Proposition 2

Let $n$ be an integer. Then every cardinal $\alpha$ such that $\alpha \le n$ is an integer. If $n \ne 0$, there exists a unique integer $m$ such that $n = m + 1$, and the relation $\alpha < n$ is equivalent to $\alpha \le m$.

If $\alpha \le n$, there exists a cardinal $\mathfrak{b}$ such that $n = \alpha + \mathfrak{b}$ (§ 3, no. 6, Proposition 13). Then $(\alpha + 1) + \mathfrak{b} = (\alpha + \mathfrak{b}) + 1 = n + 1$ (§ 3, no. 3, Proposition 5, Corollary); and since $n \ne n + 1$, we have

Hence $\alpha + 1 \ne \alpha$, which means that $\alpha$ is an integer. If $n \ne 0$, we have $n \ge 1$ (§ 3, no. 2), and therefore there exists a unique cardinal $m$ such that $n = m + 1$ (§ 3, no. 6, Proposition 13 and no. 4, Proposition 8). Since $m \leqslant n$, $m$ is an integer, from what has already been proved. Finally, if an integer $a$ is such that $a < n$, we have $n = a + b$, with $b \neq 0$ (§ 3, no. 6, Proposition 13); since $b$ is an integer, we have $b = c + 1$ and $n = m + 1 = (a + c) + 1$. It follows that $m = a + c$ (§ 3, no. 4, Proposition 8), hence $a \leqslant m$. Conversely, if $a \leqslant m$, we have

and if $a = n = m + 1$, we would have $a > m$, contrary to hypothesis.