\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{t33\_funct\_1 (TMbSAq8LsygQq3DAc4eu3PrtemBfFKQ3WPT)}
\maketitle
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.
Let $ k10\_xtuple\_0 : {{\iota}{\unicdbdjdcd}{\iota}}$ be given.
Let $ k9\_xtuple\_0 : {{\iota}{\unicdbdjdcd}{\iota}}$ be given.
Let $ k2\_funct\_1 : {{\iota}{\unicdbdjdcd}{\iota}}$ be given.
Let $ k2\_relat\_1 : {{\iota}{\unicdbdjdcd}{\iota}}$ be given.
Assume the following.
\begin{equation}
\begin{array}{c}{\unicdcdadad} X0 .  (v1\_relat\_1~X0){\unicdbdjdcd}((k10\_xtuple\_0~X0=k9\_xtuple\_0~(k2\_relat\_1~\\
X0)){\unicdcdcdhd}(k9\_xtuple\_0~X0=k10\_xtuple\_0~(k2\_relat\_1~X0)))\end{array}
\label{hyp:ehcdadpfchfgmgbgehpfbd}
\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}\\
(k2\_funct\_1~X0=k2\_relat\_1~X0))\end{array}
\label{hyp:egfdpfggfhogdgehpfbd}
\end{equation}
\begin{theorem}\label{thm:ehddddpfggfhogdgehpfbd}
$$\begin{array}{c}{\unicdcdadad} X0 .  ((v1\_relat\_1~X0){\unicdcdcdhd}(v1\_funct\_1~X0)){\unicdbdjdcd}((v2\_funct\_1~X0){\unicdbdjdcd}\\
((k10\_xtuple\_0~X0=k9\_xtuple\_0~(k2\_funct\_1~X0)){\unicdcdcdhd}(k9\_xtuple\_0~\\
X0=k10\_xtuple\_0~(k2\_funct\_1~X0))))\end{array}$$
\end{theorem}
\end{document}
