\documentclass{article}
\newtheorem{axiom}{Axiom}
\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}
\newcommand\unicdcdadad{\forall}
\newcommand\unicdbdjdcd{\Rightarrow}
\newcommand\unicdcdadjd{\notin}
\newcommand\unicdbdjded{\Leftrightarrow}
\newcommand\unicdcdcdid{\lor}
\newcommand\unicdcdcdhd{\land}
\newcommand\uniddcgcg{\lambda}
\newcommand\unicdcdaddd{\exists}
\newcommand\unicdcdidid{\not\subseteq}
\newcommand\preoh[1]{\lnot #1}
\newcommand\mname[1]{{\mathsf{#1}}}
\begin{document}
\title{l2\_substut2 (TMF6GHWnpWcHptdxt3RWa5UpoPnijLmiKS7)}
\maketitle
Let $ m1\_qc\_lang1 : {{\iota}{\unicdbdjdcd}{o}}$ be given.
Let $ m1\_subset\_1 : {{\iota}{\unicdbdjdcd}{{\iota}{\unicdbdjdcd}{o}}}$ be given.
Let $ k5\_numbers : {\iota}$ be given.
Let $ m2\_subset\_1 : {{\iota}{\unicdbdjdcd}{{\iota}{\unicdbdjdcd}{{\iota}{\unicdbdjdcd}{o}}}}$ be given.
Let $ k6\_qc\_lang1 : {{\iota}{\unicdbdjdcd}{\iota}}$ be given.
Let $ k8\_qc\_lang1 : {{\iota}{\unicdbdjdcd}{{\iota}{\unicdbdjdcd}{\iota}}}$ be given.
Let $ k7\_qc\_lang1 : {{\iota}{\unicdbdjdcd}{{\iota}{\unicdbdjdcd}{\iota}}}$ be given.
Assume the following.
\begin{equation}
\begin{array}{c}{\unicdcdadad} X0 .  (m1\_qc\_lang1~X0){\unicdbdjdcd}({\unicdcdadad} X1 .  (m1\_subset\_1~X1~k5\_numbers){\unicdbdjdcd}\\
({\unicdcdadad} X2 .  (m2\_subset\_1~X2~(k6\_qc\_lang1~X0)~(k8\_qc\_lang1~X0~X1)){\unicdbdjdcd}\\
(k7\_qc\_lang1~X0~X2=X1)))\end{array}
\label{hyp:ehbdbdpfbhdgpfmgbgoghgbd}
\end{equation}
\begin{theorem}\label{thm:mgcdpfdhfhcgdhehfhehcd}
$$\begin{array}{c}{\unicdcdadad} X0 .  (m1\_qc\_lang1~X0){\unicdbdjdcd}({\unicdcdadad} X1 .  (m1\_subset\_1~X1~k5\_numbers){\unicdbdjdcd}\\
({\unicdcdadad} X2 .  (m2\_subset\_1~X2~(k6\_qc\_lang1~X0)~(k8\_qc\_lang1~X0~X1)){\unicdbdjdcd}\\
({\unicdcdadad} X3 .  (m1\_subset\_1~X3~k5\_numbers){\unicdbdjdcd}({\unicdcdadad} X4 .  (m1\_subset\_1~\\
X4~k5\_numbers){\unicdbdjdcd}(((m2\_subset\_1~X2~(k6\_qc\_lang1~X0)~(k8\_qc\_lang1~\\
X0~X3)){\unicdcdcdhd}(m2\_subset\_1~X2~(k6\_qc\_lang1~X0)~(k8\_qc\_lang1~X0~X4))){\unicdbdjdcd}\\
(X3=X4))))))\end{array}$$
\end{theorem}
\end{document}
