\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}
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\newcommand\unicdbdjdcd{\Rightarrow}
\newcommand\unicdcdadjd{\notin}
\newcommand\unicdbdjded{\Leftrightarrow}
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\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{t79\_seq\_4 (TMRT9pDr6CyqKc3hebWqXpPytra4QbvAnAx)}
\maketitle
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 $ m2\_finseq\_2 : {{\iota}{\unicdbdjdcd}{{\iota}{\unicdbdjdcd}{{\iota}{\unicdbdjdcd}{o}}}}$ be given.
Let $ k2\_numbers : {\iota}$ be given.
Let $ k14\_seq\_4 : {{\iota}{\unicdbdjdcd}{\iota}}$ be given.
Let $ k19\_seq\_4 : {{\iota}{\unicdbdjdcd}{{\iota}{\unicdbdjdcd}{{\iota}{\unicdbdjdcd}{\iota}}}}$ be given.
Let $ k15\_seq\_4 : {{\iota}{\unicdbdjdcd}{{\iota}{\unicdbdjdcd}{{\iota}{\unicdbdjdcd}{\iota}}}}$ be given.
Let $ k17\_seq\_4 : {{\iota}{\unicdbdjdcd}{\iota}}$ be given.
Assume the following.
\begin{equation}
\begin{array}{c}{\unicdcdadad} X0 .  (m2\_subset\_1~X0~k1\_numbers~k5\_numbers){\unicdbdjdcd}({\unicdcdadad} X1 .  \\
(m2\_finseq\_2~X1~k2\_numbers~(k14\_seq\_4~X0)){\unicdbdjdcd}({\unicdcdadad} X2 .  (m2\_finseq\_2~\\
X2~k2\_numbers~(k14\_seq\_4~X0)){\unicdbdjdcd}({\unicdcdadad} X3 .  (m2\_finseq\_2~X3~k2\_numbers~\\
(k14\_seq\_4~X0)){\unicdbdjdcd}(k15\_seq\_4~X0~X1~(k19\_seq\_4~X0~X2~X3)=k19\_seq\_4~\\
X0~(k15\_seq\_4~X0~X1~X2)~X3))))\end{array}
\label{hyp:ehhdgdpfdhfgbhpfed}
\end{equation}
Assume the following.
\begin{equation}
\begin{array}{c}{\unicdcdadad} X0 .  (m2\_subset\_1~X0~k1\_numbers~k5\_numbers){\unicdbdjdcd}({\unicdcdadad} X1 .  \\
(m2\_finseq\_2~X1~k2\_numbers~(k14\_seq\_4~X0)){\unicdbdjdcd}(k19\_seq\_4~X0~X1~X1=\\
k17\_seq\_4~X0))\end{array}
\label{hyp:ehhdddpfdhfgbhpfed}
\end{equation}
Assume the following.
\begin{equation}
\begin{array}{c}{\unicdcdadad} X0 .  (m2\_subset\_1~X0~k1\_numbers~k5\_numbers){\unicdbdjdcd}({\unicdcdadad} X1 .  \\
(m2\_finseq\_2~X1~k2\_numbers~(k14\_seq\_4~X0)){\unicdbdjdcd}((k15\_seq\_4~X0~X1~(\\
k17\_seq\_4~X0)=X1){\unicdcdcdhd}(X1=k15\_seq\_4~X0~(k17\_seq\_4~X0)~X1)))\end{array}
\label{hyp:ehfdjdpfdhfgbhpfed}
\end{equation}
\begin{theorem}\label{thm:ehhdjdpfdhfgbhpfed}
$$\begin{array}{c}{\unicdcdadad} X0 .  (m2\_subset\_1~X0~k1\_numbers~k5\_numbers){\unicdbdjdcd}({\unicdcdadad} X1 .  \\
(m2\_finseq\_2~X1~k2\_numbers~(k14\_seq\_4~X0)){\unicdbdjdcd}({\unicdcdadad} X2 .  (m2\_finseq\_2~\\
X2~k2\_numbers~(k14\_seq\_4~X0)){\unicdbdjdcd}(X1=k19\_seq\_4~X0~(k15\_seq\_4~X0~X1~\\
X2)~X2)))\end{array}$$
\end{theorem}
\end{document}
