\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{t12\_xboolean (TMUPK2AKuqriK56XVJsq6EHeg3chaL2NHvz)}
\maketitle
Let $ v1\_xboolean : {{\iota}{\unicdbdjdcd}{o}}$ be given.
Let $ k5\_xboolean : {{\iota}{\unicdbdjdcd}{{\iota}{\unicdbdjdcd}{\iota}}}$ be given.
Let $ k4\_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}(k4\_xboolean~\\
X0~(k5\_xboolean~X0~X1)=X0))\end{array}
\label{hyp:ehgdpfihcgpgpgmgfgbgog}
\end{equation}
Assume the following.
\begin{equation}
\begin{array}{c}{\unicdcdadad} X0 .  (v1\_xboolean~X0){\unicdbdjdcd}({\unicdcdadad} X1 .  (v1\_xboolean~X1){\unicdbdjdcd}(k5\_xboolean~\\
X0~(k4\_xboolean~X0~X1)=X0))\end{array}
\label{hyp:ehfdpfihcgpgpgmgfgbgog}
\end{equation}
Assume the following.
\begin{equation}
\begin{array}{c}{\unicdcdadad} X0 .  (v1\_xboolean~X0){\unicdbdjdcd}({\unicdcdadad} X1 .  (v1\_xboolean~X1){\unicdbdjdcd}(k4\_xboolean~\\
X0~(k5\_xboolean~(k3\_xboolean~X0)~X1)=k4\_xboolean~X0~X1))\end{array}
\label{hyp:ehbdbdpfihcgpgpgmgfgbgog}
\end{equation}
Assume the following.
\begin{equation}
{{\unicdcdadad} X0 .  {{({{v1\_xboolean}~{X0}})}{\unicdbdjdcd}{({{{k3\_xboolean}~{({{k3\_xboolean}~{X0}})}}={X0}})}}}\label{hyp:jgogghpgmgfhehjgghfgogfgdhdhpflgddpfihcgpgpgmgfgbgog}
\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 .  {\unicdcdadad} X1 .  ((v1\_xboolean~X0){\unicdcdcdhd}(v1\_xboolean~X1)){\unicdbdjdcd}(\\
v1\_xboolean~(k4\_xboolean~X0~X1))\end{array}
\label{hyp:ggdgddpfihcgpgpgmgfgbgog}
\end{equation}
Assume the following.
\begin{equation}
{{\unicdcdadad} X0 .  {{({{v1\_xboolean}~{X0}})}{\unicdbdjdcd}{({{v1\_xboolean}~{({{k3\_xboolean}~{X0}})}})}}}\label{hyp:egehpflgddpfihcgpgpgmgfgbgog}
\end{equation}
Assume the following.
\begin{equation}
\begin{array}{c}{\unicdcdadad} X0 .  (v1\_xboolean~X0){\unicdbdjdcd}({\unicdcdadad} X1 .  (v1\_xboolean~X1){\unicdbdjdcd}(k5\_xboolean~\\
X0~X1=k3\_xboolean~(k4\_xboolean~(k3\_xboolean~X0)~(k3\_xboolean~\\
X1))))\end{array}
\label{hyp:eggdpfihcgpgpgmgfgbgog}
\end{equation}
Assume the following.
\begin{equation}
\begin{array}{c}{\unicdcdadad} X0 .  {\unicdcdadad} X1 .  ((v1\_xboolean~X0){\unicdcdcdhd}(v1\_xboolean~X1)){\unicdbdjdcd}(\\
k5\_xboolean~X0~X1=k5\_xboolean~X1~X0)\end{array}
\label{hyp:dgpgngngfhehbgehjgghjgehjhpflgfdpfihcgpgpgmgfgbgog}
\end{equation}
Assume the following.
\begin{equation}
\begin{array}{c}{\unicdcdadad} X0 .  {\unicdcdadad} X1 .  ((v1\_xboolean~X0){\unicdcdcdhd}(v1\_xboolean~X1)){\unicdbdjdcd}(\\
k4\_xboolean~X0~X1=k4\_xboolean~X1~X0)\end{array}
\label{hyp:dgpgngngfhehbgehjgghjgehjhpflgedpfihcgpgpgmgfgbgog}
\end{equation}
\begin{theorem}\label{thm:ehbdcdpfihcgpgpgmgfgbgog}
$$\begin{array}{c}{\unicdcdadad} X0 .  (v1\_xboolean~X0){\unicdbdjdcd}({\unicdcdadad} X1 .  (v1\_xboolean~X1){\unicdbdjdcd}(k5\_xboolean~\\
X0~(k4\_xboolean~(k3\_xboolean~X0)~X1)=k5\_xboolean~X0~X1))\end{array}$$
\end{theorem}
\end{document}
