\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{t94\_qc\_lang2 (TMHutWTf9ynK7RoQGpWQEnbK3RRBRAdX2zF)}
\maketitle
Let $ m1\_qc\_lang1 : {{\iota}{\unicdbdjdcd}{o}}$ be given.
Let $ m1\_subset\_1 : {{\iota}{\unicdbdjdcd}{{\iota}{\unicdbdjdcd}{o}}}$ be given.
Let $ k9\_qc\_lang1 : {{\iota}{\unicdbdjdcd}{\iota}}$ be given.
Let $ k12\_qc\_lang1 : {{\iota}{\unicdbdjdcd}{\iota}}$ be given.
Let $ v2\_qc\_lang1 : {{\iota}{\unicdbdjdcd}{{\iota}{\unicdbdjdcd}{o}}}$ be given.
Let $ k15\_qc\_lang2 : {{\iota}{\unicdbdjdcd}{{\iota}{\unicdbdjdcd}{\iota}}}$ be given.
Let $ k1\_tarski : {{\iota}{\unicdbdjdcd}{\iota}}$ be given.
Let $ v3\_qc\_lang1 : {{\iota}{\unicdbdjdcd}{{\iota}{\unicdbdjdcd}{o}}}$ be given.
Let $ v4\_qc\_lang1 : {{\iota}{\unicdbdjdcd}{{\iota}{\unicdbdjdcd}{o}}}$ be given.
Let $ v5\_qc\_lang1 : {{\iota}{\unicdbdjdcd}{{\iota}{\unicdbdjdcd}{o}}}$ be given.
Let $ k5\_numbers : {\iota}$ be given.
Let $ m2\_subset\_1 : {{\iota}{\unicdbdjdcd}{{\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 $ 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 $ k10\_qc\_lang1 : {{\iota}{\unicdbdjdcd}{{\iota}{\unicdbdjdcd}{{\iota}{\unicdbdjdcd}{\iota}}}}$ be given.
Let $ k15\_qc\_lang1 : {{\iota}{\unicdbdjdcd}{{\iota}{\unicdbdjdcd}{{\iota}{\unicdbdjdcd}{\iota}}}}$ be given.
Let $ k21\_qc\_lang1 : {{\iota}{\unicdbdjdcd}{{\iota}{\unicdbdjdcd}{\iota}}}$ be given.
Let $ k22\_qc\_lang1 : {{\iota}{\unicdbdjdcd}{{\iota}{\unicdbdjdcd}{\iota}}}$ be given.
Let $ r1\_qc\_lang2 : {{\iota}{\unicdbdjdcd}{{\iota}{\unicdbdjdcd}{{\iota}{\unicdbdjdcd}{o}}}}$ be given.
Let $ r3\_qc\_lang2 : {{\iota}{\unicdbdjdcd}{{\iota}{\unicdbdjdcd}{{\iota}{\unicdbdjdcd}{o}}}}$ be given.
Let $ k3\_qc\_lang1 : {{\iota}{\unicdbdjdcd}{\iota}}$ be given.
Let $ k14\_qc\_lang1 : {{\iota}{\unicdbdjdcd}{{\iota}{\unicdbdjdcd}{{\iota}{\unicdbdjdcd}{\iota}}}}$ be given.
Let $ k13\_qc\_lang1 : {{\iota}{\unicdbdjdcd}{{\iota}{\unicdbdjdcd}{\iota}}}$ be given.
Let $ k19\_qc\_lang1 : {{\iota}{\unicdbdjdcd}{{\iota}{\unicdbdjdcd}{\iota}}}$ be given.
Let $ k20\_qc\_lang1 : {{\iota}{\unicdbdjdcd}{{\iota}{\unicdbdjdcd}{\iota}}}$ be given.
Let $ k18\_qc\_lang1 : {{\iota}{\unicdbdjdcd}{{\iota}{\unicdbdjdcd}{\iota}}}$ be given.
Let $ r2\_qc\_lang2 : {{\iota}{\unicdbdjdcd}{{\iota}{\unicdbdjdcd}{{\iota}{\unicdbdjdcd}{o}}}}$ be given.
Assume the following.
\begin{equation}
\begin{array}{c}{\unicdcdadad} X0 .  (m1\_qc\_lang1~X0){\unicdbdjdcd}({\unicdcdadad} X1 .  (m1\_subset\_1~X1~(k9\_qc\_lang1~\\
X0)){\unicdbdjdcd}(\preoh (X1{\unicdcdgdad}k12\_qc\_lang1~X0){\unicdcdcdhd}((\preoh v2\_qc\_lang1~X1~X0){\unicdcdcdhd}((\preoh v3\_qc\_lang1~\\
X1~X0){\unicdcdcdhd}((\preoh v4\_qc\_lang1~X1~X0){\unicdcdcdhd}(\preoh v5\_qc\_lang1~X1~X0))))))\end{array}
\label{hyp:ehjdpfbhdgpfmgbgoghgbd}
\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 .  ((v3\_card\_1~X3~X1){\unicdcdcdhd}(m2\_finseq\_1~X3~(k2\_qc\_lang1~X0))){\unicdbdjdcd}\\
(k15\_qc\_lang2~X0~(k10\_qc\_lang1~X0~X2~X3)=k1\_tarski~(k10\_qc\_lang1~\\
X0~X2~X3)))))\end{array}
\label{hyp:ehidgdpfbhdgpfmgbgoghgcd}
\end{equation}
Assume the following.
\begin{equation}
\begin{array}{c}{\unicdcdadad} X0 .  (m1\_qc\_lang1~X0){\unicdbdjdcd}(k15\_qc\_lang2~X0~(k12\_qc\_lang1~X0)=\\
k1\_tarski~(k12\_qc\_lang1~X0))\end{array}
\label{hyp:ehidfdpfbhdgpfmgbgoghgcd}
\end{equation}
Assume the following.
\begin{equation}
\begin{array}{c}{\unicdcdadad} X0 .  (m1\_qc\_lang1~X0){\unicdbdjdcd}({\unicdcdadad} X1 .  (m1\_subset\_1~X1~(k9\_qc\_lang1~\\
X0)){\unicdbdjdcd}((v5\_qc\_lang1~X1~X0){\unicdbdjdcd}(X1=k15\_qc\_lang1~X0~(k21\_qc\_lang1~\\
X0~X1)~(k22\_qc\_lang1~X0~X1))))\end{array}
\label{hyp:ehgdpfbhdgpfmgbgoghgcd}
\end{equation}
Assume the following.
\begin{equation}
\begin{array}{c}{\unicdcdadad} X0 .  (m1\_qc\_lang1~X0){\unicdbdjdcd}({\unicdcdadad} X1 .  (m1\_subset\_1~X1~(k9\_qc\_lang1~\\
X0)){\unicdbdjdcd}({\unicdcdadad} X2 .  (m1\_subset\_1~X2~(k9\_qc\_lang1~X0)){\unicdbdjdcd}((r1\_qc\_lang2~\\
X0~X1~X2){\unicdbdjdcd}(r3\_qc\_lang2~X0~X1~X2))))\end{array}
\label{hyp:ehfdddpfbhdgpfmgbgoghgcd}
\end{equation}
Assume the following.
\begin{equation}
\begin{array}{c}{\unicdcdadad} X0 .  (m1\_qc\_lang1~X0){\unicdbdjdcd}({\unicdcdadad} X1 .  (m1\_subset\_1~X1~(k9\_qc\_lang1~\\
X0)){\unicdbdjdcd}({\unicdcdadad} X2 .  (m1\_subset\_1~X2~(k9\_qc\_lang1~X0)){\unicdbdjdcd}({\unicdcdadad} X3 .  \\
(m2\_subset\_1~X3~(k2\_qc\_lang1~X0)~(k3\_qc\_lang1~X0)){\unicdbdjdcd}((r1\_qc\_lang2~\\
X0~X1~(k15\_qc\_lang1~X0~X3~X2)){\unicdbdjded}(X1=X2)))))\end{array}
\label{hyp:ehedgdpfbhdgpfmgbgoghgcd}
\end{equation}
Assume the following.
\begin{equation}
\begin{array}{c}{\unicdcdadad} X0 .  (m1\_qc\_lang1~X0){\unicdbdjdcd}({\unicdcdadad} X1 .  (m1\_subset\_1~X1~(k9\_qc\_lang1~\\
X0)){\unicdbdjdcd}({\unicdcdadad} X2 .  (m1\_subset\_1~X2~(k9\_qc\_lang1~X0)){\unicdbdjdcd}({\unicdcdadad} X3 .  \\
(m1\_subset\_1~X3~(k9\_qc\_lang1~X0)){\unicdbdjdcd}((r1\_qc\_lang2~X0~X1~(k14\_qc\_lang1~\\
X0~X2~X3)){\unicdbdjded}((X1=X2){\unicdcdcdid}(X1=X3))))))\end{array}
\label{hyp:ehedfdpfbhdgpfmgbgoghgcd}
\end{equation}
Assume the following.
\begin{equation}
\begin{array}{c}{\unicdcdadad} X0 .  (m1\_qc\_lang1~X0){\unicdbdjdcd}({\unicdcdadad} X1 .  (m1\_subset\_1~X1~(k9\_qc\_lang1~\\
X0)){\unicdbdjdcd}({\unicdcdadad} X2 .  (m1\_subset\_1~X2~(k9\_qc\_lang1~X0)){\unicdbdjdcd}((r1\_qc\_lang2~\\
X0~X1~(k13\_qc\_lang1~X0~X2)){\unicdbdjded}(X1=X2))))\end{array}
\label{hyp:ehedddpfbhdgpfmgbgoghgcd}
\end{equation}
Assume the following.
\begin{equation}
\begin{array}{c}{\unicdcdadad} X0 .  (m1\_qc\_lang1~X0){\unicdbdjdcd}({\unicdcdadad} X1 .  (m1\_subset\_1~X1~(k9\_qc\_lang1~\\
X0)){\unicdbdjdcd}((v4\_qc\_lang1~X1~X0){\unicdbdjdcd}(X1=k14\_qc\_lang1~X0~(k19\_qc\_lang1~\\
X0~X1)~(k20\_qc\_lang1~X0~X1))))\end{array}
\label{hyp:ehddpfbhdgpfmgbgoghgcd}
\end{equation}
Assume the following.
\begin{equation}
\begin{array}{c}{\unicdcdadad} X0 .  {\unicdcdadad} X1 .  ((m1\_qc\_lang1~X0){\unicdcdcdhd}(m1\_subset\_1~X1~(k9\_qc\_lang1~\\
X0))){\unicdbdjdcd}(m1\_subset\_1~(k22\_qc\_lang1~X0~X1)~(k9\_qc\_lang1~X0))\end{array}
\label{hyp:egehpflgcdcdpfbhdgpfmgbgoghgbd}
\end{equation}
Assume the following.
\begin{equation}
\begin{array}{c}{\unicdcdadad} X0 .  {\unicdcdadad} X1 .  ((m1\_qc\_lang1~X0){\unicdcdcdhd}(m1\_subset\_1~X1~(k9\_qc\_lang1~\\
X0))){\unicdbdjdcd}(m2\_subset\_1~(k21\_qc\_lang1~X0~X1)~(k2\_qc\_lang1~X0)~(k3\_qc\_lang1~\\
X0))\end{array}
\label{hyp:egehpflgcdbdpfbhdgpfmgbgoghgbd}
\end{equation}
Assume the following.
\begin{equation}
\begin{array}{c}{\unicdcdadad} X0 .  {\unicdcdadad} X1 .  ((m1\_qc\_lang1~X0){\unicdcdcdhd}(m1\_subset\_1~X1~(k9\_qc\_lang1~\\
X0))){\unicdbdjdcd}(m1\_subset\_1~(k20\_qc\_lang1~X0~X1)~(k9\_qc\_lang1~X0))\end{array}
\label{hyp:egehpflgcdadpfbhdgpfmgbgoghgbd}
\end{equation}
Assume the following.
\begin{equation}
\begin{array}{c}{\unicdcdadad} X0 .  {\unicdcdadad} X1 .  ((m1\_qc\_lang1~X0){\unicdcdcdhd}(m1\_subset\_1~X1~(k9\_qc\_lang1~\\
X0))){\unicdbdjdcd}(m1\_subset\_1~(k19\_qc\_lang1~X0~X1)~(k9\_qc\_lang1~X0))\end{array}
\label{hyp:egehpflgbdjdpfbhdgpfmgbgoghgbd}
\end{equation}
Assume the following.
\begin{equation}
\begin{array}{c}{\unicdcdadad} X0 .  {\unicdcdadad} X1 .  ((m1\_qc\_lang1~X0){\unicdcdcdhd}(m1\_subset\_1~X1~(k9\_qc\_lang1~\\
X0))){\unicdbdjdcd}(m1\_subset\_1~(k18\_qc\_lang1~X0~X1)~(k9\_qc\_lang1~X0))\end{array}
\label{hyp:egehpflgbdidpfbhdgpfmgbgoghgbd}
\end{equation}
Assume the following.
\begin{equation}
\begin{array}{c}{\unicdcdadad} X0 .  (m1\_qc\_lang1~X0){\unicdbdjdcd}({\unicdcdadad} X1 .  (m1\_subset\_1~X1~(k9\_qc\_lang1~\\
X0)){\unicdbdjdcd}((v3\_qc\_lang1~X1~X0){\unicdbdjdcd}({\unicdcdadad} X2 .  (m1\_subset\_1~X2~(k9\_qc\_lang1~\\
X0)){\unicdbdjdcd}((X2=k18\_qc\_lang1~X0~X1){\unicdbdjded}(X1=k13\_qc\_lang1~X0~X2)))))\end{array}
\label{hyp:egcdedpfbhdgpfmgbgoghgbd}
\end{equation}
Assume the following.
\begin{equation}
\begin{array}{c}{\unicdcdadad} X0 .  (m1\_qc\_lang1~X0){\unicdbdjdcd}({\unicdcdadad} X1 .  (m1\_subset\_1~X1~(k9\_qc\_lang1~\\
X0)){\unicdbdjdcd}({\unicdcdadad} X2 .  (X2=k15\_qc\_lang2~X0~X1){\unicdbdjded}({\unicdcdadad} X3 .  (X3 \in X2){\unicdbdjded}\\
({\unicdcdaddd} X4 .  (m1\_subset\_1~X4~(k9\_qc\_lang1~X0)){\unicdcdcdhd}((X4=X3){\unicdcdcdhd}(r2\_qc\_lang2~\\
X0~X4~X1))))))\end{array}
\label{hyp:egcdcdpfbhdgpfmgbgoghgcd}
\end{equation}
Assume the following.
\begin{equation}
\begin{array}{c}{\unicdcdadad} X0 .  (m1\_qc\_lang1~X0){\unicdbdjdcd}({\unicdcdadad} X1 .  (m1\_subset\_1~X1~(k9\_qc\_lang1~\\
X0)){\unicdbdjdcd}({\unicdcdadad} X2 .  (m1\_subset\_1~X2~(k9\_qc\_lang1~X0)){\unicdbdjdcd}((r3\_qc\_lang2~\\
X0~X1~X2){\unicdbdjded}((r2\_qc\_lang2~X0~X1~X2){\unicdcdcdhd}(X1{\unicdcdgdad}X2)))))\end{array}
\label{hyp:egcdbdpfbhdgpfmgbgoghgcd}
\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 .  (m1\_qc\_lang1~X0){\unicdbdjdcd}({\unicdcdadad} X1 .  (m1\_subset\_1~X1~(k9\_qc\_lang1~\\
X0)){\unicdbdjdcd}((v2\_qc\_lang1~X1~X0){\unicdbdjded}({\unicdcdaddd} X2 .  (m1\_subset\_1~X2~k5\_numbers){\unicdcdcdhd}\\
({\unicdcdaddd} X3 .  (m2\_subset\_1~X3~(k6\_qc\_lang1~X0)~(k8\_qc\_lang1~X0~X2)){\unicdcdcdhd}\\
({\unicdcdaddd} X4 .  ((v3\_card\_1~X4~X2){\unicdcdcdhd}(m2\_finseq\_1~X4~(k2\_qc\_lang1~X0))){\unicdcdcdhd}\\
(X1=k10\_qc\_lang1~X0~X3~X4))))))\end{array}
\label{hyp:egbdidpfbhdgpfmgbgoghgbd}
\end{equation}
\begin{theorem}\label{thm:ehjdedpfbhdgpfmgbgoghgcd}
$$\begin{array}{c}{\unicdcdadad} X0 .  (m1\_qc\_lang1~X0){\unicdbdjdcd}({\unicdcdadad} X1 .  (m1\_subset\_1~X1~(k9\_qc\_lang1~\\
X0)){\unicdbdjdcd}(((X1=k12\_qc\_lang1~X0){\unicdcdcdid}(v2\_qc\_lang1~X1~X0)){\unicdbdjded}(k15\_qc\_lang2~\\
X0~X1=k1\_tarski~X1)))\end{array}$$
\end{theorem}
\end{document}
