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  "body": "\n## Proof\n\n1. Let $(\\Omega, \\mathcal{F}, P)$ be a [[probability space|0ZYBZ34RT2]] and let $A, B \\in \\mathcal{F}$ be [[events|00SQGJNRTB]].\n2. The sets $A \\cup B$ and $A \\cap B = (A^c \\cup B^c)^c$ are in $\\mathcal{F}$, because $\\mathcal{F}$ is a [[σ-algebra|04WSW8T3SY]].\n3. We have $A \\subseteq A \\cup B$. By [[monotonicity of probability|0464Z0JP2E]], $P((A \\cup B) \\setminus A) = P(A \\cup B) - P(A)$.\n4. A point of $A \\cup B$ that is not in $A$ lies in $B$. So $(A \\cup B) \\setminus A = B \\setminus A$.\n5. We have $A \\cap B \\subseteq B$. By monotonicity of probability again, $P(B \\setminus (A \\cap B)) = P(B) - P(A \\cap B)$.\n6. A point of $B$ lies outside $A \\cap B$ exactly when it lies outside $A$. So $B \\setminus (A \\cap B) = B \\setminus A$.\n7. By steps 3 to 6, both differences equal $P(B \\setminus A)$. So $P(A \\cup B) - P(A) = P(B) - P(A \\cap B)$.\n8. Adding $P(A) + P(A \\cap B)$ to both sides gives $P(A \\cup B) + P(A \\cap B) = P(A) + P(B)$. Subtracting $P(A \\cap B)$ from both sides gives the equivalent form.\n\n## Idea\n\nThe part of $B$ outside $A$ can be measured in two ways. It is what $A \\cup B$ adds to $A$, and it is what remains of $B$ after removing $A \\cap B$. Equating the two measurements gives the rule.\n",
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