\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}
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\newcommand\unicdbdjdcd{\Rightarrow}
\newcommand\unicdcdadjd{\notin}
\newcommand\unicdbdjded{\Leftrightarrow}
\newcommand\unicdcdcdid{\lor}
\newcommand\unicdcdcdhd{\land}
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\newcommand\unicdcdaddd{\exists}
\newcommand\unicdcdidid{\not\subseteq}
\newcommand\preoh[1]{\lnot #1}
\newcommand\mname[1]{{\mathsf{#1}}}
\begin{document}
\title{t22\_cqc\_sim1 (TMGsHaAtWxEdHLUyXvUnNEKZWeSCc4qYha5)}
\maketitle
Let $ m1\_qc\_lang1 : {{\iota}{\unicdbdjdcd}{o}}$ be given.
Let $ k13\_cqc\_sim1 : {{\iota}{\unicdbdjdcd}{{\iota}{\unicdbdjdcd}{\iota}}}$ be given.
Let $ k5\_cqc\_lang : {{\iota}{\unicdbdjdcd}{\iota}}$ be given.
Let $ k26\_qc\_lang1 : {{\iota}{\unicdbdjdcd}{\iota}}$ be given.
Let $ v6\_qc\_lang1 : {{\iota}{\unicdbdjdcd}{{\iota}{\unicdbdjdcd}{o}}}$ be given.
Let $ k12\_qc\_lang1 : {{\iota}{\unicdbdjdcd}{\iota}}$ be given.
Let $ k1\_qc\_lang2 : {{\iota}{\unicdbdjdcd}{\iota}}$ be given.
Let $ m2\_subset\_1 : {{\iota}{\unicdbdjdcd}{{\iota}{\unicdbdjdcd}{{\iota}{\unicdbdjdcd}{o}}}}$ 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}((v6\_qc\_lang1~(k12\_qc\_lang1~X0)~\\
X0){\unicdcdcdhd}(v6\_qc\_lang1~(k1\_qc\_lang2~X0)~X0))\end{array}
\label{hyp:ehcdadpfbhdgpfmgbgoghgdd}
\end{equation}
Assume the following.
\begin{equation}
\begin{array}{c}{\unicdcdadad} X0 .  (m1\_qc\_lang1~X0){\unicdbdjdcd}({\unicdcdadad} X1 .  (m2\_subset\_1~X1~(k9\_qc\_lang1~\\
X0)~(k3\_cqc\_lang~X0)){\unicdbdjdcd}((k13\_cqc\_sim1~X0~X1=k26\_qc\_lang1~X0){\unicdbdjded}\\
(v6\_qc\_lang1~X1~X0)))\end{array}
\label{hyp:ehcdadpfdgbhdgpfdhjgngbd}
\end{equation}
Assume the following.
\begin{equation}
{{\unicdcdadad} X0 .  {{({{m1\_qc\_lang1}~{X0}})}{\unicdbdjdcd}{({{{k5\_cqc\_lang}~{X0}}={{k12\_qc\_lang1}~{X0}}})}}}\label{hyp:chfgegfgggjgogjgehjgpgogpflgfdpfdgbhdgpfmgbgoghg}
\end{equation}
Assume the following.
\begin{equation}
\begin{array}{c}{\unicdcdadad} X0 .  (m1\_qc\_lang1~X0){\unicdbdjdcd}(m2\_subset\_1~(k5\_cqc\_lang~X0)~(k9\_qc\_lang1~\\
X0)~(k3\_cqc\_lang~X0))\end{array}
\label{hyp:egehpflgfdpfdgbhdgpfmgbgoghg}
\end{equation}
\begin{theorem}\label{thm:ehcdcdpfdgbhdgpfdhjgngbd}
$$\begin{array}{c}{\unicdcdadad} X0 .  (m1\_qc\_lang1~X0){\unicdbdjdcd}(k13\_cqc\_sim1~X0~(k5\_cqc\_lang~X0)=\\
k26\_qc\_lang1~X0)\end{array}$$
\end{theorem}
\end{document}
