Chapter 13: Boolean Algebra Class 12 Computer Science NCERT Solutions

Boolean Algebra is a branch of algebra that deals with binary variables and logical operations. It is the foundation of digital logic design and is used to analyze and simplify digital circuits. Chapter 13 introduces the concepts of Boolean variables, logic gates, truth tables, and simplification techniques essential for understanding computer logic and circuit design.

What You Will Learn in Chapter 13

This chapter covers the fundamental concepts of Boolean algebra, laws, identities, truth tables, canonical forms, and circuit design using logic gates. It also focuses on simplification using Boolean laws and Karnaugh Maps (K-Maps).

Key Topics Covered

Boolean Operators

  • AND (·) – Output is true only if both inputs are true
  • OR (+) – Output is true if at least one input is true
  • NOT (¬ or ‘) – Inverts the value (1 becomes 0, and 0 becomes 1)

Truth Tables

A | B | A + B | A · B | A\'
---------------------------
0 | 0 |   0   |   0   |  1
0 | 1 |   1   |   0   |  1
1 | 0 |   1   |   0   |  0
1 | 1 |   1   |   1   |  0

Basic Logic Gates

  • AND Gate – Output: 1 only if both inputs are 1
  • OR Gate – Output: 1 if any input is 1
  • NOT Gate – Output is inverse of input
  • NAND, NOR, XOR, XNOR Gates – Derived logic gates used in circuits

Boolean Laws and Properties

  • Commutative Law: A + B = B + A ; A · B = B · A
  • Associative Law: A + (B + C) = (A + B) + C
  • Distributive Law: A · (B + C) = A·B + A·C
  • Identity Law: A + 0 = A ; A · 1 = A
  • Complement Law: A + A\’ = 1 ; A · A\’ = 0

Duality Principle

Interchanging AND ↔ OR and 0 ↔ 1 in a Boolean expression gives its dual. Example:
Original: A + 0 = A → Dual: A · 1 = A

Canonical Forms

  • SOP (Sum of Products): OR of multiple AND terms
  • POS (Product of Sums): AND of multiple OR terms
Example SOP: A·B + A\'·C
Example POS: (A + B)·(A\' + C)

Minterms and Maxterms

  • Minterm: A product (AND) term representing a unique combination of inputs that results in output = 1
  • Maxterm: A sum (OR) term representing a unique combination of inputs that results in output = 0

Karnaugh Map (K-Map)

K-Map is a visual method of simplifying Boolean expressions using grouping of 1s (for SOP) or 0s (for POS).

2-Variable K-Map Example

AB | 0 | 1
--------------
  0  | 1 | 0
  1  | 1 | 1

Minimized Expression: A + A\’·B\’

Logic Circuit Design

  • Draw logic circuits for a given Boolean expression using logic gates
  • Simplify expressions before circuit construction to reduce gate usage

Standard Questions in Board Exams

  • Simplify Boolean expressions using laws
  • Draw logic circuits and truth tables
  • Convert expressions to canonical forms
  • Simplify expressions using K-Map
  • Draw simplified circuit diagrams

Important Symbols and Representation

  • ‘ (prime) or ¬ used for NOT
  • · used for AND (can be omitted)
  • + used for OR

Common Mistakes to Avoid

  • Confusing SOP and POS forms
  • Using incorrect K-Map grouping
  • Overusing logic gates due to lack of simplification

Download PDF – Boolean Algebra Solutions

Includes:

  • NCERT exercise solutions with logic gate diagrams
  • Simplified Boolean expressions with laws applied
  • K-Map practice questions with answers
  • Board exam sample questions

Preparation Tips for Boolean Algebra

  • Memorize Boolean laws and identities
  • Practice simplification and K-Map problems
  • Draw logic circuits neatly with correct gate symbols
  • Understand rather than memorize – logic matters!

Additional Resources

  • K-Map Solvers: Tools to verify your manual work
  • Logic Gate Simulators: Online tools to build and simulate circuits
  • Sample Worksheets: Practice SOP, POS, and simplification

Conclusion

Boolean Algebra is essential for understanding how computers make decisions using logic circuits. This chapter builds a strong foundation for students pursuing careers in engineering, programming, electronics, and more. Regular practice of laws, K-Map simplifications, and circuit diagrams ensures success in board exams and real-world applications.

Class 12 Computer Science (C++) – Chapter 13 Boolean Algebra Language

TOPIC-1 Basics of Boolean Algebra Very Short Answer Type Questions [1 mark each]

Question 1: Which gates are known as universal gates? Why?

Universal gates are the ones which can be used to implement any gate like AND, OR, and NOT, or any combination of these basic gates. NAND and NOR gates are known as universal gates.

Question 2: Draw the equivalent logic circuit for the following Boolean expression:

[Logic Circuit Representation]

Question 3: Express the OR operator in terms of AND and NOT operators.

(A . B)’ = A̅ + B̅ (A̅ + B̅)’ = A + B

Question 4: Specify which axioms/theorems are being used in the following Boolean reductions:

(a) (be)’ + be = 1 → Complementary law
(b) xyz + zx = xz → Absorption law

Question 5: State and verify Associative law using a truth table.

Associative Law: (A + B) + C = A + (B + C) (A . B) . C = A . (B . C)

Truth Table Verification for (A + B) + C = A + (B + C)

A B C (A + B) + C A + (B + C)
0 0 0 0 0
0 0 1 1 1
0 1 0 1 1
0 1 1 1 1
1 0 0 1 1
1 0 1 1 1
1 1 0 1 1
1 1 1 1 1
Conclusion: The law (A + B) + C = A + (B + C) holds true for all combinations.

Truth Table Verification for (A . B) . C = A . (B . C)

A B C (A . B) . C A . (B . C)
0 0 0 0 0
0 0 1 0 0
0 1 0 0 0
0 1 1 0 0
1 0 0 0 0
1 0 1 0 0
1 1 0 0 0
1 1 1 1 1
Conclusion: The law (A . B) . C = A . (B . C) also holds true for all combinations.

Short Answer Type Questions-I [2 mark each]

Question 1: Correct the following Boolean statements:


X + 1 = 1            (Correct form of X + 1 = X)
((A′)′) = A          (Correct form of (A′)′ = A′)
A + A′ = 1           (Correct form, originally A + A′ = 0)
(A + B)′ = A′ . B′   (Correct form of (A + B)′ = A . B)

Question 2: Write the POS form of a Boolean Function F, which is represented in a truth table as follows:

Answer: F = (P + Q + R) · (P′ + Q + R) · (P′ + Q′ + R)

Short Answer Type Questions – II [3 Marks Each]

Question 1: Correct the following Boolean statements:


X + 1 = 1            (Correct form of X + 1 = X)
((A′)′) = A          (Correct form of (A′)′ = A′)
A + A′ = 1           (Correct form, originally A + A′ = 0)
(A + B)′ = A′ . B′   (Correct form of (A + B)′ = A . B)

Question 2: Write the POS form of a Boolean Function F, which is represented in a truth table as follows:

Answer: F = (P + Q + R) · (P′ + Q + R) · (P′ + Q′ + R)

Short Answer Type Questions – II [3 Marks Each]

Question 1: State and Verify Absorption law algebraically.

Absorption Law: A + AB = A A · (A + B) = A Algebraic Verification of A + AB = A: 

A + AB = (A · 1) + (A · B)   [Using Identity]
       = A · (1 + B)         [Distributive Law]
       = A · 1               [1 + B = 1]
       = A

Question 2: State and define the principle of duality. Why is it important in Boolean Algebra?

Principle of Duality: From every Boolean expression, another valid expression (dual) can be derived by:
  • Replacing each + with .
  • Replacing each . with +
Example: Dual of A + A′B is A · (A′ + B) Importance: It allows us to derive new theorems from existing ones, cutting the effort to prove expressions in half.

Question 3: Name the law shown below & verify it using a truth table: X + X′ · Y = X + Y

This is known as the Absorption Law or Redundancy Law.

Question 4: Logic Circuit for Boolean Expression: ab + c · d′


   A ───┐
        ├── AND ─┐
   B ───┘        │
                 ├── OR ──── OUTPUT
   C ───┐        │
        ├── AND ┘
   D ──|> NOT

Question 5: Write the SOP form of a Boolean function F, which is represented in a truth table as follows:

Answer: F = A′B′C + A′BC + AB′C + AB′C

Question 6: Logic Circuit for Boolean Expression: (U + V) · W + Z


   U ──┐
       ├── OR ─┐
   V ──┘       │
               ├── AND ─┐
   W ──────────┘        │
                        ├── OR ──── OUTPUT
   Z ───────────────────┘

Question 7: Verify the following using Boolean Laws: U′ + V = U′V + LP·V + U·V

Answer: 

LHS = U′ + V
    = U′ · (V + V) + V · (LP + U)       [Using Distribution Law]
    = U′ · V + LP · V + U · V + U · V
    = U′ · V + LP · V + U · V           [Combining U·V terms]
    = RHS

Question 8: Draw the Logic Circuit for the following Boolean Expression: (X′ + Y) · Z + W′


X ──|> NOT ─┐
            ├── OR ─┐
Y ──────────┘       │
                    ├── AND ─┐
Z ──────────────────────────┘
                             ├── OR ──── OUTPUT
W ──|> NOT ──────────────────┘

Question 9: Write the equivalent Boolean expression for the following logic circuit:

Answer: 

((X′ · Y)′ + (X · Y′)′)′

Question 10: Write the equivalent Boolean Expression for the following Logic Circuit:

Given: Z = (A + B) · (B′ + C) 

Z = (A + B) · (B′ + C)
  = A · B′ + A · C + B · B′ + B · C       [Distributive Law]
  = A · B′ + A · C + 0 + B · C            [B · B′ = 0]
  = A · B′ + A · C + B · C

Question 11: Obtain the Boolean Expression for the logic circuit shown below:


F = (A′ · B) + (C + D′)
  = (A′ · B) + C′ · D

Question 12: Name the law shown below & verify it using a truth table: A + B · C = (A + B) · (A + C)

This law is called the “Distributive Law”.

Truth Table:

A B C B·C A + B·C A + B A + C (A + B)(A + C)
0 0 0 0 0 0 0 0
0 0 1 0 0 0 1 0
0 1 0 0 0 1 0 0
0 1 1 1 1 1 1 1
1 0 0 0 1 1 1 1
1 0 1 0 1 1 1 1
1 1 0 0 1 1 1 1
1 1 1 1 1 1 1 1
Conclusion: The truth table confirms A + B·C = (A + B)(A + C).

Question 13: Obtain the Boolean Expression for the logic circuit shown below:

Answer: 

F = (X · Y′) + (Z′ + W)
F = X′ + Y + Z′ + W

Question 14: State DeMorgan’s law. Verify one of them using a truth table:

DeMorgan’s Laws:
  • (i) (A · B)’ = A’ + B’
  • (ii) (A + B)’ = A’ · B’

Verification of (A · B)’ = A’ + B’ using a Truth Table:

A B A · B (A · B)’ A’ B’ A’ + B’
0 0 0 1 1 1 1
0 1 0 1 1 0 1
1 0 0 1 0 1 1
1 1 1 0 0 0 0
Conclusion: The truth table verifies that (A · B)’ = A’ + B’

Question 15: Draw a logic Circuit for the Boolean expression: A · B′ + (C + B′) · A′


A ──┐              ┌────────── AND ─┐
    │              │               │
    └── AND ┐      │               │
B ──|> NOT ┘        └─────┐        │
                          │        │
C ─────┐                 OR ──┐    │
       └────────────┐         │    │
B ──|> NOT ────────┘          │    │
                             AND ─┘
A ──┐
    └─|> NOT ────────────────┘
                    │
                    ▼
                OUTPUT

Question 16: Obtain the Boolean Expression for the logic circuit shown below:


F = P′ · Q + (Q + R′)
F = Q · (P′ + R′)

Question 17: Verify the following using Boolean Laws: X + Z = X + X′ · Z + Y · Z

Proof: 

RHS = X + X′ · Z + Y · Z
    = (X + X′) · (X + Z) + Y · Z       [Distribution]
    = 1 · (X + Z) + Y · Z              [X + X′ = 1]
    = X + Z + Y · Z
    = X + Z · (1 + Y)
    = X + Z                           [1 + Y = 1, 1 · Z = Z]
Hence, LHS = RHS → Verified

Question 18: Verify the following using Boolean Laws: A + C = A + A · C + B · C


RHS = A + A′ · C + B · C
    = (A + A) · (A + C) + B · C         [Distributive Law]
    = 1 · (A + C) + B · C
    = A + C + B · C
    = A + C · (1 + B)
    = A + C · 1
    = A + C
Hence, LHS = RHS → Verified

Question 19: Obtain the Boolean Expression for the logic circuit shown below:


F = (X′ · Y) + (Y + Z′)
  = (X′ + 1) · Y + Z′
  = Y + Z′          [Since (X′ + 1) = 1]

Question 20: Verify the following using truth table:

X X′ X · X′ X + 1
0 1 0 1
1 0 0 1
Conclusion: (i) X · X′ = 0 is verified (ii) X + 1 = 1 is verified

Question 21: Write the equivalent Boolean expression for the following logic circuit:


Y = U · V′ + U′ · W′

Question 22: Write the equivalent Boolean expression for the following logic circuit:


Y = (U + V′) · (U + W′)

Question 23: Verify the following using truth table:

X X + 0 X′ X + X′
0 0 1 1
1 1 0 1
Conclusion: (i) X + 0 = X is verified (ii) X + X′ = 1 is verified

Question 24: Derive a Canonical SOP expression for a Boolean function F, represented by the following truth table:


F(A, B, C) = A′B′C + A′BC + AB′C + ABC
Or using Σ notation:
F(A, B, C) = ∑(0, 3, 4, 7)

Question 25: Derive a Canonical POS expression for a Boolean function F, represented by the following truth table:


F(P, Q, R) = (P + Q + R′) · (P + Q + R) · (P′ + Q + R′) · (P′ + Q′ + R)
Or using Π notation:
F(P, Q, R) = Π(1, 2, 5, 6)

Question 26: Obtain a simplified form for the Boolean expression: F(U, V, W, Z) = ∏(0,1,3,5,6,7,15)


Simplified expression:
F = (U + V + W) · (U + Z′) · (V′ + W′) · (U′ + W′ + Z)

Question 27: Reduce the following Boolean expression to its simplest form using K-map: F(X, Y, Z, W) = Σ(0,1,6,8,9,10,11,12,15)


Simplified expression:
F = XY + Y′Z′ + XZ′W′ + XZW + X′YZW′

Question 28: Reduce the following Boolean expression to its simplest form using K-map: F(X, Y, Z, W) = Σ(0,1,4,5,6,7,8,9,11,15)


Simplified expression:
F = Y′Z′ + XY + XZW

Question 29: Verify the following using Boolean laws: X + Y′ = X·Y + X·Y + X′·Y


LHS: X + Y′
= X·(Y + Y′) + (X + X′)·Y′
= X·1 + Y′·1
= X + Y′

RHS: X·Y + X·Y′ + X′·Y′
= X·(Y + Y′) + X′·Y′
= X·1 + X′·Y′
= X + X′·Y′
= X + Y′  [Since X + X′·Y′ = X + Y′]

Hence, verified.

Question 30: State the Distributive law and verify it using a truth table.

Distributive Law:
  • (i) X · (Y + Z) = X · Y + X · Z
  • (ii) X + (Y · Z) = (X + Y) · (X + Z)

Truth Table for (i): X · (Y + Z) = X · Y + X · Z

X Y Z Y+Z X·(Y+Z) X·Y X·Z X·Y + X·Z
0 0 0 0 0 0 0 0
0 0 1 1 0 0 0 0
0 1 0 1 0 0 0 0
0 1 1 1 0 0 0 0
1 0 0 0 0 0 0 0
1 0 1 1 1 0 1 1
1 1 0 1 1 1 0 1
1 1 1 1 1 1 1 1
Conclusion: Verified that X · (Y + Z) = X · Y + X · Z

Question 31: Reduce the following Boolean expression using K-map: F(A, B, C, D) = Σ(0,1,3,5,6,7,9,11,13,14,15)


Simplified expression:
F = A′B′C′ + D + B·C
 

TOPIC-2 Karnaugh Map Minimization and Applications of Boolean Algebra Very Short Answer Type Questions [1 mark each]

Question 1: Write the Product of Sum expression of the function F(a, b, c, d) from the given truth table.


F(a, b, c, d) =
(a + b + c + d) ·
(a + b + c + d′) ·
(a + b′ + c + d) ·
(a + b′ + c′ + d′) ·
(a′ + b + c + d) ·
(a′ + b + c + d′) ·
(a′ + b′ + c + d) ·
(a′ + b′ + c + d′) ·
(a′ + b′ + c′ + d)

Question 2: Convert the following Boolean expression into its equivalent Canonical Sum of Products form (SOP): (U + V + W) · (U + V + W′) · (U′ + V + W) · (U′ + V′ + W′)


π(0, 1, 4, 7) → F = Σ(2, 3, 5, 6)
Canonical SOP = U′V W′ + U′V W + U V′W + U V W′

Question 3: Write the Sum of Product form of the function F(R, Q, R) for the following truth table representation of F.

Answer: (Based on specific truth table values) 

Example SOP (if F=1 at minterms 0, 2, 4):
F(R, Q, R) = R′Q′R + R′QR′ + RQ′R′

Question 4: Write the Product of Sum form of the function F(X, Y, Z) for the following truth table representation of F.

Answer: (Assuming F=0 at maxterms 0, 1, 4) 

F(X, Y, Z) = (X + Y + Z) · (X + Y + Z′) · (X + Y′ + Z)

Question 5: Write the Product of Sum form of the function G(U, V, W) for the following truth table representation of G.

Answer: (Assuming G=0 at maxterms 1, 3, 5) 

G(U, V, W) = (U + V + W′) · (U + V′ + W′) · (U′ + V + W′)

Question 6: Write the Product of Sum form of function G(U, V, W) for the following truth table representation of G.

Answer: (Assuming G=0 at maxterms 2, 4, 6) 

G(U, V, W) = (U + V′ + W) · (U′ + V + W′) · (U′ + V′ + W)

Question 7: Write the Sum of Product form of the function F(A, B, C) for the following truth table representation of F.


SOP = A′B C′ + A′B C + A B′C′ + A B C

Question 8: Write the SOP form of a Boolean function F, which is represented in a truth table as follows:


F(X, Y, Z) = X′Y′Z′ + X′Y Z′ + X Y′Z′ + X Y Z

Question 9: Write the POS form of Boolean function G, which is represented in a truth table as follows:


G(A, B, C) = (A + B + C) · (A + B′ + C′) · (A′ + B + C) · (A′ + B + C′)
 

Short Answer Type Questions-II

Question 1: Obtain the minimal SOP form for the following Boolean expression using K-Map: F(A,B,C,D) = ∑(0,2,3,5,7,8,10,11,14,15)


Quad 1 = m0 + m2 + m8 + m10 = B′D′
Quad 2 = m3 + m7 + m15 + m11 = C·D
Quad 3 = m5 + m7 + m15 + m13 = B·D

Minimal SOP: F = B′D′ + CD + BD

Question 2: Reduce the following Boolean expression using K-Map: F(A,B,C,D) = π(0,1,2,4,5,6,8,10)


Minimal POS: F = (A + C)(A + D)(B + D)

Question 3: Reduce the following using K-Map: F(A,B,C,D) = ∑(1,3,4,5,6,7,12,13)


Simplified SOP: F = A′C + A′D + BC

Question 4: Reduce the following Boolean expression using K-map: F(P, Q, R, S) = ∑(0,2,4,5,6,7,8,10,13,15)


Simplified SOP: F = P′Q + Q′S′ + Q·S

Question 5: Reduce the following Boolean expression using K-Map: F(P, Q, R, S) = ∑(1,2,3,4,5,6,7,8,10)


Simplified SOP: F = P′Q + P′S + P′R′S′ + PQ′S′

Question 6: Reduce the following Boolean expression using K-Map: F(A, B, C, D) = ∑(2,3,4,5,6,7,8,10,11)


Simplified SOP: F = A′B + A′C + B′C + A·B·D′
 

Long Answer Type Questions [4 marks each]

Question 1: Verify the following using Boolean Laws:


A′ + B′C = A′B′C′ + A′BC′ + A′BC + A′B′C + AB′C

Step-by-step:
Group A′ terms: A′C′(B′ + B) + A′C
               = A′C′ · 1 + A′C
               = A′C′ + A′C
               = A′(C′ + C)
               = A′ · 1
               = A′

Now consider AB′C (outside A′ group):
Final: A′ + AB′C
=> RHS = A′ + B′C
=> Hence, Verified.

Question 2: Write the Boolean Expression for the result of the Logic Circuit:


F = (u + v′) · (u + w) · (v + w′)

Question 3: Derive a Canonical POS expression for a Boolean function F, represented by the following truth table:


F = Π(0, 3, 4, 5)
POS = (P + Q + R) · (P + Q′ + R′) · (P′ + Q + R) · (P′ + Q + R′)

Question 4: Reduce the following Boolean Expression to its simplest form using K-Map:
F(X, Y, Z, W) = ∑(2, 6, 7, 8, 9, 10, 11, 13, 14, 15)


Simplified SOP: F = XY′ + ZW′ + XW + YZ

Question 5: Write the Boolean Expression for the result of the Logic Circuit:


G = PQ′ + PR + QR′

Question 6: Derive a Canonical SOP expression for a Boolean function G, represented by the following truth table:


G = ∑(0, 2, 6, 7)
SOP = A′B′C′ + A′BC′ + ABC′ + ABC

Question 7: Verify the following using Boolean Laws:


X′ + Y′Z = X′Y′Z′ + X′YZ′ + X′YZ + X′Y′Z + XY′Z

Group terms:
X′Z′(Y′ + Y) + X′Z(Y′ + Y) + XY′Z
= X′Z′ + X′Z + XY′Z
= X′(Z′ + Z) + XY′Z
= X′ + XY′Z
=> X′ + Y′Z (verified)

Question 8: Reduce the following Boolean Expression to its simplest form using K-Map:
F(P, Q, R, S) = ∑(0, 4, 5, 8, 9, 10, 11, 12, 13, 15)


Simplified SOP: F = R′S′ + PQ′ + QR′ + PS