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\begin{document}
\title{t37\_fib\_num3 (TMGx4GZcXjaTZsWz1HTma1vCGf4YuaEM4Z3)}
\maketitle
Let $ m1\_subset\_1 : {{\iota}{\unicdbdjdcd}{{\iota}{\unicdbdjdcd}{o}}}$ be given.
Let $ k5\_numbers : {\iota}$ be given.
Let $ k3\_fib\_num3 : {{\iota}{\unicdbdjdcd}{{\iota}{\unicdbdjdcd}{{\iota}{\unicdbdjdcd}{\iota}}}}$ be given.
Let $ k6\_numbers : {\iota}$ be given.
Let $ np\_\_1 : {\iota}$ be given.
Let $ k1\_pre\_ff : {{\iota}{\unicdbdjdcd}{\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 $ np\_\_2 : {\iota}$ be given.
Let $ k1\_xboole\_0 : {\iota}$ be given.
Let $ k4\_ordinal1 : {\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.
Assume the following.
\begin{equation}
\begin{array}{c}{\unicdcdadad} X0 .  (v7\_ordinal1~X0){\unicdbdjdcd}({\unicdcdadad} X1 .  (v7\_ordinal1~X1){\unicdbdjdcd}({\unicdcdadad} X2 .  \\
(v7\_ordinal1~X2){\unicdbdjdcd}(k2\_nat\_1~(k3\_fib\_num3~X0~X1~X2)~(k3\_fib\_num3~\\
X0~X1~(k1\_nat\_1~X2~np\_\_1))=k3\_fib\_num3~X0~X1~(k1\_nat\_1~X2~np\_\_2))))\end{array}
\label{hyp:ehddedpfggjgcgpfogfhngdd}
\end{equation}
Assume the following.
\begin{equation}
\begin{array}{c}{\unicdcdadad} X0 .  (v7\_ordinal1~X0){\unicdbdjdcd}({\unicdcdadad} X1 .  (v7\_ordinal1~X1){\unicdbdjdcd}((\\
k3\_fib\_num3~X0~X1~k6\_numbers=X0){\unicdcdcdhd}((k3\_fib\_num3~X0~X1~np\_\_1=X1){\unicdcdcdhd}\\
({\unicdcdadad} X2 .  (v7\_ordinal1~X2){\unicdbdjdcd}(k3\_fib\_num3~X0~X1~(k2\_nat\_1~(k1\_nat\_1~\\
X2~np\_\_1)~np\_\_1)=k2\_nat\_1~(k3\_fib\_num3~X0~X1~X2)~(k3\_fib\_num3~X0~\\
X1~(k1\_nat\_1~X2~np\_\_1)))))))\end{array}
\label{hyp:ehddcdpfggjgcgpfogfhngdd}
\end{equation}
Assume the following.
\begin{equation}
\begin{array}{c}{\unicdcdadad} X0 .  (v7\_ordinal1~X0){\unicdbdjdcd}(k1\_pre\_ff~(k1\_nat\_1~X0~np\_\_2)=k2\_nat\_1~\\
(k1\_pre\_ff~X0)~(k1\_pre\_ff~(k1\_nat\_1~X0~np\_\_1)))\end{array}
\label{hyp:ehcdedpfggjgcgpfogfhngcd}
\end{equation}
Assume the following.
\begin{equation}
\begin{array}{c}(k1\_pre\_ff~k6\_numbers=k6\_numbers){\unicdcdcdhd}((k1\_pre\_ff~np\_\_1=np\_\_1){\unicdcdcdhd}\\
({\unicdcdadad} X0 .  (v7\_ordinal1~X0){\unicdbdjdcd}(k1\_pre\_ff~(k2\_nat\_1~(k1\_nat\_1~\\
X0~np\_\_1)~np\_\_1)=k2\_nat\_1~(k1\_pre\_ff~X0)~(k1\_pre\_ff~(k1\_nat\_1~\\
X0~np\_\_1)))))\end{array}
\label{hyp:ehbdpfahchfgpfgggg}
\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}((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}
\begin{array}{c}{\unicdcdadad} X0 : \iota{\unicdbdjdcd}o .  ((X0~k6\_numbers){\unicdcdcdhd}((X0~np\_\_1){\unicdcdcdhd}({\unicdcdadad} X1 .  \\
(v7\_ordinal1~X1){\unicdbdjdcd}(((X0~X1){\unicdcdcdhd}(X0~(k1\_nat\_1~X1~np\_\_1))){\unicdbdjdcd}(X0~(k1\_nat\_1~\\
X1~np\_\_2)))))){\unicdbdjdcd}({\unicdcdadad} X1 .  (v7\_ordinal1~X1){\unicdbdjdcd}(X0~X1))\end{array}
\label{hyp:dhbdpfggjgcgpfogfhng}
\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}
{{\unicdcdadad} X0 .  {{({{{m1\_subset\_1}~{X0}}~{k4\_ordinal1}})}{\unicdbdjdcd}{({{v7\_ordinal1}~{X0}})}}}\label{hyp:dgdgidpfpgchegjgogbgmgbd}
\end{equation}
\begin{theorem}\label{thm:ehddhdpfggjgcgpfogfhngdd}
$$\begin{array}{c}{\unicdcdadad} X0 .  (m1\_subset\_1~X0~k5\_numbers){\unicdbdjdcd}(k3\_fib\_num3~k6\_numbers~\\
np\_\_1~X0=k1\_pre\_ff~X0)\end{array}$$
\end{theorem}
\end{document}
