Induction on the integers

Statement

Let $P(z)$ be a property of integers $z$. Suppose that $P(0)$ holds. Suppose that for each natural number $n$, $P(n)$ implies $P(n + 1)$. Suppose that for each natural number $n$, $P(-n)$ implies $P(-n - 1)$. Then $P(z)$ holds for every integer $z$.

Remarks

The proof applies ordinary induction two times: one time to $P(n)$ and one time to $P(-n)$, for natural numbers $n$. Each integer is $n$ or $-n$ for some natural number $n$, thus the two inductions cover all of $\mathbb{Z}$.

Explanations

L1 human

You start at $0$ and you walk in both directions on the number line. One induction goes up through $1, 2, 3, \dots$, and the other goes down through $-1, -2, -3, \dots$.