\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}
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\newcommand\unicdcdadjd{\notin}
\newcommand\unicdbdjded{\Leftrightarrow}
\newcommand\unicdcdcdid{\lor}
\newcommand\unicdcdcdhd{\land}
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\newcommand\unicdcdidid{\not\subseteq}
\newcommand\preoh[1]{\lnot #1}
\newcommand\mname[1]{{\mathsf{#1}}}
\begin{document}
\title{t10\_necklace (TMGJy8WFjoofCV6WB1znxu9cLMBqNghqhDJ)}
\maketitle
Let $ k2\_funct\_1 : {{\iota}{\unicdbdjdcd}{\iota}}$ be given.
Let $ k4\_funct\_4 : {{\iota}{\unicdbdjdcd}{{\iota}{\unicdbdjdcd}{{\iota}{\unicdbdjdcd}{{\iota}{\unicdbdjdcd}{\iota}}}}}$ be given.
Let $ k16\_funcop\_1 : {{\iota}{\unicdbdjdcd}{{\iota}{\unicdbdjdcd}{\iota}}}$ be given.
Let $ k10\_xtuple\_0 : {{\iota}{\unicdbdjdcd}{\iota}}$ be given.
Let $ k2\_tarski : {{\iota}{\unicdbdjdcd}{{\iota}{\unicdbdjdcd}{\iota}}}$ be given.
Let $ k1\_funct\_1 : {{\iota}{\unicdbdjdcd}{{\iota}{\unicdbdjdcd}{\iota}}}$ be given.
Let $ k9\_xtuple\_0 : {{\iota}{\unicdbdjdcd}{\iota}}$ be given.
Let $ r1\_tarski : {{\iota}{\unicdbdjdcd}{{\iota}{\unicdbdjdcd}{o}}}$ be given.
Let $ v1\_relat\_1 : {{\iota}{\unicdbdjdcd}{o}}$ be given.
Let $ v1\_funct\_1 : {{\iota}{\unicdbdjdcd}{o}}$ be given.
Let $ v2\_funct\_1 : {{\iota}{\unicdbdjdcd}{o}}$ be given.
Assume the following.
\begin{equation}
\begin{array}{c}{\unicdcdadad} X0 .  {\unicdcdadad} X1 .  k2\_funct\_1~(k16\_funcop\_1~X0~X1)=k16\_funcop\_1~\\
X1~X0\end{array}
\label{hyp:ehjdpfogfgdglgmgbgdgfg}
\end{equation}
Assume the following.
\begin{equation}
\begin{array}{c}{\unicdcdadad} X0 .  {\unicdcdadad} X1 .  {\unicdcdadad} X2 .  k4\_funct\_4~X0~X0~X1~X2=k16\_funcop\_1~\\
X0~X2\end{array}
\label{hyp:ehidbdpfggfhogdgehpfed}
\end{equation}
Assume the following.
\begin{equation}
\begin{array}{c}{\unicdcdadad} X0 .  {\unicdcdadad} X1 .  {\unicdcdadad} X2 .  {\unicdcdadad} X3 .  (X0{\unicdcdgdad}X1){\unicdbdjdcd}(k10\_xtuple\_0~\\
(k4\_funct\_4~X0~X1~X2~X3)=k2\_tarski~X2~X3)\end{array}
\label{hyp:ehgdedpfggfhogdgehpfed}
\end{equation}
Assume the following.
\begin{equation}
\begin{array}{c}{\unicdcdadad} X0 .  {\unicdcdadad} X1 .  {\unicdcdadad} X2 .  {\unicdcdadad} X3 .  ((X0{\unicdcdgdad}X1){\unicdbdjdcd}(k1\_funct\_1~\\
(k4\_funct\_4~X0~X1~X2~X3)~X0=X2)){\unicdcdcdhd}(k1\_funct\_1~(k4\_funct\_4~X0~X1~X2~\\
X3)~X1=X3)\end{array}
\label{hyp:ehgdddpfggfhogdgehpfed}
\end{equation}
Assume the following.
\begin{equation}
\begin{array}{c}{\unicdcdadad} X0 .  {\unicdcdadad} X1 .  {\unicdcdadad} X2 .  {\unicdcdadad} X3 .  (k9\_xtuple\_0~(k4\_funct\_4~\\
X0~X1~X2~X3)=k2\_tarski~X0~X1){\unicdcdcdhd}(r1\_tarski~(k10\_xtuple\_0~(k4\_funct\_4~\\
X0~X1~X2~X3))~(k2\_tarski~X2~X3))\end{array}
\label{hyp:ehgdcdpfggfhogdgehpfed}
\end{equation}
Assume the following.
\begin{equation}
\begin{array}{c}{\unicdcdadad} X0 .  ((v1\_relat\_1~X0){\unicdcdcdhd}(v1\_funct\_1~X0)){\unicdbdjdcd}((v2\_funct\_1~X0){\unicdbdjdcd}\\
({\unicdcdadad} X1 .  ((v1\_relat\_1~X1){\unicdcdcdhd}(v1\_funct\_1~X1)){\unicdbdjdcd}((X1=k2\_funct\_1~\\
X0){\unicdbdjded}((k9\_xtuple\_0~X1=k10\_xtuple\_0~X0){\unicdcdcdhd}({\unicdcdadad} X2 .  {\unicdcdadad} X3 .  \\
(((X2 \in k10\_xtuple\_0~X0){\unicdcdcdhd}(X3=k1\_funct\_1~X1~X2)){\unicdbdjdcd}((X3 \in k9\_xtuple\_0~\\
X0){\unicdcdcdhd}(X2=k1\_funct\_1~X0~X3))){\unicdcdcdhd}(((X3 \in k9\_xtuple\_0~X0){\unicdcdcdhd}(X2=k1\_funct\_1~\\
X0~X3)){\unicdbdjdcd}((X2 \in k10\_xtuple\_0~X0){\unicdcdcdhd}(X3=k1\_funct\_1~X1~X2))))))))\end{array}
\label{hyp:ehddcdpfggfhogdgehpfbd}
\end{equation}
Assume the following.
\begin{equation}
\begin{array}{c}{\unicdcdadad} X0 .  {\unicdcdadad} X1 .  {\unicdcdadad} X2 .  {\unicdcdadad} X3 .  (v1\_relat\_1~(k4\_funct\_4~\\
X0~X1~X2~X3)){\unicdcdcdhd}(v1\_funct\_1~(k4\_funct\_4~X0~X1~X2~X3))\end{array}
\label{hyp:ggdgbdpfggfhogdgehpfed}
\end{equation}
Assume the following.
\begin{equation}
\begin{array}{c}{\unicdcdadad} X0 .  ((v1\_relat\_1~X0){\unicdcdcdhd}(v1\_funct\_1~X0)){\unicdbdjdcd}((v2\_funct\_1~X0){\unicdbdjded}\\
({\unicdcdadad} X1 .  {\unicdcdadad} X2 .  ((X1 \in k9\_xtuple\_0~X0){\unicdcdcdhd}((X2 \in k9\_xtuple\_0~\\
X0){\unicdcdcdhd}(k1\_funct\_1~X0~X1=k1\_funct\_1~X0~X2))){\unicdbdjdcd}(X1=X2)))\end{array}
\label{hyp:egedpfggfhogdgehpfbd}
\end{equation}
Assume the following.
\begin{equation}
\begin{array}{c}{\unicdcdadad} X0 .  {\unicdcdadad} X1 .  {\unicdcdadad} X2 .  (X2=k2\_tarski~X0~X1){\unicdbdjded}({\unicdcdadad} X3 .  \\
(X3 \in X2){\unicdbdjded}((X3=X0){\unicdcdcdid}(X3=X1)))\end{array}
\label{hyp:egcdpfehbgchdhlgjg}
\end{equation}
\begin{theorem}\label{thm:ehbdadpfogfgdglgmgbgdgfg}
$$\begin{array}{c}{\unicdcdadad} X0 .  {\unicdcdadad} X1 .  {\unicdcdadad} X2 .  {\unicdcdadad} X3 .  \preoh ((X0=X1){\unicdbdjdcd}(X2=X3)){\unicdcdcdhd}\\
(((X2=X3){\unicdbdjdcd}(X0=X1)){\unicdcdcdhd}(k2\_funct\_1~(k4\_funct\_4~X0~X1~X2~X3){\unicdcdgdad}k4\_funct\_4~\\
X2~X3~X0~X1))\end{array}$$
\end{theorem}
\end{document}
