Result:
inverse_hyperoperator_arithmetic
Source:
\documentclass[]{article}
\usepackage{amsmath}
\usepackage{tikz}
\usepackage{listings}
\newcommand{\HO}[1]{%
\begin{tikzpicture}[#1]%
\draw (0,1.3ex) — (3ex,1.3ex);%
\draw (1.5ex,1ex) — (1.5ex,1.6ex);%
\end{tikzpicture}%
}
\newcommand{\IHO}[1]{%
\begin{tikzpicture}[#1]%
\draw (0,0.4ex) — (3ex,0.4ex);%
\draw (1.5ex,0.025ex) — (1.5ex,0.05ex);%
\end{tikzpicture}%
}
\begin{document}
\section{Notation}
\subsection{Hyperoperation}
\begin{equation}
a+b = a {\overset{1}{\HO{scale=2}}} b
\end{equation}
\begin{equation}
\label{hyperoperator}
a {\overset{n}{\HO{scale=2}}} a {\overset{n}{\HO{scale=2}}} … a = a {\overset{n+1}{\HO{scale=2}}} b
\end{equation}
\subsection{Inverse-hyperoperation}
\begin{equation}
c \overset{1}{\IHO{scale=2}} b =c-b
\end{equation}
\begin{equation}
\label{hyperoperator}
c {\overset{n}{\IHO{scale=2}}} b = a
\end{equation}
\end{document}