\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{t39\_fomodel0 (TMMUe4ysup8CMaV1C857iMGdLVRc3pyV4Nj)}
\maketitle
Let $ v1\_xboolean : {{\iota}{\unicdbdjdcd}{o}}$ be given.
Let $ np\_\_1 : {\iota}$ be given.
Let $ k1\_xboole\_0 : {\iota}$ be given.
Let $ v1\_xboole\_0 : {{\iota}{\unicdbdjdcd}{o}}$ be given.
Let $ k6\_numbers : {\iota}$ be given.
Let $ k1\_xboolean : {\iota}$ be given.
Let $ k2\_xboolean : {\iota}$ be given.
Assume the following.
\begin{equation}
{\preoh{{{v1\_xboole\_0}~{np\_\_1}}}}\label{hyp:dhahdgbdpfcgpgpgmgfg}
\end{equation}
Assume the following.
\begin{equation}
{{k6\_numbers}={k1\_xboole\_0}}\label{hyp:chfgegfgggjgogjgehjgpgogpflggdpfogfhngcgfgchdh}
\end{equation}
Assume the following.
\begin{equation}
{{v1\_xboole\_0}~{k1\_xboole\_0}}\label{hyp:ggdgbdpfihcgpgpgmgfgpfad}
\end{equation}
Assume the following.
\begin{equation}
{{\unicdcdadad} X0 .  {{({{v1\_xboolean}~{X0}})}{\unicdbdjded}{({{({{X0}={k1\_xboolean}})}{\unicdcdcdid}{({{X0}={k2\_xboolean}})}})}}}\label{hyp:egddpfihcgpgpgmgfgbgog}
\end{equation}
Assume the following.
\begin{equation}
{{k2\_xboolean}={np\_\_1}}\label{hyp:egcdpfihcgpgpgmgfgbgog}
\end{equation}
Assume the following.
\begin{equation}
{{k1\_xboolean}={k6\_numbers}}\label{hyp:egbdpfihcgpgpgmgfgbgog}
\end{equation}
\begin{theorem}\label{thm:ehddjdpfggpgngpgegfgmgad}
${{\unicdcdadad} X0 .  {{({{v1\_xboolean}~{X0}})}{\unicdbdjdcd}{({{({{X0}={np\_\_1}})}{\unicdbdjded}{({{X0}{\unicdcdgdad}{k1\_xboole\_0}})}})}}}$.
\end{theorem}
\end{document}
