First define what we mean by 2, 4, etc, using the successor function \(s(n)\) that maps each number \(n\) onto the number immediately after it, so:

\[\begin{align} s(1) & = 2 \label{a}\tag{Eq.1} \\ s(2) & = 3 \label{b}\tag{Eq.2} \\ s(3) & = 4 \label{c}\tag{Eq.3} \\ \end{align}\]

And so on. Secondly define what we mean by addition:

\[\begin{align} n + 1 & = s(n) \label{d}\tag{Eq.4} \\ n + s(k) & = s(n + k) \label{e}\tag{Eq.5} \end{align}\]

Take \(\ref{d}\) and substitute \(n = 2\):

\[ 2 + 1 = s(2) \]

From \(\ref{b}\) we can substitute \(s(2)\) for \(3\) to get

\[ 2 + 1 = 3 \label{f}\tag{Eq.6} \]

Take \(\ref{e}\) and substitute \(n = 2\), \(k = 1\):

\[ 2 + s(1) = s(2 + 1) \]

From \(\ref{a}\) we can substitute \(s(1)\) for \(2\), and from \(\ref{f}\) we can substitute \(2 + 1\) for \(3\), therefore:

\[ 2 + 2 = s(3) \]

From \(\ref{c}\) we can substitute \(s(3)\) for \(4\), therefore:

\[ 2 + 2 = 4 \]