\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}
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\newcommand\unicdcdaddd{\exists}
\newcommand\unicdcdidid{\not\subseteq}
\newcommand\preoh[1]{\lnot #1}
\newcommand\mname[1]{{\mathsf{#1}}}
\begin{document}
\title{l24\_fib\_num4 (TMYQqQpVWufUSnk3aSNruSxfErofRqDoicd)}
\maketitle
Let $ k3\_square\_1 : {{\iota}{\unicdbdjdcd}{\iota}}$ be given.
Let $ k2\_fib\_num : {\iota}$ be given.
Let $ k13\_complex1 : {{\iota}{\unicdbdjdcd}{{\iota}{\unicdbdjdcd}{\iota}}}$ be given.
Let $ k6\_xcmplx\_0 : {{\iota}{\unicdbdjdcd}{{\iota}{\unicdbdjdcd}{\iota}}}$ be given.
Let $ np\_\_3 : {\iota}$ be given.
Let $ k7\_square\_1 : {{\iota}{\unicdbdjdcd}{\iota}}$ be given.
Let $ np\_\_5 : {\iota}$ be given.
Let $ np\_\_2 : {\iota}$ be given.
Let $ v1\_xreal\_0 : {{\iota}{\unicdbdjdcd}{o}}$ be given.
Let $ k3\_power : {{\iota}{\unicdbdjdcd}{{\iota}{\unicdbdjdcd}{\iota}}}$ 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}k3\_power~k2\_fib\_num~np\_\_2=k13\_complex1~(k6\_xcmplx\_0~np\_\_3~(k7\_square\_1~\\
np\_\_5))~np\_\_2\end{array}
\label{hyp:mgbdbdpfggjgcgpfogfhnged}
\end{equation}
Assume the following.
\begin{equation}
{{v1\_xreal\_0}~{k2\_fib\_num}}\label{hyp:egehpflgcdpfggjgcgpfogfhng}
\end{equation}
\begin{theorem}\label{thm:mgcdedpfggjgcgpfogfhnged}
$$\begin{array}{c}k3\_square\_1~k2\_fib\_num=k13\_complex1~(k6\_xcmplx\_0~np\_\_3~(k7\_square\_1~\\
np\_\_5))~np\_\_2\end{array}$$
\end{theorem}
\end{document}
