\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{t23\_int\_4 (TMWtZGvQMPX14zGsh47GGZ4zxNu25bBBGes)}
\maketitle
Let $ v7\_ordinal1 : {{\iota}{\unicdbdjdcd}{o}}$ be given.
Let $ r1\_xxreal\_0 : {{\iota}{\unicdbdjdcd}{{\iota}{\unicdbdjdcd}{o}}}$ be given.
Let $ k6\_numbers : {\iota}$ be given.
Let $ k4\_nat\_d : {{\iota}{\unicdbdjdcd}{{\iota}{\unicdbdjdcd}{\iota}}}$ be given.
Let $ r1\_int\_1 : {{\iota}{\unicdbdjdcd}{{\iota}{\unicdbdjdcd}{o}}}$ be given.
Let $ k6\_xcmplx\_0 : {{\iota}{\unicdbdjdcd}{{\iota}{\unicdbdjdcd}{\iota}}}$ be given.
Let $ v1\_int\_1 : {{\iota}{\unicdbdjdcd}{o}}$ be given.
Let $ k6\_int\_1 : {{\iota}{\unicdbdjdcd}{{\iota}{\unicdbdjdcd}{\iota}}}$ be given.
Assume the following.
\begin{equation}
\begin{array}{c}{\unicdcdadad} X0 .  (v7\_ordinal1~X0){\unicdbdjdcd}((\preoh r1\_xxreal\_0~X0~k6\_numbers){\unicdbdjdcd}(\\
{\unicdcdadad} X1 .  (v1\_int\_1~X1){\unicdbdjdcd}((k6\_int\_1~X1~X0=k6\_numbers){\unicdbdjded}(r1\_int\_1~\\
X0~X1))))\end{array}
\label{hyp:ehgdcdpfjgogehpfbd}
\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}((\preoh r1\_xxreal\_0~X2~k6\_numbers){\unicdbdjdcd}((k6\_int\_1~(\\
k6\_xcmplx\_0~X0~X1)~X2=k6\_numbers){\unicdbdjded}(k4\_nat\_d~X0~X2=k4\_nat\_d~X1~\\
X2)))))\end{array}
\label{hyp:ehcdcdpfjgogehpfed}
\end{equation}
Assume the following.
\begin{equation}
\begin{array}{c}{\unicdcdadad} X0 .  {\unicdcdadad} X1 .  ((v1\_int\_1~X0){\unicdcdcdhd}(v1\_int\_1~X1)){\unicdbdjdcd}(v1\_int\_1~\\
(k6\_xcmplx\_0~X0~X1))\end{array}
\label{hyp:ggdgedpfjgogehpfbd}
\end{equation}
Assume the following.
\begin{equation}
{{\unicdcdadad} X0 .  {{({{v7\_ordinal1}~{X0}})}{\unicdbdjdcd}{({{v1\_int\_1}~{X0}})}}}\label{hyp:dgdgcdpfjgogehpfbd}
\end{equation}
\begin{theorem}\label{thm:ehcdddpfjgogehpfed}
$$\begin{array}{c}{\unicdcdadad} X0 .  (v7\_ordinal1~X0){\unicdbdjdcd}({\unicdcdadad} X1 .  (v7\_ordinal1~X1){\unicdbdjdcd}({\unicdcdadad} X2 .  \\
(v7\_ordinal1~X2){\unicdbdjdcd}((\preoh r1\_xxreal\_0~X2~k6\_numbers){\unicdbdjdcd}((k4\_nat\_d~X0~\\
X2=k4\_nat\_d~X1~X2){\unicdbdjded}(r1\_int\_1~X2~(k6\_xcmplx\_0~X0~X1))))))\end{array}$$
\end{theorem}
\end{document}
