Sets

$n(A ∪ B) = n(A) + n(B) − n(A ∩ B)$

Counts elements in at least one of A or B; subtracts the overlap counted twice.

$n(A∪B∪C) = n(A)+n(B)+n(C) − n(A∩B) − n(B∩C) − n(A∩C) + n(A∩B∩C)$

$n(A') = n(U) − n(A)$

The complement A' contains everything in U not in A.

$(A ∪ B)' = A' ∩ B'$

The complement of a union equals the intersection of the complements.

$(A ∩ B)' = A' ∪ B'$

$|P(A)| = 2ⁿ where n = |A|$

$A − B = {x : x ∈ A and x ∉ B} = A ∩ B'$

Elements in A but not in B. Note: A−B ≠ B−A in general.

$A × B = {(a,b) : a ∈ A, b ∈ B}; n(A×B) = n(A)·n(B)$

$Exactly one = Σn(A) − 2Σn(A∩B) + 3n(A∩B∩C); Exactly two = Σn(A∩B) − 3n(A∩B∩C); Exactly three = n(A∩B∩C)$

Splits the $7$ regions of $a three-circle$ Venn diagram by HOW MANY of A,B,C an element belongs to — essential for 'exactly k' survey questions, distinct from the plain union formula.

$A Δ (B Δ C) = (A Δ B) Δ C; A Δ ∅ = A; A Δ A = ∅; A Δ B = A Δ C ⟹ B = C$

Under Δ, the power set P(U) forms an abelian group: Δ is associative and commutative, ∅ is the identity element, and every set is its own inverse $(A Δ A = ∅).$ This is why, unlike ∪ or ∩, Δ obeys a CANCELLATION LAW.

$Swap ∪ ↔ ∩ and ∅ ↔ U throughout any true set identity to obtain another true identity$

Because ∪ and ∩ (and ∅, U) play symmetric roles in the axioms of set algebra, every proven law has a 'mirror' law obtained by this swap — halves the memorisation load.