\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{l4\_bvfunc14 (TMdgoBgzBTHsx7UAaLvvkm4E3bFuXhf1C7J)}
\maketitle
Let $ v1\_relat\_1 : {{\iota}{\unicdbdjdcd}{o}}$ be given.
Let $ v1\_funct\_1 : {{\iota}{\unicdbdjdcd}{o}}$ be given.
Let $ k1\_funct\_1 : {{\iota}{\unicdbdjdcd}{{\iota}{\unicdbdjdcd}{\iota}}}$ be given.
Let $ k1\_funct\_4 : {{\iota}{\unicdbdjdcd}{{\iota}{\unicdbdjdcd}{\iota}}}$ be given.
Let $ k16\_funcop\_1 : {{\iota}{\unicdbdjdcd}{{\iota}{\unicdbdjdcd}{\iota}}}$ be given.
Assume the following.
\begin{equation}
\begin{array}{c}{\unicdcdadad} X0 .  ((v1\_relat\_1~X0){\unicdcdcdhd}(v1\_funct\_1~X0)){\unicdbdjdcd}({\unicdcdadad} X1 .  {\unicdcdadad} X2 .  \\
k1\_funct\_1~(k1\_funct\_4~X0~(k16\_funcop\_1~X1~X2))~X1=X2)\end{array}
\label{hyp:ehjdedpfggfhogdgehpfhd}
\end{equation}
Assume the following.
\begin{equation}
\begin{array}{c}{\unicdcdadad} X0 .  ((v1\_relat\_1~X0){\unicdcdcdhd}(v1\_funct\_1~X0)){\unicdbdjdcd}({\unicdcdadad} X1 .  {\unicdcdadad} X2 .  \\
{\unicdcdadad} X3 .  (X1{\unicdcdgdad}X3){\unicdbdjdcd}(k1\_funct\_1~(k1\_funct\_4~X0~(k16\_funcop\_1~\\
X1~X2))~X3=k1\_funct\_1~X0~X3))\end{array}
\label{hyp:ehidddpfggfhogdgehpfed}
\end{equation}
Assume the following.
\begin{equation}
\begin{array}{c}{\unicdcdadad} X0 .  {\unicdcdadad} X1 .  (v1\_relat\_1~(k16\_funcop\_1~X0~X1)){\unicdcdcdhd}(v1\_funct\_1~\\
(k16\_funcop\_1~X0~X1))\end{array}
\label{hyp:ggdgbdbdpfggfhogdgpgahpfbd}
\end{equation}
Assume the following.
\begin{equation}
\begin{array}{c}{\unicdcdadad} X0 .  {\unicdcdadad} X1 .  (((v1\_relat\_1~X0){\unicdcdcdhd}(v1\_funct\_1~X0)){\unicdcdcdhd}((\\
v1\_relat\_1~X1){\unicdcdcdhd}(v1\_funct\_1~X1))){\unicdbdjdcd}((v1\_relat\_1~(k1\_funct\_4~X0~\\
X1)){\unicdcdcdhd}(v1\_funct\_1~(k1\_funct\_4~X0~X1)))\end{array}
\label{hyp:egehpflgbdpfggfhogdgehpfed}
\end{equation}
\begin{theorem}\label{thm:mgedpfcgghggfhogdgbded}
$$\begin{array}{c}{\unicdcdadad} X0 .  ((v1\_relat\_1~X0){\unicdcdcdhd}(v1\_funct\_1~X0)){\unicdbdjdcd}({\unicdcdadad} X1 .  {\unicdcdadad} X2 .  \\
{\unicdcdadad} X3 .  {\unicdcdadad} X4 .  (X1{\unicdcdgdad}X2){\unicdbdjdcd}(k1\_funct\_1~(k1\_funct\_4~(k1\_funct\_4~\\
X0~(k16\_funcop\_1~X1~X3))~(k16\_funcop\_1~X2~X4))~X1=X3))\end{array}$$
\end{theorem}
\end{document}
