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\newtheorem{conjecture}{Conjecture}
\newcommand\unicdcdgdad{\not=}
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\begin{document}
\title{t34\_qc\_lang3 (TMb4SfKkK8fKW2utgm8qyubRYaeLqyWJfP8)}
\maketitle
Let $ m1\_qc\_lang1 : {{\iota}{\unicdbdjdcd}{o}}$ 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 $ k5\_qc\_lang1 : {{\iota}{\unicdbdjdcd}{\iota}}$ be given.
Let $ k3\_qc\_lang1 : {{\iota}{\unicdbdjdcd}{\iota}}$ be given.
Let $ m1\_subset\_1 : {{\iota}{\unicdbdjdcd}{{\iota}{\unicdbdjdcd}{o}}}$ be given.
Let $ k5\_numbers : {\iota}$ be given.
Let $ k3\_qc\_lang3 : {{\iota}{\unicdbdjdcd}{{\iota}{\unicdbdjdcd}{\iota}}}$ be given.
Let $ k1\_qc\_lang1 : {{\iota}{\unicdbdjdcd}{\iota}}$ be given.
Let $ k2\_qc\_lang3 : {{\iota}{\unicdbdjdcd}{{\iota}{\unicdbdjdcd}{\iota}}}$ be given.
Let $ k4\_tarski : {{\iota}{\unicdbdjdcd}{{\iota}{\unicdbdjdcd}{\iota}}}$ be given.
Let $ k2\_tarski : {{\iota}{\unicdbdjdcd}{{\iota}{\unicdbdjdcd}{\iota}}}$ be given.
Let $ k1\_tarski : {{\iota}{\unicdbdjdcd}{\iota}}$ be given.
Let $ np\_\_6 : {\iota}$ be given.
Let $ np\_\_4 : {\iota}$ be given.
Assume the following.
\begin{equation}
\begin{array}{c}{\unicdcdadad} X0 .  (m1\_qc\_lang1~X0){\unicdbdjdcd}({\unicdcdadad} X1 .  (m2\_subset\_1~X1~(k2\_qc\_lang1~\\
X0)~(k5\_qc\_lang1~X0)){\unicdbdjdcd}({\unicdcdaddd} X2 .  (m1\_subset\_1~X2~k5\_numbers){\unicdcdcdhd}\\
(k3\_qc\_lang3~X0~X2=X1)))\end{array}
\label{hyp:ehddbdpfbhdgpfmgbgoghgdd}
\end{equation}
Assume the following.
\begin{equation}
\begin{array}{c}{\unicdcdadad} X0 .  (m1\_qc\_lang1~X0){\unicdbdjdcd}({\unicdcdadad} X1 .  (m2\_subset\_1~X1~(k2\_qc\_lang1~\\
X0)~(k3\_qc\_lang1~X0)){\unicdbdjdcd}({\unicdcdaddd} X2 .  (m1\_subset\_1~X2~(k1\_qc\_lang1~\\
X0)){\unicdcdcdhd}(k2\_qc\_lang3~X0~X2=X1)))\end{array}
\label{hyp:ehddadpfbhdgpfmgbgoghgdd}
\end{equation}
Assume the following.
\begin{equation}
\begin{array}{c}{\unicdcdadad} X0 .  {\unicdcdadad} X1 .  {\unicdcdadad} X2 .  {\unicdcdadad} X3 .  (k4\_tarski~X0~X1=\\
k4\_tarski~X2~X3){\unicdbdjdcd}((X0=X2){\unicdcdcdhd}(X1=X3))\end{array}
\label{hyp:ehbdpfihehfhahmgfgpfad}
\end{equation}
Assume the following.
\begin{equation}
\begin{array}{c}{\unicdcdadad} X0 .  {\unicdcdadad} X1 .  k4\_tarski~X0~X1=k2\_tarski~(k2\_tarski~X0~\\
X1)~(k1\_tarski~X0)\end{array}
\label{hyp:egfdpfehbgchdhlgjg}
\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}\\
(k3\_qc\_lang3~X0~X1=k4\_tarski~np\_\_6~X1))\end{array}
\label{hyp:egddpfbhdgpfmgbgoghgdd}
\end{equation}
Assume the following.
\begin{equation}
\begin{array}{c}{\unicdcdadad} X0 .  (m1\_qc\_lang1~X0){\unicdbdjdcd}({\unicdcdadad} X1 .  (m1\_subset\_1~X1~(k1\_qc\_lang1~\\
X0)){\unicdbdjdcd}(k2\_qc\_lang3~X0~X1=k4\_tarski~np\_\_4~X1))\end{array}
\label{hyp:egcdpfbhdgpfmgbgoghgdd}
\end{equation}
Assume the following.
\begin{equation}
{{\unicdcdadad} X0 .  {{\unicdcdadad} X1 .  {{{{k2\_tarski}~{X0}}~{X1}}={{{k2\_tarski}~{X1}}~{X0}}}}}\label{hyp:dgpgngngfhehbgehjgghjgehjhpflgcdpfehbgchdhlgjg}
\end{equation}
\begin{theorem}\label{thm:ehddedpfbhdgpfmgbgoghgdd}
$$\begin{array}{c}{\unicdcdadad} X0 .  (m1\_qc\_lang1~X0){\unicdbdjdcd}({\unicdcdadad} X1 .  (m2\_subset\_1~X1~(k2\_qc\_lang1~\\
X0)~(k5\_qc\_lang1~X0)){\unicdbdjdcd}({\unicdcdadad} X2 .  (m2\_subset\_1~X2~(k2\_qc\_lang1~\\
X0)~(k3\_qc\_lang1~X0)){\unicdbdjdcd}(X1{\unicdcdgdad}X2)))\end{array}$$
\end{theorem}
\end{document}
