\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{t28\_sublemma (TMNbKshqRxPyzAnea1up1oP8HKmVcQpbeCN)}
\maketitle
Let $ m1\_qc\_lang1 : {{\iota}{\unicdbdjdcd}{o}}$ be given.
Let $ m2\_subset\_1 : {{\iota}{\unicdbdjdcd}{{\iota}{\unicdbdjdcd}{{\iota}{\unicdbdjdcd}{o}}}}$ be given.
Let $ k16\_substut1 : {{\iota}{\unicdbdjdcd}{\iota}}$ be given.
Let $ k38\_substut1 : {{\iota}{\unicdbdjdcd}{\iota}}$ be given.
Let $ v7\_substut1 : {{\iota}{\unicdbdjdcd}{{\iota}{\unicdbdjdcd}{o}}}$ be given.
Let $ k39\_substut1 : {{\iota}{\unicdbdjdcd}{{\iota}{\unicdbdjdcd}{\iota}}}$ be given.
Let $ k11\_sublemma : {{\iota}{\unicdbdjdcd}{{\iota}{\unicdbdjdcd}{{\iota}{\unicdbdjdcd}{\iota}}}}$ be given.
Let $ k10\_sublemma : {{\iota}{\unicdbdjdcd}{{\iota}{\unicdbdjdcd}{\iota}}}$ be given.
Let $ m1\_subset\_1 : {{\iota}{\unicdbdjdcd}{{\iota}{\unicdbdjdcd}{o}}}$ be given.
Let $ k37\_substut1 : {{\iota}{\unicdbdjdcd}{{\iota}{\unicdbdjdcd}{\iota}}}$ be given.
Let $ k36\_substut1 : {{\iota}{\unicdbdjdcd}{{\iota}{\unicdbdjdcd}{{\iota}{\unicdbdjdcd}{\iota}}}}$ be given.
Let $ k31\_substut1 : {{\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 $ k9\_qc\_lang1 : {{\iota}{\unicdbdjdcd}{\iota}}$ be given.
Let $ k3\_cqc\_lang : {{\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~(k16\_substut1~\\
X0)){\unicdbdjdcd}((v7\_substut1~X1~X0){\unicdbdjdcd}(k37\_substut1~X0~X1=k36\_substut1~X0~\\
X1~(k37\_substut1~X0~(k31\_substut1~X0~X1)))))\end{array}
\label{hyp:ehddcdpfdhfhcgdhehfhehbd}
\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}
\begin{array}{c}{\unicdcdadad} X0 .  {\unicdcdadad} X1 .  ((m1\_qc\_lang1~X0){\unicdcdcdhd}(m1\_subset\_1~X1~(k38\_substut1~\\
X0))){\unicdbdjdcd}(k39\_substut1~X0~X1=k37\_substut1~X0~X1)\end{array}
\label{hyp:chfgegfgggjgogjgehjgpgogpflgddjdpfdhfhcgdhehfhehbd}
\end{equation}
Assume the following.
\begin{equation}
{{\unicdcdadad} X0 .  {{({{m1\_qc\_lang1}~{X0}})}{\unicdbdjdcd}{({\preoh{{{v1\_xboole\_0}~{({{k38\_substut1}~{X0}})}}}})}}}\label{hyp:ggdghdpfdhfhcgdhehfhehbd}
\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){\unicdbdjdcd}(m1\_subset\_1~X2~X0))\end{array}
\label{hyp:egehpfngcdpfdhfhcgdhfgehpfbd}
\end{equation}
Assume the following.
\begin{equation}
\begin{array}{c}{\unicdcdadad} X0 .  {\unicdcdadad} X1 .  ((m1\_qc\_lang1~X0){\unicdcdcdhd}(m1\_subset\_1~X1~(k38\_substut1~\\
X0))){\unicdbdjdcd}(m2\_subset\_1~(k39\_substut1~X0~X1)~(k9\_qc\_lang1~X0)~(k3\_cqc\_lang~\\
X0))\end{array}
\label{hyp:egehpflgddjdpfdhfhcgdhehfhehbd}
\end{equation}
Assume the following.
\begin{equation}
\begin{array}{c}{\unicdcdadad} X0 .  (m1\_qc\_lang1~X0){\unicdbdjdcd}(m1\_subset\_1~(k38\_substut1~X0)~(\\
k1\_zfmisc\_1~(k16\_substut1~X0)))\end{array}
\label{hyp:egehpflgddidpfdhfhcgdhehfhehbd}
\end{equation}
Assume the following.
\begin{equation}
\begin{array}{c}{\unicdcdadad} X0 .  {\unicdcdadad} X1 .  ((m1\_qc\_lang1~X0){\unicdcdcdhd}(m1\_subset\_1~X1~(k38\_substut1~\\
X0))){\unicdbdjdcd}(m2\_subset\_1~(k10\_sublemma~X0~X1)~(k16\_substut1~X0)~(k38\_substut1~\\
X0))\end{array}
\label{hyp:egehpflgbdadpfdhfhcgmgfgngngbg}
\end{equation}
Assume the following.
\begin{equation}
\begin{array}{c}{\unicdcdadad} X0 .  (m1\_qc\_lang1~X0){\unicdbdjdcd}({\unicdcdadad} X1 .  (m2\_subset\_1~X1~(k16\_substut1~\\
X0)~(k38\_substut1~X0)){\unicdbdjdcd}({\unicdcdadad} X2 .  (m2\_subset\_1~X2~(k9\_qc\_lang1~\\
X0)~(k3\_cqc\_lang~X0)){\unicdbdjdcd}(((v7\_substut1~X1~X0){\unicdcdcdhd}(X2=k39\_substut1~\\
X0~(k10\_sublemma~X0~X1))){\unicdbdjdcd}(k11\_sublemma~X0~X1~X2=k36\_substut1~\\
X0~X1~X2))))\end{array}
\label{hyp:eghdpfdhfhcgmgfgngngbg}
\end{equation}
Assume the following.
\begin{equation}
\begin{array}{c}{\unicdcdadad} X0 .  (m1\_qc\_lang1~X0){\unicdbdjdcd}({\unicdcdadad} X1 .  (m2\_subset\_1~X1~(k16\_substut1~\\
X0)~(k38\_substut1~X0)){\unicdbdjdcd}((v7\_substut1~X1~X0){\unicdbdjdcd}(k10\_sublemma~X0~\\
X1=k31\_substut1~X0~X1)))\end{array}
\label{hyp:eggdpfdhfhcgmgfgngngbg}
\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:ehcdidpfdhfhcgmgfgngngbg}
$$\begin{array}{c}{\unicdcdadad} X0 .  (m1\_qc\_lang1~X0){\unicdbdjdcd}({\unicdcdadad} X1 .  (m2\_subset\_1~X1~(k16\_substut1~\\
X0)~(k38\_substut1~X0)){\unicdbdjdcd}((v7\_substut1~X1~X0){\unicdbdjdcd}(k39\_substut1~X0~\\
X1=k11\_sublemma~X0~X1~(k39\_substut1~X0~(k10\_sublemma~X0~X1)))))\end{array}$$
\end{theorem}
\end{document}
