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  "body": "\n## Proof\n\nWe show that $\\neg a \\wedge \\neg b$ is a complement of $a \\vee b$, and then use that complements are unique.\nAll axioms quoted are those of the definition of a [[Boolean algebra|06GKBTAA5E]] $A$.\n\n1. Let $u \\in A$. By commutativity and the identity axiom, $0 \\vee u = u$ and $1 \\wedge u = u$.\n2. Let $u \\in A$. Absorption gives $0 \\wedge (0 \\vee u) = 0$. By step 1, $0 \\vee u = u$, so $0 \\wedge u = 0$.\n3. In the same way, absorption gives $1 \\vee (1 \\wedge u) = 1$. By step 1, $1 \\wedge u = u$, so $1 \\vee u = 1$.\n4. Uniqueness of complements. Let $u, x \\in A$ with $u \\vee x = 1$ and $u \\wedge x = 0$. We claim $x = \\neg u$.\n5. By the identity, complement and distributive axioms, $x = x \\wedge 1 = x \\wedge (u \\vee \\neg u) = (x \\wedge u) \\vee (x \\wedge \\neg u)$.\n6. By commutativity, $x \\wedge u = u \\wedge x = 0$. So by step 1, step 5 gives $x = x \\wedge \\neg u$.\n7. In the same way, $\\neg u = \\neg u \\wedge (u \\vee x) = (\\neg u \\wedge u) \\vee (\\neg u \\wedge x) = 0 \\vee (\\neg u \\wedge x) = \\neg u \\wedge x$. Here $\\neg u \\wedge u = u \\wedge \\neg u = 0$.\n8. By commutativity, $x \\wedge \\neg u = \\neg u \\wedge x$. So steps 6 and 7 give $x = \\neg u$, which proves the claim of step 4.\n9. Now fix $a, b \\in A$. By associativity and commutativity of $\\vee$, $(a \\vee b) \\vee \\neg a = (a \\vee \\neg a) \\vee b = 1 \\vee b$. By step 3 this is $1$.\n10. In the same way, $(a \\vee b) \\vee \\neg b = a \\vee (b \\vee \\neg b) = a \\vee 1 = 1 \\vee a = 1$.\n11. By distributivity, $(a \\vee b) \\vee (\\neg a \\wedge \\neg b) = ((a \\vee b) \\vee \\neg a) \\wedge ((a \\vee b) \\vee \\neg b)$. By steps 9 and 10 this is $1 \\wedge 1 = 1$.\n12. By associativity and commutativity of $\\wedge$, $(\\neg a \\wedge \\neg b) \\wedge a = (a \\wedge \\neg a) \\wedge \\neg b = 0 \\wedge \\neg b$. By step 2 this is $0$.\n13. In the same way, $(\\neg a \\wedge \\neg b) \\wedge b = \\neg a \\wedge (b \\wedge \\neg b) = \\neg a \\wedge 0 = 0 \\wedge \\neg a = 0$.\n14. By commutativity and distributivity, $(a \\vee b) \\wedge (\\neg a \\wedge \\neg b) = ((\\neg a \\wedge \\neg b) \\wedge a) \\vee ((\\neg a \\wedge \\neg b) \\wedge b)$. By steps 12 and 13 this is $0 \\vee 0 = 0$.\n15. By steps 11 and 14, the element $x = \\neg a \\wedge \\neg b$ satisfies the hypotheses of step 4 with $u = a \\vee b$. So $\\neg(a \\vee b) = \\neg a \\wedge \\neg b$.\n16. For the second law, we check that $\\neg a \\vee \\neg b$ is a complement of $a \\wedge b$. By commutativity and distributivity, $(a \\wedge b) \\vee (\\neg a \\vee \\neg b) = ((\\neg a \\vee \\neg b) \\vee a) \\wedge ((\\neg a \\vee \\neg b) \\vee b)$.\n17. By associativity and commutativity, $(\\neg a \\vee \\neg b) \\vee a = (a \\vee \\neg a) \\vee \\neg b = 1 \\vee \\neg b = 1$, using step 3.\n18. In the same way, $(\\neg a \\vee \\neg b) \\vee b = \\neg a \\vee (b \\vee \\neg b) = \\neg a \\vee 1 = 1 \\vee \\neg a = 1$.\n19. By steps 16 to 18, $(a \\wedge b) \\vee (\\neg a \\vee \\neg b) = 1 \\wedge 1 = 1$.\n20. By distributivity, $(a \\wedge b) \\wedge (\\neg a \\vee \\neg b) = ((a \\wedge b) \\wedge \\neg a) \\vee ((a \\wedge b) \\wedge \\neg b)$.\n21. By associativity and commutativity, $(a \\wedge b) \\wedge \\neg a = (a \\wedge \\neg a) \\wedge b = 0 \\wedge b = 0$, using step 2.\n22. In the same way, $(a \\wedge b) \\wedge \\neg b = a \\wedge (b \\wedge \\neg b) = a \\wedge 0 = 0 \\wedge a = 0$.\n23. By steps 20 to 22, $(a \\wedge b) \\wedge (\\neg a \\vee \\neg b) = 0 \\vee 0 = 0$.\n24. By steps 19 and 23 and step 4 with $u = a \\wedge b$, we get $\\neg(a \\wedge b) = \\neg a \\vee \\neg b$. This proves both laws for all $a, b \\in A$.\n\n## Idea\n\nAn element has only one complement: one element that joins with it to $1$ and meets it in $0$.\nSo to find $\\neg(a \\vee b)$ it is enough to guess a candidate and check the two complement equations.\nThe candidate $\\neg a \\wedge \\neg b$ passes both checks by distributivity, and the second law works the same way with the roles of $\\wedge$ and $\\vee$ swapped.\n",
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