\documentclass{article}
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\newtheorem{proposition}{Proposition}
\newtheorem{definition}{Definition}
\newtheorem{example}{Example}
\newtheorem{conjecture}{Conjecture}
\newcommand\unicdcdgdad{\not=}
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\begin{document}
\title{t15\_fib\_num3 (TMGWgwWaYVAdZhNev3kuPaaobqfoDY85sYu)}
\maketitle
Let $ k1\_fib\_num3 : {{\iota}{\unicdbdjdcd}{\iota}}$ be given.
Let $ np\_\_3 : {\iota}$ be given.
Let $ np\_\_4 : {\iota}$ be given.
Let $ np\_\_2 : {\iota}$ be given.
Let $ k6\_numbers : {\iota}$ be given.
Let $ np\_\_1 : {\iota}$ be given.
Let $ v7\_ordinal1 : {{\iota}{\unicdbdjdcd}{o}}$ be given.
Let $ k2\_nat\_1 : {{\iota}{\unicdbdjdcd}{{\iota}{\unicdbdjdcd}{\iota}}}$ be given.
Let $ k1\_nat\_1 : {{\iota}{\unicdbdjdcd}{{\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 $ k2\_xcmplx\_0 : {{\iota}{\unicdbdjdcd}{{\iota}{\unicdbdjdcd}{\iota}}}$ be given.
Let $ k4\_ordinal1 : {\iota}$ be given.
Assume the following.
\begin{equation}
{{{k1\_fib\_num3}~{np\_\_2}}={np\_\_3}}\label{hyp:ehbdedpfggjgcgpfogfhngdd}
\end{equation}
Assume the following.
\begin{equation}
\begin{array}{c}(k1\_fib\_num3~k6\_numbers=np\_\_2){\unicdcdcdhd}((k1\_fib\_num3~np\_\_1=np\_\_1){\unicdcdcdhd}\\
({\unicdcdadad} X0 .  (v7\_ordinal1~X0){\unicdbdjdcd}(k1\_fib\_num3~(k2\_nat\_1~(k1\_nat\_1~\\
X0~np\_\_1)~np\_\_1)=k2\_nat\_1~(k1\_fib\_num3~X0)~(k1\_fib\_num3~(k1\_nat\_1~\\
X0~np\_\_1)))))\end{array}
\label{hyp:ehbdbdpfggjgcgpfogfhngdd}
\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}
\begin{array}{c}((v2\_xxreal\_0~np\_\_2){\unicdcdcdhd}(m2\_subset\_1~np\_\_2~k1\_numbers~k5\_numbers)){\unicdcdcdhd}\\
((m1\_subset\_1~np\_\_2~k5\_numbers){\unicdcdcdhd}(m1\_subset\_1~np\_\_2~k1\_numbers))\end{array}
\label{hyp:dhahdgcdpfogfhngfgchbgmgdh}
\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}
{{{{k2\_xcmplx\_0}~{np\_\_3}}~{np\_\_1}}={np\_\_4}}\label{hyp:chbhcffgbgmgbeegegpfpflgcdpfihdgngahmgihpfadpfpfchddpfchbdpfched}
\end{equation}
Assume the following.
\begin{equation}
{{{{k2\_xcmplx\_0}~{np\_\_2}}~{np\_\_1}}={np\_\_3}}\label{hyp:chbhcffgbgmgbeegegpfpflgcdpfihdgngahmgihpfadpfpfchcdpfchbdpfchdd}
\end{equation}
Assume the following.
\begin{equation}
{{{{k2\_xcmplx\_0}~{np\_\_1}}~{np\_\_1}}={np\_\_2}}\label{hyp:chbhcffgbgmgbeegegpfpflgcdpfihdgngahmgihpfadpfpfchbdpfchbdpfchcd}
\end{equation}
Assume the following.
\begin{equation}
{{k5\_numbers}={k4\_ordinal1}}\label{hyp:chfgegfgggjgogjgehjgpgogpflgfdpfogfhngcgfgchdh}
\end{equation}
Assume the following.
\begin{equation}
\begin{array}{c}{\unicdcdadad} X0 .  {\unicdcdadad} X1 .  ((m1\_subset\_1~X0~k5\_numbers){\unicdcdcdhd}(v7\_ordinal1~\\
X1)){\unicdbdjdcd}(k2\_nat\_1~X0~X1=k2\_xcmplx\_0~X0~X1)\end{array}
\label{hyp:chfgegfgggjgogjgehjgpgogpflgcdpfogbgehpfbd}
\end{equation}
Assume the following.
\begin{equation}
\begin{array}{c}{\unicdcdadad} X0 .  {\unicdcdadad} X1 .  ((v7\_ordinal1~X0){\unicdcdcdhd}(m1\_subset\_1~X1~k5\_numbers)){\unicdbdjdcd}\\
(k1\_nat\_1~X0~X1=k2\_xcmplx\_0~X0~X1)\end{array}
\label{hyp:chfgegfgggjgogjgehjgpgogpflgbdpfogbgehpfbd}
\end{equation}
Assume the following.
\begin{equation}
\begin{array}{c}{\unicdcdadad} X0 .  {\unicdcdadad} X1 .  ((m1\_subset\_1~X0~k5\_numbers){\unicdcdcdhd}(v7\_ordinal1~\\
X1)){\unicdbdjdcd}(k2\_nat\_1~X0~X1=k2\_nat\_1~X1~X0)\end{array}
\label{hyp:dgpgngngfhehbgehjgghjgehjhpflgcdpfogbgehpfbd}
\end{equation}
Assume the following.
\begin{equation}
{{\unicdcdadad} X0 .  {{({{{m1\_subset\_1}~{X0}}~{k4\_ordinal1}})}{\unicdbdjdcd}{({{v7\_ordinal1}~{X0}})}}}\label{hyp:dgdgidpfpgchegjgogbgmgbd}
\end{equation}
\begin{theorem}\label{thm:ehbdfdpfggjgcgpfogfhngdd}
${{{k1\_fib\_num3}~{np\_\_3}}={np\_\_4}}$.
\end{theorem}
\end{document}
