\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}
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\newcommand\unicdcdaddd{\exists}
\newcommand\unicdcdidid{\not\subseteq}
\newcommand\preoh[1]{\lnot #1}
\newcommand\mname[1]{{\mathsf{#1}}}
\begin{document}
\title{t36\_xboolean (TMLHf6164Q1izfB6JxUohF2SE7QhEybLgVT)}
\maketitle
Let $ v1\_xboolean : {{\iota}{\unicdbdjdcd}{o}}$ be given.
Let $ k9\_xboolean : {{\iota}{\unicdbdjdcd}{{\iota}{\unicdbdjdcd}{\iota}}}$ be given.
Let $ k5\_xboolean : {{\iota}{\unicdbdjdcd}{{\iota}{\unicdbdjdcd}{\iota}}}$ be given.
Let $ k3\_xboolean : {{\iota}{\unicdbdjdcd}{\iota}}$ be given.
Assume the following.
\begin{equation}
\begin{array}{c}{\unicdcdadad} X0 .  (v1\_xboolean~X0){\unicdbdjdcd}({\unicdcdadad} X1 .  (v1\_xboolean~X1){\unicdbdjdcd}(k5\_xboolean~\\
X0~(k5\_xboolean~X0~X1)=k5\_xboolean~X0~X1))\end{array}
\label{hyp:ehedpfihcgpgpgmgfgbgog}
\end{equation}
Assume the following.
\begin{equation}
\begin{array}{c}{\unicdcdadad} X0 .  {\unicdcdadad} X1 .  ((v1\_xboolean~X0){\unicdcdcdhd}(v1\_xboolean~X1)){\unicdbdjdcd}(\\
v1\_xboolean~(k5\_xboolean~X0~X1))\end{array}
\label{hyp:ggdgedpfihcgpgpgmgfgbgog}
\end{equation}
Assume the following.
\begin{equation}
\begin{array}{c}{\unicdcdadad} X0 .  (v1\_xboolean~X0){\unicdbdjdcd}({\unicdcdadad} X1 .  (v1\_xboolean~X1){\unicdbdjdcd}(k9\_xboolean~\\
X0~X1=k3\_xboolean~(k5\_xboolean~X0~X1)))\end{array}
\label{hyp:egbdadpfihcgpgpgmgfgbgog}
\end{equation}
\begin{theorem}\label{thm:ehddgdpfihcgpgpgmgfgbgog}
$$\begin{array}{c}{\unicdcdadad} X0 .  (v1\_xboolean~X0){\unicdbdjdcd}({\unicdcdadad} X1 .  (v1\_xboolean~X1){\unicdbdjdcd}(k9\_xboolean~\\
X0~(k5\_xboolean~X0~X1)=k9\_xboolean~X0~X1))\end{array}$$
\end{theorem}
\end{document}
