\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{t24\_qc\_lang3 (TMGEoArXmmafYcKSmC9oF3ws9twDXkffL9d)}
\maketitle
Let $ m1\_qc\_lang1 : {{\iota}{\unicdbdjdcd}{o}}$ be given.
Let $ m2\_subset\_1 : {{\iota}{\unicdbdjdcd}{{\iota}{\unicdbdjdcd}{{\iota}{\unicdbdjdcd}{o}}}}$ be given.
Let $ k2\_qc\_lang1 : {{\iota}{\unicdbdjdcd}{\iota}}$ be given.
Let $ k3\_qc\_lang1 : {{\iota}{\unicdbdjdcd}{\iota}}$ be given.
Let $ m1\_subset\_1 : {{\iota}{\unicdbdjdcd}{{\iota}{\unicdbdjdcd}{o}}}$ be given.
Let $ k9\_qc\_lang1 : {{\iota}{\unicdbdjdcd}{\iota}}$ be given.
Let $ v6\_qc\_lang1 : {{\iota}{\unicdbdjdcd}{{\iota}{\unicdbdjdcd}{o}}}$ be given.
Let $ k15\_qc\_lang1 : {{\iota}{\unicdbdjdcd}{{\iota}{\unicdbdjdcd}{{\iota}{\unicdbdjdcd}{\iota}}}}$ be given.
Let $ r1\_tarski : {{\iota}{\unicdbdjdcd}{{\iota}{\unicdbdjdcd}{o}}}$ be given.
Let $ k1\_xboole\_0 : {\iota}$ be given.
Let $ k24\_qc\_lang1 : {{\iota}{\unicdbdjdcd}{{\iota}{\unicdbdjdcd}{\iota}}}$ be given.
Let $ k1\_tarski : {{\iota}{\unicdbdjdcd}{\iota}}$ be given.
Assume the following.
\begin{equation}
{{\unicdcdadad} X0 .  {{{r1\_tarski}~{k1\_xboole\_0}}~{X0}}}\label{hyp:ehcdpfihcgpgpgmgfgpfbd}
\end{equation}
Assume the following.
\begin{equation}
\begin{array}{c}{\unicdcdadad} X0 .  (m1\_qc\_lang1~X0){\unicdbdjdcd}({\unicdcdadad} X1 .  (m2\_subset\_1~X1~(k2\_qc\_lang1~\\
X0)~(k3\_qc\_lang1~X0)){\unicdbdjdcd}({\unicdcdadad} X2 .  (m1\_subset\_1~X2~(k9\_qc\_lang1~\\
X0)){\unicdbdjdcd}((v6\_qc\_lang1~(k15\_qc\_lang1~X0~X1~X2)~X0){\unicdbdjded}(r1\_tarski~(k24\_qc\_lang1~\\
X0~X2)~(k1\_tarski~X1)))))\end{array}
\label{hyp:ehcdddpfbhdgpfmgbgoghgdd}
\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}((v6\_qc\_lang1~X1~X0){\unicdbdjded}(k24\_qc\_lang1~X0~X1=k1\_xboole\_0)))\end{array}
\label{hyp:egddbdpfbhdgpfmgbgoghgbd}
\end{equation}
\begin{theorem}\label{thm:ehcdedpfbhdgpfmgbgoghgdd}
$$\begin{array}{c}{\unicdcdadad} X0 .  (m1\_qc\_lang1~X0){\unicdbdjdcd}({\unicdcdadad} X1 .  (m2\_subset\_1~X1~(k2\_qc\_lang1~\\
X0)~(k3\_qc\_lang1~X0)){\unicdbdjdcd}({\unicdcdadad} X2 .  (m1\_subset\_1~X2~(k9\_qc\_lang1~\\
X0)){\unicdbdjdcd}((v6\_qc\_lang1~X2~X0){\unicdbdjdcd}(v6\_qc\_lang1~(k15\_qc\_lang1~X0~X1~X2)~\\
X0))))\end{array}$$
\end{theorem}
\end{document}
