\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{t18\_qc\_lang4 (TMbBvJkdgUHuaeDmdKQqyjTMX9PwEhu6d2k)}
\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 $ m1\_trees\_1 : {{\iota}{\unicdbdjdcd}{{\iota}{\unicdbdjdcd}{o}}}$ be given.
Let $ k9\_xtuple\_0 : {{\iota}{\unicdbdjdcd}{\iota}}$ be given.
Let $ k2\_qc\_lang4 : {{\iota}{\unicdbdjdcd}{{\iota}{\unicdbdjdcd}{\iota}}}$ be given.
Let $ k3\_trees\_2 : {{\iota}{\unicdbdjdcd}{{\iota}{\unicdbdjdcd}{{\iota}{\unicdbdjdcd}{\iota}}}}$ be given.
Let $ r3\_xboole\_0 : {{\iota}{\unicdbdjdcd}{{\iota}{\unicdbdjdcd}{o}}}$ be given.
Let $ r2\_xboole\_0 : {{\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 .  (m1\_trees\_1~X2~(k9\_xtuple\_0~(k2\_qc\_lang4~X0~\\
X1))){\unicdbdjdcd}({\unicdcdadad} X3 .  (m1\_trees\_1~X3~(k9\_xtuple\_0~(k2\_qc\_lang4~X0~\\
X1))){\unicdbdjdcd}(\preoh (r2\_xboole\_0~X2~X3){\unicdcdcdhd}(k3\_trees\_2~(k9\_qc\_lang1~X0)~(k2\_qc\_lang4~\\
X0~X1)~X3=k3\_trees\_2~(k9\_qc\_lang1~X0)~(k2\_qc\_lang4~X0~X1)~X2)))))\end{array}
\label{hyp:ehbdfdpfbhdgpfmgbgoghged}
\end{equation}
Assume the following.
\begin{equation}
\begin{array}{c}{\unicdcdadad} X0 .  {\unicdcdadad} X1 .  (\preoh (\preoh r2\_xboole\_0~X0~X1){\unicdcdcdhd}((X0{\unicdcdgdad}X1){\unicdcdcdhd}(\preoh r2\_xboole\_0~\\
X1~X0))){\unicdbdjded}(r3\_xboole\_0~X0~X1)\end{array}
\label{hyp:ehbdadedpfihcgpgpgmgfgpfbd}
\end{equation}
\begin{theorem}\label{thm:ehbdidpfbhdgpfmgbgoghged}
$$\begin{array}{c}{\unicdcdadad} X0 .  (m1\_qc\_lang1~X0){\unicdbdjdcd}({\unicdcdadad} X1 .  (m1\_subset\_1~X1~(k9\_qc\_lang1~\\
X0)){\unicdbdjdcd}({\unicdcdadad} X2 .  (m1\_trees\_1~X2~(k9\_xtuple\_0~(k2\_qc\_lang4~X0~\\
X1))){\unicdbdjdcd}({\unicdcdadad} X3 .  (m1\_trees\_1~X3~(k9\_xtuple\_0~(k2\_qc\_lang4~X0~\\
X1))){\unicdbdjdcd}(\preoh (X2{\unicdcdgdad}X3){\unicdcdcdhd}((k3\_trees\_2~(k9\_qc\_lang1~X0)~(k2\_qc\_lang4~\\
X0~X1)~X2=k3\_trees\_2~(k9\_qc\_lang1~X0)~(k2\_qc\_lang4~X0~X1)~X3){\unicdcdcdhd}(\\
r3\_xboole\_0~X2~X3))))))\end{array}$$
\end{theorem}
\end{document}
