\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}
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\newcommand\unicdbdjdcd{\Rightarrow}
\newcommand\unicdcdadjd{\notin}
\newcommand\unicdbdjded{\Leftrightarrow}
\newcommand\unicdcdcdid{\lor}
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\newcommand\unicdcdaddd{\exists}
\newcommand\unicdcdidid{\not\subseteq}
\newcommand\preoh[1]{\lnot #1}
\newcommand\mname[1]{{\mathsf{#1}}}
\begin{document}
\title{t51\_qc\_lang3 (TMZCWFWeR8uGrzfqTFiT7syb5THo4H5mCyk)}
\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 $ v1\_xboole\_0 : {{\iota}{\unicdbdjdcd}{o}}$ be given.
Let $ k1\_zfmisc\_1 : {{\iota}{\unicdbdjdcd}{\iota}}$ be given.
Let $ k2\_qc\_lang1 : {{\iota}{\unicdbdjdcd}{\iota}}$ be given.
Let $ v4\_qc\_lang2 : {{\iota}{\unicdbdjdcd}{{\iota}{\unicdbdjdcd}{o}}}$ be given.
Let $ k4\_qc\_lang3 : {{\iota}{\unicdbdjdcd}{{\iota}{\unicdbdjdcd}{{\iota}{\unicdbdjdcd}{\iota}}}}$ be given.
Let $ k18\_qc\_lang1 : {{\iota}{\unicdbdjdcd}{{\iota}{\unicdbdjdcd}{\iota}}}$ be given.
Let $ k22\_qc\_lang1 : {{\iota}{\unicdbdjdcd}{{\iota}{\unicdbdjdcd}{\iota}}}$ be given.
Let $ v5\_qc\_lang1 : {{\iota}{\unicdbdjdcd}{{\iota}{\unicdbdjdcd}{o}}}$ be given.
Let $ v3\_qc\_lang1 : {{\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 .  ((\preoh v1\_xboole\_0~X2){\unicdcdcdhd}(m1\_subset\_1~X2~(k1\_zfmisc\_1~\\
(k2\_qc\_lang1~X0)))){\unicdbdjdcd}((v5\_qc\_lang1~X1~X0){\unicdbdjdcd}(k4\_qc\_lang3~X0~X2~X1=\\
k4\_qc\_lang3~X0~X2~(k22\_qc\_lang1~X0~X1)))))\end{array}
\label{hyp:ehedddpfbhdgpfmgbgoghgdd}
\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 .  ((\preoh v1\_xboole\_0~X2){\unicdcdcdhd}(m1\_subset\_1~X2~(k1\_zfmisc\_1~\\
(k2\_qc\_lang1~X0)))){\unicdbdjdcd}((v3\_qc\_lang1~X1~X0){\unicdbdjdcd}(k4\_qc\_lang3~X0~X2~X1=\\
k4\_qc\_lang3~X0~X2~(k18\_qc\_lang1~X0~X1)))))\end{array}
\label{hyp:ehddidpfbhdgpfmgbgoghgdd}
\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}((v4\_qc\_lang2~X1~X0){\unicdbdjdcd}((v3\_qc\_lang1~X1~X0){\unicdcdcdhd}((v5\_qc\_lang1~\\
(k18\_qc\_lang1~X0~X1)~X0){\unicdcdcdhd}(v3\_qc\_lang1~(k22\_qc\_lang1~X0~(k18\_qc\_lang1~\\
X0~X1))~X0)))))\end{array}
\label{hyp:ehddgdpfbhdgpfmgbgoghgcd}
\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}(m1\_subset\_1~(k22\_qc\_lang1~X0~X1)~(k9\_qc\_lang1~X0))\end{array}
\label{hyp:egehpflgcdcdpfbhdgpfmgbgoghgbd}
\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}(m1\_subset\_1~(k18\_qc\_lang1~X0~X1)~(k9\_qc\_lang1~X0))\end{array}
\label{hyp:egehpflgbdidpfbhdgpfmgbgoghgbd}
\end{equation}
\begin{theorem}\label{thm:ehfdbdpfbhdgpfmgbgoghgdd}
$$\begin{array}{c}{\unicdcdadad} X0 .  (m1\_qc\_lang1~X0){\unicdbdjdcd}({\unicdcdadad} X1 .  (m1\_subset\_1~X1~(k9\_qc\_lang1~\\
X0)){\unicdbdjdcd}({\unicdcdadad} X2 .  ((\preoh v1\_xboole\_0~X2){\unicdcdcdhd}(m1\_subset\_1~X2~(k1\_zfmisc\_1~\\
(k2\_qc\_lang1~X0)))){\unicdbdjdcd}((v4\_qc\_lang2~X1~X0){\unicdbdjdcd}(k4\_qc\_lang3~X0~X2~X1=\\
k4\_qc\_lang3~X0~X2~(k18\_qc\_lang1~X0~(k22\_qc\_lang1~X0~(k18\_qc\_lang1~\\
X0~X1)))))))\end{array}$$
\end{theorem}
\end{document}
