\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{l6\_bvfunc14 (TMKvk2QBLe1YKKRLpR42xjo3sW9F79EHXwv)}
\maketitle
Let $ v1\_relat\_1 : {{\iota}{\unicdbdjdcd}{o}}$ be given.
Let $ v1\_funct\_1 : {{\iota}{\unicdbdjdcd}{o}}$ be given.
Let $ k1\_funct\_4 : {{\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 $ k1\_enumset1 : {{\iota}{\unicdbdjdcd}{{\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.
Assume the following.
\begin{equation}
\begin{array}{c}{\unicdcdadad} X0 .  {\unicdcdadad} X1 .  {\unicdcdadad} X2 .  {\unicdcdadad} X3 .  {\unicdcdadad} X4 .  {\unicdcdadad} X5 .  \\
k9\_xtuple\_0~(k1\_funct\_4~(k1\_funct\_4~(k16\_funcop\_1~X0~X3)~(k16\_funcop\_1~\\
X1~X4))~(k16\_funcop\_1~X2~X5))=k1\_enumset1~X0~X1~X2\end{array}
\label{hyp:mgddpfcgghggfhogdgbded}
\end{equation}
Assume the following.
\begin{equation}
\begin{array}{c}{\unicdcdadad} X0 .  ((v1\_relat\_1~X0){\unicdcdcdhd}(v1\_funct\_1~X0)){\unicdbdjdcd}({\unicdcdadad} X1 .  (X1=\\
k10\_xtuple\_0~X0){\unicdbdjded}({\unicdcdadad} X2 .  (X2 \in X1){\unicdbdjded}({\unicdcdaddd} X3 .  (X3 \in k9\_xtuple\_0~\\
X0){\unicdcdcdhd}(X2=k1\_funct\_1~X0~X3))))\end{array}
\label{hyp:egddpfggfhogdgehpfbd}
\end{equation}
Assume the following.
\begin{equation}
\begin{array}{c}{\unicdcdadad} X0 .  {\unicdcdadad} X1 .  {\unicdcdadad} X2 .  {\unicdcdadad} X3 .  (X3=k1\_enumset1~\\
X0~X1~X2){\unicdbdjded}({\unicdcdadad} X4 .  (X4 \in X3){\unicdbdjded}(\preoh (X4{\unicdcdgdad}X0){\unicdcdcdhd}((X4{\unicdcdgdad}X1){\unicdcdcdhd}(X4{\unicdcdgdad}X2))))\end{array}
\label{hyp:egbdpffgogfhngdhfgehbd}
\end{equation}
\begin{theorem}\label{thm:mggdpfcgghggfhogdgbded}
$$\begin{array}{c}{\unicdcdadad} X0 .  {\unicdcdadad} X1 .  {\unicdcdadad} X2 .  {\unicdcdadad} X3 .  {\unicdcdadad} X4 .  {\unicdcdadad} X5 .  \\
{\unicdcdadad} X6 .  ((v1\_relat\_1~X6){\unicdcdcdhd}(v1\_funct\_1~X6)){\unicdbdjdcd}((X6=k1\_funct\_4~\\
(k1\_funct\_4~(k16\_funcop\_1~X0~X3)~(k16\_funcop\_1~X1~X4))~(k16\_funcop\_1~\\
X2~X5)){\unicdbdjdcd}(k10\_xtuple\_0~X6=k1\_enumset1~(k1\_funct\_1~X6~X0)~(k1\_funct\_1~\\
X6~X1)~(k1\_funct\_1~X6~X2)))\end{array}$$
\end{theorem}
\end{document}
