\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=}
\newcommand\unicdcdcdbg{\cup}
\newcommand\unicdcdbdgd{\setminus}
\newcommand\unicdcdcdjd{\cap}
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\newcommand\unicdbdjdcd{\Rightarrow}
\newcommand\unicdcdadjd{\notin}
\newcommand\unicdbdjded{\Leftrightarrow}
\newcommand\unicdcdcdid{\lor}
\newcommand\unicdcdcdhd{\land}
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\newcommand\unicdcdaddd{\exists}
\newcommand\unicdcdidid{\not\subseteq}
\newcommand\preoh[1]{\lnot #1}
\newcommand\mname[1]{{\mathsf{#1}}}
\begin{document}
\title{t50\_nat\_4 (TMKPG9RHZyoy8NT5tzU1xBCvadx7ZwVEGD1)}
\maketitle
Let $ v7\_ordinal1 : {{\iota}{\unicdbdjdcd}{o}}$ be given.
Let $ v1\_int\_2 : {{\iota}{\unicdbdjdcd}{o}}$ be given.
Let $ m1\_trees\_4 : {{\iota}{\unicdbdjdcd}{{\iota}{\unicdbdjdcd}{{\iota}{\unicdbdjdcd}{o}}}}$ be given.
Let $ k1\_numbers : {\iota}$ be given.
Let $ k5\_numbers : {\iota}$ be given.
Let $ k1\_xboole\_0 : {\iota}$ be given.
Let $ k1\_nat\_4 : {{\iota}{\unicdbdjdcd}{{\iota}{\unicdbdjdcd}{\iota}}}$ be given.
Let $ v1\_xboole\_0 : {{\iota}{\unicdbdjdcd}{o}}$ be given.
Let $ v1\_relat\_1 : {{\iota}{\unicdbdjdcd}{o}}$ be given.
Let $ v1\_funct\_1 : {{\iota}{\unicdbdjdcd}{o}}$ be given.
Let $ v1\_finseq\_1 : {{\iota}{\unicdbdjdcd}{o}}$ be given.
Let $ k4\_finseq\_1 : {{\iota}{\unicdbdjdcd}{\iota}}$ be given.
Let $ k9\_xtuple\_0 : {{\iota}{\unicdbdjdcd}{\iota}}$ be given.
Let $ m1\_finseq\_1 : {{\iota}{\unicdbdjdcd}{{\iota}{\unicdbdjdcd}{o}}}$ be given.
Let $ v4\_relat\_1 : {{\iota}{\unicdbdjdcd}{{\iota}{\unicdbdjdcd}{o}}}$ be given.
Let $ v5\_relat\_1 : {{\iota}{\unicdbdjdcd}{{\iota}{\unicdbdjdcd}{o}}}$ be given.
Let $ v1\_finset\_1 : {{\iota}{\unicdbdjdcd}{o}}$ be given.
Let $ k3\_finseq\_1 : {{\iota}{\unicdbdjdcd}{\iota}}$ be given.
Let $ k1\_recdef\_1 : {{\iota}{\unicdbdjdcd}{{\iota}{\unicdbdjdcd}{\iota}}}$ be given.
Let $ k11\_nat\_3 : {{\iota}{\unicdbdjdcd}{{\iota}{\unicdbdjdcd}{\iota}}}$ be given.
Assume the following.
\begin{equation}
{{\unicdcdadad} X0 .  {{\unicdcdadad} X1 .  {{\preoh{{({{X0} \in {X1}})}}}{\unicdcdcdhd}{({{v1\_xboole\_0}~{X1}})}}}}\label{hyp:ehhdpfcgpgpgmgfg}
\end{equation}
Assume the following.
\begin{equation}
{{\unicdcdadad} X0 .  {{({{v1\_xboole\_0}~{X0}})}{\unicdbdjdcd}{({{X0}={k1\_xboole\_0}})}}}\label{hyp:ehgdpfcgpgpgmgfg}
\end{equation}
Assume the following.
\begin{equation}
\begin{array}{c}{\unicdcdadad} X0 .  ((v1\_relat\_1~X0){\unicdcdcdhd}((v1\_funct\_1~X0){\unicdcdcdhd}(v1\_finseq\_1~X0))){\unicdbdjdcd}\\
(k4\_finseq\_1~X0=k9\_xtuple\_0~X0)\end{array}
\label{hyp:chfgegfgggjgogjgehjgpgogpflgedpfggjgogdhfgbhpfbd}
\end{equation}
Assume the following.
\begin{equation}
\begin{array}{c}{\unicdcdadad} X0 .  {\unicdcdaddd} X1 .  (m1\_finseq\_1~X1~X0){\unicdcdcdhd}((v1\_relat\_1~X1){\unicdcdcdhd}(\\
(v4\_relat\_1~X1~k5\_numbers){\unicdcdcdhd}((v5\_relat\_1~X1~X0){\unicdcdcdhd}((v1\_funct\_1~\\
X1){\unicdcdcdhd}((v1\_xboole\_0~X1){\unicdcdcdhd}((v1\_finset\_1~X1){\unicdcdcdhd}(v1\_finseq\_1~X1)))))))\end{array}
\label{hyp:chdgcdpfggjgogdhfgbhpfbd}
\end{equation}
Assume the following.
\begin{equation}
{{\unicdcdadad} X0 .  {{({{v1\_xboole\_0}~{X0}})}{\unicdbdjdcd}{({{v1\_xboole\_0}~{({{k9\_xtuple\_0}~{X0}})}})}}}\label{hyp:ggdgbdadpfchfgmgbgehpfbd}
\end{equation}
Assume the following.
\begin{equation}
\begin{array}{c}{\unicdcdadad} X0 .  (m1\_trees\_4~X0~k1\_numbers~k5\_numbers){\unicdbdjdcd}({\unicdcdadad} X1 .  \\
((v7\_ordinal1~X1){\unicdcdcdhd}(v1\_int\_2~X1)){\unicdbdjdcd}({\unicdcdadad} X2 .  (m1\_trees\_4~X2~\\
k1\_numbers~k5\_numbers){\unicdbdjdcd}((X2=k1\_nat\_4~X0~X1){\unicdbdjded}((k3\_finseq\_1~X2=\\
k3\_finseq\_1~X0){\unicdcdcdhd}({\unicdcdadad} X3 .  (X3 \in k4\_finseq\_1~X2){\unicdbdjdcd}(k1\_recdef\_1~\\
X2~X3=k11\_nat\_3~(k1\_recdef\_1~X0~X3)~X1))))))\end{array}
\label{hyp:egbdpfogbgehpfed}
\end{equation}
\begin{theorem}\label{thm:ehfdadpfogbgehpfed}
$$\begin{array}{c}{\unicdcdadad} X0 .  ((v7\_ordinal1~X0){\unicdcdcdhd}(v1\_int\_2~X0)){\unicdbdjdcd}({\unicdcdadad} X1 .  (m1\_trees\_4~\\
X1~k1\_numbers~k5\_numbers){\unicdbdjdcd}((X1=k1\_xboole\_0){\unicdbdjdcd}(k1\_nat\_4~X1~X0=\\
k1\_xboole\_0)))\end{array}$$
\end{theorem}
\end{document}
