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  "body": "\n## Proof\n\nWe show that each side has the same elements as the other. Two [[sets|0SJ3RX9TXA]] with the same elements are equal, by the [[axiom of extensionality|0TB16T2C97]].\n\n1. Let $U$ be a set, let $A, B \\subseteq U$, and write $X^c = U \\setminus X$ for the [[complement|0B1FSS1VZW]] of $X \\subseteq U$.\n2. Let $x$ be an object. By the definitions of complement and [[union|07KVB4JSC7]], $x \\in (A \\cup B)^c$ holds exactly when $x \\in U$ and $\\neg (x \\in A \\lor x \\in B)$.\n3. By the first of [[De Morgan's laws|09PKJKA6J0]], $\\neg (x \\in A \\lor x \\in B)$ is equivalent to $x \\notin A \\land x \\notin B$.\n4. So $x \\in (A \\cup B)^c$ holds exactly when $x \\in U$, $x \\notin A$ and $x \\notin B$. This says $x \\in A^c$ and $x \\in B^c$.\n5. By the definition of [[intersection|0HBRW4E6G1]], step 4 says that $x \\in (A \\cup B)^c$ holds exactly when $x \\in A^c \\cap B^c$. Hence $(A \\cup B)^c = A^c \\cap B^c$.\n6. By the definitions of complement and intersection, $x \\in (A \\cap B)^c$ holds exactly when $x \\in U$ and $\\neg (x \\in A \\land x \\in B)$.\n7. By the second of De Morgan's laws, $\\neg (x \\in A \\land x \\in B)$ is equivalent to $x \\notin A \\lor x \\notin B$.\n8. So $x \\in (A \\cap B)^c$ holds exactly when $x \\in U$, and $x \\notin A$ or $x \\notin B$. This says $x \\in A^c$ or $x \\in B^c$.\n9. By the definition of union, step 8 says that $x \\in (A \\cap B)^c$ holds exactly when $x \\in A^c \\cup B^c$. Hence $(A \\cap B)^c = A^c \\cup B^c$.\n\n## Idea\n\nComplement means \"not\", union means \"or\" and intersection means \"and\". Written for one element, each set law becomes one of De Morgan's laws of logic.\n",
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