\documentclass{article}
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\newtheorem{lemma}{Lemma}
\newtheorem{proposition}{Proposition}
\newtheorem{definition}{Definition}
\newtheorem{example}{Example}
\newtheorem{conjecture}{Conjecture}
\newcommand\unicdcdgdad{\not=}
\newcommand\unicdcdcdbg{\cup}
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\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{t20\_qc\_lang1 (TMKQbXWUQnUPqBjqdPgTCiVv45FyX9bVh36)}
\maketitle
Let $ m1\_qc\_lang1 : {{\iota}{\unicdbdjdcd}{o}}$ be given.
Let $ v2\_qc\_lang1 : {{\iota}{\unicdbdjdcd}{{\iota}{\unicdbdjdcd}{o}}}$ be given.
Let $ k12\_qc\_lang1 : {{\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 $ m1\_subset\_1 : {{\iota}{\unicdbdjdcd}{{\iota}{\unicdbdjdcd}{o}}}$ be given.
Let $ k9\_qc\_lang1 : {{\iota}{\unicdbdjdcd}{\iota}}$ be given.
Let $ v1\_xboole\_0 : {{\iota}{\unicdbdjdcd}{o}}$ be given.
Let $ k1\_xboole\_0 : {\iota}$ be given.
Let $ k1\_xtuple\_0 : {{\iota}{\unicdbdjdcd}{\iota}}$ be given.
Let $ k1\_funct\_1 : {{\iota}{\unicdbdjdcd}{{\iota}{\unicdbdjdcd}{\iota}}}$ be given.
Let $ k11\_qc\_lang1 : {{\iota}{\unicdbdjdcd}{{\iota}{\unicdbdjdcd}{\iota}}}$ be given.
Let $ np\_\_1 : {\iota}$ be given.
Let $ k6\_numbers : {\iota}$ be given.
Let $ np\_\_2 : {\iota}$ be given.
Let $ np\_\_3 : {\iota}$ 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 $ np\_\_0 : {\iota}$ be given.
Assume the following.
\begin{equation}
{{\unicdcdadad} X0 .  {{({{v1\_xboole\_0}~{X0}})}{\unicdbdjdcd}{({{X0}={k1\_xboole\_0}})}}}\label{hyp:ehgdpfcgpgpgmgfg}
\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){\unicdbdjdcd}((k1\_xtuple\_0~(k1\_funct\_1~(k11\_qc\_lang1~\\
X0~X1)~np\_\_1){\unicdcdgdad}k6\_numbers){\unicdcdcdhd}((k1\_xtuple\_0~(k1\_funct\_1~(k11\_qc\_lang1~\\
X0~X1)~np\_\_1){\unicdcdgdad}np\_\_1){\unicdcdcdhd}((k1\_xtuple\_0~(k1\_funct\_1~(k11\_qc\_lang1~\\
X0~X1)~np\_\_1){\unicdcdgdad}np\_\_2){\unicdcdcdhd}(k1\_xtuple\_0~(k1\_funct\_1~(k11\_qc\_lang1~\\
X0~X1)~np\_\_1){\unicdcdgdad}np\_\_3))))))\end{array}
\label{hyp:ehbdjdpfbhdgpfmgbgoghgbd}
\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}((k1\_xtuple\_0~(k1\_funct\_1~(k11\_qc\_lang1~X0~(k12\_qc\_lang1~\\
X0))~np\_\_1)=k6\_numbers){\unicdcdcdhd}((\preoh (v2\_qc\_lang1~X1~X0){\unicdcdcdhd}({\unicdcdadad} X2 .  \\
(m1\_subset\_1~X2~k5\_numbers){\unicdbdjdcd}(\preoh m2\_subset\_1~(k1\_funct\_1~(k11\_qc\_lang1~\\
X0~X1)~np\_\_1)~(k6\_qc\_lang1~X0)~(k8\_qc\_lang1~X0~X2)))){\unicdcdcdhd}(((v3\_qc\_lang1~\\
X1~X0){\unicdbdjdcd}(k1\_xtuple\_0~(k1\_funct\_1~(k11\_qc\_lang1~X0~X1)~np\_\_1)=np\_\_1)){\unicdcdcdhd}\\
(((v4\_qc\_lang1~X1~X0){\unicdbdjdcd}(k1\_xtuple\_0~(k1\_funct\_1~(k11\_qc\_lang1~\\
X0~X1)~np\_\_1)=np\_\_2)){\unicdcdcdhd}((v5\_qc\_lang1~X1~X0){\unicdbdjdcd}(k1\_xtuple\_0~(k1\_funct\_1~\\
(k11\_qc\_lang1~X0~X1)~np\_\_1)=np\_\_3)))))))\end{array}
\label{hyp:ehbdidpfbhdgpfmgbgoghgbd}
\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}(((k1\_xtuple\_0~(k1\_funct\_1~(k11\_qc\_lang1~X0~X1)~np\_\_1)=\\
k6\_numbers){\unicdbdjdcd}(X1=k12\_qc\_lang1~X0)){\unicdcdcdhd}(((k1\_xtuple\_0~(k1\_funct\_1~\\
(k11\_qc\_lang1~X0~X1)~np\_\_1)=np\_\_1){\unicdbdjdcd}(v3\_qc\_lang1~X1~X0)){\unicdcdcdhd}(((k1\_xtuple\_0~\\
(k1\_funct\_1~(k11\_qc\_lang1~X0~X1)~np\_\_1)=np\_\_2){\unicdbdjdcd}(v4\_qc\_lang1~\\
X1~X0)){\unicdcdcdhd}(((k1\_xtuple\_0~(k1\_funct\_1~(k11\_qc\_lang1~X0~X1)~np\_\_1)=\\
np\_\_3){\unicdbdjdcd}(v5\_qc\_lang1~X1~X0)){\unicdcdcdhd}(({\unicdcdaddd} X2 .  (m1\_subset\_1~X2~k5\_numbers){\unicdcdcdhd}\\
(m2\_subset\_1~(k1\_funct\_1~(k11\_qc\_lang1~X0~X1)~np\_\_1)~(k6\_qc\_lang1~\\
X0)~(k8\_qc\_lang1~X0~X2))){\unicdbdjdcd}(v2\_qc\_lang1~X1~X0)))))))\end{array}
\label{hyp:ehbdcdpfbhdgpfmgbgoghgbd}
\end{equation}
Assume the following.
\begin{equation}
{{v1\_xboole\_0}~{np\_\_0}}\label{hyp:dhahdgadpfcgpgpgmgfg}
\end{equation}
Assume the following.
\begin{equation}
{{k6\_numbers}={k1\_xboole\_0}}\label{hyp:chfgegfgggjgogjgehjgpgogpflggdpfogfhngcgfgchdh}
\end{equation}
Assume the following.
\begin{equation}
\begin{array}{c}{\unicdcdadad} X0 .  (m1\_qc\_lang1~X0){\unicdbdjdcd}(m1\_subset\_1~(k12\_qc\_lang1~X0)~(\\
k9\_qc\_lang1~X0))\end{array}
\label{hyp:egehpflgbdcdpfbhdgpfmgbgoghgbd}
\end{equation}
Assume the following.
\begin{equation}
{{k1\_xboole\_0}={{the}~{({{\uniddcgcg} X0 : {\iota} .  {{v1\_xboole\_0}~{X0}}})}}}\label{hyp:egcdpfihcgpgpgmgfgpfad}
\end{equation}
\begin{theorem}\label{thm:ehcdadpfbhdgpfmgbgoghgbd}
$$\begin{array}{c}{\unicdcdadad} X0 .  (m1\_qc\_lang1~X0){\unicdbdjdcd}((\preoh v2\_qc\_lang1~(k12\_qc\_lang1~X0)~\\
X0){\unicdcdcdhd}((\preoh v3\_qc\_lang1~(k12\_qc\_lang1~X0)~X0){\unicdcdcdhd}((\preoh v4\_qc\_lang1~(k12\_qc\_lang1~\\
X0)~X0){\unicdcdcdhd}((\preoh v5\_qc\_lang1~(k12\_qc\_lang1~X0)~X0){\unicdcdcdhd}({\unicdcdadad} X1 .  (m1\_subset\_1~\\
X1~(k9\_qc\_lang1~X0)){\unicdbdjdcd}((\preoh (v2\_qc\_lang1~X1~X0){\unicdcdcdhd}(v3\_qc\_lang1~X1~X0)){\unicdcdcdhd}\\
((\preoh (v2\_qc\_lang1~X1~X0){\unicdcdcdhd}(v4\_qc\_lang1~X1~X0)){\unicdcdcdhd}((\preoh (v2\_qc\_lang1~\\
X1~X0){\unicdcdcdhd}(v5\_qc\_lang1~X1~X0)){\unicdcdcdhd}((\preoh (v3\_qc\_lang1~X1~X0){\unicdcdcdhd}(v4\_qc\_lang1~\\
X1~X0)){\unicdcdcdhd}((\preoh (v3\_qc\_lang1~X1~X0){\unicdcdcdhd}(v5\_qc\_lang1~X1~X0)){\unicdcdcdhd}(\preoh (v4\_qc\_lang1~\\
X1~X0){\unicdcdcdhd}(v5\_qc\_lang1~X1~X0))))))))))))\end{array}$$
\end{theorem}
\end{document}
