\documentclass{article}
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\newtheorem{theorem}{Theorem}
\newtheorem{lemma}{Lemma}
\newtheorem{proposition}{Proposition}
\newtheorem{definition}{Definition}
\newtheorem{example}{Example}
\newtheorem{conjecture}{Conjecture}
\newcommand\unicdcdgdad{\not=}
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\newcommand\unicdcdadjd{\notin}
\newcommand\unicdbdjded{\Leftrightarrow}
\newcommand\unicdcdcdid{\lor}
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\newcommand\unicdcdidid{\not\subseteq}
\newcommand\preoh[1]{\lnot #1}
\newcommand\mname[1]{{\mathsf{#1}}}
\begin{document}
\title{t14\_cqc\_sim1 (TMamNXCRkUf4WQfQegvcqcjWCu9vQ7nmpB7)}
\maketitle
Let $ m1\_qc\_lang1 : {{\iota}{\unicdbdjdcd}{o}}$ be given.
Let $ k7\_cqc\_sim1 : {{\iota}{\unicdbdjdcd}{{\iota}{\unicdbdjdcd}{\iota}}}$ be given.
Let $ k5\_cqc\_lang : {{\iota}{\unicdbdjdcd}{\iota}}$ be given.
Let $ k6\_numbers : {\iota}$ be given.
Let $ v1\_xboole\_0 : {{\iota}{\unicdbdjdcd}{o}}$ be given.
Let $ m1\_subset\_1 : {{\iota}{\unicdbdjdcd}{{\iota}{\unicdbdjdcd}{o}}}$ be given.
Let $ k1\_zfmisc\_1 : {{\iota}{\unicdbdjdcd}{\iota}}$ be given.
Let $ m2\_subset\_1 : {{\iota}{\unicdbdjdcd}{{\iota}{\unicdbdjdcd}{{\iota}{\unicdbdjdcd}{o}}}}$ be given.
Let $ k3\_cqc\_lang : {{\iota}{\unicdbdjdcd}{\iota}}$ be given.
Let $ k5\_numbers : {\iota}$ be given.
Let $ k9\_qc\_lang1 : {{\iota}{\unicdbdjdcd}{\iota}}$ be given.
Let $ v1\_funct\_1 : {{\iota}{\unicdbdjdcd}{o}}$ be given.
Let $ v1\_funct\_2 : {{\iota}{\unicdbdjdcd}{{\iota}{\unicdbdjdcd}{{\iota}{\unicdbdjdcd}{o}}}}$ be given.
Let $ k2\_zfmisc\_1 : {{\iota}{\unicdbdjdcd}{{\iota}{\unicdbdjdcd}{\iota}}}$ be given.
Let $ k1\_recdef\_1 : {{\iota}{\unicdbdjdcd}{{\iota}{\unicdbdjdcd}{\iota}}}$ be given.
Let $ k2\_qc\_lang1 : {{\iota}{\unicdbdjdcd}{\iota}}$ be given.
Let $ k3\_qc\_lang1 : {{\iota}{\unicdbdjdcd}{\iota}}$ be given.
Let $ v5\_relat\_1 : {{\iota}{\unicdbdjdcd}{{\iota}{\unicdbdjdcd}{o}}}$ be given.
Let $ v3\_card\_1 : {{\iota}{\unicdbdjdcd}{{\iota}{\unicdbdjdcd}{o}}}$ be given.
Let $ m2\_finseq\_1 : {{\iota}{\unicdbdjdcd}{{\iota}{\unicdbdjdcd}{o}}}$ be given.
Let $ k6\_qc\_lang1 : {{\iota}{\unicdbdjdcd}{\iota}}$ be given.
Let $ k8\_qc\_lang1 : {{\iota}{\unicdbdjdcd}{{\iota}{\unicdbdjdcd}{\iota}}}$ be given.
Let $ k4\_cqc\_lang : {{\iota}{\unicdbdjdcd}{{\iota}{\unicdbdjdcd}{{\iota}{\unicdbdjdcd}{{\iota}{\unicdbdjdcd}{\iota}}}}}$ be given.
Let $ k6\_cqc\_lang : {{\iota}{\unicdbdjdcd}{{\iota}{\unicdbdjdcd}{\iota}}}$ be given.
Let $ k7\_cqc\_lang : {{\iota}{\unicdbdjdcd}{{\iota}{\unicdbdjdcd}{{\iota}{\unicdbdjdcd}{\iota}}}}$ be given.
Let $ k2\_nat\_1 : {{\iota}{\unicdbdjdcd}{{\iota}{\unicdbdjdcd}{\iota}}}$ be given.
Let $ k11\_cqc\_lang : {{\iota}{\unicdbdjdcd}{{\iota}{\unicdbdjdcd}{{\iota}{\unicdbdjdcd}{\iota}}}}$ be given.
Let $ np\_\_1 : {\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}
{{\unicdcdadad} X0 .  {{({{m1\_qc\_lang1}~{X0}})}{\unicdbdjdcd}{({\preoh{{{v1\_xboole\_0}~{({{k3\_cqc\_lang}~{X0}})}}}})}}}\label{hyp:ggdgcdpfdgbhdgpfmgbgoghg}
\end{equation}
Assume the following.
\begin{equation}
\begin{array}{c}{\unicdcdadad} X0 .  {\unicdcdadad} X1 .  ((m1\_qc\_lang1~X0){\unicdcdcdhd}(m1\_subset\_1~X1~(k3\_cqc\_lang~\\
X0))){\unicdbdjdcd}(m1\_subset\_1~(k7\_cqc\_sim1~X0~X1)~k5\_numbers)\end{array}
\label{hyp:egehpflghdpfdgbhdgpfdhjgngbd}
\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}
Assume the following.
\begin{equation}
\begin{array}{c}{\unicdcdadad} X0 .  (m1\_qc\_lang1~X0){\unicdbdjdcd}(m1\_subset\_1~(k3\_cqc\_lang~X0)~(k1\_zfmisc\_1~\\
(k9\_qc\_lang1~X0)))\end{array}
\label{hyp:egehpflgddpfdgbhdgpfmgbgoghg}
\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}({\unicdcdadad} X2 .  (m1\_subset\_1~X2~k5\_numbers){\unicdbdjdcd}\\
((X2=k7\_cqc\_sim1~X0~X1){\unicdbdjded}({\unicdcdaddd} X3 .  ((v1\_funct\_1~X3){\unicdcdcdhd}((v1\_funct\_2~\\
X3~(k3\_cqc\_lang~X0)~k5\_numbers){\unicdcdcdhd}(m1\_subset\_1~X3~(k1\_zfmisc\_1~\\
(k2\_zfmisc\_1~(k3\_cqc\_lang~X0)~k5\_numbers))))){\unicdcdcdhd}((X2=k1\_recdef\_1~\\
X3~X1){\unicdcdcdhd}((k1\_recdef\_1~X3~(k5\_cqc\_lang~X0)=k6\_numbers){\unicdcdcdhd}({\unicdcdadad} X4 .  \\
(m2\_subset\_1~X4~(k9\_qc\_lang1~X0)~(k3\_cqc\_lang~X0)){\unicdbdjdcd}({\unicdcdadad} X5 .  \\
(m2\_subset\_1~X5~(k9\_qc\_lang1~X0)~(k3\_cqc\_lang~X0)){\unicdbdjdcd}({\unicdcdadad} X6 .  \\
(m2\_subset\_1~X6~(k2\_qc\_lang1~X0)~(k3\_qc\_lang1~X0)){\unicdbdjdcd}({\unicdcdadad} X7 .  \\
(m1\_subset\_1~X7~k5\_numbers){\unicdbdjdcd}({\unicdcdadad} X8 .  ((v5\_relat\_1~X8~(k3\_qc\_lang1~\\
X0)){\unicdcdcdhd}((v3\_card\_1~X8~X7){\unicdcdcdhd}(m2\_finseq\_1~X8~(k2\_qc\_lang1~X0)))){\unicdbdjdcd}\\
({\unicdcdadad} X9 .  (m2\_subset\_1~X9~(k6\_qc\_lang1~X0)~(k8\_qc\_lang1~X0~X7)){\unicdbdjdcd}\\
((k1\_recdef\_1~X3~(k4\_cqc\_lang~X7~X0~X9~X8)=k6\_numbers){\unicdcdcdhd}((k1\_recdef\_1~\\
X3~(k6\_cqc\_lang~X0~X4)=k1\_recdef\_1~X3~X4){\unicdcdcdhd}((k1\_recdef\_1~X3~(k7\_cqc\_lang~\\
X0~X4~X5)=k2\_nat\_1~(k1\_recdef\_1~X3~X4)~(k1\_recdef\_1~X3~X5)){\unicdcdcdhd}(k1\_recdef\_1~\\
X3~(k11\_cqc\_lang~X0~X6~X4)=k2\_nat\_1~(k1\_recdef\_1~X3~X4)~np\_\_1))))))))))))))))\end{array}
\label{hyp:eggdpfdgbhdgpfdhjgngbd}
\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:ehbdedpfdgbhdgpfdhjgngbd}
$$\begin{array}{c}{\unicdcdadad} X0 .  (m1\_qc\_lang1~X0){\unicdbdjdcd}(k7\_cqc\_sim1~X0~(k5\_cqc\_lang~X0)=\\
k6\_numbers)\end{array}$$
\end{theorem}
\end{document}
