Set Identities

By Daniel Machado Updated: 10/01/2026

The laws of the algebra of sets (also known as set identities) are mathematical equations involving set operations that always hold true for any sets within a universal set. These properties allow complex expressions involving union, intersection, and complement to be simplified into equivalent, simpler forms without relying on Venn diagrams or element-by-element verification.

Table of Contents

Table of Properties

Below are the primary laws governing operations on sets, considering sets A, B, and C within a universal set U, the empty set ∅, and the complement denoted as A':

Law Name	Mathematical Expression	Description
Identity laws	A ∪ ∅ = A A ∩ U = A	Union with the empty set or intersection with the universal set leaves the original set unchanged.
Domination laws	A ∪ U = U A ∩ ∅ = ∅	Union with the universal set yields the universal set, and intersection with the empty set always yields the empty set.
Idempotent laws	A ∪ A = A A ∩ A = A	Operating a set with itself under union or intersection returns the same set.
Complement laws	A ∪ A' = U A ∩ A' = ∅	The union of a set with its complement covers the entire universal set, while their intersection shares no elements.
Universal and empty set complement laws	U' = ∅ ∅' = U	The complement of the universal set is the empty set, and the complement of the empty set is the universal set.
Double complement law	(A')' = A	The complement of the complement of a set equals the original set.
Commutative laws	A ∪ B = B ∪ A A ∩ B = B ∩ A	The order of the operands does not affect the result of a union or intersection.
Associative laws	(A ∪ B) ∪ C = A ∪ (B ∪ C) (A ∩ B) ∩ C = A ∩ (B ∩ C)	When chaining operations of the same type, grouping the sets differently does not change the final result.
Distributive laws	A ∩ (B ∪ C) = (A ∩ B) ∪ (A ∩ C) A ∪ (B ∩ C) = (A ∪ B) ∩ (A ∪ C)	Intersection distributes over union, and union distributes over intersection in an analogous manner.
De Morgan's laws	(A ∪ B)' = A' ∩ B' (A ∩ B)' = A' ∪ B'	The complement of a union is the intersection of the complements, and the complement of an intersection is the union of the complements.
Absorption laws	A ∪ (A ∩ B) = A A ∩ (A ∪ B) = A A ∪ (A' ∩ B) = A ∪ B A ∩ (A' ∪ B) = A ∩ B	A combined operation of union and intersection where one set appears both inside and outside the parentheses reduces redundant terms; if the inner set is complemented, that complement is eliminated.
Set difference law	A − B = A ∩ B'	The difference between two sets equals the intersection of the first set with the complement of the second.
Basic difference laws	A − A = ∅ A − ∅ = A ∅ − A = ∅	Subtracting a set from itself yields the empty set, while subtracting the empty set leaves the set unchanged.
Symmetric difference laws	A Δ B = (A − B) ∪ (B − A) A Δ B = (A ∪ B) − (A ∩ B) A Δ ∅ = A A Δ A = ∅	The symmetric difference collects elements that belong to either set, but not to both simultaneously.

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How to Prove Set Identities

To verify that an equality between sets holds, the double containment method (mutual inclusion) is used. Two sets A and B are equal if and only if they contain the exact same elements, which requires proving two separate conditions: first that A ⊆ B, and then that B ⊆ A.

In practice, this procedure consists of taking an arbitrary element x and translating the set operations into equivalent logical statements. Each set connective corresponds to a logical operator: union represents a disjunction (∨), intersection represents a conjunction (∧), and complement represents a negation (¬). Below are two step-by-step proofs using this reasoning.

Proof of One of De Morgan's Laws

Consider the identity (A ∪ B)' = A' ∩ B'. To verify it, we prove both subset inclusions by analyzing the membership of an arbitrary element x.

First, we prove the forward inclusion, namely (A ∪ B)' ⊆ A' ∩ B':

Assume that x ∈ (A ∪ B)'. By the definition of complement, this means that x ∉ (A ∪ B).
Not belonging to the union is logically equivalent to stating that x is in neither set: ¬(x ∈ A ∨ x ∈ B).
Applying De Morgan's laws of propositional logic, this expression transforms into: x ∉ A ∧ x ∉ B.
By the definition of complement for each set individually, we deduce that x ∈ A' ∧ x ∈ B'.
Because x belongs to both complement sets simultaneously, we conclude that x ∈ A' ∩ B', completing the forward inclusion.

Next, we prove the reverse inclusion, namely A' ∩ B' ⊆ (A ∪ B)':

Now consider an element satisfying x ∈ A' ∩ B'. By the definition of intersection, it follows that x ∈ A' ∧ x ∈ B'.
Applying the definition of complement to each part yields x ∉ A ∧ x ∉ B.
Combining both negations using the corresponding logical equivalence, we write: ¬(x ∈ A ∨ x ∈ B).
This negation indicates that the element is not a member of the union: x ∉ (A ∪ B).
By the definition of complement, we finally obtain x ∈ (A ∪ B)'.

Having established that each set is a subset of the other, the equality (A ∪ B)' = A' ∩ B' is proven.

Proof of the Distributive Law

Next, we analyze the distribution of intersection over union: A ∩ (B ∪ C) = (A ∩ B) ∪ (A ∩ C). Again, we divide the argument into two parts to cover both inclusions.

To prove that A ∩ (B ∪ C) ⊆ (A ∩ B) ∪ (A ∩ C), we follow this sequence of deductions:

Start with an element satisfying x ∈ A ∩ (B ∪ C). By the definition of intersection, this is expressed as x ∈ A ∧ x ∈ (B ∪ C).
Translate the inner union into its logical form: x ∈ A ∧ (x ∈ B ∨ x ∈ C).
Apply the distributive law of propositional logic (conjunction over disjunction), obtaining: (x ∈ A ∧ x ∈ B) ∨ (x ∈ A ∧ x ∈ C).
Rewrite each parenthetical expression according to the definition of intersection: x ∈ (A ∩ B) ∨ x ∈ (A ∩ C).
By the definition of set union, we conclude that x ∈ (A ∩ B) ∪ (A ∩ C).

To complete the proof, we verify the reverse inclusion, (A ∩ B) ∪ (A ∩ C) ⊆ A ∩ (B ∪ C):

Assume that x ∈ (A ∩ B) ∪ (A ∩ C). By the definition of union, we have (x ∈ A ∩ B) ∨ (x ∈ A ∩ C).
Break down the intersections into logical conjunctions: (x ∈ A ∧ x ∈ B) ∨ (x ∈ A ∧ x ∈ C).
Factor out x ∈ A using logical distributivity: x ∈ A ∧ (x ∈ B ∨ x ∈ C).
Convert the disjunction inside the parentheses back into a set union: x ∈ A ∧ x ∈ (B ∪ C).
By the definition of intersection, we deduce that x ∈ A ∩ (B ∪ C).