\documentclass{article}
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\newtheorem{lemma}{Lemma}
\newtheorem{proposition}{Proposition}
\newtheorem{definition}{Definition}
\newtheorem{example}{Example}
\newtheorem{conjecture}{Conjecture}
\newcommand\unicdcdgdad{\not=}
\newcommand\unicdcdcdbg{\cup}
\newcommand\unicdcdbdgd{\setminus}
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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{t38\_waybel26 (TMGWaC3g499kQ9EhVbo88kErKqbRWmbi9KU)}
\maketitle
Let $ v1\_relat\_1 : {{\iota}{\unicdbdjdcd}{o}}$ be given.
Let $ v1\_funct\_1 : {{\iota}{\unicdbdjdcd}{o}}$ be given.
Let $ k4\_tarski : {{\iota}{\unicdbdjdcd}{{\iota}{\unicdbdjdcd}{\iota}}}$ be given.
Let $ k5\_waybel26 : {{\iota}{\unicdbdjdcd}{\iota}}$ be given.
Let $ k9\_xtuple\_0 : {{\iota}{\unicdbdjdcd}{\iota}}$ be given.
Let $ k1\_funct\_1 : {{\iota}{\unicdbdjdcd}{{\iota}{\unicdbdjdcd}{\iota}}}$ be given.
Let $ k3\_card\_3 : {{\iota}{\unicdbdjdcd}{\iota}}$ be given.
Let $ k2\_card\_3 : {{\iota}{\unicdbdjdcd}{\iota}}$ be given.
Let $ k2\_xtuple\_0 : {{\iota}{\unicdbdjdcd}{\iota}}$ be given.
Let $ k1\_xtuple\_0 : {{\iota}{\unicdbdjdcd}{\iota}}$ be given.
Let $ k2\_relat\_1 : {{\iota}{\unicdbdjdcd}{\iota}}$ be given.
Let $ k2\_tarski : {{\iota}{\unicdbdjdcd}{{\iota}{\unicdbdjdcd}{\iota}}}$ be given.
Let $ k1\_tarski : {{\iota}{\unicdbdjdcd}{\iota}}$ be given.
Assume the following.
\begin{equation}
\begin{array}{c}{\unicdcdadad} X0 .  {\unicdcdadad} X1 .  ((v1\_relat\_1~X1){\unicdcdcdhd}(v1\_funct\_1~X1)){\unicdbdjdcd}((X0 \in \\
k3\_card\_3~(k2\_card\_3~X1)){\unicdbdjded}((k2\_xtuple\_0~X0 \in k9\_xtuple\_0~X1){\unicdcdcdhd}\\
((k1\_xtuple\_0~X0 \in k1\_funct\_1~X1~(k2\_xtuple\_0~X0)){\unicdcdcdhd}(X0=k4\_tarski~\\
(k1\_xtuple\_0~X0)~(k2\_xtuple\_0~X0)))))\end{array}
\label{hyp:ehcdcdpfdgbgchegpfdd}
\end{equation}
Assume the following.
\begin{equation}
{{\unicdcdadad} X0 .  {{\unicdcdadad} X1 .  {{{k2\_xtuple\_0}~{({{{k4\_tarski}~{X0}}~{X1}})}}={X1}}}}\label{hyp:chegcdpfihehfhahmgfgpfad}
\end{equation}
Assume the following.
\begin{equation}
{{\unicdcdadad} X0 .  {{\unicdcdadad} X1 .  {{{k1\_xtuple\_0}~{({{{k4\_tarski}~{X0}}~{X1}})}}={X0}}}}\label{hyp:chegbdpfihehfhahmgfgpfad}
\end{equation}
Assume the following.
\begin{equation}
\begin{array}{c}{\unicdcdadad} X0 .  ((v1\_relat\_1~X0){\unicdcdcdhd}(v1\_funct\_1~X0)){\unicdbdjdcd}(v1\_relat\_1~(k3\_card\_3~\\
(k2\_card\_3~X0)))\end{array}
\label{hyp:ggdgfdpfhhbgjhcgfgmgcdgd}
\end{equation}
Assume the following.
\begin{equation}
\begin{array}{c}{\unicdcdadad} X0 .  ((v1\_relat\_1~X0){\unicdcdcdhd}(v1\_funct\_1~X0)){\unicdbdjdcd}(v1\_relat\_1~(k5\_waybel26~\\
X0))\end{array}
\label{hyp:egehpflgfdpfhhbgjhcgfgmgcdgd}
\end{equation}
Assume the following.
\begin{equation}
\begin{array}{c}{\unicdcdadad} X0 .  (v1\_relat\_1~X0){\unicdbdjdcd}({\unicdcdadad} X1 .  (v1\_relat\_1~X1){\unicdbdjdcd}((X1=\\
k2\_relat\_1~X0){\unicdbdjded}({\unicdcdadad} X2 .  {\unicdcdadad} X3 .  (k4\_tarski~X2~X3 \in X1){\unicdbdjded}(\\
k4\_tarski~X3~X2 \in X0))))\end{array}
\label{hyp:eghdpfchfgmgbgehpfbd}
\end{equation}
Assume the following.
\begin{equation}
\begin{array}{c}{\unicdcdadad} X0 .  {\unicdcdadad} X1 .  k4\_tarski~X0~X1=k2\_tarski~(k2\_tarski~X0~\\
X1)~(k1\_tarski~X0)\end{array}
\label{hyp:egfdpfehbgchdhlgjg}
\end{equation}
Assume the following.
\begin{equation}
\begin{array}{c}{\unicdcdadad} X0 .  ((v1\_relat\_1~X0){\unicdcdcdhd}(v1\_funct\_1~X0)){\unicdbdjdcd}(k5\_waybel26~X0=\\
k2\_relat\_1~(k3\_card\_3~(k2\_card\_3~X0)))\end{array}
\label{hyp:egedpfhhbgjhcgfgmgcdgd}
\end{equation}
Assume the following.
\begin{equation}
{{\unicdcdadad} X0 .  {{\unicdcdadad} X1 .  {{{{k2\_tarski}~{X0}}~{X1}}={{{k2\_tarski}~{X1}}~{X0}}}}}\label{hyp:dgpgngngfhehbgehjgghjgehjhpflgcdpfehbgchdhlgjg}
\end{equation}
\begin{theorem}\label{thm:ehddidpfhhbgjhcgfgmgcdgd}
$$\begin{array}{c}{\unicdcdadad} X0 .  {\unicdcdadad} X1 .  {\unicdcdadad} X2 .  ((v1\_relat\_1~X2){\unicdcdcdhd}(v1\_funct\_1~\\
X2)){\unicdbdjdcd}((k4\_tarski~X0~X1 \in k5\_waybel26~X2){\unicdbdjded}((X0 \in k9\_xtuple\_0~X2){\unicdcdcdhd}\\
(X1 \in k1\_funct\_1~X2~X0)))\end{array}$$
\end{theorem}
\end{document}
