\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}
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\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{t111\_seq\_4 (TMXVP1g4DvDBourLMVHPCK5d8PfnESTBZ6c)}
\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 $ m1\_subset\_1 : {{\iota}{\unicdbdjdcd}{{\iota}{\unicdbdjdcd}{o}}}$ 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 $ r1\_xxreal\_0 : {{\iota}{\unicdbdjdcd}{{\iota}{\unicdbdjdcd}{o}}}$ be given.
Let $ k6\_numbers : {\iota}$ be given.
Let $ k23\_seq\_4 : {{\iota}{\unicdbdjdcd}{{\iota}{\unicdbdjdcd}{{\iota}{\unicdbdjdcd}{\iota}}}}$ be given.
Let $ k22\_seq\_4 : {{\iota}{\unicdbdjdcd}{\iota}}$ be given.
Let $ k19\_seq\_4 : {{\iota}{\unicdbdjdcd}{{\iota}{\unicdbdjdcd}{{\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}((k22\_seq\_4~(k19\_seq\_4~X0~X1~X2)=\\
k6\_numbers){\unicdbdjded}(X1=X2))))\end{array}
\label{hyp:ehbdadcdpfdhfgbhpfed}
\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}({\unicdcdadad} X2 .  (m1\_subset\_1~\\
X2~k1\_numbers){\unicdbdjdcd}(k23\_seq\_4~X0~X1~X2=ReplSep~(toset~({\uniddcgcg} X3 : \iota .  \\
m2\_finseq\_2~X3~k2\_numbers~(k14\_seq\_4~X0)))~({\uniddcgcg} X3 : \iota .  \preoh r1\_xxreal\_0~\\
X2~(k22\_seq\_4~(k19\_seq\_4~X0~X3~X1)))~({\uniddcgcg} X3 : \iota .  X3))))\end{array}
\label{hyp:egbdgdpfdhfgbhpfed}
\end{equation}
\begin{theorem}\label{thm:ehbdbdbdpfdhfgbhpfed}
$$\begin{array}{c}{\unicdcdadad} X0 .  (m2\_subset\_1~X0~k1\_numbers~k5\_numbers){\unicdbdjdcd}({\unicdcdadad} X1 .  \\
(m1\_subset\_1~X1~k1\_numbers){\unicdbdjdcd}({\unicdcdadad} X2 .  (m2\_finseq\_2~X2~k2\_numbers~\\
(k14\_seq\_4~X0)){\unicdbdjdcd}((\preoh r1\_xxreal\_0~X1~k6\_numbers){\unicdbdjdcd}(X2 \in k23\_seq\_4~\\
X0~X2~X1))))\end{array}$$
\end{theorem}
\end{document}
