\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{t44\_fib\_num3 (TMdBfd4yBo2PLNF5hVgGcuK1vzVcaF3irtv)}
\maketitle
Let $ m1\_subset\_1 : {{\iota}{\unicdbdjdcd}{{\iota}{\unicdbdjdcd}{o}}}$ be given.
Let $ k5\_numbers : {\iota}$ be given.
Let $ k2\_nat\_1 : {{\iota}{\unicdbdjdcd}{{\iota}{\unicdbdjdcd}{\iota}}}$ be given.
Let $ k1\_fib\_num3 : {{\iota}{\unicdbdjdcd}{\iota}}$ be given.
Let $ np\_\_3 : {\iota}$ be given.
Let $ k4\_nat\_1 : {{\iota}{\unicdbdjdcd}{{\iota}{\unicdbdjdcd}{\iota}}}$ be given.
Let $ np\_\_2 : {\iota}$ be given.
Assume the following.
\begin{equation}
\begin{array}{c}{\unicdcdadad} X0 .  (m1\_subset\_1~X0~k5\_numbers){\unicdbdjdcd}(k4\_nat\_1~np\_\_2~(k1\_fib\_num3~\\
(k2\_nat\_1~X0~np\_\_2))=k2\_nat\_1~(k1\_fib\_num3~X0)~(k1\_fib\_num3~(\\
k2\_nat\_1~X0~np\_\_3)))\end{array}
\label{hyp:ehbdjdpfggjgcgpfogfhngdd}
\end{equation}
\begin{theorem}\label{thm:ehededpfggjgcgpfogfhngdd}
$$\begin{array}{c}{\unicdcdadad} X0 .  (m1\_subset\_1~X0~k5\_numbers){\unicdbdjdcd}(k2\_nat\_1~(k1\_fib\_num3~\\
X0)~(k1\_fib\_num3~(k2\_nat\_1~X0~np\_\_3))=k4\_nat\_1~np\_\_2~(k1\_fib\_num3~\\
(k2\_nat\_1~X0~np\_\_2)))\end{array}$$
\end{theorem}
\end{document}
