De Morgan's Law - Theorem, Proofs, Formula & Examples

De Morgan's Law in Set Theory

De Morgan's law in set theory defines the relationship between the union, intersection, and complements of the sets, and is given for both the complement of the union and the intersection of two sets. In set theory, there are two De Morgan's Laws that are:

• First De Morgan's Law
• Second De Morgan's Law

Let's understand these laws in detail as below:

First De Morgan's Law

First De Morgan's law states that "The complement of the union of two sets is equal to the intersection of the complements of each set."

Let A and B be two sets, then mathematically First De Morgan's Law is given as:

(A ∪ B)' = A' ∩ B'

Where 

• U represents the Union operation between sets,
• ∩ represents the intersection operation between sets, and 
• ' represents complement operation on a set.

It is also called De Morgan's Law of Union.

Detail the Proof of De Morgan's Law

Proof Using Algebra of Sets

We need to prove, (A ∪ B)' = A' ∩ B'
Let X = (A ∪ B)' and Y = A' ∩ B'
Let p be any element of X, then p ∈ X ⇒ p ∈ (A ∪ B)'
⇒ p ∉ (A ∪ B)
⇒ p ∉ A or p ∉ B
⇒ p ∈ A' and p ∈ B'
⇒ p ∈ A' ∩ B'
⇒ p ∈ Y

∴ X ⊂ Y              . . . (i)

Again, let q be any element of Y, then q ∈ Y ⇒ q ∈ A' ∩ B'
⇒ q ∈ A' and q ∈ B'
⇒ q ∉ A and q ∉ B
⇒ q ∉ (A ∪ B)
⇒ q ∈ (A ∪ B)'
⇒ q ∈ X

∴ Y ⊂ X              . . . (ii)

From (i) and (ii) X = Y

(A ∪ B)' = A' ∩ B'

Also Read -Proof of De-Morgan’s laws in boolean algebra

Proof Using a Venn Diagram

Venn Diagram for (A ∪ B)'

Venn Diagram for A' ∩ B'

From both Diagrams, we can clearly say,

(A ∪ B)' = A' ∩ B'

That is the First De Morgan's Law.

Second De Morgan's Law

Second, De Morgan's law states that "The complement of intersection of two sets is equal to the union of the complements of each set."

Let A and B be two sets, then mathematically First De Morgan's Law is given as:

(A ∩ B)' = A' ∪ B'

Where 

• U represents the Union operation between sets,
• ∩ represents the intersection operation between sets, and 
• ' represents complement operation on a set.

It is also called De Morgan's Law of Intersection.

Proof Using Algebra of Sets

Second De Morgan's law: (A ∩ B)' = A' ∪ B'

Let X = (A ∩ B)' and Y = A' ∪ B'
Let p be any element of X, then p ∈ X ⇒ p ∈ (A ∩ B)'
⇒ p ∉ (A ∩ B)
⇒ p ∉ A or p ∉ B
⇒ p ∈ A' or p ∈ B'
⇒ p ∈ A' ∪ B'
⇒ p ∈ Y

∴ X ⊂ Y --------------(i)

Again, let q be any element of Y, then q ∈ Y ⇒ q ∈ A' ∪ B'

⇒ q ∈ A' or q ∈ B'
⇒ q ∉ A and q ∉ B
⇒ q ∉ (A ∩ B)
⇒ q ∈ (A ∩ B)'
⇒ q ∈ X

∴ Y ⊂ X --------------(ii)

From (i) and (ii) X = Y

(A ∩ B)' = A' ∪ B'

Proof Using Venn Diagram

Venn Diagram for (A ∩ B)'

Venn diagram for A' ∪ B'

From both diagrams, we can clearly say

(A ∩ B)' = A' ∪ B'

That is the Second De Morgan's Law.

De Morgan's Law in Boolean Algebra

De Morgan's Law Boolean Algebra defines the relation between the OR, AND, and the complements of variables, and is given for both the complement of AND and OR of two values. In Boolean Algebra there are two De Morgan's Laws that are:

• First De Morgan's Law
• Second De Morgan's Law

Let's understand these laws in detail as below:

First De Morgan's Law in Boolean Algebra

First De Morgan's law states that "The complement of OR of two or more variables is equal to the AND of the complement of each variable."

Let A and B be two variables, then mathematically First De Morgan's Law is given as:

(A + B)' = A' . B'

Where 

• + represents the OR operator between variables,
• . represents AND operator between variables, and 
• ' represents complement operation on the variable.

First De Morgan's Law Logic Gates

In context to logic gates and Boolean Algebra, De Morgan's Law states that "Both the logic gate circuits i.e., NOT gate is added to the output of OR gate, and NOT gate is added to the input of AND gate, are equivalent. These two logic gate circuits are given as follows:

👁 l2

First De Morgan's Law Truth Table

The truth table for first De Morgan's Law is given as follows:

Second De Morgan's Law in Boolean Algebra

Second De Morgan's law states that "The complement of AND of two or more variables is equal to the OR of the complement of each variable."

Let A and B be two variables, then mathematically Second De Morgan's Law is given as:

(A . B)' = A' + B'

Where 

• + represents the OR operator between variables,
• . represents AND operator between variables, and 
• ' represents complement operation on variable.

Second De Morgan's Law Logic Gates

In context to logic gates and Boolean Algebra, De Morgan's Law states that "Both the logic gate circuits i.e., NOT gate is added to the output of AND gate, and NOT gate is added to the input of OR gate, are equivalent. These two logic gate circuits are given as follows:

👁 Untitled-Diagram-(1)

Second De Morgan's Law Truth Table

The truth table for the second De Morgan's Law is given as follows:

De Morgan's Law of Logic

In De Morgan's law for logic the below propositions are tautology:

∼ (a ∧ b) ≡ ∼ a ∨ ∼ b

∼ (a ∨ b) ≡ ∼ a ∧ ∼ b

Where,

• ∧ represents the conjunction of statements,
• ∨ represents the disjunction of statements,
• ~ represents the negation of a statement, and 
• ≡ represents the equivalence of statements.

De Morgan's Law Formula

Let's compile all the formulas for De Morgan's Law in the following list.

For set theory:

• (A ∪ B)' = A' ∩ B'
• (A ∩ B)' = A' ∪ B'

For Boolean Algebra:

• (A + B)' = A' . B'
• (A . B)' = A' + B'

For Logic:

• ∼ (a ∧ b) ≡ ∼ a ∨ ∼ b
• ∼ (a ∨ b) ≡ ∼ a ∧ ∼ b

De Morgan's Law Practical Examples

Example 1: Pizza Toppings

Imagine you're at a pizza party, and you're told you can choose any toppings except for both mushrooms and olives together.

• Using De Morgan's Law: This means if you don't want both mushrooms and olives (Not (Mushrooms and Olives)), you can either not have mushrooms (Not Mushrooms) or not have olives (Not Olives) on your pizza. So, you could have a pizza with just mushrooms, just olives, or neither!

Example 2: Library Books

Your teacher says you cannot bring books about wizards or dragons into the classroom.

• Using De Morgan's Law: This means if you're not allowed books about wizards or dragons (Not (Wizards or Dragons)), you can't bring books about wizards (Not Wizards) and you can't bring books about dragons (Not Dragons). So, books about space or animals are still okay!

Example 3: Playing Outside

Your mom says you can't play outside if it's raining and cold at the same time.

• Using De Morgan's Law: This means if you're not going out because it's raining and cold (Not (Raining and Cold)), you wouldn't go out if it's just raining (Not Raining) or just cold (Not Cold). But if it's sunny and warm, you're good to go!

Example 4: Choosing a Movie

Your friend says they don't want to watch a movie that is scary or boring.

• Using De Morgan's Law: This means if your friend doesn't want a movie that's scary or boring (Not (Scary or Boring)), they don't want a scary movie (Not Scary) and they don't want a boring movie (Not Boring). So, a funny or exciting movie would be perfect!

Solved Question on De Morgan's Law

Question 1: Given that U = {2, 3, 7, 8, 9}, A = {2, 7} and B = {2, 3, 9}. Prove De Morgan's Second Law.

Solution:

U = {2, 3, 7, 8, 9}, A = {2, 7} and B = {2, 3, 9}
To Prove: (A ∩ B)' = A' ∪ B'
(A ∩ B) = {2}
(A ∩ B)' = U - (A ∩ B) = {2, 3, 7, 8, 9} - {2}
(A ∩ B)' = {3, 7, 8, 9}
A' = U - A = {2, 3, 7, 8, 9} - {2, 7}
A' = {3, 8, 9}
B' = U - B = {2, 3, 7, 8, 9} - {2, 3, 9}
B' = {7, 8}
A' ∪ B' = {3, 8, 9} ∪ {7, 8}
A' ∪ B' = {3, 7, 8, 9}

(A ∩ B)' = A' ∪ B'

Question 2: Given that U = {1, 4, 6, 8, 9}, A = {1, 9} and B = {4, 6, 9}. Prove De Morgan's First Law.

Solution:

U = {1, 4, 6, 8, 9}, A = {1, 9} and B = {4, 6, 9}
To Prove: (A ∪ B)' = A' ∩ B'
(A ∪ B) = {1, 4, 6, 9}
(A ∪ B)' = U - (A ∪ B) = {1, 4, 6, 8, 9} - {1, 4, 6, 9}
(A ∪ B)' = {8}
A' = U - A = {1, 4, 6, 8, 9} - {1, 9}
A' = {4, 6, 8}
B' = U - B = {1, 4, 6, 8, 9} - {4, 6, 9}
B' = {1, 8}
A' ∩ B' = {4, 6, 8} ∩ {1, 8}
A' ∩ B' = {8}
(A ∪ B)' = A' ∩ B'

Hence Proved

Question 3: Simplify the Boolean Expression: Y = [(A + B).C]'

Solution:

Y = [(A + B).C]'
Applying De Morgan's law (A . B)' = A' + B'
Y = (A + B)' + C'
Applying De Morgan's law (A + B)' = A'. B'
Y = A'. B' + C'

Question 4: Simplify the Boolean Expression: X = [(A + B)' + C]'

Solution:

X = [(A + B)' + C]'
Applying De Morgan's law (A + B)' = A'. B'
X = [(A + B)']' . C'
X = (A + B). C'

Check this sources for more:

Showcase Examples of De Morgan's Law

Logic Applications of De Morgan's Law