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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]]. Let $A$, $B$ and $C$ be sets.\n\n1. Let $x \\in A \\cap (B \\cup C)$. By the definition of [[intersection|0HBRW4E6G1]], $x \\in A$ and $x \\in B \\cup C$. By the definition of [[union|07KVB4JSC7]], $x \\in B$ or $x \\in C$.\n2. In step 1, if $x \\in B$, then $x \\in A \\cap B$. If $x \\in C$, then $x \\in A \\cap C$. In both cases $x \\in (A \\cap B) \\cup (A \\cap C)$.\n3. Conversely, let $x \\in (A \\cap B) \\cup (A \\cap C)$. Then $x \\in A \\cap B$ or $x \\in A \\cap C$.\n4. In step 3, if $x \\in A \\cap B$, then $x \\in A$ and $x \\in B \\subseteq B \\cup C$. If $x \\in A \\cap C$, then $x \\in A$ and $x \\in C \\subseteq B \\cup C$. In both cases $x \\in A \\cap (B \\cup C)$.\n5. By steps 2 and 4, the sets $A \\cap (B \\cup C)$ and $(A \\cap B) \\cup (A \\cap C)$ have the same elements. So they are equal.\n6. Let $x \\in A \\cup (B \\cap C)$. Then $x \\in A$ or $x \\in B \\cap C$.\n7. In step 6, if $x \\in A$, then $x \\in A \\cup B$ and $x \\in A \\cup C$. If $x \\in B \\cap C$, then $x \\in B$, so $x \\in A \\cup B$, and $x \\in C$, so $x \\in A \\cup C$.\n8. By step 7, in both cases $x \\in (A \\cup B) \\cap (A \\cup C)$.\n9. Conversely, let $x \\in (A \\cup B) \\cap (A \\cup C)$. Then $x \\in A \\cup B$ and $x \\in A \\cup C$. By the [[law of excluded middle|0RYYK8PG4S]], either $x \\in A$ or $x \\notin A$.\n10. In step 9, if $x \\in A$, then $x \\in A \\cup (B \\cap C)$.\n11. In step 9, if $x \\notin A$, then $x \\in A \\cup B$ forces $x \\in B$, and $x \\in A \\cup C$ forces $x \\in C$. So $x \\in B \\cap C \\subseteq A \\cup (B \\cap C)$.\n12. By steps 8, 10 and 11, the sets $A \\cup (B \\cap C)$ and $(A \\cup B) \\cap (A \\cup C)$ have the same elements. So they are equal.\n\n## Idea\n\nMembership in a union is an \"or\", and membership in an intersection is an \"and\". So each set law reduces to a statement about an element, checked by splitting into cases. For the second law, the useful split is whether or not $x$ lies in $A$.\n",
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