_1~X2~k5\_numbers)){\unicdcdcdhd}(X0=k12\_zf\_lang~X1~X2))){\unicdbdjdcd}(v10\_zf\_lang~\\
X0)){\unicdcdcdhd}((\preoh (v11\_zf\_lang~X0){\unicdcdcdhd}({\unicdcdadad} X1 .  (m2\_subset\_1~X1~k5\_numbers~\\
k1\_zf\_lang){\unicdbdjdcd}({\unicdcdadad} X2 .  ((v1\_zf\_lang~X2){\unicdcdcdhd}(m2\_finseq\_1~X2~k5\_numbers)){\unicdbdjdcd}\\
(X0{\unicdcdgdad}k13\_zf\_lang~X1~X2)))){\unicdcdcdhd}(({\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=k13\_zf\_lang~X1~X2))){\unicdbdjdcd}(v11\_zf\_lang~X0)))))))))\end{array}
\label{hyp:ehgdpfkhggpfmgbgoghg}
\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}((v11\_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}(k13\_zf\_lang~X0~X3{\unicdcdgdad}X1))){\unicdcdcdhd}((({\unicdcdaddd} X3 .  ((v1\_zf\_lang~\\
X3){\unicdcdcdhd}(m2\_finseq\_1~X3~k5\_numbers)){\unicdcdcdhd}(k13\_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}(k13\_zf\_lang~X3~X2{\unicdcdgdad}X1))){\unicdcdcdhd}(({\unicdcdaddd} X3 .  \\
(m2\_subset\_1~X3~k5\_numbers~k1\_zf\_lang){\unicdcdcdhd}(k13\_zf\_lang~X3~X2=X1)){\unicdbdjdcd}\\
(X2=k24\_zf\_lang~X1))))))))\end{array}
\label{hyp:ehedddpfkhggpfmgbgoghg}
\end{equation}
Assume the following.
\begin{equation}
\begin{array}{c}{\unicdcdadad} X0 .  {\unicdcdadad} X1 .  ((\preoh v1\_xboole\_0~X0){\unicdcdcdhd}((\preoh v1\_xboole\_0~X1){\unicdcdcdhd}\\
(m1\_subset\_1~X1~(k1\_zfmisc\_1~X0)))){\unicdbdjdcd}({\unicdcdadad} X2 .  (m2\_subset\_1~\\
X2~X0~X1){\unicdbdjded}(m1\_subset\_1~X2~X1))\end{array}
\label{hyp:chfgegfgggjgogjgehjgpgogpfngcdpfdhfhcgdhfgehpfbd}
\end{equation}
Assume the following.
\begin{equation}
{{\unicdcdadad} X0 .  {{\unicdcdadad} X1 .  {{({{{m2\_finseq\_1}~{X1}}~{X0}})}{\unicdbdjded}{({{{m1\_finseq\_1}~{X1}}~{X0}})}}}}\label{hyp:chfgegfgggjgogjgehjgpgogpfngcdpfggjgogdhfgbhpfbd}
\end{equation}
Assume the following.
\begin{equation}
{{k5\_numbers}={k4\_ordinal1}}\label{hyp:chfgegfgggjgogjgehjgpgogpflgfdpfogfhngcgfgchdh}
\end{equation}
Assume the following.
\begin{equation}
{\preoh{{{v1\_xboole\_0}~{k1\_zf\_lang}}}}\label{hyp:ggdgbdpfkhggpfmgbgoghg}
\end{equation}
Assume the following.
\begin{equation}
{{{m1\_subset\_1}~{k1\_zf\_lang}}~{({{k1\_zfmisc\_1}~{k5\_numbers}})}}\label{hyp:egehpflgbdpfkhggpfmgbgoghg}
\end{equation}
Assume the following.
\begin{equation}
\begin{array}{c}{\unicdcdadad} X0 .  {\unicdcdadad} X1 .  ((m1\_subset\_1~X0~k1\_zf\_lang){\unicdcdcdhd}((v1\_zf\_lang~\\
X1){\unicdcdcdhd}(m1\_finseq\_1~X1~k5\_numbers))){\unicdbdjdcd}((v1\_zf\_lang~(k13\_zf\_lang~\\
X0~X1)){\unicdcdcdhd}(m2\_finseq\_1~(k13\_zf\_lang~X0~X1)~k5\_numbers))\end{array}
\label{hyp:egehpflgbdddpfkhggpfmgbgoghg}
\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}
\begin{array}{c}{\unicdcdadad} X0 .  (v1\_xboole\_0~X0){\unicdbdjdcd}({\unicdcdadad} X1 .  (m1\_subset\_1~X1~(k1\_zfmisc\_1~\\
X0)){\unicdbdjdcd}(v1\_xboole\_0~X1))\end{array}
\label{hyp:dgdgbdpfdhfhcgdhfgehpfbd}
\end{equation}
\begin{theorem}\label{thm:ehjdpfkhggpfmgbgoghgbd}
$$\begin{array}{c}{\unicdcdadad} X0 .  ((v1\_zf\_lang~X0){\unicdcdcdhd}(m2\_finseq\_1~X0~k5\_numbers)){\unicdbdjdcd}({\unicdcdadad} X1 .  \\
(m2\_subset\_1~X1~k5\_numbers~k1\_zf\_lang){\unicdbdjdcd}((k23\_zf\_lang~(k13\_zf\_lang~\\
X1~X0)=X1){\unicdcdcdhd}(k24\_zf\_lang~(k13\_zf\_lang~X1~X0)=X0)))\end{array}$$
\end{theorem}
\end{document}
