\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{t30\_qc\_lang3 (TMRjNStq4NEJ4mRxLpvMN3hs4SHoC5ETaoN)}
\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 $ k1\_qc\_lang1 : {{\iota}{\unicdbdjdcd}{\iota}}$ be given.
Let $ k2\_qc\_lang3 : {{\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 $ k2\_zfmisc\_1 : {{\iota}{\unicdbdjdcd}{{\iota}{\unicdbdjdcd}{\iota}}}$ be given.
Let $ k1\_tarski : {{\iota}{\unicdbdjdcd}{\iota}}$ be given.
Let $ np\_\_4 : {\iota}$ be given.
Let $ k4\_tarski : {{\iota}{\unicdbdjdcd}{{\iota}{\unicdbdjdcd}{\iota}}}$ be given.
Assume the following.
\begin{equation}
\begin{array}{c}{\unicdcdadad} X0 .  {\unicdcdadad} X1 .  (m1\_subset\_1~X0~X1){\unicdbdjdcd}((v1\_xboole\_0~X1){\unicdcdcdid}\\
(X0 \in X1))\end{array}
\label{hyp:ehcdpfdhfhcgdhfgeh}
\end{equation}
Assume the following.
\begin{equation}
{{\unicdcdadad} X0 .  {{\unicdcdadad} X1 .  {{({{X0} \in {X1}})}{\unicdbdjdcd}{({{{m1\_subset\_1}~{X0}}~{X1}})}}}}\label{hyp:ehbdpfdhfhcgdhfgeh}
\end{equation}
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}
{{\unicdcdadad} X0 .  {{({{m1\_qc\_lang1}~{X0}})}{\unicdbdjdcd}{({\preoh{{{v1\_xboole\_0}~{({{k3\_qc\_lang1}~{X0}})}}}})}}}\label{hyp:ggdgddpfbhdgpfmgbgoghgbd}
\end{equation}
Assume the following.
\begin{equation}
\begin{array}{c}{\unicdcdadad} X0 .  (m1\_qc\_lang1~X0){\unicdbdjdcd}(m1\_subset\_1~(k3\_qc\_lang1~X0)~(k1\_zfmisc\_1~\\
(k2\_qc\_lang1~X0)))\end{array}
\label{hyp:egehpflgddpfbhdgpfmgbgoghgbd}
\end{equation}
Assume the following.
\begin{equation}
\begin{array}{c}{\unicdcdadad} X0 .  (m1\_qc\_lang1~X0){\unicdbdjdcd}(k3\_qc\_lang1~X0=k2\_zfmisc\_1~(k1\_tarski~\\
np\_\_4)~(k1\_qc\_lang1~X0))\end{array}
\label{hyp:egedpfbhdgpfmgbgoghgbd}
\end{equation}
Assume the following.
\begin{equation}
\begin{array}{c}{\unicdcdadad} X0 .  {\unicdcdadad} X1 .  {\unicdcdadad} X2 .  (X2=k2\_zfmisc\_1~X0~X1){\unicdbdjded}({\unicdcdadad} X3 .  \\
(X3 \in X2){\unicdbdjded}({\unicdcdaddd} X4 .  {\unicdcdaddd} X5 .  (X4 \in X0){\unicdcdcdhd}((X5 \in X1){\unicdcdcdhd}(X3=k4\_tarski~\\
X4~X5))))\end{array}
\label{hyp:egcdpfkhggngjgdhdgpfbd}
\end{equation}
Assume the following.
\begin{equation}
\begin{array}{c}{\unicdcdadad} X0 .  (m1\_qc\_lang1~X0){\unicdbdjdcd}({\unicdcdadad} X1 .  (m1\_subset\_1~X1~(k1\_qc\_lang1~\\
X0)){\unicdbdjdcd}(k2\_qc\_lang3~X0~X1=k4\_tarski~np\_\_4~X1))\end{array}
\label{hyp:egcdpfbhdgpfmgbgoghgdd}
\end{equation}
Assume the following.
\begin{equation}
\begin{array}{c}{\unicdcdadad} X0 .  {\unicdcdadad} X1 .  (X1=k1\_tarski~X0){\unicdbdjded}({\unicdcdadad} X2 .  (X2 \in X1){\unicdbdjded}\\
(X2=X0))\end{array}
\label{hyp:egbdpfehbgchdhlgjg}
\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:ehddadpfbhdgpfmgbgoghgdd}
$$\begin{array}{c}{\unicdcdadad} X0 .  (m1\_qc\_lang1~X0){\unicdbdjdcd}({\unicdcdadad} X1 .  (m2\_subset\_1~X1~(k2\_qc\_lang1~\\
X0)~(k3\_qc\_lang1~X0)){\unicdbdjdcd}({\unicdcdaddd} X2 .  (m1\_subset\_1~X2~(k1\_qc\_lang1~\\
X0)){\unicdcdcdhd}(k2\_qc\_lang3~X0~X2=X1)))\end{array}$$
\end{theorem}
\end{document}
