Proof of induction on the integers

Proof

We apply ordinary induction two times, once upward and once downward. Let $P$ be a property of integers that satisfies the three hypotheses.

1. For each natural number $n$, let $Q_1(n)$ be the claim "$P(n)$ holds".
2. $Q_1(0)$ holds, because $P(0)$ holds. If $Q_1(n)$ holds, then $P(n + 1)$ holds by the second hypothesis, so $Q_1(n + 1)$ holds.
3. By induction, $P(n)$ holds for every natural number $n$.
4. For each natural number $n$, let $Q_2(n)$ be the claim "$P(-n)$ holds".
5. $Q_2(0)$ holds, because $-0 = 0$ and $P(0)$ holds.
6. If $Q_2(n)$ holds, then $P(-n - 1)$ holds by the third hypothesis. Since $-n - 1 = -(n + 1)$, $Q_2(n + 1)$ holds.
7. By induction, $P(-n)$ holds for every natural number $n$.
8. Each integer $z$ is $n$ or $-n$ for some natural number $n$. So $P(z)$ holds by step 3 or by step 7.