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  "body": "\n## Proof\n\nWe argue by contradiction, using a minimal counterexample.\n\n1. Let $R$ be a [[well-founded relation|0230ECX1Q5]] on a set $A$ and let $P(x)$ be a [[predicate|0T3DPBG981]] on $A$.\n2. Assume the hypothesis: for each $x \\in A$, if $P(y)$ holds for each $y \\in A$ with $y \\mathrel{R} x$, then $P(x)$ holds.\n3. Let $S = \\set{x \\in A \\mid P(x) \\text{ is false}}$. Suppose, for a contradiction, that $S$ is nonempty.\n4. Since $R$ is well-founded and $S$ is a nonempty subset of $A$, there is $m \\in S$ such that no $s \\in S$ satisfies $s \\mathrel{R} m$.\n5. Let $y \\in A$ with $y \\mathrel{R} m$. By step 4, $y \\notin S$. By the definition of $S$, $P(y)$ is not false. By the [[law of excluded middle|0RYYK8PG4S]], $P(y)$ holds.\n6. By step 5, $P(y)$ holds for each $y \\in A$ with $y \\mathrel{R} m$. By step 2 with $x = m$, $P(m)$ holds.\n7. But $m \\in S$ by step 4, so $P(m)$ is false. This contradicts step 6.\n8. Hence $S$ is empty. So for each $x \\in A$, $P(x)$ is not false. By the law of excluded middle again, $P(x)$ holds.\n\n## Idea\n\nIf $P$ failed somewhere, well-foundedness would give a minimal place where it fails. Below that place $P$ holds everywhere, so the hypothesis forces $P$ to hold there too. This contradiction shows that $P$ never fails.\n",
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