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  "body": "\n## Proof\n\nWe apply ordinary [[induction|0ASJVEFA9J]] two times, once upward and once downward.\nLet $P$ be a property of [[integers|0C4JS9RF5M]] that satisfies the three hypotheses.\n\n1. For each natural number $n$, let $Q_1(n)$ be the claim \"$P(n)$ holds\".\n2. $Q_1(0)$ 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.\n3. By induction, $P(n)$ holds for every natural number $n$.\n4. For each natural number $n$, let $Q_2(n)$ be the claim \"$P(-n)$ holds\".\n5. $Q_2(0)$ holds, because $-0 = 0$ and $P(0)$ holds.\n6. If $Q_2(n)$ holds, then $P(-n - 1)$ holds by the third hypothesis. Since $-n - 1 = -(n + 1)$, $Q_2(n + 1)$ holds.\n7. By induction, $P(-n)$ holds for every natural number $n$.\n8. Each integer $z$ is $n$ or $-n$ for some natural number $n$. So $P(z)$ holds by step 3 or by step 7.\n\n## Idea\n\nThe integers are two copies of the natural numbers glued at $0$: one going up and one going down.\nOrdinary induction covers each copy, starting from the shared base case $P(0)$.\n",
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