\subsection{Natural Number Proofs}
To prove $\forall n \in \N. P$, we of course again use induction:

\shade{blue}{Base Case} Show $P[n \mapsto 0]$

\shade{green}{Step Case} Let $m \in \N$ be arbitrary and not free in $P$. We then assume that $P[n \mapsto m]$ and show that $P[n \mapsto m + 1]$

Or the same as a natural deduction rule:
\[
    \begin{prooftree}
        \hypo{\Gamma \vdash P[n \mapsto 0]}
        \hypo{\Gamma, P[n \mapsto m] \vdash P[n \mapsto m + 1]}
        \infer2[$m$ not free in $\Gamma, P$]{\Gamma \vdash \forall n \in \N. P}
    \end{prooftree}
\]
