\documentclass{article}
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\newtheorem{conjecture}{Conjecture}
\newcommand\unicdcdgdad{\not=}
\newcommand\unicdcdcdbg{\cup}
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\newcommand\unicdbdjdcd{\Rightarrow}
\newcommand\unicdcdadjd{\notin}
\newcommand\unicdbdjded{\Leftrightarrow}
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\newcommand\unicdcdidid{\not\subseteq}
\newcommand\preoh[1]{\lnot #1}
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\begin{document}
\title{t46\_cqc\_sim1 (TMdxmML72PgT5dZvURK9kCEG3cdsMBx6tyh)}
\maketitle
Let $ m1\_qc\_lang1 : {{\iota}{\unicdbdjdcd}{o}}$ be given.
Let $ m2\_subset\_1 : {{\iota}{\unicdbdjdcd}{{\iota}{\unicdbdjdcd}{{\iota}{\unicdbdjdcd}{o}}}}$ be given.
Let $ k9\_qc\_lang1 : {{\iota}{\unicdbdjdcd}{\iota}}$ be given.
Let $ k3\_cqc\_lang : {{\iota}{\unicdbdjdcd}{\iota}}$ be given.
Let $ r2\_cqc\_sim1 : {{\iota}{\unicdbdjdcd}{{\iota}{\unicdbdjdcd}{{\iota}{\unicdbdjdcd}{o}}}}$ be given.
Let $ k14\_cqc\_sim1 : {{\iota}{\unicdbdjdcd}{{\iota}{\unicdbdjdcd}{\iota}}}$ be given.
Assume the following.
\begin{equation}
\begin{array}{c}{\unicdcdadad} X0 .  (m1\_qc\_lang1~X0){\unicdbdjdcd}({\unicdcdadad} X1 .  (m2\_subset\_1~X1~(k9\_qc\_lang1~\\
X0)~(k3\_cqc\_lang~X0)){\unicdbdjdcd}({\unicdcdadad} X2 .  (m2\_subset\_1~X2~(k9\_qc\_lang1~\\
X0)~(k3\_cqc\_lang~X0)){\unicdbdjdcd}((r2\_cqc\_sim1~X0~X1~X2){\unicdbdjded}(k14\_cqc\_sim1~X0~\\
X1=k14\_cqc\_sim1~X0~X2))))\end{array}
\label{hyp:egbdedpfdgbhdgpfdhjgngbd}
\end{equation}
\begin{theorem}\label{thm:ehedgdpfdgbhdgpfdhjgngbd}
$$\begin{array}{c}{\unicdcdadad} X0 .  (m1\_qc\_lang1~X0){\unicdbdjdcd}({\unicdcdadad} X1 .  (m2\_subset\_1~X1~(k9\_qc\_lang1~\\
X0)~(k3\_cqc\_lang~X0)){\unicdbdjdcd}({\unicdcdadad} X2 .  (m2\_subset\_1~X2~(k9\_qc\_lang1~\\
X0)~(k3\_cqc\_lang~X0)){\unicdbdjdcd}({\unicdcdadad} X3 .  (m2\_subset\_1~X3~(k9\_qc\_lang1~\\
X0)~(k3\_cqc\_lang~X0)){\unicdbdjdcd}(((r2\_cqc\_sim1~X0~X1~X2){\unicdcdcdhd}(r2\_cqc\_sim1~X0~\\
X2~X3)){\unicdbdjdcd}(r2\_cqc\_sim1~X0~X1~X3)))))\end{array}$$
\end{theorem}
\end{document}
