\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}
\newcommand\unicdcdadad{\forall}
\newcommand\unicdbdjdcd{\Rightarrow}
\newcommand\unicdcdadjd{\notin}
\newcommand\unicdbdjded{\Leftrightarrow}
\newcommand\unicdcdcdid{\lor}
\newcommand\unicdcdcdhd{\land}
\newcommand\uniddcgcg{\lambda}
\newcommand\unicdcdaddd{\exists}
\newcommand\unicdcdidid{\not\subseteq}
\newcommand\preoh[1]{\lnot #1}
\newcommand\mname[1]{{\mathsf{#1}}}
\begin{document}
\title{l51\_fib\_num4 (TMZCWpD1ALJGCHkSZ6UFo3UnJiuyXyGAoLW)}
\maketitle
Let $ r1\_xxreal\_0 : {{\iota}{\unicdbdjdcd}{{\iota}{\unicdbdjdcd}{o}}}$ be given.
Let $ k4\_power : {{\iota}{\unicdbdjdcd}{{\iota}{\unicdbdjdcd}{\iota}}}$ be given.
Let $ np\_\_2 : {\iota}$ be given.
Let $ k3\_square\_1 : {{\iota}{\unicdbdjdcd}{\iota}}$ be given.
Let $ k6\_xcmplx\_0 : {{\iota}{\unicdbdjdcd}{{\iota}{\unicdbdjdcd}{\iota}}}$ be given.
Let $ np\_\_1 : {\iota}$ be given.
Let $ k7\_square\_1 : {{\iota}{\unicdbdjdcd}{\iota}}$ be given.
Let $ np\_\_5 : {\iota}$ be given.
Let $ v1\_xreal\_0 : {{\iota}{\unicdbdjdcd}{o}}$ be given.
Let $ k3\_power : {{\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 $ k4\_ordinal1 : {\iota}$ be given.
Let $ v1\_int\_1 : {{\iota}{\unicdbdjdcd}{o}}$ be given.
Let $ k1\_pepin : {{\iota}{\unicdbdjdcd}{\iota}}$ be given.
Let $ v7\_ordinal1 : {{\iota}{\unicdbdjdcd}{o}}$ be given.
Assume the following.
\begin{equation}
{{\unicdcdadad} X0 .  {{({{v1\_xreal\_0}~{X0}})}{\unicdbdjdcd}{({{{{k3\_power}~{X0}}~{np\_\_2}}={{k3\_square\_1}~{X0}}})}}}\label{hyp:ehedgdpfahpghhfgch}
\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}
{{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~k1\_numbers){\unicdcdcdhd}(m1\_subset\_1~\\
X1~k1\_numbers)){\unicdbdjdcd}(k4\_power~X0~X1=k3\_power~X0~X1)\end{array}
\label{hyp:chfgegfgggjgogjgehjgpgogpflgedpfahpghhfgch}
\end{equation}
Assume the following.
\begin{equation}
{{\unicdcdadad} X0 .  {{({{v1\_int\_1}~{X0}})}{\unicdbdjdcd}{({{{k1\_pepin}~{X0}}={{k3\_square\_1}~{X0}}})}}}\label{hyp:chfgegfgggjgogjgehjgpgogpflgbdpfahfgahjgog}
\end{equation}
Assume the following.
\begin{equation}
\begin{array}{c}\preoh r1\_xxreal\_0~(k1\_pepin~np\_\_2)~(k3\_square\_1~(k6\_xcmplx\_0~np\_\_1~\\
(k7\_square\_1~np\_\_5)))\end{array}
\label{hyp:mgfdadpfggjgcgpfogfhnged}
\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}{({{v1\_int\_1}~{X0}})}}}\label{hyp:dgdgcdpfjgogehpfbd}
\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}
\begin{theorem}\label{thm:mgfdbdpfggjgcgpfogfhnged}
$$\begin{array}{c}\preoh r1\_xxreal\_0~(k4\_power~np\_\_2~np\_\_2)~(k3\_square\_1~(k6\_xcmplx\_0~\\
np\_\_1~(k7\_square\_1~np\_\_5)))\end{array}$$
\end{theorem}
\end{document}
