\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}
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\newcommand\unicdcdadjd{\notin}
\newcommand\unicdbdjded{\Leftrightarrow}
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\newcommand\unicdcdidid{\not\subseteq}
\newcommand\preoh[1]{\lnot #1}
\newcommand\mname[1]{{\mathsf{#1}}}
\begin{document}
\title{l22\_numbers (TMHhxtimbwjf9zeQ9KPKD8g35ZX7bMtqvN4)}
\maketitle
Let $ k9\_arytm\_3 : {{\iota}{\unicdbdjdcd}{{\iota}{\unicdbdjdcd}{\iota}}}$ be given.
Let $ k12\_arytm\_3 : {\iota}$ be given.
Let $ c2\_\_numbers : {\iota}$ be given.
Let $ v1\_xboole\_0 : {{\iota}{\unicdbdjdcd}{o}}$ be given.
Let $ k1\_xboole\_0 : {\iota}$ be given.
Let $ v3\_ordinal1 : {{\iota}{\unicdbdjdcd}{o}}$ be given.
Let $ m1\_subset\_1 : {{\iota}{\unicdbdjdcd}{{\iota}{\unicdbdjdcd}{o}}}$ be given.
Let $ k5\_arytm\_3 : {\iota}$ be given.
Let $ k8\_ordinal3 : {{\iota}{\unicdbdjdcd}{{\iota}{\unicdbdjdcd}{\iota}}}$ be given.
Let $ k10\_ordinal2 : {{\iota}{\unicdbdjdcd}{{\iota}{\unicdbdjdcd}{\iota}}}$ be given.
Let $ k1\_ordinal1 : {{\iota}{\unicdbdjdcd}{\iota}}$ be given.
Let $ v2\_xxreal\_0 : {{\iota}{\unicdbdjdcd}{o}}$ be given.
Let $ np\_\_1 : {\iota}$ be given.
Let $ m2\_subset\_1 : {{\iota}{\unicdbdjdcd}{{\iota}{\unicdbdjdcd}{{\iota}{\unicdbdjdcd}{o}}}}$ be given.
Let $ k1\_numbers : {\iota}$ be given.
Let $ k5\_numbers : {\iota}$ be given.
Let $ np\_\_0 : {\iota}$ be given.
Let $ np\_\_2 : {\iota}$ be given.
Let $ v7\_ordinal1 : {{\iota}{\unicdbdjdcd}{o}}$ be given.
Let $ k4\_ordinal1 : {\iota}$ be given.
Let $ k1\_arytm\_3 : {\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 .  ((v3\_ordinal1~X0){\unicdcdcdhd}(m1\_subset\_1~X0~k5\_arytm\_3)){\unicdbdjdcd}(\\
{\unicdcdadad} X1 .  ((v3\_ordinal1~X1){\unicdcdcdhd}(m1\_subset\_1~X1~k5\_arytm\_3)){\unicdbdjdcd}(\\
k9\_arytm\_3~X0~X1=k8\_ordinal3~X0~X1))\end{array}
\label{hyp:ehfdidpfbgchjhehngpfdd}
\end{equation}
Assume the following.
\begin{equation}
\begin{array}{c}{\unicdcdadad} X0 .  (v3\_ordinal1~X0){\unicdbdjdcd}({\unicdcdadad} X1 .  (v3\_ordinal1~X1){\unicdbdjdcd}(k10\_ordinal2~\\
X0~(k1\_ordinal1~X1)=k1\_ordinal1~(k10\_ordinal2~X0~X1)))\end{array}
\label{hyp:ehcdidpfpgchegjgogbgmgcd}
\end{equation}
Assume the following.
\begin{equation}
{{\unicdcdadad} X0 .  {{({{v3\_ordinal1}~{X0}})}{\unicdbdjdcd}{({{{{k10\_ordinal2}~{X0}}~{k1\_xboole\_0}}={X0}})}}}\label{hyp:ehcdhdpfpgchegjgogbgmgcd}
\end{equation}
Assume the following.
\begin{equation}
\begin{array}{c}((v2\_xxreal\_0~np\_\_1){\unicdcdcdhd}(m2\_subset\_1~np\_\_1~k1\_numbers~k5\_numbers)){\unicdcdcdhd}\\
((m1\_subset\_1~np\_\_1~k5\_numbers){\unicdcdcdhd}(m1\_subset\_1~np\_\_1~k1\_numbers))\end{array}
\label{hyp:dhahdgbdpfogfhngfgchbgmgdh}
\end{equation}
Assume the following.
\begin{equation}
{{v1\_xboole\_0}~{np\_\_0}}\label{hyp:dhahdgadpfcgpgpgmgfg}
\end{equation}
Assume the following.
\begin{equation}
{{{k1\_ordinal1}~{np\_\_1}}={np\_\_2}}\label{hyp:chbhdffhdgdgpfpflgbdpfpgchegjgogbgmgbdpfpfchbdpfchcd}
\end{equation}
Assume the following.
\begin{equation}
{{{k1\_ordinal1}~{np\_\_0}}={np\_\_1}}\label{hyp:chbhdffhdgdgpfpflgbdpfpgchegjgogbgmgbdpfpfchadpfchbd}
\end{equation}
Assume the following.
\begin{equation}
\begin{array}{c}{\unicdcdadad} X0 .  {\unicdcdadad} X1 .  (((v3\_ordinal1~X0){\unicdcdcdhd}(v7\_ordinal1~X0)){\unicdcdcdhd}\\
((v3\_ordinal1~X1){\unicdcdcdhd}(v7\_ordinal1~X1))){\unicdbdjdcd}(k8\_ordinal3~X0~X1=k10\_ordinal2~\\
X0~X1)\end{array}
\label{hyp:chfgegfgggjgogjgehjgpgogpflgidpfpgchegjgogbgmgdd}
\end{equation}
Assume the following.
\begin{equation}
{{k5\_numbers}={k4\_ordinal1}}\label{hyp:chfgegfgggjgogjgehjgpgogpflgfdpfogfhngcgfgchdh}
\end{equation}
Assume the following.
\begin{equation}
{{k12\_arytm\_3}={k1\_arytm\_3}}\label{hyp:chfgegfgggjgogjgehjgpgogpflgbdcdpfbgchjhehngpfdd}
\end{equation}
Assume the following.
\begin{equation}
\begin{array}{c}(\preoh v1\_xboole\_0~k12\_arytm\_3){\unicdcdcdhd}((v3\_ordinal1~k12\_arytm\_3){\unicdcdcdhd}(m1\_subset\_1~\\
k12\_arytm\_3~k5\_arytm\_3))\end{array}
\label{hyp:egehpflgbdcdpfbgchjhehngpfdd}
\end{equation}
Assume the following.
\begin{equation}
{{c2\_\_numbers}={np\_\_2}}\label{hyp:egfgpfdgcdpfpfogfhngcgfgchdh}
\end{equation}
Assume the following.
\begin{equation}
{{k1\_arytm\_3}={np\_\_1}}\label{hyp:egbdpfbgchjhehngpfdd}
\end{equation}
Assume the following.
\begin{equation}
{{\unicdcdadad} X0 .  {{({{{m1\_subset\_1}~{X0}}~{k4\_ordinal1}})}{\unicdbdjdcd}{({{v7\_ordinal1}~{X0}})}}}\label{hyp:dgdgidpfpgchegjgogbgmgbd}
\end{equation}
Assume the following.
\begin{equation}
{{\unicdcdadad} X0 .  {{({{v7\_ordinal1}~{X0}})}{\unicdbdjdcd}{({{v3\_ordinal1}~{X0}})}}}\label{hyp:dgdggdpfpgchegjgogbgmgbd}
\end{equation}
Assume the following.
\begin{equation}
{{\unicdcdadad} X0 .  {{({{v1\_xboole\_0}~{X0}})}{\unicdbdjdcd}{({{v3\_ordinal1}~{X0}})}}}\label{hyp:dgdgddpfpgchegjgogbgmgbd}
\end{equation}
\begin{theorem}\label{thm:mgcdcdpfogfhngcgfgchdh}
${{{{k9\_arytm\_3}~{k12\_arytm\_3}}~{k12\_arytm\_3}}={c2\_\_numbers}}$.
\end{theorem}
\end{document}
