\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{t57\_bvfunc\_1 (TMcKSjsQzGHYHWtYN62rCQAc3AgBP4CVbRd)}
\maketitle
Let $ v1\_xboolean : {{\iota}{\unicdbdjdcd}{o}}$ be given.
Let $ k7\_xboolean : {{\iota}{\unicdbdjdcd}{{\iota}{\unicdbdjdcd}{\iota}}}$ be given.
Let $ k7\_margrel1 : {\iota}$ be given.
Let $ k3\_xboolean : {{\iota}{\unicdbdjdcd}{\iota}}$ be given.
Let $ k10\_xboolean : {{\iota}{\unicdbdjdcd}{{\iota}{\unicdbdjdcd}{\iota}}}$ be given.
Let $ k1\_xboolean : {\iota}$ be given.
Let $ m1\_subset\_1 : {{\iota}{\unicdbdjdcd}{{\iota}{\unicdbdjdcd}{o}}}$ be given.
Let $ k6\_margrel1 : {\iota}$ be given.
Assume the following.
\begin{equation}
{{\unicdcdadad} X0 .  {{({{v1\_xboolean}~{X0}})}{\unicdbdjdcd}{({{{{k10\_xboolean}~{k7\_margrel1}}~{X0}}={X0}})}}}\label{hyp:ehidpfcgjgogbgchjgehig}
\end{equation}
Assume the following.
\begin{equation}
{{k7\_margrel1}={k1\_xboolean}}\label{hyp:chfgegfgggjgogjgehjgpgogpflghdpfngbgchhgchfgmgbd}
\end{equation}
Assume the following.
\begin{equation}
{{{m1\_subset\_1}~{k7\_margrel1}}~{k6\_margrel1}}\label{hyp:egehpflghdpfngbgchhgchfgmgbd}
\end{equation}
Assume the following.
\begin{equation}
\begin{array}{c}{\unicdcdadad} X0 .  (v1\_xboolean~X0){\unicdbdjdcd}({\unicdcdadad} X1 .  (v1\_xboolean~X1){\unicdbdjdcd}(k7\_xboolean~\\
X0~X1=k3\_xboolean~(k10\_xboolean~X0~X1)))\end{array}
\label{hyp:egbdjdpfcgghggfhogdgpfbd}
\end{equation}
Assume the following.
\begin{equation}
\begin{array}{c}{\unicdcdadad} X0 .  {\unicdcdadad} X1 .  ((v1\_xboolean~X0){\unicdcdcdhd}(v1\_xboolean~X1)){\unicdbdjdcd}(\\
k7\_xboolean~X0~X1=k7\_xboolean~X1~X0)\end{array}
\label{hyp:dgpgngngfhehbgehjgghjgehjhpflghdpfihcgpgpgmgfgbgog}
\end{equation}
Assume the following.
\begin{equation}
{{\unicdcdadad} X0 .  {{({{{m1\_subset\_1}~{X0}}~{k6\_margrel1}})}{\unicdbdjdcd}{({{v1\_xboolean}~{X0}})}}}\label{hyp:dgdgcdpfngbgchhgchfgmgbd}
\end{equation}
\begin{theorem}\label{thm:ehfdhdpfcgghggfhogdgpfbd}
$$\begin{array}{c}{\unicdcdadad} X0 .  (v1\_xboolean~X0){\unicdbdjdcd}(k7\_xboolean~k7\_margrel1~X0=k3\_xboolean~\\
X0)\end{array}$$
\end{theorem}
\end{document}
