\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\unicdbdjdcd{\Rightarrow}
\newcommand\unicdcdadjd{\notin}
\newcommand\unicdbdjded{\Leftrightarrow}
\newcommand\unicdcdcdid{\lor}
\newcommand\unicdcdcdhd{\land}
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\newcommand\unicdcdaddd{\exists}
\newcommand\unicdcdidid{\not\subseteq}
\newcommand\preoh[1]{\lnot #1}
\newcommand\mname[1]{{\mathsf{#1}}}
\begin{document}
\title{l146\_jordan (TMT4WEcixtkahseVa3ZuGnW5deqejrDwJdi)}
\maketitle
Let $ k7\_square\_1 : {{\iota}{\unicdbdjdcd}{\iota}}$ be given.
Let $ np\_\_9 : {\iota}$ be given.
Let $ np\_\_3 : {\iota}$ be given.
Let $ v1\_xboole\_0 : {{\iota}{\unicdbdjdcd}{o}}$ be given.
Let $ k1\_xboole\_0 : {\iota}$ be given.
Let $ v1\_xreal\_0 : {{\iota}{\unicdbdjdcd}{o}}$ be given.
Let $ r1\_xxreal\_0 : {{\iota}{\unicdbdjdcd}{{\iota}{\unicdbdjdcd}{o}}}$ be given.
Let $ k6\_numbers : {\iota}$ be given.
Let $ k6\_square\_1 : {{\iota}{\unicdbdjdcd}{\iota}}$ be given.
Let $ k3\_square\_1 : {{\iota}{\unicdbdjdcd}{\iota}}$ be given.
Let $ v2\_xxreal\_0 : {{\iota}{\unicdbdjdcd}{o}}$ 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 $ m1\_subset\_1 : {{\iota}{\unicdbdjdcd}{{\iota}{\unicdbdjdcd}{o}}}$ be given.
Let $ np\_\_0 : {\iota}$ be given.
Let $ k3\_xcmplx\_0 : {{\iota}{\unicdbdjdcd}{{\iota}{\unicdbdjdcd}{\iota}}}$ be given.
Let $ v1\_xcmplx\_0 : {{\iota}{\unicdbdjdcd}{o}}$ 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 .  (v1\_xreal\_0~X0){\unicdbdjdcd}((r1\_xxreal\_0~k6\_numbers~X0){\unicdbdjdcd}(k6\_square\_1~\\
(k3\_square\_1~X0)=X0))\end{array}
\label{hyp:ehcdcdpfdhbhfhbgchfgpfbd}
\end{equation}
Assume the following.
\begin{equation}
\begin{array}{c}((v2\_xxreal\_0~np\_\_9){\unicdcdcdhd}(m2\_subset\_1~np\_\_9~k1\_numbers~k5\_numbers)){\unicdcdcdhd}\\
((m1\_subset\_1~np\_\_9~k5\_numbers){\unicdcdcdhd}(m1\_subset\_1~np\_\_9~k1\_numbers))\end{array}
\label{hyp:dhahdgjdpfogfhngfgchbgmgdh}
\end{equation}
Assume the following.
\begin{equation}
\begin{array}{c}((v2\_xxreal\_0~np\_\_3){\unicdcdcdhd}(m2\_subset\_1~np\_\_3~k1\_numbers~k5\_numbers)){\unicdcdcdhd}\\
((m1\_subset\_1~np\_\_3~k5\_numbers){\unicdcdcdhd}(m1\_subset\_1~np\_\_3~k1\_numbers))\end{array}
\label{hyp:dhahdgddpfogfhngfgchbgmgdh}
\end{equation}
Assume the following.
\begin{equation}
{{v1\_xboole\_0}~{np\_\_0}}\label{hyp:dhahdgadpfcgpgpgmgfg}
\end{equation}
Assume the following.
\begin{equation}
{{{{k3\_xcmplx\_0}~{np\_\_3}}~{np\_\_3}}={np\_\_9}}\label{hyp:chbhcffgbgmgnefhmgehpfpflgddpfihdgngahmgihpfadpfpfchddpfchddpfchjd}
\end{equation}
Assume the following.
\begin{equation}
{{{r1\_xxreal\_0}~{np\_\_0}}~{np\_\_3}}\label{hyp:chbhmefgdhdhpechfebhfhbgmgpfpfchbdpfihihchfgbgmgpfadpfpfchadpfchdd}
\end{equation}
Assume the following.
\begin{equation}
\begin{array}{c}{\unicdcdadad} X0 .  (m1\_subset\_1~X0~k1\_numbers){\unicdbdjdcd}(k7\_square\_1~X0=k6\_square\_1~\\
X0)\end{array}
\label{hyp:chfgegfgggjgogjgehjgpgogpflghdpfdhbhfhbgchfgpfbd}
\end{equation}
Assume the following.
\begin{equation}
{{k6\_numbers}={k1\_xboole\_0}}\label{hyp:chfgegfgggjgogjgehjgpgogpflggdpfogfhngcgfgchdh}
\end{equation}
Assume the following.
\begin{equation}
{{\unicdcdadad} X0 .  {{({{v1\_xcmplx\_0}~{X0}})}{\unicdbdjdcd}{({{{k3\_square\_1}~{X0}}={{{k3\_xcmplx\_0}~{X0}}~{X0}}})}}}\label{hyp:egbdpfdhbhfhbgchfgpfbd}
\end{equation}
Assume the following.
\begin{equation}
{{\unicdcdadad} X0 .  {{({{{m1\_subset\_1}~{X0}}~{k1\_numbers}})}{\unicdbdjdcd}{({{v1\_xreal\_0}~{X0}})}}}\label{hyp:dgdgbdpfihchfgbgmgpfad}
\end{equation}
Assume the following.
\begin{equation}
{{\unicdcdadad} X0 .  {{({{{m1\_subset\_1}~{X0}}~{k1\_numbers}})}{\unicdbdjdcd}{({{v1\_xcmplx\_0}~{X0}})}}}\label{hyp:dgdgbdpfihdgngahmgihpfad}
\end{equation}
\begin{theorem}\label{thm:mgbdedgdpfkgpgchegbgog}
${{{k7\_square\_1}~{np\_\_9}}={np\_\_3}}$.
\end{theorem}
\end{document}
