\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}
\newcommand\unicdcdcdjd{\cap}
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\newcommand\unicdcdadjd{\notin}
\newcommand\unicdbdjded{\Leftrightarrow}
\newcommand\unicdcdcdid{\lor}
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\newcommand\unicdcdaddd{\exists}
\newcommand\unicdcdidid{\not\subseteq}
\newcommand\preoh[1]{\lnot #1}
\newcommand\mname[1]{{\mathsf{#1}}}
\begin{document}
\title{t10\_arytm\_0 (TMR8co3VyVvrogdMLcjZUgftKbFwtoXnqN1)}
\maketitle
Let $ m1\_subset\_1 : {{\iota}{\unicdbdjdcd}{{\iota}{\unicdbdjdcd}{o}}}$ be given.
Let $ k1\_numbers : {\iota}$ be given.
Let $ k5\_arytm\_0 : {{\iota}{\unicdbdjdcd}{{\iota}{\unicdbdjdcd}{\iota}}}$ be given.
Let $ k5\_funct\_4 : {{\iota}{\unicdbdjdcd}{{\iota}{\unicdbdjdcd}{{\iota}{\unicdbdjdcd}{{\iota}{\unicdbdjdcd}{{\iota}{\unicdbdjdcd}{\iota}}}}}}$ be given.
Let $ k6\_numbers : {\iota}$ be given.
Let $ np\_\_1 : {\iota}$ be given.
Let $ k4\_funct\_4 : {{\iota}{\unicdbdjdcd}{{\iota}{\unicdbdjdcd}{{\iota}{\unicdbdjdcd}{{\iota}{\unicdbdjdcd}{\iota}}}}}$ be given.
Let $ v1\_xboole\_0 : {{\iota}{\unicdbdjdcd}{o}}$ be given.
Let $ k1\_xboole\_0 : {\iota}$ be given.
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}(\preoh k5\_funct\_4~k1\_numbers~k6\_numbers~np\_\_1~X0~X1 \in \\
k1\_numbers))\end{array}
\label{hyp:ehidpfbgchjhehngpfad}
\end{equation}
Assume the following.
\begin{equation}
\begin{array}{c}{\unicdcdadad} X0 .  {\unicdcdadad} X1 .  {\unicdcdadad} X2 .  {\unicdcdadad} X3 .  {\unicdcdadad} X4 .  {\unicdcdadad} X5 .  \\
(k4\_funct\_4~X0~X1~X2~X3=k4\_funct\_4~X0~X1~X4~X5){\unicdbdjdcd}((X0=X1){\unicdcdcdid}((X2=X4){\unicdcdcdhd}\\
(X3=X5)))\end{array}
\label{hyp:ehgdidpfggfhogdgehpfed}
\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}
{\preoh{{{v1\_xboole\_0}~{np\_\_1}}}}\label{hyp:dhahdgbdpfcgpgpgmgfg}
\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 .  {\unicdcdadad} X1 .  {\unicdcdadad} X2 .  {\unicdcdadad} X3 .  {\unicdcdadad} X4 .  ((\preoh v1\_xboole\_0~\\
X0){\unicdcdcdhd}((m1\_subset\_1~X3~X0){\unicdcdcdhd}(m1\_subset\_1~X4~X0))){\unicdbdjdcd}(k5\_funct\_4~X0~\\
X1~X2~X3~X4=k4\_funct\_4~X1~X2~X3~X4)\end{array}
\label{hyp:chfgegfgggjgogjgehjgpgogpflgfdpfggfhogdgehpfed}
\end{equation}
Assume the following.
\begin{equation}
{{v1\_xboole\_0}~{k1\_xboole\_0}}\label{hyp:ggdgbdpfihcgpgpgmgfgpfad}
\end{equation}
Assume the following.
\begin{equation}
{\preoh{{{v1\_xboole\_0}~{k1\_numbers}}}}\label{hyp:ggdgbdpfogfhngcgfgchdh}
\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}
\begin{theorem}\label{thm:ehbdadpfbgchjhehngpfad}
$$\begin{array}{c}{\unicdcdadad} X0 .  (m1\_subset\_1~X0~k1\_numbers){\unicdbdjdcd}({\unicdcdadad} X1 .  (m1\_subset\_1~\\
X1~k1\_numbers){\unicdbdjdcd}({\unicdcdadad} X2 .  (m1\_subset\_1~X2~k1\_numbers){\unicdbdjdcd}({\unicdcdadad} X3 .  \\
(m1\_subset\_1~X3~k1\_numbers){\unicdbdjdcd}((k5\_arytm\_0~X0~X1=k5\_arytm\_0~X2~\\
X3){\unicdbdjdcd}((X0=X2){\unicdcdcdhd}(X1=X3))))))\end{array}$$
\end{theorem}
\end{document}
