Laws of Boolean Algebra MCQ Quiz - Objective Question with Answer for Laws of Boolean Algebra - Download Free PDF

Explanation:

Dual of a Boolean Expression

Definition: The dual of a Boolean expression is derived by replacing all AND (×) operations with OR (+) operations, all OR (+) operations with AND (×) operations, and swapping the constants 1 and 0 in the original expression. The variables in the Boolean expression remain unchanged.

To determine the dual of a given Boolean expression, follow these steps:

1. Identify all the AND (×) and OR (+) operations in the expression.
2. Replace AND (×) with OR (+) and vice versa.
3. Swap 1 with 0 wherever they appear in the expression.

Given Boolean Expression:

X + YZ = (X + Y) × (X + Z)

Let’s determine the dual of this expression:

• In the given expression, the OR (+) operation between X and YZ will be replaced with an AND (×) operation.
• The AND (×) operations within YZ and (X + Y) × (X + Z) will be replaced with OR (+) operations.
• Since there are no constants 1 or 0 in the expression, we do not need to swap them.

Replacing operations step by step:

• The left-hand side, X + YZ, becomes X × (Y + Z).
• The right-hand side, (X + Y) × (X + Z), becomes (X × Y) + (X × Z).

Thus, the dual of the given Boolean expression is:

X × (Y + Z) = (X × Y) + (X × Z)

Correct Option Analysis:

The correct option is:

Option 1: X × (Y + Z) = XY + YZ

This is the correct dual of the given Boolean expression. As derived above, the left-hand side X + YZ becomes X × (Y + Z), and the right-hand side (X + Y) × (X + Z) becomes (X × Y) + (X × Z). Hence, the correct dual is accurately represented in Option 1.

Additional Information

To further understand the analysis, let’s evaluate the other options:

Option 2: X.(Y + Z) = X.Y + X.Z

This option is not the correct dual of the given Boolean expression. While it might seem similar, it actually represents the distributive property of Boolean algebra and does not match the dual derived from the original expression.

Option 3: X + (Y + Z) = X.Y + Z

This option introduces an additional OR (+) operation within (Y + Z) that was not present in the original expression. Moreover, the right-hand side, X.Y + Z, does not correspond to the structure of the dual derived from the given Boolean expression. Therefore, it is incorrect.

Option 4: X + (YZ) = X + Y + Z

This option does not align with the dual derived from the given Boolean expression. It incorrectly simplifies X + (YZ) to X + Y + Z, which is not mathematically valid based on Boolean algebra rules. Hence, this option is incorrect.

Conclusion:

By understanding the concept of duality in Boolean algebra, we can accurately derive the dual of any given Boolean expression. The dual is obtained by swapping AND (×) and OR (+) operations and interchanging the constants 1 and 0. For the given expression X + YZ = (X + Y) × (X + Z), the correct dual is X × (Y + Z) = (X × Y) + (X × Z), which is represented in Option 1.

The correct answer is option 4: A + A̅B = A + B

Key Points

• A + A̅B = A + B is a valid Boolean identity derived using the Distributive Law and the Law of Complementation.
• It is a known simplification rule in Boolean algebra.
• Proof:
  • A + A̅B = (A + A̅)(A + B) (Distributive Law)
  • (A + A̅) = 1 (Complement Law)
  • ⇒ 1.(A + B) = A + B

Additional Information

• Option 1 – A.A̅ = 1: ❌ Incorrect. A.A̅ = 0 (by Complement Law)
• Option 2 – A + AB = A + B: ❌ Incorrect. A + AB = A (by Absorption Law)
• Option 3 – A(A + B) = B: ❌ Incorrect. A(A + B) = A (by Absorption Law)

Hence, the correct answer is: option 4: A + A̅B = A + B

The correct answer is Option 4.

Key Points

• The correct Boolean algebra rule is A + A̅B = A + B.
  • This rule can be derived using the Distributive Law and the Complement Law in Boolean algebra.
  • First, we use the Distributive Law: A + A̅B = (A + A̅)(A + B).
  • According to the Complement Law, (A + A̅) = 1.
  • So, (A + A̅)(A + B) = 1.(A + B) = A + B.
  • Therefore, the expression A + A̅B simplifies to A + B.

Additional Information

• Boolean algebra is a branch of algebra that deals with true or false values, typically denoted as 1 and 0 respectively.
• It is fundamental in the design of digital circuits and computer algorithms.
• Basic operations in Boolean algebra include AND, OR, and NOT.
• Boolean algebra helps in simplifying complex logical expressions and is used extensively in computer science and electrical engineering.

Clearly this is the code of a Half adder where:

• the Sum bit is the XOR of the two input bits
• the carry bit is simply the AND of the two input bits

A half adder circuit is basically made up of an AND gate and an XOR gate as shown below:

• A half adder is also known as XOR gate because XOR is applied to both inputs to produce the sum
• Half adder can add only two bits (A and B) and has nothing to do with the carry
• If the input to a half adder has a carry, then it will neglect it and adds only the A and B bits
• That means the binary addition process is not complete and that's why it is called a half adder

Sum (S) = A⊕B, Carry = A.B

The correct answer is 4

Concept:

From the figure we may conclude that both the input of the XNOR gate are the same

The XNOR gate output is 1 when both inputs are the same.

Truth Table:

Output Equation: \(Y={\overline{A\oplus B}}\)

Key Points

1) If B is always Low, the output is the inverted value of the other input A, i.e. A̅.

2) The output is low when both inputs are different.

3) The output is high when both inputs are the same.

4) The XNOR gate produces an output only when the two inputs are the same.

X = [(A + B̅) (B + C)] B

= (AB + AC + 0 +  B̅C)B

= AB + ABC

= AB(1 + C)

= AB

Concept:

Consensus Law is one of the most powerful theorems used in digital electronics for the minimization of Boolean function or equation either in the successive reduction method or in the K-Map method.

Statement:

• The consensus theorem states that the consensus term of a disjunction is defined when the terms in function are reciprocals to each other (such as A and A̅).
• The consensus theorem is defined in two statements (normal form and it's dual). They are
• AB + ĀC+BC = AB+ĀC
• (A+B)(Ā+C)(B+C) = (A+B)( Ā+C)

Calculation:

Y = P̅Q + QR + PR

Y = P̅Q + PR + QR (P̅ + P)

Y = P̅Q + PR + QRP̅ + QRP

Y = P̅Q(1 + R) + PR(1 + Q)

Y = P̅Q + PR where (1 + A = 1) according to Boolean algebra. 

Analysis:

Y = AB + A(B + C) + B(B + C)

= AB + AB + AC + B + BC

Since AB + AB = AB, we get:

Y = AB + AC + B (1 + C)

Since 1 + X (any variable) = X, we get:

Y = AB + AC + B     

Y = B(1 + A) + AC  

Y = B + AC

Key Points

All Boolean algebra laws are shown below

Laws in Boolean Algebra

1.) Distributive law

A + A̅B = (A + A̅) (A+B)

Since, (A + A̅) = 1

A + A̅B = A + B

2.) Commutative law

A + B = B + A

3.) Associative law

A + B + C = A + (B + C) = (A + B) + C 

4.) Absorptive law

A(A + B) = A A + (AB) 

Since, AA = 1

A(A + B) = A

5.) Identity law:
A.1 = A
A + 0 = A

\(Y = ABC + AB\bar C + \bar A\bar BC + \bar ABC\)

\(Y = AB\left( {C + \bar C} \right) + \bar AC\left( {B + \bar B} \right)\)

\(= AB + \bar AC\)

To find the complement of Y.

\(\bar Y = \overline {AB + \bar AC}\)

\(= \left( {\overline {AB} } \right) \cdot \left(\overline {\bar AC} \right)\)

\(= \left( {\bar A + \bar B} \right)\left( {\overline {\bar A} + \bar C} \right)\)

Concept:

De Morgan’s law states that:

\(\overline {\left( {{A_1}.{A_2} \ldots {A_n}} \right)} = \left( {\overline {{A_1}} + \overline {{A_2}} + \ldots + \overline {{A_n}} } \right)\)

\(\overline {\left( {{A_1} + {A_2} + \ldots + {A_n}} \right)} = \left( {\overline {{A_1}} \;.\;\overline {{A_2}} \;.\;..\;\overline {{A_n}} } \right)\)

Application:

\(f = \overline {AB} + \overline {A + B} \)

This can be written as:

f = A̅ + B̅ + A̅ B̅ 

f = A̅ (1 + B̅) + B̅

f = A̅ + B̅ 

Again using De-Morgan's property, we get:

\(f=\overline {AB} \)

We have,

X = AB + A (B + C) + B (B + C)

or, X = AB + AB + AC + B + BC

or, X = AB + AC + B + BC

or, X = AB + AC + B(1 + C)  Since, (1 + A = 1)

or, X = AB + AC + B = B + AB + AC

or, X = B(1 + A) + AC = B + AC

Given:

Y = AC̅ + ABC̅ 

Y = AC̅(1 + B)

Y = AC̅

Additional Information

Laws of Boolean Algebra:

All Boolean algebra laws are shown below

A(BC) = (AB)C is associative law.

A+B = B + A is commutative law.

The expression for absorption law is given by,

A + AB = A,

A.(A+B) = A is absorption law.

AB' + B = A + B is also absorption law.

The commutative law of Boolean algebra for two input variables is defined as:

AND form: AB = BA

OR Form: A + B = B + A

Important Points