\documentclass{article}
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\newtheorem{lemma}{Lemma}
\newtheorem{proposition}{Proposition}
\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}
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\newcommand\unicdcdidid{\not\subseteq}
\newcommand\preoh[1]{\lnot #1}
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\begin{document}
\title{t18\_qc\_lang3 (TMSdw1UMLWXUNTtQuSNC83ozbkp4KRgVfxJ)}
\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 $ k24\_qc\_lang1 : {{\iota}{\unicdbdjdcd}{{\iota}{\unicdbdjdcd}{\iota}}}$ be given.
Let $ k4\_qc\_lang2 : {{\iota}{\unicdbdjdcd}{{\iota}{\unicdbdjdcd}{{\iota}{\unicdbdjdcd}{\iota}}}}$ be given.
Let $ k4\_subset\_1 : {{\iota}{\unicdbdjdcd}{{\iota}{\unicdbdjdcd}{{\iota}{\unicdbdjdcd}{\iota}}}}$ be given.
Let $ k3\_qc\_lang1 : {{\iota}{\unicdbdjdcd}{\iota}}$ be given.
Let $ k13\_qc\_lang2 : {{\iota}{\unicdbdjdcd}{{\iota}{\unicdbdjdcd}{\iota}}}$ be given.
Let $ k14\_qc\_lang2 : {{\iota}{\unicdbdjdcd}{{\iota}{\unicdbdjdcd}{\iota}}}$ be given.
Let $ k19\_qc\_lang1 : {{\iota}{\unicdbdjdcd}{{\iota}{\unicdbdjdcd}{\iota}}}$ be given.
Let $ k2\_qc\_lang2 : {{\iota}{\unicdbdjdcd}{{\iota}{\unicdbdjdcd}{{\iota}{\unicdbdjdcd}{\iota}}}}$ be given.
Let $ k20\_qc\_lang1 : {{\iota}{\unicdbdjdcd}{{\iota}{\unicdbdjdcd}{\iota}}}$ be given.
Let $ v3\_qc\_lang2 : {{\iota}{\unicdbdjdcd}{{\iota}{\unicdbdjdcd}{o}}}$ 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\_subset\_1~X2~(k9\_qc\_lang1~X0)){\unicdbdjdcd}((k13\_qc\_lang2~\\
X0~(k4\_qc\_lang2~X0~X1~X2)=X1){\unicdcdcdhd}((k14\_qc\_lang2~X0~(k4\_qc\_lang2~X0~\\
X1~X2)=X2){\unicdcdcdhd}((k19\_qc\_lang1~X0~(k4\_qc\_lang2~X0~X1~X2)=k2\_qc\_lang2~\\
X0~X1~X2){\unicdcdcdhd}(k20\_qc\_lang1~X0~(k4\_qc\_lang2~X0~X1~X2)=k2\_qc\_lang2~X0~\\
X2~X1))))))\end{array}
\label{hyp:ehddbdpfbhdgpfmgbgoghgcd}
\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}((v3\_qc\_lang2~X1~X0){\unicdbdjdcd}(k24\_qc\_lang1~X0~X1=k4\_subset\_1~(k3\_qc\_lang1~\\
X0)~(k24\_qc\_lang1~X0~(k13\_qc\_lang2~X0~X1))~(k24\_qc\_lang1~X0~(k14\_qc\_lang2~\\
X0~X1)))))\end{array}
\label{hyp:ehbdhdpfbhdgpfmgbgoghgdd}
\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\_subset\_1~X2~(k9\_qc\_lang1~X0)))){\unicdbdjdcd}(m1\_subset\_1~\\
(k4\_qc\_lang2~X0~X1~X2)~(k9\_qc\_lang1~X0))\end{array}
\label{hyp:egehpflgedpfbhdgpfmgbgoghgcd}
\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}((v3\_qc\_lang2~X1~X0){\unicdbdjded}({\unicdcdaddd} X2 .  (m1\_subset\_1~X2~(k9\_qc\_lang1~\\
X0)){\unicdcdcdhd}({\unicdcdaddd} X3 .  (m1\_subset\_1~X3~(k9\_qc\_lang1~X0)){\unicdcdcdhd}(X1=k4\_qc\_lang2~\\
X0~X2~X3)))))\end{array}
\label{hyp:egbdcdpfbhdgpfmgbgoghgcd}
\end{equation}
\begin{theorem}\label{thm:ehbdidpfbhdgpfmgbgoghgdd}
$$\begin{array}{c}{\unicdcdadad} X0 .  (m1\_qc\_lang1~X0){\unicdbdjdcd}({\unicdcdadad} X1 .  (m1\_subset\_1~X1~(k9\_qc\_lang1~\\
X0)){\unicdbdjdcd}({\unicdcdadad} X2 .  (m1\_subset\_1~X2~(k9\_qc\_lang1~X0)){\unicdbdjdcd}(k24\_qc\_lang1~\\
X0~(k4\_qc\_lang2~X0~X1~X2)=k4\_subset\_1~(k3\_qc\_lang1~X0)~(k24\_qc\_lang1~\\
X0~X1)~(k24\_qc\_lang1~X0~X2))))\end{array}$$
\end{theorem}
\end{document}
