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276 LearnersLast updated on November 20, 2025

De Morgan’s Law is a pair of rules in boolean algebra, logic, and set theory. It relates the intersection and union of sets through complements. De Morgan’s Law is applied in the fields of mathematics, computer science, and design. In this article, we will learn more about De Morgan’s Law.
De Morgan’s laws are two key transformation rules in Boolean algebra and set theory. They show how the union and intersection of sets relate to their complements. These laws help simplify complex expressions, making calculations and Boolean operations easier to handle.
According to De Morgan’s Laws:
These relationships can be clearly understood with the help of Venn diagrams. In this lesson, we will explore the statements of De Morgan’s laws, see how they are proven, and understand their applications with examples.
For example,
Let:
A = {1, 2, 3, 4}
B = {3, 4, 5, 6}
Universal set U = {1, 2, 3, 4, 5, 6}
A ∪ B = {1, 2, 3, 4, 5, 6}
Its complement \((A ∪ B)’ = ∅\)
A’ = {5, 6}
B’ = {1, 2}
Now,
\(A' \cap B' = \{ \} = \varnothing \)
So,
\(\ (A \cup B)' = A' \cap B' \ \)
This verifies the first De Morgan’s law.
De Morgan’s laws are used in both Boolean algebra and set theory to simplify expressions involving unions, intersections, and complements. Let A and B be two subsets of a universal set U. The complements of A and B are written as A’ and B’ respectively. The symbol ∩ represents intersection, and ∪ means union.
Using these notations, De Morgan’s laws are stated as follows:
1. De Morgan’s Law of Union
The complement of the union of two sets A and B is equal to the intersection of their complements.
In symbolic form:
\((A ∪ B)’ = A’ ∩ B’\)
This rule can be extended to multiple sets. For a collection of sets \(\ \{A_1, A_2, \ldots, A_n\} \ \):
\(\ \left( \bigcup_{i=1}^{n} A_i \right)' = \bigcap_{i=1}^{n} A_i' \ \)
In the Venn diagram, the orange region shows A ∪ B, while the blue shaded region represents (A ∪ B)’.
2. De Morgan’s Law of Intersection
The complement of the intersection of sets A and B is equal to the union of their complements.
Symbolically:
\((A ∩ B)’ = A’ ∪ B’\)
This law also generalizes to multiple sets:
\(\ \left( \bigcap_{i=1}^{n} A_i \right)' = \bigcup_{i=1}^{n} A_i' \ \)
In the Venn diagram, the orange region indicates A ∩ B, while the blue area shows (A ∩ B)’.
In set theory, De Morgan’s laws show that when taking the complement of a union or intersection, the operations switch places. These laws can be proven using mathematical reasoning or truth tables.
1. Proof of \((A ∪ B)’ = A’ ∩ B’\)
Let \(R = (A ∪ B)’\) and \(S = A’ ∩ B’\).
To show \(R = S\), we must prove both \(R ⊂ S\) and \(S ⊂ R\).
Step 1: Proving \(R ⊂ S\)
Take any element \(y ∈ R.\)
\(y ∈ (A ∪ B)’\)
⇒ \(y ∉ (A ∪ B)\)
⇒ \( y ∉ A\) and \(y ∉ B\)
⇒ \(y ∈ A’\) and \(y ∈ B’\)
⇒ \(y ∈ A’ ∩ B’\)
⇒ \(y ∈ S\)
Thus, \(R ⊂ S. …(1)\)
Step 2: Proving \( S ⊂ R\)
Take any element \(z ∈ S\).
\(z ∈ A’ ∩ B’\)
⇒ \(z ∈ A’\) and \(z ∈ B’\)
⇒ \(z ∉ A \) and \(z ∉ B\)
⇒ \(z ∉ (A ∪ B)\)
⇒ \(xz ∈ (A ∪ B)’\)
⇒\( z ∈ R\)
Thus, \(S ⊂ R. …(2)\)
From (1) and (2), we conclude:
\((A ∪ B)’ = A’ ∩ B’\)
2. Proof of \((A ∩ B)’ = A’ ∪ B’\)
Let\( G = (A ∩ B)’ \) and \(H = A’ ∪ B’\).
To show G = H, we again prove \( G ⊂ H\) and \(H ⊂ G\).
Step 1: Proving \(G ⊂ H\)
Take any element \( y ∈ G\).
\(y ∈ (A ∩ B)’\)
⇒ \(y ∉ (A ∩ B)\)
⇒ \( y ∉ A \) or \(y ∉ B\)
⇒ \(y ∈ A’\) or \(y ∈ B’\)
⇒ \(y ∈ A’ ∪ B’\)
⇒ \(y ∈ H\)
Thus, \(G ⊂ H. …(1)\)
\( G ⊂ H. …(1)\)
Step 2: Proving \(H ⊂ G\)
Take any element \(z ∈ H\).
\(z ∈ A’ ∪ B’\)
⇒\( z ∈ A’ or z ∈ B’\)
⇒ \(z ∉ A or z ∉ B\)
⇒ \(z ∉ (A ∩ B)\)
⇒ \(z ∈ (A ∩ B)’\)
⇒ \(z ∈ G\)
Thus, \(H ⊂ G. …(2)\)
Combining (1) and (2), we get:
\((A ∩ B)’ = A’ ∪ B’.\)


De Morgan’s laws describe how logical operations behave under negation. In Boolean algebra, they are written as:
Here, A and B are binary input variables that take values 0 or 1. Logical operations such as\( AND (A • B)\), \(OR (A + B)\), and NOT (negation) are represented using truth tables. These laws can be proved using these truth tables.
1. First De Morgan’s Law
Statement:
When two or more variables are OR’ed and then negated, the result is equal to the AND of their complements.
\(⇁(A+B)=⇁A⇁B\)
Truth Table Proof
| A | A+B | ¬(A + B) | ¬A | ¬B | ¬A • ¬B |
| 0 | 0 | 1 | 1 | 1 | 1 |
| 1 | 1 | 0 | 0 | 1 | 0 |
| 0 | 1 | 0 | 1 | 0 | 0 |
| 1 | 1 | 0 | 0 | 0 | 0 |
The last two columns are identical, proving:
\(¬(A+B)=¬A⋅¬B\)
2. Second De Morgan’s Law
Statement:
When two or more variables are AND’ed and then negated, the result is equal to the OR of their complements.
\(¬(A⋅B)=¬A+¬B\)
Truth Table Proof
| A | A • B | ¬(A • B) | ¬A | ¬B | ¬A + ¬B |
| 0 | 0 | 1 | 1 | 1 | 1 |
| 1 | 0 | 1 | 0 | 1 | 1 |
| 0 | 0 | 1 | 1 | 0 | 1 |
| 1 | 1 | 0 | 0 | 0 | 0 |
Again, the last two columns match, proving:
\(¬(A⋅B)=¬A+¬B\)
Let us learn more about De Morgan’s formulas in set theory, boolean algebra, and logic.
In set theory:
Where:
In Logic:
Where:
De Morgan's law helps simplify complex logical expressions in math, programming, and circuits. Regular practice makes applying the rules faster and more accurate.
De Morgan’s law can be quite overwhelming as there are many rules to remember. This may lead to students making mistakes. Here are a few common mistakes students make and ways to avoid them:
De Morgan’s law is widely used in several fields, mainly computer science. Here are some real-world uses of De Morgan’s law:
Simplify the logical expression ¬ ( A ∧ B ).
\(¬ A ∨ ¬ B\).
De Morgan’s Law states that the negation of an AND is equivalent to the OR of the negations: \(¬ (A ∧ B) = ¬ A ∨ ¬ B\).
Express the complement of the intersection of sets A and B in terms of Ac and Bc
\((A ∩ B )^c = A^c ∪ B^c.\)
De Morgan’s Law states that the complement of an intersection is the union of the complements.
Express the complement of the union of sets A and B in terms of A^c and B^c
\((A∪B)^c = A^c ∩ B^c.\)
In sets, the complement of a union is given by the intersection of the complements.
Express the complement of the union of sets A, B, and C in terms of their complements.
\((A ∪ B ∪ C)^c = A^c ∩ B^c ∩ C^c.\)
De Morgan’s law generalizes to multiple sets: the complement of a union equals the intersection of the complements.
Simplify the Boolean expression ¬((X ∧ Y) ∨ Z).
\( ( ¬ X ∨ ¬ Y) ∧ ¬ Z.\)
First, apply De Morgan’s Law to the outer OR:
\(¬((X ∧ Y) ∨ Z) = ¬(X ∧ Y) ∧ ¬Z ¬((X ∧ Y) ∨ Z) = ¬(X ∧ Y) ∧ ¬Z.\)
Then, simplify \(¬(X ∧ Y) = ¬X ∨ ¬Y ¬(X ∧ Y) = ¬X ∨ ¬Y\).
Thus, the final result is: \( ( ¬ X ∨ ¬ Y) ∧ ¬ Z\).
Jaipreet Kour Wazir is a data wizard with over 5 years of expertise in simplifying complex data concepts. From crunching numbers to crafting insightful visualizations, she turns raw data into compelling stories. Her journey from analytics to education ref
: She compares datasets to puzzle games—the more you play with them, the clearer the picture becomes!






