\documentclass{article}
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\newtheorem{lemma}{Lemma}
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\newtheorem{definition}{Definition}
\newtheorem{example}{Example}
\newtheorem{conjecture}{Conjecture}
\newcommand\unicdcdgdad{\not=}
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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}
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\begin{document}
\title{t37\_qc\_lang4 (TMVvEcrAeH9ZVgqnz8aGC8YK7UsSh558Mv9)}
\maketitle
Let $ m1\_qc\_lang1 : {{\iota}{\unicdbdjdcd}{o}}$ be given.
Let $ m1\_subset\_1 : {{\iota}{\unicdbdjdcd}{{\iota}{\unicdbdjdcd}{o}}}$ be given.
Let $ k9\_qc\_lang1 : {{\iota}{\unicdbdjdcd}{\iota}}$ be given.
Let $ m1\_qc\_lang4 : {{\iota}{\unicdbdjdcd}{{\iota}{\unicdbdjdcd}{{\iota}{\unicdbdjdcd}{o}}}}$ be given.
Let $ m2\_qc\_lang4 : {{\iota}{\unicdbdjdcd}{{\iota}{\unicdbdjdcd}{{\iota}{\unicdbdjdcd}{{\iota}{\unicdbdjdcd}{o}}}}}$ be given.
Let $ m1\_trees\_1 : {{\iota}{\unicdbdjdcd}{{\iota}{\unicdbdjdcd}{o}}}$ be given.
Let $ k9\_xtuple\_0 : {{\iota}{\unicdbdjdcd}{\iota}}$ be given.
Let $ k2\_qc\_lang4 : {{\iota}{\unicdbdjdcd}{{\iota}{\unicdbdjdcd}{\iota}}}$ be given.
Let $ k3\_qc\_lang4 : {{\iota}{\unicdbdjdcd}{{\iota}{\unicdbdjdcd}{{\iota}{\unicdbdjdcd}{\iota}}}}$ be given.
Let $ k8\_finseq\_1 : {{\iota}{\unicdbdjdcd}{{\iota}{\unicdbdjdcd}{{\iota}{\unicdbdjdcd}{\iota}}}}$ be given.
Let $ k5\_numbers : {\iota}$ be given.
Let $ k4\_qc\_lang4 : {{\iota}{\unicdbdjdcd}{{\iota}{\unicdbdjdcd}{{\iota}{\unicdbdjdcd}{\iota}}}}$ be given.
Assume the following.
\begin{equation}
\begin{array}{c}{\unicdcdadad} X0 .  (m1\_qc\_lang1~X0){\unicdbdjdcd}({\unicdcdadad} X1 .  (m1\_subset\_1~X1~(k9\_qc\_lang1~\\
X0)){\unicdbdjdcd}({\unicdcdadad} X2 .  (m1\_qc\_lang4~X2~X0~X1){\unicdbdjdcd}(k4\_qc\_lang4~X0~X1~X2=\\
ReplSep~(toset~({\uniddcgcg} X3 : \iota .  m2\_qc\_lang4~X3~X0~X1~X2))~({\uniddcgcg} X3 : \iota .  \\
X3=X3)~({\uniddcgcg} X3 : \iota .  X3))))\end{array}
\label{hyp:ehddgdpfbhdgpfmgbgoghged}
\end{equation}
Assume the following.
\begin{equation}
\begin{array}{c}{\unicdcdadad} X0 .  (m1\_qc\_lang1~X0){\unicdbdjdcd}({\unicdcdadad} X1 .  (m1\_subset\_1~X1~(k9\_qc\_lang1~\\
X0)){\unicdbdjdcd}({\unicdcdadad} X2 .  (m1\_qc\_lang4~X2~X0~X1){\unicdbdjdcd}({\unicdcdadad} X3 .  (m2\_qc\_lang4~\\
X3~X0~X1~X2){\unicdbdjdcd}(X3 \in k4\_qc\_lang4~X0~X1~X2))))\end{array}
\label{hyp:ehddfdpfbhdgpfmgbgoghged}
\end{equation}
Assume the following.
\begin{equation}
\begin{array}{c}{\unicdcdadad} X0 .  (m1\_qc\_lang1~X0){\unicdbdjdcd}({\unicdcdadad} X1 .  (m1\_subset\_1~X1~(k9\_qc\_lang1~\\
X0)){\unicdbdjdcd}({\unicdcdadad} X2 .  (m1\_subset\_1~X2~(k9\_qc\_lang1~X0)){\unicdbdjdcd}({\unicdcdadad} X3 .  \\
(m1\_subset\_1~X3~(k9\_qc\_lang1~X0)){\unicdbdjdcd}({\unicdcdadad} X4 .  (m1\_trees\_1~X4~\\
(k9\_xtuple\_0~(k2\_qc\_lang4~X0~X1))){\unicdbdjdcd}({\unicdcdadad} X5 .  (m1\_trees\_1~X5~\\
(k9\_xtuple\_0~(k2\_qc\_lang4~X0~X2))){\unicdbdjdcd}(((X4 \in k3\_qc\_lang4~X0~X1~X2){\unicdcdcdhd}\\
(X5 \in k3\_qc\_lang4~X0~X2~X3)){\unicdbdjdcd}(k8\_finseq\_1~k5\_numbers~X4~X5 \in k3\_qc\_lang4~\\
X0~X1~X3)))))))\end{array}
\label{hyp:ehcdhdpfbhdgpfmgbgoghged}
\end{equation}
Assume the following.
\begin{equation}
\begin{array}{c}{\unicdcdadad} X0 .  {\unicdcdadad} X1 .  {\unicdcdadad} X2 .  ((m1\_qc\_lang1~X0){\unicdcdcdhd}((m1\_subset\_1~\\
X1~(k9\_qc\_lang1~X0)){\unicdcdcdhd}(m1\_qc\_lang4~X2~X0~X1))){\unicdbdjdcd}({\unicdcdadad} X3 .  (m2\_qc\_lang4~\\
X3~X0~X1~X2){\unicdbdjdcd}(m1\_trees\_1~X3~(k9\_xtuple\_0~(k2\_qc\_lang4~X0~X1))))\end{array}
\label{hyp:egehpfngcdpfbhdgpfmgbgoghged}
\end{equation}
Assume the following.
\begin{equation}
\begin{array}{c}{\unicdcdadad} X0 .  {\unicdcdadad} X1 .  ((m1\_qc\_lang1~X0){\unicdcdcdhd}(m1\_subset\_1~X1~(k9\_qc\_lang1~\\
X0))){\unicdbdjdcd}({\unicdcdadad} X2 .  (m1\_qc\_lang4~X2~X0~X1){\unicdbdjdcd}(m1\_subset\_1~X2~(k9\_qc\_lang1~\\
X0)))\end{array}
\label{hyp:egehpfngbdpfbhdgpfmgbgoghged}
\end{equation}
Assume the following.
\begin{equation}
\begin{array}{c}{\unicdcdadad} X0 .  (m1\_qc\_lang1~X0){\unicdbdjdcd}({\unicdcdadad} X1 .  (m1\_subset\_1~X1~(k9\_qc\_lang1~\\
X0)){\unicdbdjdcd}({\unicdcdadad} X2 .  (m1\_qc\_lang4~X2~X0~X1){\unicdbdjdcd}(k4\_qc\_lang4~X0~X1~X2=\\
k3\_qc\_lang4~X0~X1~X2)))\end{array}
\label{hyp:eggdpfbhdgpfmgbgoghged}
\end{equation}
\begin{theorem}\label{thm:ehddhdpfbhdgpfmgbgoghged}
$$\begin{array}{c}{\unicdcdadad} X0 .  (m1\_qc\_lang1~X0){\unicdbdjdcd}({\unicdcdadad} X1 .  (m1\_subset\_1~X1~(k9\_qc\_lang1~\\
X0)){\unicdbdjdcd}({\unicdcdadad} X2 .  (m1\_qc\_lang4~X2~X0~X1){\unicdbdjdcd}({\unicdcdadad} X3 .  (m1\_qc\_lang4~\\
X3~X0~X1){\unicdbdjdcd}({\unicdcdadad} X4 .  (m2\_qc\_lang4~X4~X0~X1~X2){\unicdbdjdcd}({\unicdcdadad} X5 .  (m1\_trees\_1~\\
X5~(k9\_xtuple\_0~(k2\_qc\_lang4~X0~X2))){\unicdbdjdcd}((X5 \in k3\_qc\_lang4~X0~X2~X3){\unicdbdjdcd}\\
(m2\_qc\_lang4~(k8\_finseq\_1~k5\_numbers~X4~X5)~X0~X1~X3)))))))\end{array}$$
\end{theorem}
\end{document}
