\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}
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\newcommand\unicdcdaddd{\exists}
\newcommand\unicdcdidid{\not\subseteq}
\newcommand\preoh[1]{\lnot #1}
\newcommand\mname[1]{{\mathsf{#1}}}
\begin{document}
\title{t29\_int\_5 (TMYkk17mykFpCfBxStTYh472eZzZcBvwTW5)}
\maketitle
Let $ v1\_int\_1 : {{\iota}{\unicdbdjdcd}{o}}$ be given.
Let $ v7\_ordinal1 : {{\iota}{\unicdbdjdcd}{o}}$ be given.
Let $ v1\_int\_2 : {{\iota}{\unicdbdjdcd}{o}}$ be given.
Let $ k3\_int\_2 : {{\iota}{\unicdbdjdcd}{{\iota}{\unicdbdjdcd}{\iota}}}$ be given.
Let $ np\_\_1 : {\iota}$ be given.
Let $ r2\_int\_1 : {{\iota}{\unicdbdjdcd}{{\iota}{\unicdbdjdcd}{{\iota}{\unicdbdjdcd}{o}}}}$ be given.
Let $ r1\_xxreal\_0 : {{\iota}{\unicdbdjdcd}{{\iota}{\unicdbdjdcd}{o}}}$ be given.
Let $ np\_\_2 : {\iota}$ be given.
Let $ k2\_int\_5 : {{\iota}{\unicdbdjdcd}{{\iota}{\unicdbdjdcd}{\iota}}}$ be given.
Let $ k6\_int\_1 : {{\iota}{\unicdbdjdcd}{{\iota}{\unicdbdjdcd}{\iota}}}$ be given.
Let $ k6\_numbers : {\iota}$ be given.
Let $ r1\_int\_5 : {{\iota}{\unicdbdjdcd}{{\iota}{\unicdbdjdcd}{o}}}$ be given.
Let $ k4\_xcmplx\_0 : {{\iota}{\unicdbdjdcd}{\iota}}$ be given.
Assume the following.
\begin{equation}
\begin{array}{c}{\unicdcdadad} X0 .  (v1\_int\_1~X0){\unicdbdjdcd}({\unicdcdadad} X1 .  (v1\_int\_1~X1){\unicdbdjdcd}({\unicdcdadad} X2 .  \\
(v1\_int\_1~X2){\unicdbdjdcd}(((k6\_int\_1~X1~X0=k6\_int\_1~X2~X0){\unicdbdjdcd}((X0=k6\_numbers){\unicdcdcdid}\\
(r2\_int\_1~X1~X2~X0))){\unicdcdcdhd}((r2\_int\_1~X1~X2~X0){\unicdbdjdcd}(k6\_int\_1~X1~X0=k6\_int\_1~\\
X2~X0)))))\end{array}
\label{hyp:ehgdedpfogbgehpfeg}
\end{equation}
Assume the following.
\begin{equation}
\begin{array}{c}{\unicdcdadad} X0 .  (v1\_int\_1~X0){\unicdbdjdcd}({\unicdcdadad} X1 .  (v1\_int\_1~X1){\unicdbdjdcd}({\unicdcdadad} X2 .  \\
(v1\_int\_1~X2){\unicdbdjdcd}((r2\_int\_1~X0~X1~X2){\unicdbdjdcd}(r2\_int\_1~X1~X0~X2))))\end{array}
\label{hyp:ehbdedpfjgogehpfbd}
\end{equation}
Assume the following.
\begin{equation}
\begin{array}{c}{\unicdcdadad} X0 .  (v1\_int\_1~X0){\unicdbdjdcd}({\unicdcdadad} X1 .  (v1\_int\_1~X1){\unicdbdjdcd}({\unicdcdadad} X2 .  \\
(v7\_ordinal1~X2){\unicdbdjdcd}(((r1\_int\_5~X0~X2){\unicdcdcdhd}(r2\_int\_1~X0~X1~X2)){\unicdbdjdcd}(r1\_int\_5~\\
X1~X2))))\end{array}
\label{hyp:ehbdbdpfjgogehpffd}
\end{equation}
Assume the following.
\begin{equation}
\begin{array}{c}{\unicdcdadad} X0 .  (v1\_int\_1~X0){\unicdbdjdcd}({\unicdcdadad} X1 .  ((v7\_ordinal1~X1){\unicdcdcdhd}(v1\_int\_2~\\
X1)){\unicdbdjdcd}(((r1\_int\_5~X0~X1){\unicdbdjdcd}((k6\_int\_1~X0~X1=k6\_numbers){\unicdcdcdid}(k2\_int\_5~\\
X0~X1=np\_\_1))){\unicdcdcdhd}((((r1\_int\_5~X0~X1){\unicdcdcdhd}(k6\_int\_1~X0~X1=k6\_numbers)){\unicdbdjdcd}\\
(k2\_int\_5~X0~X1=k6\_numbers)){\unicdcdcdhd}(\preoh (\preoh (r1\_int\_5~X0~X1){\unicdcdcdhd}(k6\_int\_1~\\
X0~X1{\unicdcdgdad}k6\_numbers)){\unicdcdcdhd}((\preoh (r1\_int\_5~X0~X1){\unicdcdcdhd}(k6\_int\_1~X0~X1=k6\_numbers)){\unicdcdcdhd}\\
(k2\_int\_5~X0~X1{\unicdcdgdad}k4\_xcmplx\_0~np\_\_1))))))\end{array}
\label{hyp:egddpfjgogehpffd}
\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:ehcdjdpfjgogehpffd}
$$\begin{array}{c}{\unicdcdadad} X0 .  (v1\_int\_1~X0){\unicdbdjdcd}({\unicdcdadad} X1 .  ((v7\_ordinal1~X1){\unicdcdcdhd}(v1\_int\_2~\\
X1)){\unicdbdjdcd}({\unicdcdadad} X2 .  (v1\_int\_1~X2){\unicdbdjdcd}(((k3\_int\_2~X0~X1=np\_\_1){\unicdcdcdhd}(r2\_int\_1~\\
X0~X2~X1)){\unicdbdjdcd}((r1\_xxreal\_0~X1~np\_\_2){\unicdcdcdid}(k2\_int\_5~X0~X1=k2\_int\_5~X2~\\
X1)))))\end{array}$$
\end{theorem}
\end{document}
