\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{t1\_qc\_lang3 (TMN3CiYDA4Y92cdoRoyMDQmdnqayNhcG8dm)}
\maketitle
Let $ m1\_qc\_lang1 : {{\iota}{\unicdbdjdcd}{o}}$ be given.
Let $ m1\_subset\_1 : {{\iota}{\unicdbdjdcd}{{\iota}{\unicdbdjdcd}{o}}}$ be given.
Let $ k6\_qc\_lang1 : {{\iota}{\unicdbdjdcd}{\iota}}$ be given.
Let $ m2\_subset\_1 : {{\iota}{\unicdbdjdcd}{{\iota}{\unicdbdjdcd}{{\iota}{\unicdbdjdcd}{o}}}}$ be given.
Let $ k8\_qc\_lang1 : {{\iota}{\unicdbdjdcd}{{\iota}{\unicdbdjdcd}{\iota}}}$ be given.
Let $ k7\_qc\_lang1 : {{\iota}{\unicdbdjdcd}{{\iota}{\unicdbdjdcd}{\iota}}}$ be given.
Let $ v1\_xboole\_0 : {{\iota}{\unicdbdjdcd}{o}}$ be given.
Let $ k1\_zfmisc\_1 : {{\iota}{\unicdbdjdcd}{\iota}}$ be given.
Let $ k5\_numbers : {\iota}$ be given.
Assume the following.
\begin{equation}
\begin{array}{c}{\unicdcdadad} X0 .  {\unicdcdadad} X1 .  ((\preoh v1\_xboole\_0~X0){\unicdcdcdhd}((\preoh v1\_xboole\_0~X1){\unicdcdcdhd}\\
(m1\_subset\_1~X1~(k1\_zfmisc\_1~X0)))){\unicdbdjdcd}({\unicdcdadad} X2 .  (m2\_subset\_1~\\
X2~X0~X1){\unicdbdjded}(m1\_subset\_1~X2~X1))\end{array}
\label{hyp:chfgegfgggjgogjgehjgpgogpfngcdpfdhfhcgdhfgehpfbd}
\end{equation}
Assume the following.
\begin{equation}
\begin{array}{c}{\unicdcdadad} X0 .  {\unicdcdadad} X1 .  ((m1\_subset\_1~X0~k5\_numbers){\unicdcdcdhd}(m1\_qc\_lang1~\\
X1)){\unicdbdjdcd}(\preoh v1\_xboole\_0~(k8\_qc\_lang1~X1~X0))\end{array}
\label{hyp:ggdghdpfbhdgpfmgbgoghgbd}
\end{equation}
Assume the following.
\begin{equation}
\begin{array}{c}{\unicdcdadad} X0 .  {\unicdcdadad} X1 .  ((m1\_qc\_lang1~X0){\unicdcdcdhd}(m1\_subset\_1~X1~k5\_numbers)){\unicdbdjdcd}\\
(m1\_subset\_1~(k8\_qc\_lang1~X0~X1)~(k1\_zfmisc\_1~(k6\_qc\_lang1~X0)))\end{array}
\label{hyp:egehpflgidpfbhdgpfmgbgoghgbd}
\end{equation}
Assume the following.
\begin{equation}
\begin{array}{c}{\unicdcdadad} X0 .  {\unicdcdadad} X1 .  ((m1\_qc\_lang1~X0){\unicdcdcdhd}(m1\_subset\_1~X1~(k6\_qc\_lang1~\\
X0))){\unicdbdjdcd}(m1\_subset\_1~(k7\_qc\_lang1~X0~X1)~k5\_numbers)\end{array}
\label{hyp:egehpflghdpfbhdgpfmgbgoghgbd}
\end{equation}
Assume the following.
\begin{equation}
\begin{array}{c}{\unicdcdadad} X0 .  (m1\_qc\_lang1~X0){\unicdbdjdcd}({\unicdcdadad} X1 .  (m1\_subset\_1~X1~k5\_numbers){\unicdbdjdcd}\\
(k8\_qc\_lang1~X0~X1=ReplSep~(toset~({\uniddcgcg} X2 : \iota .  m1\_subset\_1~X2~\\
(k6\_qc\_lang1~X0)))~({\uniddcgcg} X2 : \iota .  k7\_qc\_lang1~X0~X2=X1)~({\uniddcgcg} X2 : \\
\iota .  X2)))\end{array}
\label{hyp:egjdpfbhdgpfmgbgoghgbd}
\end{equation}
Assume the following.
\begin{equation}
\begin{array}{c}{\unicdcdadad} X0 .  {\unicdcdadad} X1 .  ((\preoh v1\_xboole\_0~X0){\unicdbdjdcd}((m1\_subset\_1~X1~X0){\unicdbdjded}\\
(X1 \in X0))){\unicdcdcdhd}((v1\_xboole\_0~X0){\unicdbdjdcd}((m1\_subset\_1~X1~X0){\unicdbdjded}(v1\_xboole\_0~\\
X1)))\end{array}
\label{hyp:egbdpfdhfhcgdhfgehpfbd}
\end{equation}
Assume the following.
\begin{equation}
\begin{array}{c}{\unicdcdadad} X0 .  (v1\_xboole\_0~X0){\unicdbdjdcd}({\unicdcdadad} X1 .  (m1\_subset\_1~X1~(k1\_zfmisc\_1~\\
X0)){\unicdbdjdcd}(v1\_xboole\_0~X1))\end{array}
\label{hyp:dgdgbdpfdhfhcgdhfgehpfbd}
\end{equation}
\begin{theorem}\label{thm:ehbdpfbhdgpfmgbgoghgdd}
$$\begin{array}{c}{\unicdcdadad} X0 .  (m1\_qc\_lang1~X0){\unicdbdjdcd}({\unicdcdadad} X1 .  (m1\_subset\_1~X1~(k6\_qc\_lang1~\\
X0)){\unicdbdjdcd}(m2\_subset\_1~X1~(k6\_qc\_lang1~X0)~(k8\_qc\_lang1~X0~(k7\_qc\_lang1~\\
X0~X1))))\end{array}$$
\end{theorem}
\end{document}
