\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{t3\_binarith (TMUtyjBE5QuqM5fTDMCJtaCUHs6riWVYK3W)}
\maketitle
Let $ v1\_xboolean : {{\iota}{\unicdbdjdcd}{o}}$ be given.
Let $ k5\_xboolean : {{\iota}{\unicdbdjdcd}{{\iota}{\unicdbdjdcd}{\iota}}}$ be given.
Let $ k7\_margrel1 : {\iota}$ be given.
Let $ k4\_xboolean : {{\iota}{\unicdbdjdcd}{{\iota}{\unicdbdjdcd}{\iota}}}$ be given.
Let $ k3\_xboolean : {{\iota}{\unicdbdjdcd}{\iota}}$ be given.
Let $ k1\_xboolean : {\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~(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}(k4\_xboolean~X0~(k3\_xboolean~X0)=\\
k1\_xboolean)\end{array}
\label{hyp:ehbdddidpfihcgpgpgmgfgbgog}
\end{equation}
Assume the following.
\begin{equation}
{{k7\_margrel1}={k1\_xboolean}}\label{hyp:chfgegfgggjgogjgehjgpgogpflghdpfngbgchhgchfgmgbd}
\end{equation}
Assume the following.
\begin{equation}
{{\unicdcdadad} X0 .  {{({{v1\_xboolean}~{X0}})}{\unicdbdjdcd}{({{v1\_xboolean}~{({{k3\_xboolean}~{X0}})}})}}}\label{hyp:egehpflgddpfihcgpgpgmgfgbgog}
\end{equation}
\begin{theorem}\label{thm:ehddpfcgjgogbgchjgehig}
${{\unicdcdadad} X0 .  {{({{v1\_xboolean}~{X0}})}{\unicdbdjdcd}{({{{{k5\_xboolean}~{X0}}~{k7\_margrel1}}={X0}})}}}$.
\end{theorem}
\end{document}
