\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{t12\_substut2 (TMYVkQDinzQbF4oSKph5WukWg5BXp5sTtwn)}
\maketitle
Let $ m1\_qc\_lang1 : {{\iota}{\unicdbdjdcd}{o}}$ 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.
Let $ m1\_subset\_1 : {{\iota}{\unicdbdjdcd}{{\iota}{\unicdbdjdcd}{o}}}$ be given.
Let $ k1\_substut1 : {{\iota}{\unicdbdjdcd}{\iota}}$ be given.
Let $ k16\_substut1 : {{\iota}{\unicdbdjdcd}{\iota}}$ be given.
Let $ k38\_substut1 : {{\iota}{\unicdbdjdcd}{\iota}}$ be given.
Let $ k2\_sublemma : {{\iota}{\unicdbdjdcd}{{\iota}{\unicdbdjdcd}{\iota}}}$ be given.
Let $ k19\_substut1 : {{\iota}{\unicdbdjdcd}{{\iota}{\unicdbdjdcd}{\iota}}}$ be given.
Let $ k7\_cqc\_lang : {{\iota}{\unicdbdjdcd}{{\iota}{\unicdbdjdcd}{{\iota}{\unicdbdjdcd}{\iota}}}}$ be given.
Let $ k6\_cqc\_lang : {{\iota}{\unicdbdjdcd}{{\iota}{\unicdbdjdcd}{\iota}}}$ be given.
Let $ k5\_numbers : {\iota}$ be given.
Let $ k6\_qc\_lang1 : {{\iota}{\unicdbdjdcd}{\iota}}$ be given.
Let $ k8\_qc\_lang1 : {{\iota}{\unicdbdjdcd}{{\iota}{\unicdbdjdcd}{\iota}}}$ be given.
Let $ v5\_relat\_1 : {{\iota}{\unicdbdjdcd}{{\iota}{\unicdbdjdcd}{o}}}$ be given.
Let $ k3\_qc\_lang1 : {{\iota}{\unicdbdjdcd}{\iota}}$ 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 $ k2\_qc\_lang1 : {{\iota}{\unicdbdjdcd}{\iota}}$ be given.
Let $ k4\_cqc\_lang : {{\iota}{\unicdbdjdcd}{{\iota}{\unicdbdjdcd}{{\iota}{\unicdbdjdcd}{{\iota}{\unicdbdjdcd}{\iota}}}}}$ be given.
Let $ k5\_cqc\_lang : {{\iota}{\unicdbdjdcd}{\iota}}$ be given.
Let $ k11\_cqc\_lang : {{\iota}{\unicdbdjdcd}{{\iota}{\unicdbdjdcd}{{\iota}{\unicdbdjdcd}{\iota}}}}$ be given.
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 .  (m2\_subset\_1~X2~(k9\_qc\_lang1~\\
X0)~(k3\_cqc\_lang~X0)){\unicdbdjdcd}((({\unicdcdadad} X3 .  (m1\_subset\_1~X3~(k1\_substut1~\\
X0)){\unicdbdjdcd}({\unicdcdaddd} X4 .  (m2\_subset\_1~X4~(k16\_substut1~X0)~(k38\_substut1~\\
X0)){\unicdcdcdhd}((k2\_sublemma~X0~X4=X1){\unicdcdcdhd}(k19\_substut1~X0~X4=X3)))){\unicdcdcdhd}({\unicdcdadad} X3 .  \\
(m1\_subset\_1~X3~(k1\_substut1~X0)){\unicdbdjdcd}({\unicdcdaddd} X4 .  (m2\_subset\_1~X4~\\
(k16\_substut1~X0)~(k38\_substut1~X0)){\unicdcdcdhd}((k2\_sublemma~X0~X4=X2){\unicdcdcdhd}\\
(k19\_substut1~X0~X4=X3))))){\unicdbdjdcd}({\unicdcdadad} X3 .  (m1\_subset\_1~X3~(k1\_substut1~\\
X0)){\unicdbdjdcd}({\unicdcdaddd} X4 .  (m2\_subset\_1~X4~(k16\_substut1~X0)~(k38\_substut1~\\
X0)){\unicdcdcdhd}((k2\_sublemma~X0~X4=k7\_cqc\_lang~X0~X1~X2){\unicdcdcdhd}(k19\_substut1~\\
X0~X4=X3)))))))\end{array}
\label{hyp:ehfdpfdhfhcgdhehfhehcd}
\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~(k1\_substut1~\\
X0)){\unicdbdjdcd}({\unicdcdaddd} X3 .  (m2\_subset\_1~X3~(k16\_substut1~X0)~(k38\_substut1~\\
X0)){\unicdcdcdhd}((k2\_sublemma~X0~X3=X1){\unicdcdcdhd}(k19\_substut1~X0~X3=X2)))){\unicdbdjdcd}({\unicdcdadad} X2 .  \\
(m1\_subset\_1~X2~(k1\_substut1~X0)){\unicdbdjdcd}({\unicdcdaddd} X3 .  (m2\_subset\_1~X3~\\
(k16\_substut1~X0)~(k38\_substut1~X0)){\unicdcdcdhd}((k2\_sublemma~X0~X3=k6\_cqc\_lang~\\
X0~X1){\unicdcdcdhd}(k19\_substut1~X0~X3=X2))))))\end{array}
\label{hyp:ehedpfdhfhcgdhehfhehcd}
\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}\\
({\unicdcdadad} X2 .  (m2\_subset\_1~X2~(k6\_qc\_lang1~X0)~(k8\_qc\_lang1~X0~X1)){\unicdbdjdcd}\\
({\unicdcdadad} X3 .  ((v5\_relat\_1~X3~(k3\_qc\_lang1~X0)){\unicdcdcdhd}((v3\_card\_1~X3~\\
X1){\unicdcdcdhd}(m2\_finseq\_1~X3~(k2\_qc\_lang1~X0)))){\unicdbdjdcd}({\unicdcdadad} X4 .  (m1\_subset\_1~\\
X4~(k1\_substut1~X0)){\unicdbdjdcd}({\unicdcdaddd} X5 .  (m2\_subset\_1~X5~(k16\_substut1~\\
X0)~(k38\_substut1~X0)){\unicdcdcdhd}((k2\_sublemma~X0~X5=k4\_cqc\_lang~X1~X0~X2~\\
X3){\unicdcdcdhd}(k19\_substut1~X0~X5=X4)))))))\end{array}
\label{hyp:ehcdpfdhfhcgdhehfhehcd}
\end{equation}
Assume the following.
\begin{equation}
\begin{array}{c}{\unicdcdadad} X0 .  (m1\_qc\_lang1~X0){\unicdbdjdcd}({\unicdcdadad} X1 .  (m1\_subset\_1~X1~(k1\_substut1~\\
X0)){\unicdbdjdcd}({\unicdcdaddd} X2 .  (m2\_subset\_1~X2~(k16\_substut1~X0)~(k38\_substut1~\\
X0)){\unicdcdcdhd}((k2\_sublemma~X0~X2=k5\_cqc\_lang~X0){\unicdcdcdhd}(k19\_substut1~X0~X2=\\
X1))))\end{array}
\label{hyp:ehbdpfdhfhcgdhehfhehcd}
\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 .  (m2\_subset\_1~X2~(k2\_qc\_lang1~\\
X0)~(k3\_qc\_lang1~X0)){\unicdbdjdcd}(({\unicdcdadad} X3 .  (m1\_subset\_1~X3~(k1\_substut1~\\
X0)){\unicdbdjdcd}({\unicdcdaddd} X4 .  (m2\_subset\_1~X4~(k16\_substut1~X0)~(k38\_substut1~\\
X0)){\unicdcdcdhd}((k2\_sublemma~X0~X4=X1){\unicdcdcdhd}(k19\_substut1~X0~X4=X3)))){\unicdbdjdcd}({\unicdcdadad} X3 .  \\
(m1\_subset\_1~X3~(k1\_substut1~X0)){\unicdbdjdcd}({\unicdcdaddd} X4 .  (m2\_subset\_1~X4~\\
(k16\_substut1~X0)~(k38\_substut1~X0)){\unicdcdcdhd}((k2\_sublemma~X0~X4=k11\_cqc\_lang~\\
X0~X2~X1){\unicdcdcdhd}(k19\_substut1~X0~X4=X3)))))))\end{array}
\label{hyp:ehbdbdpfdhfhcgdhehfhehcd}
\end{equation}
Assume the following.
\begin{equation}
\begin{array}{c}{\unicdcdadad} X0 : \iota{\unicdbdjdcd}o .  {\unicdcdadad} X1 .  ({\unicdcdadad} X2 .  (m2\_subset\_1~X2~(\\
k9\_qc\_lang1~X1)~(k3\_cqc\_lang~X1)){\unicdbdjdcd}({\unicdcdadad} X3 .  (m2\_subset\_1~X3~\\
(k9\_qc\_lang1~X1)~(k3\_cqc\_lang~X1)){\unicdbdjdcd}({\unicdcdadad} X4 .  (m2\_subset\_1~\\
X4~(k2\_qc\_lang1~X1)~(k3\_qc\_lang1~X1)){\unicdbdjdcd}({\unicdcdadad} X5 .  (m1\_subset\_1~\\
X5~k5\_numbers){\unicdbdjdcd}({\unicdcdadad} X6 .  ((v5\_relat\_1~X6~(k3\_qc\_lang1~X1)){\unicdcdcdhd}\\
((v3\_card\_1~X6~X5){\unicdcdcdhd}(m2\_finseq\_1~X6~(k2\_qc\_lang1~X1)))){\unicdbdjdcd}({\unicdcdadad} X7 .  \\
(m2\_subset\_1~X7~(k6\_qc\_lang1~X1)~(k8\_qc\_lang1~X1~X5)){\unicdbdjdcd}((X0~(k5\_cqc\_lang~\\
X1)){\unicdcdcdhd}((X0~(k4\_cqc\_lang~X5~X1~X7~X6)){\unicdcdcdhd}(((X0~X2){\unicdbdjdcd}(X0~(k6\_cqc\_lang~\\
X1~X2))){\unicdcdcdhd}((((X0~X2){\unicdcdcdhd}(X0~X3)){\unicdbdjdcd}(X0~(k7\_cqc\_lang~X1~X2~X3))){\unicdcdcdhd}((X0~\\
X2){\unicdbdjdcd}(X0~(k11\_cqc\_lang~X1~X4~X2))))))))))))){\unicdbdjdcd}({\unicdcdadad} X2 .  (m2\_subset\_1~\\
X2~(k9\_qc\_lang1~X1)~(k3\_cqc\_lang~X1)){\unicdbdjdcd}(X0~X2))\end{array}
\label{hyp:dhbdpfdgbhdgpfmgbgoghg}
\end{equation}
\begin{theorem}\label{thm:ehbdcdpfdhfhcgdhehfhehcd}
$$\begin{array}{c}{\unicdcdadad} X0 .  (m1\_qc\_lang1~X0){\unicdbdjdcd}({\unicdcdadad} X1 .  (m2\_subset\_1~X1~(k9\_qc\_lang1~\\
X0)~(k3\_cqc\_lang~X0)){\unicdbdjdcd}({\unicdcdadad} X2 .  (m1\_subset\_1~X2~(k1\_substut1~\\
X0)){\unicdbdjdcd}({\unicdcdaddd} X3 .  (m2\_subset\_1~X3~(k16\_substut1~X0)~(k38\_substut1~\\
X0)){\unicdcdcdhd}((k2\_sublemma~X0~X3=X1){\unicdcdcdhd}(k19\_substut1~X0~X3=X2)))))\end{array}$$
\end{theorem}
\end{document}
