De-Morgan's Laws

What are De-Morgan's Laws in Sets?

In set theory, the relationship between union, complements, and intersection is provided by De-Morgan's law. It provides the relationship between AND, OR, and the variable's complements in Boolean algebra; in logic, it provides the relationship between AND, OR, or the statement's negation. De Morgan's Law allows us to optimise different Boolean circuits that use logic gates to do the same task with a minimum amount of equipment.

According to De-Morgan's Laws:

• The complement of the union of two sets is equal to the intersection of their individual complements.
• Additionally, the complement of the intersection of two sets is equal to the union of their individual complements.

De-Morgan's Law of Sets Formula

De Morgan's Law Formula outlines two fundamental rules in set theory that connect the operations of union, intersection, and complement. These laws are essential for simplifying set expressions and solving problems in logic, Boolean algebra, and mathematics.

First De-Morgan's Law: Law of Union

The first law of De Morgan states that the complement of the union of two sets is equal to the intersection of their individual complements. This principle is widely applied in set theory problems, Venn diagrams, and logic gate circuits.

If $A$ and $B$ are two sets, then:

$(A \cup B)' = A' \cap B'$

Where:

• $\cup$ denotes the union of sets
• $\cap$ denotes the intersection of sets
• $'$ denotes the complement of a set

This is also known as De-Morgan's Law of Union.

Second De-Morgan's Law: Law of Intersection

The second law of De Morgan states that the complement of the intersection of two sets is equal to the union of their individual complements. This law helps in transforming and simplifying logical and mathematical statements involving negations.

If $A$ and $B$ are two sets, then:

$(A \cap B)' = A' \cup B'$

Where:

• $\cap$ denotes the intersection of sets
• $\cup$ denotes the union of sets
• $'$ denotes the complement of a set

This is also referred to as De-Morgan's Law of Intersection.

De Morgan’s Laws Proof in Set Theory

De Morgan’s Laws are fundamental identities in set theory, logic, Boolean algebra, and computer science. They describe how the complement of a union and the complement of an intersection behave under set operations. These laws are especially useful for simplifying complex set expressions, solving Venn diagram problems, and transforming logical statements.

In set theory, De Morgan’s Laws establish a precise relationship between union, intersection, and complement, making them essential for both theoretical understanding and exam problem-solving.

Proof of De Morgan’s Law for Union

Given below is the detailed proof of the De Morgan's law of union along with the statement:

Statement of De Morgan’s Law (Union)

The De Morgan’s Law of Union states that the complement of the union of two sets is equal to the intersection of their complements:

$(A \cup B)' = A' \cap B'$

This means that an element outside both sets $A$ and $B$ must lie in the complement of each set individually.

Proof

Let $x \in (A \cup B)'$.

By definition of complement, $x \notin (A \cup B)$.
This implies that $x \notin A$ and $x \notin B$.
Hence, $x \in A'$ and $x \in B'$.
Therefore, $x \in (A' \cap B')$.

Since every element of $(A \cup B)'$ belongs to $A' \cap B'$, and vice versa, we conclude that: $(A \cup B)' = A' \cap B'$

Thus, the complement of a union is equal to the intersection of complements.

Generalised Form of De Morgan’s Law (Union)

For any finite number of sets $A_1, A_2, \dots, A_n$, the generalized form is:

$\left( \bigcup_{i=1}^{n} A_i \right)' = \bigcap_{i=1}^{n} A_i'$

This shows that the law holds true not just for two sets, but for any number of sets.

Proof of De Morgan’s Law for Intersection

We have given below the detailed proof and statement of De-Morgan's Law of intersection:

Statement of De Morgan’s Law (Intersection)

The De Morgan’s Law of Intersection states that the complement of the intersection of two sets is equal to the union of their complements:

$(A \cap B)' = A' \cup B'$

This means that an element not common to both sets must belong to at least one of their complements.

Proof

Let $x \in (A \cap B)'$.

By definition of complement, $x \notin (A \cap B)$.
This implies that $x \notin A$ or $x \notin B$.
Hence, $x \in A'$ or $x \in B'$.
Therefore, $x \in (A' \cup B')$.

Since every element of $(A \cap B)'$ belongs to $A' \cup B'$, and vice versa, we conclude that:

$(A \cap B)' = A' \cup B'$

Thus, the complement of an intersection is equal to the union of complements.

Generalised Form of De Morgan’s Law (Intersection)

For any finite number of sets $A_1, A_2, \dots, A_n$, the generalized form is:

$\left( \bigcap_{i=1}^{n} A_i \right)' = \bigcup_{i=1}^{n} A_i'$

This confirms that De Morgan’s Law applies universally to intersections of multiple sets.

Importance of De Morgan’s Laws

These proofs confirm that De Morgan’s Laws in set theory are logically sound and universally valid. They are widely used to:

• Simplify set expressions

• Solve Venn diagram problems

• Convert logical statements

• Apply Boolean algebra rules

• Analyze probability events

Understanding the proof of De Morgan’s Laws not only strengthens conceptual clarity but also improves accuracy and speed in competitive exams and mathematical reasoning problems.

De-Morgan's Law Truth Table

De-Morgan's Laws describe how complements interact with unions and intersections. These logical equivalences can be verified using truth tables.

De-Morgan's First Law Statement and Proof

Statement:
The complement of the union of two sets is equal to the intersection of their individual complements.

$(A \cup B)' = A' \cap B'$

Truth Table:

De-Morgan's Second Law Statement and Proof

Statement:
The complement of the intersection of two sets is equal to the union of their individual complements.

$(A \cap B)' = A' \cup B'$

Truth Table:

De-Morgan's Law Examples

Let us understand De-Morgan’s Laws using a simple example.

Let the universal set be: $U = \{7, 8, 9, 10, 11, 12, 13\}$

Let the two subsets be: $A = \{11, 12, 13\}$, $B = \{7, 8\}$

De-Morgan’s Law of Union example

We have: $A \cup B = \{7, 8, 11, 12, 13\}$
Therefore,
$(A \cup B)' = U - (A \cup B) = {9, 10}$

Also, $A' = U - A = \{7, 8, 9, 10\}$
$B' = U - B = \{9, 10, 11, 12, 13\}$
So, $A' \cap B' = \{9, 10\}$

Hence, $(A \cup B)' = A' \cap B'$

De-Morgan’s Law of Intersection example

We have:
$A \cap B = \phi$
So, $(A \cap B)' = U - \phi = U = \{7, 8, 9, 10, 11, 12, 13\}$

Also, $A' = \{7, 8, 9, 10\}$
$B' = \{9, 10, 11, 12, 13\}$
So, $A' \cup B' = \{7, 8, 9, 10, 11, 12, 13\}$

Thus, $(A \cap B)' = A' \cup B'$

Generalised De Morgan’s Law for n Sets

In set theory and discrete mathematics, De Morgan’s Laws are not limited to just two sets. They extend naturally to any finite number of sets, which is known as the generalised De Morgan’s Laws. This generalisation is extremely important when dealing with multiple sets, large datasets, probability problems, and logical expressions.

Generalised Law for Union

The complement of the union of n sets is equal to the intersection of their complements:

$(A_1 \cup A_2 \cup \dots \cup A_n)' = A_1' \cap A_2' \cap \dots \cap A_n'$

This means that if an element does not belong to any of the sets $A_1, A_2, \dots, A_n$, then it must lie in the complement of each individual set.

Generalised Law for Intersection

The complement of the intersection of n sets is equal to the union of their complements:

$(A_1 \cap A_2 \cap \dots \cap A_n)' = A_1' \cup A_2' \cup \dots \cup A_n'$

This means that if an element fails to belong to at least one of the sets, it will appear in the union of their complements.

Why the Generalised Form Matters

The generalised De Morgan’s Laws are widely used in:

• Set operations involving many sets

• Probability and inclusion–exclusion problems

• Logical reasoning and predicate logic

• Database queries and filtering

• Boolean algebra and digital circuit design

They allow complex expressions to be simplified systematically.

Logical Interpretation of De Morgan’s Laws

To truly understand De Morgan’s Laws, it helps to interpret them using logical operators like AND, OR, and NOT. This logical view is especially important in mathematical logic, computer science, and programming.

Logical Meaning

• Union ($\cup$) corresponds to OR

• Intersection ($\cap$) corresponds to AND

• Complement ($'$) corresponds to NOT

Using this idea:

• NOT (A OR B) = (NOT A) AND (NOT B)

• NOT (A AND B) = (NOT A) OR (NOT B)

This shows that negation flips OR into AND and AND into OR.

Everyday Logic Example

If someone says,
“I am not going to the mall or the cinema,”

Logically, it means:
“I am not going to the mall and I am not going to the cinema.”

This real-life reasoning is exactly what De Morgan’s Laws formalize in mathematics.

Importance of Logical Interpretation

Understanding this logical perspective helps in:

• Simplifying logical expressions

• Writing efficient conditions in programming

• Solving Boolean algebra problems

• Analyzing truth tables and circuits

De Morgan’s Laws in Venn Diagram Form

Venn diagrams provide a powerful visual way to understand De Morgan’s Laws in set theory. They make it immediately clear how complements, unions, and intersections interact.

Complement of Union Using Venn Diagram

For $(A \cup B)'$:

• First, shade the region representing $A \cup B$

• Then take its complement

• The shaded region lies outside both sets A and B

This visually confirms: $(A \cup B)' = A' \cap B'$

Complement of Intersection Using Venn Diagram

For $(A \cap B)'$:

• First, identify the overlapping region of $A$ and $B$

• Take its complement

• The shaded area includes everything except the overlap

This confirms: $(A \cap B)' = A' \cup B'$

Solved Examples of De-Morgan's Law

Example 1: If $(A \cup B)=P$, then evaluate $P^{\prime}$
1) $A^{\prime} \cup B$
2) $A \cap B^{\prime}$
3) $A^{\prime} \cup B^{\prime}$
4) $A^{\prime} \cap B^{\prime}$

Solution:
Using De-Morgan's Law:
$P^{\prime}=(A \cup B)^{\prime}=A^{\prime} \cap B^{\prime}$
Hence, the answer is option 4.

Example 2: Which of the following is not a property of a union of sets?
1) $A \cup(B \cup C)=(A \cup B) \cup C$
2) $A \cup B=B \cup A$
3) $(A \cup B)^c=A^c \cup B^c$
4) $(A \cap B)^c=A^c \cup B^c$

Solution:
Let $A$ and $B$ be any two sets. The union of $A$ and $B$ is the set which consists of all the elements of $A$ and all the elements of $B$, the common elements being taken only once. The symbol ' $u$ ' is used to denote the union.

Symbolically, we write $A \cup B=\{x: x \in A$ or $x \in B\}$.
De Morgan's Law -
$(A \cup B)^{\prime}=A^{\prime} \cap B^{\prime}$
Hence, option 3 is incorrect.
Hence, the answer is option 3.

Example 3: If $(A \cap B \cap C)=P$. Then evaluate $P^{\prime}$
1) $A^{\prime} \cap B^{\prime} \cap C^{\prime}$
2) $A^{\prime} \cup B^{\prime} \cup C^{\prime}$
3) $A^{\prime} \cup B^{\prime} \cap C^{\prime}$
4) $A^{\prime} \cup B^{\prime} \cap C^{\prime}$

Solution:
$P' = (A \cap B \cap C)'$

$= ((A \cap B) \cap C)'$

$= (A \cap B)' \cup C' \quad \text{(by De Morgan's Law)}$

$= A' \cup B' \cup C'$

Hence, the answer is option 2.

Example 4: If the set $A^{\prime}=\{3,5,7\}$ and $B^{\prime}=\{1,5,9\}$, then the set $(A \cup B)^{\prime}=$
1) $\{1,3,5,7,9\}$
2) $\{5,7\}$
3) $\{5\}$
4) $\{1,5,9\}$

Solution:
De Morgan's Law:
$(A \cup B)^{\prime}=A^{\prime} \cap B^{\prime}$
The intersection of $\mathrm{A}^{\prime}$ and $\mathrm{B}^{\prime}$ is $\{5\}$.
Hence, the answer is option 3.

Example 5: If $A-B=X$ and $A-C=Y$. then the simplification of $A-(B \cup C)$ is
1) $X \cap Y$
2) $X \cup Y$
3) $X-Y$
4) $Y-X$

Solution:
$P-Q=P \cap Q^{\prime}$
So,
$A - (B \cup C) = A \cap (B \cup C)'$

$A \cap (B' \cap C') = (A \cap B') \cap (A \cap C') = (A - B) \cap (A - C)$

$X \cap Y$
Hence, the answer is option 1.

List of Topics Related to De-Morgan's Laws

To fully understand De-Morgan's Laws in set theory, it's important to first grasp the foundational set concepts they rely on. This section outlines key related topics like the roster and set-builder forms, types of sets, subsets, and operations such as difference and union. Mastery of these topics will help build the logical base needed to apply De-Morgan’s Laws accurately.

NCERT Resources

Access curated NCERT resources to strengthen your understanding of Sets. This section includes detailed solutions, revision notes, and exemplar problems from Class 11 Chapter 1, helping you build a strong foundation for exams through structured and NCERT-aligned preparation.

NCERT Solutions for Class 11 Chapter 1 Sets

NCERT Notes for Class 11 Chapter 1 Sets

NCERT Exemplar for Class 11 Chapter 1 Sets

Practice Questions on De-Morgan's Law in Sets

Sharpen your grasp of De-Morgan's Laws in Sets with targeted practice questions. This section offers a variety of MCQs designed to test your understanding of definitions, properties, and logical applications. You’ll also find practice sets on related foundational topics to strengthen your overall command of set theory.

Practice Questions here: De Morgan's Laws - Practice Question MCQ

You can practice some important topics of Sets below: