\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=}
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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}
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\begin{document}
\title{t35\_lukasi\_1 (TMXhzkfNcjQihNHY1oqFzv5psCCMDReZmGi)}
\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 $ k8\_cqc\_lang : {{\iota}{\unicdbdjdcd}{{\iota}{\unicdbdjdcd}{{\iota}{\unicdbdjdcd}{\iota}}}}$ be given.
Let $ k6\_cqc\_lang : {{\iota}{\unicdbdjdcd}{{\iota}{\unicdbdjdcd}{\iota}}}$ be given.
Let $ k4\_cqc\_the1 : {{\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}((k8\_cqc\_lang~X0~X1~X2 \in k4\_cqc\_the1~X0){\unicdbdjded}\\
(k8\_cqc\_lang~X0~(k6\_cqc\_lang~X0~X2)~(k6\_cqc\_lang~X0~X1) \in k4\_cqc\_the1~\\
X0))))\end{array}
\label{hyp:ehddedpfmgfhlgbgdhjgpfbd}
\end{equation}
\begin{theorem}\label{thm:ehddfdpfmgfhlgbgdhjgpfbd}
$$\begin{array}{c}{\unicdcdadad} X0 .  (m1\_qc\_lang1~X0){\unicdbdjdcd}({\unicdcdadad} X1 .  (m2\_subset\_1~X1~(k9\_qc\_lang1~\\
X0)~(k3\_cqc\_lang~X0)){\unicdbdjdcd}({\unicdcdadad} X2 .  (m2\_subset\_1~X2~(k9\_qc\_lang1~\\
X0)~(k3\_cqc\_lang~X0)){\unicdbdjdcd}((k8\_cqc\_lang~X0~(k6\_cqc\_lang~X0~X1)~(k6\_cqc\_lang~\\
X0~X2) \in k4\_cqc\_the1~X0){\unicdbdjdcd}(k8\_cqc\_lang~X0~X2~X1 \in k4\_cqc\_the1~X0))))\end{array}$$
\end{theorem}
\end{document}
