\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{t37\_funct\_3 (TMTUzF4ttBCyAapmb87qPQiKj3r5dPkSSWx)}
\maketitle
Let $ k6\_subset\_1 : {{\iota}{\unicdbdjdcd}{{\iota}{\unicdbdjdcd}{\iota}}}$ be given.
Let $ k1\_funct\_1 : {{\iota}{\unicdbdjdcd}{{\iota}{\unicdbdjdcd}{\iota}}}$ be given.
Let $ k4\_funct\_3 : {{\iota}{\unicdbdjdcd}{{\iota}{\unicdbdjdcd}{\iota}}}$ be given.
Let $ k1\_xboole\_0 : {\iota}$ be given.
Let $ k4\_xboole\_0 : {{\iota}{\unicdbdjdcd}{{\iota}{\unicdbdjdcd}{\iota}}}$ be given.
Let $ v1\_relat\_1 : {{\iota}{\unicdbdjdcd}{o}}$ be given.
Let $ v1\_funct\_1 : {{\iota}{\unicdbdjdcd}{o}}$ be given.
Let $ k9\_xtuple\_0 : {{\iota}{\unicdbdjdcd}{\iota}}$ be given.
Let $ np\_\_1 : {\iota}$ be given.
Assume the following.
\begin{equation}
{{\unicdcdadad} X0 .  {{\unicdcdadad} X1 .  {{{{k6\_subset\_1}~{X0}}~{X1}}={{{k4\_xboole\_0}~{X0}}~{X1}}}}}\label{hyp:chfgegfgggjgogjgehjgpgogpflggdpfdhfhcgdhfgehpfbd}
\end{equation}
Assume the following.
\begin{equation}
\begin{array}{c}{\unicdcdadad} X0 .  {\unicdcdadad} X1 .  (v1\_relat\_1~(k4\_funct\_3~X0~X1)){\unicdcdcdhd}(v1\_funct\_1~\\
(k4\_funct\_3~X0~X1))\end{array}
\label{hyp:egehpflgedpfggfhogdgehpfdd}
\end{equation}
Assume the following.
\begin{equation}
\begin{array}{c}{\unicdcdadad} X0 .  {\unicdcdadad} X1 .  {\unicdcdadad} X2 .  (X2=k4\_xboole\_0~X0~X1){\unicdbdjded}({\unicdcdadad} X3 .  \\
(X3 \in X2){\unicdbdjded}((X3 \in X0){\unicdcdcdhd}(\preoh X3 \in X1)))\end{array}
\label{hyp:egfdpfihcgpgpgmgfgpfad}
\end{equation}
Assume the following.
\begin{equation}
\begin{array}{c}{\unicdcdadad} X0 .  {\unicdcdadad} X1 .  {\unicdcdadad} X2 .  ((v1\_relat\_1~X2){\unicdcdcdhd}(v1\_funct\_1~\\
X2)){\unicdbdjdcd}((X2=k4\_funct\_3~X0~X1){\unicdbdjded}((k9\_xtuple\_0~X2=X1){\unicdcdcdhd}({\unicdcdadad} X3 .  \\
(X3 \in X1){\unicdbdjdcd}(((X3 \in X0){\unicdbdjdcd}(k1\_funct\_1~X2~X3=np\_\_1)){\unicdcdcdhd}((\preoh X3 \in X0){\unicdbdjdcd}(k1\_funct\_1~\\
X2~X3=k1\_xboole\_0))))))\end{array}
\label{hyp:egddpfggfhogdgehpfdd}
\end{equation}
\begin{theorem}\label{thm:ehddhdpfggfhogdgehpfdd}
$$\begin{array}{c}{\unicdcdadad} X0 .  {\unicdcdadad} X1 .  {\unicdcdadad} X2 .  (X0 \in k6\_subset\_1~X1~X2){\unicdbdjdcd}(k1\_funct\_1~\\
(k4\_funct\_3~X2~X1)~X0=k1\_xboole\_0)\end{array}$$
\end{theorem}
\end{document}
