\documentclass{article}
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\newtheorem{conjecture}{Conjecture}
\newcommand\unicdcdgdad{\not=}
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\begin{document}
\title{t97\_qc\_lang2 (TMRmp1doPT4bnQ9gtcCLqBWLY3z2CXmut5a)}
\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 $ v5\_qc\_lang1 : {{\iota}{\unicdbdjdcd}{{\iota}{\unicdbdjdcd}{o}}}$ be given.
Let $ k15\_qc\_lang2 : {{\iota}{\unicdbdjdcd}{{\iota}{\unicdbdjdcd}{\iota}}}$ be given.
Let $ k2\_xboole\_0 : {{\iota}{\unicdbdjdcd}{{\iota}{\unicdbdjdcd}{\iota}}}$ be given.
Let $ k22\_qc\_lang1 : {{\iota}{\unicdbdjdcd}{{\iota}{\unicdbdjdcd}{\iota}}}$ be given.
Let $ k1\_tarski : {{\iota}{\unicdbdjdcd}{\iota}}$ be given.
Let $ m2\_subset\_1 : {{\iota}{\unicdbdjdcd}{{\iota}{\unicdbdjdcd}{{\iota}{\unicdbdjdcd}{o}}}}$ be given.
Let $ k2\_qc\_lang1 : {{\iota}{\unicdbdjdcd}{\iota}}$ be given.
Let $ k3\_qc\_lang1 : {{\iota}{\unicdbdjdcd}{\iota}}$ be given.
Let $ k15\_qc\_lang1 : {{\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 .  (m1\_subset\_1~X1~(k9\_qc\_lang1~\\
X0)){\unicdbdjdcd}({\unicdcdadad} X2 .  (m2\_subset\_1~X2~(k2\_qc\_lang1~X0)~(k3\_qc\_lang1~\\
X0)){\unicdbdjdcd}(k15\_qc\_lang2~X0~(k15\_qc\_lang1~X0~X2~X1)=k2\_xboole\_0~(k15\_qc\_lang2~\\
X0~X1)~(k1\_tarski~(k15\_qc\_lang1~X0~X2~X1)))))\end{array}
\label{hyp:ehjdadpfbhdgpfmgbgoghgcd}
\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}({\unicdcdadad} X2 .  (m1\_subset\_1~X2~(k9\_qc\_lang1~\\
X0)){\unicdbdjdcd}((X2=k22\_qc\_lang1~X0~X1){\unicdbdjded}({\unicdcdaddd} X3 .  (m2\_subset\_1~X3~(k2\_qc\_lang1~\\
X0)~(k3\_qc\_lang1~X0)){\unicdcdcdhd}(X1=k15\_qc\_lang1~X0~X3~X2))))))\end{array}
\label{hyp:egcdidpfbhdgpfmgbgoghgbd}
\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){\unicdbdjded}({\unicdcdaddd} X2 .  (m2\_subset\_1~X2~(k2\_qc\_lang1~\\
X0)~(k3\_qc\_lang1~X0)){\unicdcdcdhd}({\unicdcdaddd} X3 .  (m1\_subset\_1~X3~(k9\_qc\_lang1~\\
X0)){\unicdcdcdhd}(X1=k15\_qc\_lang1~X0~X2~X3)))))\end{array}
\label{hyp:egcdbdpfbhdgpfmgbgoghgbd}
\end{equation}
Assume the following.
\begin{equation}
{{\unicdcdadad} X0 .  {{\unicdcdadad} X1 .  {{{{k2\_xboole\_0}~{X0}}~{X1}}={{{k2\_xboole\_0}~{X1}}~{X0}}}}}\label{hyp:dgpgngngfhehbgehjgghjgehjhpflgcdpfihcgpgpgmgfgpfad}
\end{equation}
\begin{theorem}\label{thm:ehjdhdpfbhdgpfmgbgoghgcd}
$$\begin{array}{c}{\unicdcdadad} X0 .  (m1\_qc\_lang1~X0){\unicdbdjdcd}({\unicdcdadad} X1 .  (m1\_subset\_1~X1~(k9\_qc\_lang1~\\
X0)){\unicdbdjdcd}((v5\_qc\_lang1~X1~X0){\unicdbdjdcd}(k15\_qc\_lang2~X0~X1=k2\_xboole\_0~(k15\_qc\_lang2~\\
X0~(k22\_qc\_lang1~X0~X1))~(k1\_tarski~X1))))\end{array}$$
\end{theorem}
\end{document}
