\documentclass{article}
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\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}
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\newcommand\unicdcdaddd{\exists}
\newcommand\unicdcdidid{\not\subseteq}
\newcommand\preoh[1]{\lnot #1}
\newcommand\mname[1]{{\mathsf{#1}}}
\begin{document}
\title{t4\_comput\_1 (TMRwNRqMxtVY7U5hX9XiR4uzyh5m14gCujN)}
\maketitle
Let $ m1\_subset\_1 : {{\iota}{\unicdbdjdcd}{{\iota}{\unicdbdjdcd}{o}}}$ be given.
Let $ k5\_numbers : {\iota}$ be given.
Let $ v1\_relat\_1 : {{\iota}{\unicdbdjdcd}{o}}$ be given.
Let $ v1\_funct\_1 : {{\iota}{\unicdbdjdcd}{o}}$ be given.
Let $ v1\_finseq\_1 : {{\iota}{\unicdbdjdcd}{o}}$ be given.
Let $ k2\_funct\_7 : {{\iota}{\unicdbdjdcd}{{\iota}{\unicdbdjdcd}{{\iota}{\unicdbdjdcd}{\iota}}}}$ be given.
Let $ k2\_finseq\_3 : {{\iota}{\unicdbdjdcd}{{\iota}{\unicdbdjdcd}{\iota}}}$ be given.
Assume the following.
\begin{equation}
\begin{array}{c}{\unicdcdadad} X0 .  (m1\_subset\_1~X0~k5\_numbers){\unicdbdjdcd}({\unicdcdadad} X1 .  {\unicdcdadad} X2 .  \\
((v1\_relat\_1~X2){\unicdcdcdhd}((v1\_funct\_1~X2){\unicdcdcdhd}(v1\_finseq\_1~X2))){\unicdbdjdcd}(k2\_finseq\_3~\\
X0~(k2\_funct\_7~X2~X0~X1)=k2\_finseq\_3~X0~X2))\end{array}
\label{hyp:ehddpfdgpgngahfhehpfbd}
\end{equation}
\begin{theorem}\label{thm:ehedpfdgpgngahfhehpfbd}
$$\begin{array}{c}{\unicdcdadad} X0 .  (m1\_subset\_1~X0~k5\_numbers){\unicdbdjdcd}({\unicdcdadad} X1 .  {\unicdcdadad} X2 .  \\
((v1\_relat\_1~X2){\unicdcdcdhd}((v1\_funct\_1~X2){\unicdcdcdhd}(v1\_finseq\_1~X2))){\unicdbdjdcd}({\unicdcdadad} X3 .  \\
((v1\_relat\_1~X3){\unicdcdcdhd}((v1\_funct\_1~X3){\unicdcdcdhd}(v1\_finseq\_1~X3))){\unicdbdjdcd}((k2\_funct\_7~\\
X2~X0~X1=k2\_funct\_7~X3~X0~X1){\unicdbdjdcd}(k2\_finseq\_3~X0~X2=k2\_finseq\_3~X0~\\
X3))))\end{array}$$
\end{theorem}
\end{document}
