\documentclass{article}
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\newtheorem{conjecture}{Conjecture}
\newcommand\unicdcdgdad{\not=}
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\begin{document}
\title{t2\_arithm (TMPvhAr63DHa8XZbihkRfrxDAvvjMVVMAz5)}
\maketitle
Let $ v1\_xcmplx\_0 : {{\iota}{\unicdbdjdcd}{o}}$ be given.
Let $ k3\_xcmplx\_0 : {{\iota}{\unicdbdjdcd}{{\iota}{\unicdbdjdcd}{\iota}}}$ be given.
Let $ k6\_numbers : {\iota}$ be given.
Let $ m1\_subset\_1 : {{\iota}{\unicdbdjdcd}{{\iota}{\unicdbdjdcd}{o}}}$ be given.
Let $ k2\_numbers : {\iota}$ be given.
Let $ k1\_numbers : {\iota}$ be given.
Let $ k5\_arytm\_0 : {{\iota}{\unicdbdjdcd}{{\iota}{\unicdbdjdcd}{\iota}}}$ be given.
Let $ k1\_zfmisc\_1 : {{\iota}{\unicdbdjdcd}{\iota}}$ be given.
Let $ r1\_tarski : {{\iota}{\unicdbdjdcd}{{\iota}{\unicdbdjdcd}{o}}}$ be given.
Let $ v1\_xboole\_0 : {{\iota}{\unicdbdjdcd}{o}}$ be given.
Let $ k1\_xboole\_0 : {\iota}$ be given.
Let $ k4\_ordinal1 : {\iota}$ be given.
Let $ k2\_arytm\_0 : {{\iota}{\unicdbdjdcd}{{\iota}{\unicdbdjdcd}{\iota}}}$ be given.
Let $ k1\_arytm\_0 : {{\iota}{\unicdbdjdcd}{{\iota}{\unicdbdjdcd}{\iota}}}$ be given.
Let $ k5\_numbers : {\iota}$ be given.
Let $ k3\_arytm\_0 : {{\iota}{\unicdbdjdcd}{\iota}}$ be given.
Let $ v3\_ordinal1 : {{\iota}{\unicdbdjdcd}{o}}$ be given.
Let $ k5\_funct\_4 : {{\iota}{\unicdbdjdcd}{{\iota}{\unicdbdjdcd}{{\iota}{\unicdbdjdcd}{{\iota}{\unicdbdjdcd}{{\iota}{\unicdbdjdcd}{\iota}}}}}}$ be given.
Let $ np\_\_1 : {\iota}$ be given.
Assume the following.
\begin{equation}
\begin{array}{c}{\unicdcdadad} X0 .  (m1\_subset\_1~X0~k2\_numbers){\unicdbdjdcd}({\unicdcdaddd} X1 .  (m1\_subset\_1~\\
X1~k1\_numbers){\unicdcdcdhd}({\unicdcdaddd} X2 .  (m1\_subset\_1~X2~k1\_numbers){\unicdcdcdhd}(X0=\\
k5\_arytm\_0~X1~X2)))\end{array}
\label{hyp:ehjdpfbgchjhehngpfad}
\end{equation}
Assume the following.
\begin{equation}
\begin{array}{c}{\unicdcdadad} X0 .  {\unicdcdadad} X1 .  {\unicdcdadad} X2 .  ((X0 \in X1){\unicdcdcdhd}(m1\_subset\_1~X1~(k1\_zfmisc\_1~\\
X2))){\unicdbdjdcd}(m1\_subset\_1~X0~X2)\end{array}
\label{hyp:ehedpfdhfhcgdhfgeh}
\end{equation}
Assume the following.
\begin{equation}
\begin{array}{c}{\unicdcdadad} X0 .  {\unicdcdadad} X1 .  (m1\_subset\_1~X0~(k1\_zfmisc\_1~X1)){\unicdbdjded}(r1\_tarski~\\
X0~X1)\end{array}
\label{hyp:ehddpfdhfhcgdhfgeh}
\end{equation}
Assume the following.
\begin{equation}
\begin{array}{c}{\unicdcdadad} X0 .  {\unicdcdadad} X1 .  (m1\_subset\_1~X0~X1){\unicdbdjdcd}((v1\_xboole\_0~X1){\unicdcdcdid}\\
(X0 \in X1))\end{array}
\label{hyp:ehcdpfdhfhcgdhfgeh}
\end{equation}
Assume the following.
\begin{equation}
{{{m1\_subset\_1}~{k1\_xboole\_0}}~{k4\_ordinal1}}\label{hyp:ehbdpfogfhngfgchbgmgdh}
\end{equation}
Assume the following.
\begin{equation}
\begin{array}{c}{\unicdcdadad} X0 .  (m1\_subset\_1~X0~k1\_numbers){\unicdbdjdcd}({\unicdcdadad} X1 .  (m1\_subset\_1~\\
X1~k1\_numbers){\unicdbdjdcd}((X1=k6\_numbers){\unicdbdjdcd}(k2\_arytm\_0~X0~X1=k6\_numbers)))\end{array}
\label{hyp:ehbdcdpfbgchjhehngpfad}
\end{equation}
Assume the following.
\begin{equation}
\begin{array}{c}{\unicdcdadad} X0 .  (m1\_subset\_1~X0~k1\_numbers){\unicdbdjdcd}({\unicdcdadad} X1 .  (m1\_subset\_1~\\
X1~k1\_numbers){\unicdbdjdcd}((X1=k6\_numbers){\unicdbdjdcd}(k1\_arytm\_0~X0~X1=X0)))\end{array}
\label{hyp:ehbdbdpfbgchjhehngpfad}
\end{equation}
Assume the following.
\begin{equation}
{{\unicdcdadad} X0 .  {{\unicdcdadad} X1 .  {{{r1\_tarski}~{X0}}~{X0}}}}\label{hyp:chfgggmgfgihjgghjgehjhpfchbdpfehbgchdhlgjg}
\end{equation}
Assume the following.
\begin{equation}
{{k6\_numbers}={k1\_xboole\_0}}\label{hyp:chfgegfgggjgogjgehjgpgogpflggdpfogfhngcgfgchdh}
\end{equation}
Assume the following.
\begin{equation}
{{k5\_numbers}={k4\_ordinal1}}\label{hyp:chfgegfgggjgogjgehjgpgogpflgfdpfogfhngcgfgchdh}
\end{equation}
Assume the following.
\begin{equation}
{{{k3\_arytm\_0}~{k6\_numbers}}={k6\_numbers}}\label{hyp:mgddpfbgchjgehigng}
\end{equation}
Assume the following.
\begin{equation}
{{({\preoh{{{v1\_xboole\_0}~{k4\_ordinal1}}}})}{\unicdcdcdhd}{({{v3\_ordinal1}~{k4\_ordinal1}})}}\label{hyp:ggdggdpfpgchegjgogbgmgbd}
\end{equation}
Assume the following.
\begin{equation}
{{{m1\_subset\_1}~{k5\_numbers}}~{({{k1\_zfmisc\_1}~{k1\_numbers}})}}\label{hyp:egehpflgfdpfogfhngcgfgchdh}
\end{equation}
Assume the following.
\begin{equation}
\begin{array}{c}{\unicdcdadad} X0 .  {\unicdcdadad} X1 .  ((m1\_subset\_1~X0~k1\_numbers){\unicdcdcdhd}(m1\_subset\_1~\\
X1~k1\_numbers)){\unicdbdjdcd}(m1\_subset\_1~(k5\_arytm\_0~X0~X1)~k2\_numbers)\end{array}
\label{hyp:egehpflgfdpfbgchjhehngpfad}
\end{equation}
Assume the following.
\begin{equation}
\begin{array}{c}{\unicdcdadad} X0 .  (v1\_xcmplx\_0~X0){\unicdbdjdcd}({\unicdcdadad} X1 .  (v1\_xcmplx\_0~X1){\unicdbdjdcd}({\unicdcdadad} X2 .  \\
(X2=k3\_xcmplx\_0~X0~X1){\unicdbdjded}({\unicdcdaddd} X3 .  (m1\_subset\_1~X3~k1\_numbers){\unicdcdcdhd}\\
({\unicdcdaddd} X4 .  (m1\_subset\_1~X4~k1\_numbers){\unicdcdcdhd}({\unicdcdaddd} X5 .  (m1\_subset\_1~\\
X5~k1\_numbers){\unicdcdcdhd}({\unicdcdaddd} X6 .  (m1\_subset\_1~X6~k1\_numbers){\unicdcdcdhd}((X0=\\
k5\_arytm\_0~X3~X4){\unicdcdcdhd}((X1=k5\_arytm\_0~X5~X6){\unicdcdcdhd}(X2=k5\_arytm\_0~(k1\_arytm\_0~\\
(k2\_arytm\_0~X3~X5)~(k3\_arytm\_0~(k2\_arytm\_0~X4~X6)))~(k1\_arytm\_0~\\
(k2\_arytm\_0~X3~X6)~(k2\_arytm\_0~X4~X5)))))))))))\end{array}
\label{hyp:egfdpfihdgngahmgihpfad}
\end{equation}
Assume the following.
\begin{equation}
\begin{array}{c}{\unicdcdadad} X0 .  (m1\_subset\_1~X0~k1\_numbers){\unicdbdjdcd}({\unicdcdadad} X1 .  (m1\_subset\_1~\\
X1~k1\_numbers){\unicdbdjdcd}(((X1=k6\_numbers){\unicdbdjdcd}(k5\_arytm\_0~X0~X1=X0)){\unicdcdcdhd}((\\
X1{\unicdcdgdad}k6\_numbers){\unicdbdjdcd}(k5\_arytm\_0~X0~X1=k5\_funct\_4~k1\_numbers~k6\_numbers~\\
np\_\_1~X0~X1))))\end{array}
\label{hyp:egfdpfbgchjhehngpfad}
\end{equation}
Assume the following.
\begin{equation}
{{\unicdcdadad} X0 .  {{({{v1\_xcmplx\_0}~{X0}})}{\unicdbdjded}{({{X0} \in {k2\_numbers}})}}}\label{hyp:egcdpfihdgngahmgihpfad}
\end{equation}
Assume the following.
\begin{equation}
{{\unicdcdadad} X0 .  {{({{{m1\_subset\_1}~{X0}}~{k2\_numbers}})}{\unicdbdjdcd}{({{v1\_xcmplx\_0}~{X0}})}}}\label{hyp:dgdgddpfihdgngahmgihpfad}
\end{equation}
\begin{theorem}\label{thm:ehcdpfbgchjgehigng}
${{\unicdcdadad} X0 .  {{({{v1\_xcmplx\_0}~{X0}})}{\unicdbdjdcd}{({{{{k3\_xcmplx\_0}~{X0}}~{k6\_numbers}}={k6\_numbers}})}}}$.
\end{theorem}
\end{document}
