\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{t134\_xcmplx\_1 (TMHefKDHSPobNZBzLvq1uoyYTf42BkAP4gF)}
\maketitle
Let $ v1\_xcmplx\_0 : {{\iota}{\unicdbdjdcd}{o}}$ be given.
Let $ k4\_xcmplx\_0 : {{\iota}{\unicdbdjdcd}{\iota}}$ be given.
Let $ k6\_numbers : {\iota}$ be given.
Assume the following.
\begin{equation}
{{{k4\_xcmplx\_0}~{k6\_numbers}}={k6\_numbers}}\label{hyp:mgcdpfbgchjgehigng}
\end{equation}
Assume the following.
\begin{equation}
{{\unicdcdadad} X0 .  {{({{v1\_xcmplx\_0}~{X0}})}{\unicdbdjdcd}{({{{k4\_xcmplx\_0}~{({{k4\_xcmplx\_0}~{X0}})}}={X0}})}}}\label{hyp:jgogghpgmgfhehjgghfgogfgdhdhpflgedpfihdgngahmgihpfad}
\end{equation}
\begin{theorem}\label{thm:ehbdddedpfihdgngahmgihpfbd}
$$\begin{array}{c}{\unicdcdadad} X0 .  (v1\_xcmplx\_0~X0){\unicdbdjdcd}((k4\_xcmplx\_0~X0=k6\_numbers){\unicdbdjdcd}(\\
X0=k6\_numbers))\end{array}$$
\end{theorem}
\end{document}
