\documentclass{article}
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\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}
\newcommand\unicdcdcdjd{\cap}
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\newcommand\unicdbdjdcd{\Rightarrow}
\newcommand\unicdcdadjd{\notin}
\newcommand\unicdbdjded{\Leftrightarrow}
\newcommand\unicdcdcdid{\lor}
\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{t98\_qc\_lang2 (TMLNwsEod463kTJX4WxCpHSRA7DcRatTrgf)}
\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 $ k15\_qc\_lang2 : {{\iota}{\unicdbdjdcd}{{\iota}{\unicdbdjdcd}{\iota}}}$ be given.
Let $ r1\_qc\_lang2 : {{\iota}{\unicdbdjdcd}{{\iota}{\unicdbdjdcd}{{\iota}{\unicdbdjdcd}{o}}}}$ be given.
Let $ r3\_qc\_lang2 : {{\iota}{\unicdbdjdcd}{{\iota}{\unicdbdjdcd}{{\iota}{\unicdbdjdcd}{o}}}}$ be given.
Let $ r2\_qc\_lang2 : {{\iota}{\unicdbdjdcd}{{\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}({\unicdcdadad} X3 .  \\
(m1\_subset\_1~X3~(k9\_qc\_lang1~X0)){\unicdbdjdcd}(((r2\_qc\_lang2~X0~X1~X2){\unicdcdcdhd}(r2\_qc\_lang2~\\
X0~X2~X3)){\unicdbdjdcd}(r2\_qc\_lang2~X0~X1~X3)))))\end{array}
\label{hyp:ehfdhdpfbhdgpfmgbgoghgcd}
\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}((r1\_qc\_lang2~\\
X0~X1~X2){\unicdbdjdcd}(r2\_qc\_lang2~X0~X1~X2))))\end{array}
\label{hyp:ehfdcdpfbhdgpfmgbgoghgcd}
\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 .  (X2=k15\_qc\_lang2~X0~X1){\unicdbdjded}({\unicdcdadad} X3 .  (X3 \in X2){\unicdbdjded}\\
({\unicdcdaddd} X4 .  (m1\_subset\_1~X4~(k9\_qc\_lang1~X0)){\unicdcdcdhd}((X4=X3){\unicdcdcdhd}(r2\_qc\_lang2~\\
X0~X4~X1))))))\end{array}
\label{hyp:egcdcdpfbhdgpfmgbgoghgcd}
\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}((r3\_qc\_lang2~\\
X0~X1~X2){\unicdbdjded}((r2\_qc\_lang2~X0~X1~X2){\unicdcdcdhd}(X1{\unicdcdgdad}X2)))))\end{array}
\label{hyp:egcdbdpfbhdgpfmgbgoghgcd}
\end{equation}
\begin{theorem}\label{thm:ehjdidpfbhdgpfmgbgoghgcd}
$$\begin{array}{c}{\unicdcdadad} X0 .  (m1\_qc\_lang1~X0){\unicdbdjdcd}({\unicdcdadad} X1 .  (m1\_subset\_1~X1~(k9\_qc\_lang1~\\
X0)){\unicdbdjdcd}({\unicdcdadad} X2 .  (m1\_subset\_1~X2~(k9\_qc\_lang1~X0)){\unicdbdjdcd}({\unicdcdadad} X3 .  \\
(m1\_subset\_1~X3~(k9\_qc\_lang1~X0)){\unicdbdjdcd}((X2 \in k15\_qc\_lang2~X0~X3){\unicdbdjdcd}(\\
((\preoh r1\_qc\_lang2~X0~X1~X2){\unicdcdcdhd}((\preoh r3\_qc\_lang2~X0~X1~X2){\unicdcdcdhd}(\preoh r2\_qc\_lang2~\\
X0~X1~X2))){\unicdcdcdid}(X1 \in k15\_qc\_lang2~X0~X3))))))\end{array}$$
\end{theorem}
\end{document}
