\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{t32\_real\_3 (TMFzScjs2XonJDjjxCG1Tv5qGFH2KJHYJt8)}
\maketitle
Let $ v7\_ordinal1 : {{\iota}{\unicdbdjdcd}{o}}$ be given.
Let $ v1\_int\_1 : {{\iota}{\unicdbdjdcd}{o}}$ be given.
Let $ r1\_xxreal\_0 : {{\iota}{\unicdbdjdcd}{{\iota}{\unicdbdjdcd}{o}}}$ be given.
Let $ np\_\_1 : {\iota}$ be given.
Let $ k3\_funct\_2 : {{\iota}{\unicdbdjdcd}{{\iota}{\unicdbdjdcd}{{\iota}{\unicdbdjdcd}{{\iota}{\unicdbdjdcd}{\iota}}}}}$ be given.
Let $ k5\_numbers : {\iota}$ be given.
Let $ k1\_numbers : {\iota}$ be given.
Let $ k4\_real\_3 : {{\iota}{\unicdbdjdcd}{\iota}}$ be given.
Let $ k10\_real\_1 : {{\iota}{\unicdbdjdcd}{{\iota}{\unicdbdjdcd}{\iota}}}$ be given.
Let $ k6\_numbers : {\iota}$ be given.
Let $ k1\_nat\_1 : {{\iota}{\unicdbdjdcd}{{\iota}{\unicdbdjdcd}{\iota}}}$ be given.
Let $ np\_\_2 : {\iota}$ be given.
Let $ k3\_real\_3 : {{\iota}{\unicdbdjdcd}{\iota}}$ be given.
Assume the following.
\begin{equation}
\begin{array}{c}{\unicdcdadad} X0 .  (v7\_ordinal1~X0){\unicdbdjdcd}({\unicdcdadad} X1 .  (v1\_int\_1~X1){\unicdbdjdcd}((\preoh r1\_xxreal\_0~\\
X1~np\_\_1){\unicdbdjdcd}((k3\_funct\_2~k5\_numbers~k1\_numbers~(k3\_real\_3~(k10\_real\_1~\\
np\_\_1~X1))~np\_\_1=X1){\unicdcdcdhd}((k3\_funct\_2~k5\_numbers~k1\_numbers~(k3\_real\_3~\\
(k10\_real\_1~np\_\_1~X1))~(k1\_nat\_1~X0~np\_\_2)=k6\_numbers){\unicdcdcdhd}((k3\_funct\_2~\\
k5\_numbers~k1\_numbers~(k4\_real\_3~(k10\_real\_1~np\_\_1~X1))~k6\_numbers=\\
k6\_numbers){\unicdcdcdhd}((k3\_funct\_2~k5\_numbers~k1\_numbers~(k4\_real\_3~(\\
k10\_real\_1~np\_\_1~X1))~np\_\_1=X1){\unicdcdcdhd}(k3\_funct\_2~k5\_numbers~k1\_numbers~\\
(k4\_real\_3~(k10\_real\_1~np\_\_1~X1))~(k1\_nat\_1~X0~np\_\_2)=k6\_numbers)))))))\end{array}
\label{hyp:mgedbdpfchfgbgmgpfdd}
\end{equation}
\begin{theorem}\label{thm:ehddcdpfchfgbgmgpfdd}
$$\begin{array}{c}{\unicdcdadad} X0 .  (v7\_ordinal1~X0){\unicdbdjdcd}({\unicdcdadad} X1 .  (v1\_int\_1~X1){\unicdbdjdcd}((\preoh r1\_xxreal\_0~\\
X1~np\_\_1){\unicdbdjdcd}((k3\_funct\_2~k5\_numbers~k1\_numbers~(k4\_real\_3~(k10\_real\_1~\\
np\_\_1~X1))~k6\_numbers=k6\_numbers){\unicdcdcdhd}((k3\_funct\_2~k5\_numbers~k1\_numbers~\\
(k4\_real\_3~(k10\_real\_1~np\_\_1~X1))~np\_\_1=X1){\unicdcdcdhd}(k3\_funct\_2~k5\_numbers~\\
k1\_numbers~(k4\_real\_3~(k10\_real\_1~np\_\_1~X1))~(k1\_nat\_1~X0~np\_\_2)=\\
k6\_numbers)))))\end{array}$$
\end{theorem}
\end{document}
