Proposition 5

If $a$ and $b$ are integers such that $a \leqslant b$, then the interval $[a, b]$ is a finite set whose number of elements is $b - a + 1$.

By Proposition 4 we may restrict ouselves to the case where $a = 0$. The proof is by induction on $b$. If $b = 0$, the result is clear. The relation $0 \leqslant x \leqslant b + 1$ is equivalent to "$0 \leqslant x < b + 1$ or $x = b + 1$", and the relation $0 \leqslant x < b + 1$ is equivalent to $0 \leqslant x \leqslant b$ (§ 4, no. 2, Proposition 2); in other words, the interval $[0, b + 1]$ is the union of $[0, b]$ and $\{b + 1\}$, and these two sets are disjoint. By the inductive hypothesis the number of elements of $[0, b + 1]$ is equal to $(b + 1) + 1$, and the Proposition is proved.