\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{l28\_seqm\_3 (TMRzECUshmimc8c99CyDJKF5T3aQXiebEND)}
\maketitle
Let $ v5\_valued\_0 : {{\iota}{\unicdbdjdcd}{o}}$ be given.
Let $ k6\_partfun1 : {{\iota}{\unicdbdjdcd}{\iota}}$ be given.
Let $ k5\_numbers : {\iota}$ be given.
Let $ v4\_valued\_0 : {{\iota}{\unicdbdjdcd}{o}}$ be given.
Let $ k4\_relat\_1 : {{\iota}{\unicdbdjdcd}{\iota}}$ be given.
Let $ m1\_subset\_1 : {{\iota}{\unicdbdjdcd}{{\iota}{\unicdbdjdcd}{o}}}$ be given.
Let $ k1\_zfmisc\_1 : {{\iota}{\unicdbdjdcd}{\iota}}$ be given.
Let $ k6\_funct\_3 : {{\iota}{\unicdbdjdcd}{{\iota}{\unicdbdjdcd}{\iota}}}$ be given.
Let $ k4\_ordinal1 : {\iota}$ be given.
Let $ k1\_numbers : {\iota}$ be given.
Let $ v6\_membered : {{\iota}{\unicdbdjdcd}{o}}$ be given.
Let $ v1\_relat\_1 : {{\iota}{\unicdbdjdcd}{o}}$ be given.
Assume the following.
\begin{equation}
{{\unicdcdadad} X0 .  {{{k6\_partfun1}~{X0}}={{k4\_relat\_1}~{X0}}}}\label{hyp:chfgegfgggjgogjgehjgpgogpflggdpfahbgchehggfhogbd}
\end{equation}
Assume the following.
\begin{equation}
\begin{array}{c}{\unicdcdadad} X0 .  {\unicdcdadad} X1 .  (m1\_subset\_1~X1~(k1\_zfmisc\_1~X0)){\unicdbdjdcd}(k6\_funct\_3~\\
X0~X1=k4\_relat\_1~X1)\end{array}
\label{hyp:chfgegfgggjgogjgehjgpgogpflggdpfggfhogdgehpfdd}
\end{equation}
Assume the following.
\begin{equation}
{{k5\_numbers}={k4\_ordinal1}}\label{hyp:chfgegfgggjgogjgehjgpgogpflgfdpfogfhngcgfgchdh}
\end{equation}
Assume the following.
\begin{equation}
\begin{array}{c}(v5\_valued\_0~(k6\_funct\_3~k1\_numbers~k5\_numbers)){\unicdcdcdhd}(v4\_valued\_0~\\
(k6\_funct\_3~k1\_numbers~k5\_numbers))\end{array}
\label{hyp:mgcdhdpfdhfgbhngpfdd}
\end{equation}
Assume the following.
\begin{equation}
{{v6\_membered}~{k4\_ordinal1}}\label{hyp:ggdggdpfngfgngcgfgchfgeg}
\end{equation}
Assume the following.
\begin{equation}
\begin{array}{c}{\unicdcdadad} X0 .  (v6\_membered~X0){\unicdbdjdcd}((v1\_relat\_1~(k4\_relat\_1~X0)){\unicdcdcdhd}(\\
v4\_valued\_0~(k4\_relat\_1~X0)))\end{array}
\label{hyp:ggdgfdedpfghbgmgfhfgegpfad}
\end{equation}
Assume the following.
\begin{equation}
{{{m1\_subset\_1}~{k5\_numbers}}~{({{k1\_zfmisc\_1}~{k1\_numbers}})}}\label{hyp:egehpflgfdpfogfhngcgfgchdh}
\end{equation}
\begin{theorem}\label{thm:mgcdidpfdhfgbhngpfdd}
$$\begin{array}{c}(v5\_valued\_0~(k6\_partfun1~k5\_numbers)){\unicdcdcdhd}(v4\_valued\_0~(k6\_partfun1~\\
k5\_numbers))\end{array}$$
\end{theorem}
\end{document}
