\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{t3\_jordan4 (TMXK5m3C4potX5bsqCcTFr8nGkBzh4FF7j1)}
\maketitle
Let $ v7\_ordinal1 : {{\iota}{\unicdbdjdcd}{o}}$ be given.
Let $ k4\_nat\_d : {{\iota}{\unicdbdjdcd}{{\iota}{\unicdbdjdcd}{\iota}}}$ be given.
Let $ k2\_xcmplx\_0 : {{\iota}{\unicdbdjdcd}{{\iota}{\unicdbdjdcd}{\iota}}}$ be given.
Let $ k6\_numbers : {\iota}$ be given.
Assume the following.
\begin{equation}
{{\unicdcdadad} X0 .  {{({{v7\_ordinal1}~{X0}})}{\unicdbdjdcd}{({{{{k4\_nat\_d}~{X0}}~{X0}}={k6\_numbers}})}}}\label{hyp:ehcdfdpfogbgehpfeg}
\end{equation}
Assume the following.
\begin{equation}
\begin{array}{c}{\unicdcdadad} X0 .  (v7\_ordinal1~X0){\unicdbdjdcd}({\unicdcdadad} X1 .  (v7\_ordinal1~X1){\unicdbdjdcd}({\unicdcdadad} X2 .  \\
(v7\_ordinal1~X2){\unicdbdjdcd}((k4\_nat\_d~X0~X1=k6\_numbers){\unicdbdjdcd}(k4\_nat\_d~(k2\_xcmplx\_0~\\
X0~X2)~X1=k4\_nat\_d~X2~X1))))\end{array}
\label{hyp:ehbdhdpfogbgehpfeg}
\end{equation}
\begin{theorem}\label{thm:ehddpfkgpgchegbgoged}
$$\begin{array}{c}{\unicdcdadad} X0 .  (v7\_ordinal1~X0){\unicdbdjdcd}({\unicdcdadad} X1 .  (v7\_ordinal1~X1){\unicdbdjdcd}(k4\_nat\_d~\\
(k2\_xcmplx\_0~X1~X1)~X1=k6\_numbers))\end{array}$$
\end{theorem}
\end{document}
