\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{t15\_pepin (TMPcQYgJ5szKtgfJ3Zjna4meYB78dzF2tCp)}
\maketitle
Let $ v7\_ordinal1 : {{\iota}{\unicdbdjdcd}{o}}$ be given.
Let $ v1\_int\_2 : {{\iota}{\unicdbdjdcd}{o}}$ be given.
Let $ k4\_nat\_d : {{\iota}{\unicdbdjdcd}{{\iota}{\unicdbdjdcd}{\iota}}}$ be given.
Let $ k4\_xcmplx\_0 : {{\iota}{\unicdbdjdcd}{\iota}}$ be given.
Let $ np\_\_1 : {\iota}$ be given.
Let $ k3\_square\_1 : {{\iota}{\unicdbdjdcd}{\iota}}$ be given.
Let $ v2\_xxreal\_0 : {{\iota}{\unicdbdjdcd}{o}}$ be given.
Let $ m2\_subset\_1 : {{\iota}{\unicdbdjdcd}{{\iota}{\unicdbdjdcd}{{\iota}{\unicdbdjdcd}{o}}}}$ be given.
Let $ k1\_numbers : {\iota}$ be given.
Let $ k5\_numbers : {\iota}$ be given.
Let $ m1\_subset\_1 : {{\iota}{\unicdbdjdcd}{{\iota}{\unicdbdjdcd}{o}}}$ be given.
Let $ v1\_xboole\_0 : {{\iota}{\unicdbdjdcd}{o}}$ be given.
Let $ k4\_ordinal1 : {\iota}$ be given.
Let $ v1\_xcmplx\_0 : {{\iota}{\unicdbdjdcd}{o}}$ be given.
Assume the following.
\begin{equation}
\begin{array}{c}((v2\_xxreal\_0~np\_\_1){\unicdcdcdhd}(m2\_subset\_1~np\_\_1~k1\_numbers~k5\_numbers)){\unicdcdcdhd}\\
((m1\_subset\_1~np\_\_1~k5\_numbers){\unicdcdcdhd}(m1\_subset\_1~np\_\_1~k1\_numbers))\end{array}
\label{hyp:dhahdgbdpfogfhngfgchbgmgdh}
\end{equation}
Assume the following.
\begin{equation}
{\preoh{{{v1\_xboole\_0}~{np\_\_1}}}}\label{hyp:dhahdgbdpfcgpgpgmgfg}
\end{equation}
Assume the following.
\begin{equation}
{{k5\_numbers}={k4\_ordinal1}}\label{hyp:chfgegfgggjgogjgehjgpgogpflgfdpfogfhngcgfgchdh}
\end{equation}
Assume the following.
\begin{equation}
\begin{array}{c}{\unicdcdadad} X0 .  ((\preoh v1\_xboole\_0~X0){\unicdcdcdhd}(v7\_ordinal1~X0)){\unicdbdjdcd}((\preoh v7\_ordinal1~\\
(k4\_xcmplx\_0~X0)){\unicdcdcdhd}(v1\_xcmplx\_0~(k4\_xcmplx\_0~X0)))\end{array}
\label{hyp:ggdgbdpfjgogehpfcd}
\end{equation}
Assume the following.
\begin{equation}
\begin{array}{c}{\unicdcdadad} X0 .  {\unicdcdadad} X1 .  ((v7\_ordinal1~X0){\unicdcdcdhd}(v7\_ordinal1~X1)){\unicdbdjdcd}(\\
m1\_subset\_1~(k4\_nat\_d~X0~X1)~k5\_numbers)\end{array}
\label{hyp:egehpflgedpfogbgehpfeg}
\end{equation}
Assume the following.
\begin{equation}
{{\unicdcdadad} X0 .  {{({{{m1\_subset\_1}~{X0}}~{k4\_ordinal1}})}{\unicdbdjdcd}{({{v7\_ordinal1}~{X0}})}}}\label{hyp:dgdgidpfpgchegjgogbgmgbd}
\end{equation}
\begin{theorem}\label{thm:ehbdfdpfahfgahjgog}
$$\begin{array}{c}{\unicdcdadad} X0 .  (v7\_ordinal1~X0){\unicdbdjdcd}({\unicdcdadad} X1 .  (v7\_ordinal1~X1){\unicdbdjdcd}((\\
(v1\_int\_2~X0){\unicdcdcdhd}(k4\_nat\_d~X1~X0=k4\_xcmplx\_0~np\_\_1)){\unicdbdjdcd}(k4\_nat\_d~\\
(k3\_square\_1~X1)~X0=np\_\_1)))\end{array}$$
\end{theorem}
\end{document}
