Boolean Algebra Question Bank Set 2

MATH 250 - INTRODUCTION TO DISCRETE MATHEMATICS - Boolean Algebra Question Bank - Set 2 Liberty University Question 1 Question Let A, B, and C be Boolean variables. Simplify the following Boolean expression: (A + B) · (A + C). Solution To simplify the given Boolean expression (A + B) · (A + C), we will use the distributive law and absorption law of Boolean algebra. Step 1: Distributive Law By distributing the terms (A + B) and (A + C), we get: (A + B) · (A + C) = A · (A + C) + B · (A + C) Step 2: Absorption Law Using the absorption law (X + X · Y = X), we simplify the expression as follows: A · (A + C) + B · (A + C) = A + A · C + B · (A + C) Step 3: Distributive Law (Once More) Now, distribute the term B into (A + C): A + A · C + B · (A + C) = A + A · C + A · B + B · C Step 4: Commutative Law and Absorption Law Applying the commutative (XY = Y X) and absorption laws, we simplify the expression further: A + A · C + A · B + B · C = A + A · (B + C) + B · C Step 5: Absorption Law Finally, we can simplify the expression using the absorption law: A + A · (B + C) + B · C = A + B · C Therefore, the simplified form of (A + B) · (A + C) is A + B · C. Question 2 Question Simplify the following Boolean expression: (A + B + C)(A + B + C)(A + B + C). Solution Step 1: Use the distributive law to multiply the first two terms together. (A + B + C)(A + B + C) = (A · A + A · B + A · C + B · A + B · B + B · C + C · A + C · B + C · C) = (A + AB + AC + BA + 0 + BC + AC + 0 + 0) = (A + AC + AB + AB + BC + AC) = (A + A(C + C) + AB + BC) = (A + A + AB + BC) = (A + AB + BC) Step 2: Now, multiply the result from step 1 with the third term and simplify. (A + AB + BC)(A + B + C) = (AA + AB + ABC + AAB + ABB + ABBC + AAC + ABC + ABBC) = (0 + AB + ABC + 0 + 0 + 0 + 0 + 0 + 0) = (AB + ABC) Therefore, the simplified Boolean expression is AB + ABC. Question 3 Question Let F (A, B, C) = (A + B)(A + B + C)(A + B + C) where A, B, and C are Boolean variables. Simplify the Boolean expression F to its simplest form using Boolean Algebra laws. 2 Solution To simplify the Boolean expression F (A, B, C) = (A+B)(A+B+C)(A+B+C), we will apply various Boolean Algebra laws. Step 1: Apply the distributive law: XY + XZ = X(Y + Z). F (A, B, C) = (A + B)(A + B + C)(A + B + C) = (A + B)(A + B + C + A + B + C) = (A + B)(A + A + B + B + C + C) = (A + B)(1 + B + C) = (A + B) Therefore, the simplified form of the Boolean expression F (A, B, C) = (A + B). Question 4 Question Simplify the following Boolean expression using Boolean Algebra: (A + B)(A + C)(B + C). Solution Step 1: Apply the Distributive Law to expand the expression. (A + B)(A + C)(B + C) = A(A + C)(B + C) + B(A + C)(B + C) = AA + AC + AB + BC + BC = A + AC + AB + BC Step 2: Apply the Idempotent Law (AA = A). A + AC + AB + BC = A + A(C + B) + BC = A + (A)(C + B) + BC = A + AC + AB + BC Step 3: Apply the Absorption Law (A + AB = A). A + AC + AB + BC = A + AC + A(C + B) = A + AC + A = A(1 + C) + A =A+A =A Therefore, the simplified expression is A. 3 Question 5 Question Let F (A, B, C) = A′ B ′ C ′ + A′ BC + AB ′ C ′ + ABC. Simplify F (A, B, C) using Boolean algebra laws. Solution To simplify the given Boolean expression F (A, B, C) = A′ B ′ C ′ + A′ BC + AB ′ C ′ + ABC, we will apply various Boolean algebra laws. Step 1: Apply the absorption law X + XY = X. F (A, B, C) = A′ B ′ C ′ + A′ B(C + C ′ ) + AB ′ C ′ + AB(C + C ′ ) = A′ B ′ C ′ + A′ B + AB ′ C ′ + AB = A′ B(1 + 1) + AB(1 + 1) = A′ B + AB Step 2: Apply the consensus theorem X + X ′ Y = X + Y . F (A, B, C) = A′ B + AB = B(A′ + A) = B(1) =B Therefore, the simplified form of F (A, B, C) is B. Question 6 Question Simplify the Boolean expression (A + B + C)(A + B)(A + C) using Boolean algebra laws and theorems. Solution We can simplify the given Boolean expression step by step using Boolean algebra laws and theorems. (A + B + C)(A + B)(A + C) = (A + B + C)(AA + AC + BA + BC) = (A + B + C)(0 + AC + BA + BC) Exp App = AC(A + B + C) + BA(A + B + C) + BC(A + B + C) = ACA + ACB + ACC + BAA + BAB + BAC + BCA + BCB + BCC = 0 + 0 + AC + 0 + 0 + BC + 0 + 0 + 0 = AC + BC 4 Appl Therefore, the simplified Boolean expression is AC + BC. Question 7 Question Simplify the following Boolean expression: (A + B)(A + B) + A. Solution To simplify the given Boolean expression, we will first expand the terms and then apply Boolean algebra rules to simplify it step by step. Step 1: Expand the terms using the distributive law. (A + B)(A + B) = AA + AB + BA + BB Step 2: Apply the idempotent law (AA = A) and null law (AB + BB = A + B). AA + AB + BA + BB = A + A + B Step 3: Simplify the expression. A+A+B =A+B Thus, the simplified form of the Boolean expression (A + B)(A + B) + A is A + B. Question 8 Question Simplify the Boolean expression (A + B + C)(A′ + B + C ′ )(A + B ′ ). Solution To simplify the Boolean expression, we will use the properties of Boolean algebra, such as idempotent law, identity law, complementary law, etc. Step 1: Use the distributive law to expand the expression. (A+B+C)(A′ +B+C ′ )(A+B ′ ) = (AA′ +AB+AC+BA′ +BB+BC ′ +CA+CB+CC ′ )(A+B ′ ) Step 2: Simplify the expression by removing terms AA′ , BB, CC ′ . (AB + AC + BA′ + BC ′ + CA + CB)(A + B ′ ) Step 3: Use the absorption law to further simplify the expression. AB + AC + BA′ + BC ′ + CA + CB 5 Step 4: Use the consensus theorem to simplify the expression. AB + AC + CA Step 5: Use the consensus theorem again to simplify the expression. AB + AC Therefore, the simplified form of (A+B+C)(A′ +B+C ′ )(A+B ′ ) is AB+AC. Question 9 Question Let F (A, B, C) = A′ BC ′ + AB ′ C + A′ B ′ C ′ . Simplify the expression using Boolean algebra properties. Solution To simplify the given expression F (A, B, C) = A′ BC ′ + AB ′ C + A′ B ′ C ′ , we will apply Boolean algebra properties such as the distributive law, complement law, idempotent law, etc. Step 1: Apply the distributive law (A(B + C) = AB + AC) to simplify the expression: F (A, B, C) = A′ BC ′ + AB ′ C + A′ B ′ C ′ = A′ C ′ (B + B ′ ) + A′ B(C + C ′ ) Step 2: Use the complement law (X + X ′ Y = X + Y ) to simplify the expression further: F (A, B, C) = A′ C ′ + A′ B Step 3: Apply the distributive law in reverse (X + XY = X) to simplify the expression: F (A, B, C) = A′ C ′ + A′ B = A′ (C ′ + B) Therefore, the simplified expression for F (A, B, C) is A′ (C ′ + B). Question 10 Question Simplify the following Boolean expression: (A + B + C)(A + B + C)(A + B + C)(A + B + C). 6 Solution Step 1: Expand the given expression using the distributive property. (A + B + C)(A + B + C)(A + B + C)(A + B + C) = A(A+B+C)(A+B+C)(A+B+C)+B(A+B+C)(A+B+C)(A+B+C)+C(A+B+C)(A+B+C)(A+B+C) Step 2: Apply the distributive property again to expand each term. (A+AB+AC)(A+B+C)(A+B+C)+(BA+B+BC)(A+B+C)(A+B+C)+(CA+CB+CC)(A+B+C)(A+B+C Step 3: Continue expanding and simplifying each term. (A + AB + AC)(A + B + C)(A + B + C) = (AA + AB + AC + AB + ABB + ABC + AAC + AC + ACC) = (A + AB + AC + AB + ABC + AC + AC) = A(1 + B + C + B + BC + C + C) = A(1) =A Step 4: Similarly simplify the other two expanded terms to get the final simplified expression. Final expression: A + B + C Therefore, (A + B + C)(A + B + C)(A + B + C)(A + B + C) simplifies to A + B + C. Question 11 Question Simplify the boolean expression (A+B)(A+C)(B+C) using the laws of Boolean Algebra. Solution We will simplify the given expression step by step using the laws of Boolean Algebra. (A + B)(A + C)(B + C) = (A + B)(AB + AC + BC + BC) Expand (A + C)(B + C) using the distributive = ABB + ABC + ABC + ACC + BCC + BCC = AB + ABC + ABC + AC + BC + BC = AB + AC + BC Simplify using XX = X and X + X = X Simplify using XX = X and X + X = X Therefore, (A + B)(A + C)(B + C) simplifies to AB + AC + BC. 7 Distribute (A + B) over (AB + AC Question 12 Question Simplify the following Boolean expression using Boolean algebra rules: (A + B + C)(A + B + C). Solution We will simplify the given Boolean expression step by step using Boolean algebra rules. Step 1: Apply the Distributive Law: (A + B + C)(A + B + C) = A(A + B + C) + B(A + B + C) + C(A + B + C). Step 2: Use the Absorption Law: AA = 0 and A+AB = A+B. Apply these laws to simplify each term: AA + AB + AC + BA + BB + BC + CA + CB + CC. Step 3: Further simplify using the Complement Law: AA = 0, AB = AB, AC = AC, BA = AB, BB = 0, BC = BC, CA = AC, CB = BC, and CC = 0. Step 4: Combine the simplified terms: AB + AC + AB + BC + AC = AB + AC + BC . Question 13 Question Simplify the following Boolean expression: (A + B + C)(A + B + C)(A + B + C)(A + B + C). Solution Step 1: Use the distributive property to expand the expression. (A + B + C)(A + B + C)(A + B + C)(A + B + C) = A(A + B + C)(A + B + C)(A + B + C) + B(A + B + C)(A + B + C)(A + B + C) + C(A + B + C)(A + B + C)(A + B + C) Step 2: Apply the distributive property again. = A(AA + AB + AC + AB + AB + BC + ABC + BB + BC) + B(AA + AB + AC + AB + AB + BC + ABC + BB + BC) + C(AA + AB + AC + AB + AB + BC + ABC + BB + BC) Step 3: Simplify the terms. 8 = A(AC + AB + BC + ABC + BB + BC) + B(AC + AB + BC + ABC + BB + BC) + C(AC + AB + BC + ABC + BB + BC) = A(AC + AB + BB + BC) + B(AC + AB + BB + BC) + C(AC + AB + BB + BC) = A(C + B) + B(C + A) + C(B + A) = AC + AB + BC Therefore, the simplified expression is AC + AB + BC. Question 14 Question Simplify the following Boolean expression using algebraic manipulation techniques: (A + B + C)(A + B + C)(A + B + C)(A + B + C). Solution 1. We start by expanding the given expression using the distributive law: (A + B + C)(A + B + C)(A + B + C)(A + B + C) = (A + B + C)(A(B + C) + B(B + C) + C(A + B)) = (A + B + C)(AB + AC + BB + BC + CA + CB) = (A + B + C)(AB + AC + BC + CA + CB) = ABA + ABC + AAC + ACB + ACC + ABC + BA + BC + CA + CB + CC = 0 + ABC + A + 0 + 0 + ABC + BA + 0 + 0 + C. 9 2. Next, simplify the expression further by combining terms: ABC + A + ABC + BA + C = A(BC + 1 + BC) + BA + C = A(1 + B + C) + BA + C = A + AB + AC + BA + C
= A 1 + B + AC + C = A + AC + C = A(1 + C) + C = A + C. Therefore, the simplified form of the given Boolean expression is A + C. Question 15 Question Simplify the following Boolean expression using algebraic manipulation: F = A′ B ′ C ′ D′ +A′ B ′ C ′ D+A′ BC ′ D+A′ BCD+AB ′ C ′ D′ +AB ′ C ′ D+AB ′ CD+ABCD Solution To simplify the given Boolean expression, we will use the laws of Boolean algebra including absorption, identity, complement, and distributive laws. Step 1: Apply the absorption law. F = A′ B ′ (C ′ D′ +C ′ D+CD)+AB ′ (C ′ D′ +C ′ D+CD)+A(B ′ C ′ D′ +B ′ C ′ D+BCD) Step 2: Apply the idempotent law XX = X. F = A′ B ′ (D′ + D) + AB ′ (D′ + D) + A(C ′ D′ + C ′ D + CD) Step 3: Apply the complement law X + X ′ Y = X + Y . F = A′ B ′ + AB ′ + AC + AD Step 4: Apply the consensus law XY + X ′ Z + Y Z = XY + X ′ Z. F = A(B ′ + B) + AC + AD Step 5: Apply the idempotent law XX = X. F = A + AC + AD 10 Step 6: Apply the absorption law. F = A(1 + C) Step 7: Apply the identity law X1 = X. F =A Therefore, the simplified Boolean expression is F = A. Question 16 Question Simplify the Boolean expression (A + B) · (A′ · B ′ ) + A · B. Solution We will simplify the given Boolean expression using the laws of Boolean algebra. Step 1: Apply the Distributive Law to expand the expression. (A + B) · (A′ · B ′ ) + A · B = A · A′ · B ′ + B · A′ · B ′ + A · B Step 2: Apply the Complement Law to simplify the expression A · A′ . A · A′ · B ′ + B · A′ · B ′ + A · B = 0 · B ′ + B · A′ · B ′ + A · B Step 3: Apply the Annulment Law to simplify the term 0 · B ′ . 0 · B ′ + B · A′ · B ′ + A · B = 0 + B · A′ · B ′ + A · B Step 4: Apply the Identity Law to remove the term 0. B · A′ · B ′ + A · B Step 5: Apply the Absorption Law to the expression B · A′ · B ′ . A·B Therefore, the simplified form of the expression (A + B) · (A′ · B ′ ) + A · B is A · B. Question 17 Question Simplify the following Boolean expression using algebraic manipulation: (A′ B + AB ′ )(A + B) + AB 11 Solution Let’s simplify the given Boolean expression step by step: Step 1: Apply the distributive law: ′ ′ (A B + AB )(A + B) + AB = A′ B(A + B) + AB ′ (A + B) + AB Step 2: Apply the distributive law again: = A′ BA + A′ BB + AB ′ A + AB ′ B + AB Step 3: Apply the idempotent law: = AB + 0 + 0 + 0 + AB Step 4: Apply the identity law: = AB + AB Step 5: Apply the idempotent law again: = AB Therefore, the simplified form of the Boolean expression is AB . Question 18 Question Simplify the following Boolean expression: (A + B)(A + C)(A + D). Solution To simplify the Boolean expression (A + B)(A + C)(A + D), we will use the following Boolean algebra identities: ˆ Complement Law: A + A = 1 ˆ Absorption Law: A + AB = A Step 1: Apply the Complement Law to A + D: (A + B)(A + C)(A + D) = (A + B)(A + C)(1) Step 2: Simplify the expression by using the Absorption Law: (A + B)(A + C)(1) = (A + B)(A + C) Step 3: Apply the Distributive Law to expand the expression: (A + B)(A + C) = AA + AC + BA + BC 12 = A + AC + BA + BC Step 4: Use the Absorption Law to simplify the expression: A + AC + BA + BC = A + AC Step 5: Apply the Distributive Law to factor out A: A + AC = A(1 + C) Step 6: Apply the Complement Law to simplify the expression: A(1 + C) = A Therefore, the simplified form of (A + B)(A + C)(A + D) is A. Question 19 Question Simplify the Boolean expression (A + B ′ )(A′ B + AB ′ ). Solution Step 1: Use the distributive property to expand the expression. (A + B ′ )(A′ B + AB ′ ) = A · A′ B + A · AB ′ + B ′ · A′ B + B ′ · AB ′ = AA′ B + AAB ′ + B ′ A′ B + B ′ AB ′ = AAB ′ + AAB ′ + AB ′ B + AB ′ B =A+A+0+0 =A Step 2: Simplify further to obtain the final simplified expression. (A + B ′ )(A′ B + AB ′ ) = A Question 20 Question Simplify the following Boolean expression: F = A′ B ′ C ′ + AB ′ C + ABC + A′ BC ′ 13 Solution To simplify the Boolean expression F = A′ B ′ C ′ + AB ′ C + ABC + A′ BC ′ , we will use Boolean algebra laws and rules. Step 1: Apply the absorption law (A + AB = A) to eliminate redundant terms. F = A′ B ′ C ′ + AB ′ C + ABC + A′ BC ′ = A′ B ′ C ′ + AB ′ C + A(BC + B ′ C ′ ) ′ ′ ′ ′ ′ = A B C + AB C + A(C(B + B )) = A′ B ′ C ′ + AB ′ C + AC (Grouping terms) (Distributive law) (Complement law: B + B ′ = 1) Step 2: Apply the consensus theorem (AB + AC + BC = AB + AC) to simplify the expression further. F = A′ B ′ C ′ + AB ′ C + AC = A′ B ′ C ′ + AB ′ C + A′ C (Consensus theorem) Step 3: Finally, we apply the consensus theorem once more to obtain the simplified expression. F = A′ B ′ C ′ + AB ′ C + A′ C = A′ C(B ′ + B) + A′ B ′ C ′ ′ ′ ′ =AC +AB C ′ ′ ′ (Complement law: B ′ + B = 1) = C(A + B ) + A B ′ C ′ ′ ′ =C +AB C ′ ′ (Distributive law) (Distributive law) (Complement law: A′ + B ′ = 1) Therefore, the simplified Boolean expression is F = C + A′ B ′ C ′ . Question 21 Question Simplify the Boolean expression (A + B + C) · (A + B) + A · B · C. Solution Step 1: Use the distributive law to expand the expression. (A + B + C) · (A + B) + A · B · C = (A · A + A · B + B · A + B · B + C · A + C · B) + A · B · C = (0 + A · B + A · B + 0 + C · A + C · B) + A · B · C = (A · B + A · B + C · A + C · B) + A · B · C 14 Step 2: Use the consensus theorem to simplify the expression. (A · B + A · B + C · A + C · B) + A · B · C = (A · B + A · B + C · A + C · B) = (A · B + A · B + C · (A + B)) = (A · B + A · B + C) Therefore, the simplified Boolean expression is A · B + A · B + C. Question 22 Question Simplify the following Boolean expression: (A + B + C)(A + B + C)(A + B + C). Solution Step 1: Use the distributive property to expand the expression. (A + B + C)(A + B + C)(A + B + C) = (A(A + B + C) + B(A + B + C) + C(A + B + C))(A + B + C) = (A + AB + AC + BA + B 2 + BC + CA + BC + CC)(A + B + C) = (A + AB + AC + A + B + BC + CA + BC + 0)(A + B + C) = (A + B + AC + BC + CA + BC)(A + B + C) Step 2: Continue simplifying by removing redundant terms. (A + B + AC + BC + CA + BC)(A + B + C) = (A(1 + C) + B(1 + C) + AC + BC)(A + B + C) = (A + B + AC + BC)(A + B + C) = (A(1 + C) + B(1 + C))(A + B + C) = (A + B)(A + B + C) = A + AB + BA + B + BC + BC = A + B + BA + BC + BC = A + B + BC + BC = A + (B + BC) + BC = A + B + BC = A + BC Therefore, the simplified expression is A + BC. 15 Question 23 Question Simplify the Boolean expression F = A′ B ′ C ′ + AB ′ C + ABC + A′ B ′ C Solution To simplify the given Boolean expression, we will use the laws of Boolean algebra. Step 1: Apply the idempotent law: X + X = X F = A′ B ′ C ′ + AB ′ C + ABC + A′ B ′ C Step 2: Group the terms which contain the same variables: F = A′ B ′ C ′ + A′ B ′ C + AB ′ C + ABC Step 3: Apply the idempotent law: X + X = X F = A′ B ′ C ′ + AB ′ C + ABC Step 4: Apply the absorption law: X + XY = X F = A′ B ′ (C ′ + C) + ABC Step 5: Apply the complement law: X + X ′ Y = X + Y F = A′ B ′ + ABC Step 6: Apply the absorption law: X + XY = X F = B(A′ + AC) Thus, the simplified form of the given Boolean expression is F = B(A′ +AC). Question 24 Question Simplify the following Boolean expression: (A + B)(A′ C + BC). Solution To simplify the given Boolean expression (A + B)(A′ C + BC), we will use the distributive property, complement property, and absorption property of Boolean algebra. Step 1: Apply the distributive property to expand the expression: (A + B)(A′ C + BC) = A · A′ C + A · BC + B · A′ C + B · BC = AA′ C + ABC + BA′ C + BBC Step 2: Simplify using the complement property AA′ = 0: AA′ C + ABC + BA′ C + BBC = 0C + ABC + 0B + B0 = 0 + AB + 0 + 0 = AB Step 3: Final answer: The simplified expression is AB. 16 Question 25 Question Simplify the following Boolean expression using algebraic manipulation techniques: F (A, B, C, D) = (A + B + C) · (A+!B + C) + (!A + B + C) · (A + B + C) Solution Step 1: Apply the Distributive Property to expand the expression. F (A, B, C, D) = (A + B + C) · (A+!B + C) + (!A + B + C) · (A + B + C) = A · A + A·!B + A · C + B · A + B·!B + B · C + C · A + C·!B + C · C + (!A) · A + (!A) · B + (!A) · C + B · A + B · B + B · C + C · A + C · B + C · C Step 2: Simplify the expanded expression by removing redundant terms and applying the Idempotent Law (X · X = X). F (A, B, C, D) = A + A·!B + A · C + B + 0 + B · C + C + 0 + 0 + 0 + 0 + 0 + A · B + B + 0 + C + 0 + 0 = A + A·!B + A · C + B + B · C + C + A · B + C =A+A·C +B+B·C +C +A·B Step 3: Apply the Absorption Law (X + X · Y = X) and the Complement Law (X+!X = 1) to simplify the expression further. F (A, B, C, D) = A + A · C + B + B · C + C + A · B =A+B+C Therefore, the simplified Boolean expression is F (A, B, C, D) = A + B + C. Question 26 Question Simplify the following Boolean expression: (A + B + C)(A + B + C)(A + B + C)(A + B + C) Solution To simplify the given Boolean expression, we will use the laws of Boolean algebra, including the distributive law, the absorption law, and the complement law. Step 1: Apply the distributive law: (A + B + C)(A + B + C)(A + B + C)(A + B + C) ⇒ (A + B)(A + B + C)(A + B + C)(A + B + C) 17 Step 2: Apply the distributive law again: (A + B)(A + B + C)(A + B + C)(A + B + C) ⇒ (A + B)(A + B + C)(A + B + C) Step 3: Apply the absorption law: (A + B)(A + B + C)(A + B + C) ⇒ (A + B)(A + B + C) Step 4: Apply the distributive law once more: (A + B)(A + B + C) ⇒ AA + AB + AC + BB + BBC Step 5: Apply the complement law: BB = 0 ⇒ AB + AC Therefore, the simplified Boolean expression is AB + AC. Question 27 Question Simplify the Boolean expression (A + B) · (A + B) · (A + B + C) using Boolean algebra laws. Solution Step 1: Apply the Distributive Law to the first two terms: (A+B)·(A+B) = A·A+A·B +B ·A+B ·B = A+AB +AB +0 = A+AB +AB Step 2: Apply the Associative Law to the result of Step 1 with the third term: (A+AB +AB)·(A+B +C) = A·(A+B +C)+AB ·(A+B +C)+AB ·(A+B +C) = A + AB + AC + AB + ABB + ABC + AB + ABB + ABC = A + AB + AC + AB + 0 + ABC + AB + 0 + ABC = A + AB + AC + AB + ABC + AB + ABC Step 3: Apply the Absorption Law to the result of Step 2: A + AB + AC + AB + ABC + AB + ABC = A(1 + B + C) + AB + ABC = A + AB + ABC = A(1 + B + BC) Therefore, the simplified expression is A(1 + B + BC). Question 28 Question Simplify the following Boolean expression: F = AB ′ + A′ B + (A + B ′ )′ 18 Solution Step 1: Apply De Morgan’s Theorem to simplify the third term. (A + B ′ )′ = A′ B ′′ Step 2: Simplify the second term using the complement law. A′ B = A · B Step 3: Substitute the simplified expressions back into the original expression. F = AB ′ + A · B + A′ B ′′ = AB ′ + AB + A′ B Step 4: Use the absorption law to simplify the expression. F = AB ′ + AB + A′ B = AB ′ + A′ B Therefore, the simplified form of the given Boolean expression is AB ′ + A′ B. Question 29 Question Simplify the Boolean expression (A + B)(A + C)(B + C) + ABC. Solution To simplify the given Boolean expression, we will first expand it using the distributive property, then use the Boolean algebra laws to simplify the resulting expression. Step 1: Expand the given expression (A + B)(A + C)(B + C) + ABC = (A + B)(A + C)B + (A + B)(A + C)C + ABC = AB + BB + AC + BC + AC + CC + ABC = AB + B + AC + BC + AC + C + ABC = AB + B + AC + BC + AC + C + AC + BC = AB + B + AC + BC + C + BC = AB + AC + BC + B + C Step 2: Apply Boolean algebra laws to simplify the expression AB + AC + BC + B + C = B(A + C) + C(A + B) + B + C =B+C +B+C =C +B 19 Therefore, the simplified form of the Boolean expression (A+B)(A+C)(B + C) + ABC is C + B. Question 30 Question Simplify the Boolean expression (A + B) · (A + B) using Boolean algebra rules. Solution 1. Let’s first expand the expression using the distributive law: (A + B) · (A + B) = A · A + A · B + B · A + B · B 2. Use the idempotent law X · X = X to simplify A · A: A·A+A·B+B·A+B·B =A+A·B+B·A+B·B 3. Use the absorption law X + X · Y = X to simplify B · B: A+A·B+B·A+B·B =A+A·B+B·A 4. Use the commutative law XY = Y X to rearrange A · B and B · A: A+A·B+B·A=A+A·B+A·B 5. Use the distribution law X + XY = X to simplify A · B + A · B: A + A · B + A · B = A + A · (B + B) 6. Use the complement law X + X ′ Y = X + Y to simplify B + B: A + A · (B + B) = A + A · 1 = A Therefore, (A + B) · (A + B) simplifies to A . Question 31 Question Simplify the Boolean expression F = A′ B + AB ′ + AC + BC ′ . 20 Solution To simplify the Boolean expression F = A′ B + AB ′ + AC + BC ′ , we will use Boolean algebra rules to simplify term by term. Step 1: Apply the absorption law: X + XY = X. We have: F = A′ B + AB ′ + AC + BC ′ = B(A′ + A) + AC + BC ′ = B + AC + BC ′ Step 2: Apply the consensus theorem: X + X ′ Z + Y Z = X + Y Z. We can rewrite the expression as: F = B + AC + BC ′ = B(1 + C ′ ) + AC = B + AC Step 3: Apply the consensus theorem again: X + X ′ Y + Z = X + Y + Z. Simplifying further, we get: F = B + AC = B(1 + C) + A(C + C ′ ) =B+A Therefore, the simplified form of the Boolean expression F = A′ B + AB ′ + AC + BC ′ is F = A + B. Question 32 Question Simplify the following Boolean expression: F = (A + B ′ )(A′ + B)(A′ + B ′ ) Solution To simplify the given Boolean expression, we will use the laws of Boolean algebra including the distributive law, complement law, and identity law. Step 1: Apply the distributive law to the expression: F = (A + B ′ )(A′ + B)(A′ + B ′ ) F = (A + B ′ )(A′ A′ + A′ B + B ′ A′ + B ′ B ′ ) F = (A + B ′ )(0 + A′ B + 0 + 0) 21 F = (A + B ′ )A′ B Step 2: Use the absorption law to simplify the expression: F = A′ B + A′ B ′ B F = A′ B Therefore, the simplified expression is F = A′ B. Question 33 Question Simplify the Boolean expression (A + B)(A + C)(B + C). Solution To simplify the given Boolean expression, we will use the following Boolean algebra identities: 1. Commutative Property: A + B = B + A and A · B = B · A 2. Associative Property: (A + B) + C = A + (B + C) and (A · B) · C = A · (B · C) 3. Distributive Property: A · (B + C) = A · B + A · C and A + (B · C) = (A + B) · (A + C) 4. Absorption Law: A + (A · B) = A and A · (A + B) = A 5. Identity Law: A + 0 = A and A · 1 = A 6. Zero Law: A · 0 = 0 and A + 1 = 1 Step 1: Apply the Distributive Property: (A + B)(A + C)(B + C) = ((A + B)A + (A + B)C)(B + C) Step 2: Apply the Distributive Property again: = ((A · A + B · A) + (A · C + B · C))(B + C) Step 3: Simplify using the Absorption Law and the Identity Law: = (A + B) + (A + C)(B + C) Step 4: Apply the Distributive Property once more: = (A + B) + AB + AC + BC Step 5: Apply the Commutative Property and the Associative Property: = A + AB + AC + BC + B Step 6: Apply the Absorption Law: = A + B + AC Therefore, the simplified Boolean expression (A + B)(A + C)(B + C) is equivalent to A + B + AC. 22 Question 34 Question Simplify the Boolean expression (A′ + B) · (A + B ′ ) + (A + B ′ ) · (A′ + B). Solution We simplify the given Boolean expression step by step using Boolean algebra laws. Step 1: Apply the Distributive Law: X · (Y + Z) = X · Y + X · Z. (A′ +B)·(A+B ′ )+(A+B ′ )·(A′ +B) = (A′ ·A+A′ ·B ′ +B·A+B·B ′ )+(A·A′ +A·B+B ′ ·A′ +B ′ ·B) Step 2: Apply the Complement Law: X · X ′ = 0. = (0 + A′ · B ′ + B · A + 0) + (0 + A · B + B ′ · A′ + 0) Step 3: Apply the Identity Law: X + 0 = X. = A′ · B ′ + B · A + A · B + B ′ · A′ Step 4: Apply the Commutative Law: X · Y = Y · X and X + Y = Y + X. = A · B + B · A + A′ · B ′ + B ′ · A′ Step 5: Apply the Idempotent Law: X + X = X and X · X = X. = A · B + A′ · B ′ Therefore, (A′ + B) · (A + B ′ ) + (A + B ′ ) · (A′ + B) simplifies to A · B + A′ · B ′ . Question 35 Question Simplify the following Boolean expression: (A + B + C)(A + B + C). Solution Step 1: Apply the distributive property to expand the expression. (A + B + C)(A + B + C) = A(A + B + C) + B(A + B + C) + C(A + B + C) Step 2: Apply the absorption law X + X = 1 to simplify the terms. = A + B(A + B + C) + C(A + B + C) = A + B + C(A + B + C) 23 Step 5: Absorption Law Finally, we can simplify the expression using the absorption law: A + A · (B + C) + B · C = A + B · C Therefore, the simplified form of (A + B) · (A + C) is A + B · C. Question 2 Question Simplify the following Boolean expression: (A + B + C)(A + B + C)(A + B + C). Solution Step 1: Use the distributive law to multiply the first two terms together. (A + B + C)(A + B + C) = (A · A + A · B + A · C + B · A + B · B + B · C + C · A + C · B + C · C) = (A + AB + AC + BA + 0 + BC + AC + 0 + 0) = (A + AC + AB + AB + BC + AC) = (A + A(C + C) + AB + BC) = (A + A + AB + BC) = (A + AB + BC) Step 2: Now, multiply the result from step 1 with the third term and simplify. (A + AB + BC)(A + B + C) = (AA + AB + ABC + AAB + ABB + ABBC + AAC + ABC + ABBC) = (0 + AB + ABC + 0 + 0 + 0 + 0 + 0 + 0) = (AB + ABC) Therefore, the simplified Boolean expression is AB + ABC. Question 3 Question Let F (A, B, C) = (A + B)(A + B + C)(A + B + C) where A, B, and C are Boolean variables. Simplify the Boolean expression F to its simplest form using Boolean Algebra laws. 2 Solution To simplify the Boolean expression F (A, B, C) = (A+B)(A+B+C)(A+B+C), we will apply various Boolean Algebra laws. Step 1: Apply the distributive law: XY + XZ = X(Y + Z). F (A, B, C) = (A + B)(A + B + C)(A + B + C) = (A + B)(A + B + C + A + B + C) = (A + B)(A + A + B + B + C + C) = (A + B)(1 + B + C) = (A + B) Therefore, the simplified form of the Boolean expression F (A, B, C) = (A + B). Question 4 Question Simplify the following Boolean expression using Boolean Algebra: (A + B)(A + C)(B + C). Solution Step 1: Apply the Distributive Law to expand the expression. (A + B)(A + C)(B + C) = A(A + C)(B + C) + B(A + C)(B + C) = AA + AC + AB + BC + BC = A + AC + AB + BC Step 2: Apply the Idempotent Law (AA = A). A + AC + AB + BC = A + A(C + B) + BC = A + (A)(C + B) + BC = A + AC + AB + BC Step 3: Apply the Absorption Law (A + AB = A). A + AC + AB + BC = A + AC + A(C + B) = A + AC + A = A(1 + C) + A =A+A =A Therefore, the simplified expression is A. 3 Question 5 Question Let F (A, B, C) = A′ B ′ C ′ + A′ BC + AB ′ C ′ + ABC. Simplify F (A, B, C) using Boolean algebra laws. Solution To simplify the given Boolean expression F (A, B, C) = A′ B ′ C ′ + A′ BC + AB ′ C ′ + ABC, we will apply various Boolean algebra laws. Step 1: Apply the absorption law X + XY = X. F (A, B, C) = A′ B ′ C ′ + A′ B(C + C ′ ) + AB ′ C ′ + AB(C + C ′ ) = A′ B ′ C ′ + A′ B + AB ′ C ′ + AB = A′ B(1 + 1) + AB(1 + 1) = A′ B + AB Step 2: Apply the consensus theorem X + X ′ Y = X + Y . F (A, B, C) = A′ B + AB = B(A′ + A) = B(1) =B Therefore, the simplified form of F (A, B, C) is B. Question 6 Question Simplify the Boolean expression (A + B + C)(A + B)(A + C) using Boolean algebra laws and theorems. Solution We can simplify the given Boolean expression step by step using Boolean algebra laws and theorems. (A + B + C)(A + B)(A + C) = (A + B + C)(AA + AC + BA + BC) = (A + B + C)(0 + AC + BA + BC) Exp App = AC(A + B + C) + BA(A + B + C) + BC(A + B + C) = ACA + ACB + ACC + BAA + BAB + BAC + BCA + BCB + BCC = 0 + 0 + AC + 0 + 0 + BC + 0 + 0 + 0 = AC + BC 4 Appl Therefore, the simplified Boolean expression is AC + BC. Question 7 Question Simplify the following Boolean expression: (A + B)(A + B) + A. Solution To simplify the given Boolean expression, we will first expand the terms and then apply Boolean algebra rules to simplify it step by step. Step 1: Expand the terms using the distributive law. (A + B)(A + B) = AA + AB + BA + BB Step 2: Apply the idempotent law (AA = A) and null law (AB + BB = A + B). AA + AB + BA + BB = A + A + B Step 3: Simplify the expression. A+A+B =A+B Thus, the simplified form of the Boolean expression (A + B)(A + B) + A is A + B. Question 8 Question Simplify the Boolean expression (A + B + C)(A′ + B + C ′ )(A + B ′ ). Solution To simplify the Boolean expression, we will use the properties of Boolean algebra, such as idempotent law, identity law, complementary law, etc. Step 1: Use the distributive law to expand the expression. (A+B+C)(A′ +B+C ′ )(A+B ′ ) = (AA′ +AB+AC+BA′ +BB+BC ′ +CA+CB+CC ′ )(A+B ′ ) Step 2: Simplify the expression by removing terms AA′ , BB, CC ′ . (AB + AC + BA′ + BC ′ + CA + CB)(A + B ′ ) Step 3: Use the absorption law to further simplify the expression. AB + AC + BA′ + BC ′ + CA + CB 5 Step 4: Use the consensus theorem to simplify the expression. AB + AC + CA Step 5: Use the consensus theorem again to simplify the expression. AB + AC Therefore, the simplified form of (A+B+C)(A′ +B+C ′ )(A+B ′ ) is AB+AC. Question 9 Question Let F (A, B, C) = A′ BC ′ + AB ′ C + A′ B ′ C ′ . Simplify the expression using Boolean algebra properties. Solution To simplify the given expression F (A, B, C) = A′ BC ′ + AB ′ C + A′ B ′ C ′ , we will apply Boolean algebra properties such as the distributive law, complement law, idempotent law, etc. Step 1: Apply the distributive law (A(B + C) = AB + AC) to simplify the expression: F (A, B, C) = A′ BC ′ + AB ′ C + A′ B ′ C ′ = A′ C ′ (B + B ′ ) + A′ B(C + C ′ ) Step 2: Use the complement law (X + X ′ Y = X + Y ) to simplify the expression further: F (A, B, C) = A′ C ′ + A′ B Step 3: Apply the distributive law in reverse (X + XY = X) to simplify the expression: F (A, B, C) = A′ C ′ + A′ B = A′ (C ′ + B) Therefore, the simplified expression for F (A, B, C) is A′ (C ′ + B). Question 10 Question Simplify the following Boolean expression: (A + B + C)(A + B + C)(A + B + C)(A + B + C). 6 Solution Step 1: Expand the given expression using the distributive property. (A + B + C)(A + B + C)(A + B + C)(A + B + C) = A(A+B+C)(A+B+C)(A+B+C)+B(A+B+C)(A+B+C)(A+B+C)+C(A+B+C)(A+B+C)(A+B+C) Step 2: Apply the distributive property again to expand each term. (A+AB+AC)(A+B+C)(A+B+C)+(BA+B+BC)(A+B+C)(A+B+C)+(CA+CB+CC)(A+B+C)(A+B+C Step 3: Continue expanding and simplifying each term. (A + AB + AC)(A + B + C)(A + B + C) = (AA + AB + AC + AB + ABB + ABC + AAC + AC + ACC) = (A + AB + AC + AB + ABC + AC + AC) = A(1 + B + C + B + BC + C + C) = A(1) =A Step 4: Similarly simplify the other two expanded terms to get the final simplified expression. Final expression: A + B + C Therefore, (A + B + C)(A + B + C)(A + B + C)(A + B + C) simplifies to A + B + C. Question 11 Question Simplify the boolean expression (A+B)(A+C)(B+C) using the laws of Boolean Algebra. Solution We will simplify the given expression step by step using the laws of Boolean Algebra. (A + B)(A + C)(B + C) = (A + B)(AB + AC + BC + BC) Expand (A + C)(B + C) using the distributive = ABB + ABC + ABC + ACC + BCC + BCC = AB + ABC + ABC + AC + BC + BC = AB + AC + BC Simplify using XX = X and X + X = X Simplify using XX = X and X + X = X Therefore, (A + B)(A + C)(B + C) simplifies to AB + AC + BC. 7 Distribute (A + B) over (AB + AC Question 12 Question Simplify the following Boolean expression using Boolean algebra rules: (A + B + C)(A + B + C). Solution We will simplify the given Boolean expression step by step using Boolean algebra rules. Step 1: Apply the Distributive Law: (A + B + C)(A + B + C) = A(A + B + C) + B(A + B + C) + C(A + B + C). Step 2: Use the Absorption Law: AA = 0 and A+AB = A+B. Apply these laws to simplify each term: AA + AB + AC + BA + BB + BC + CA + CB + CC. Step 3: Further simplify using the Complement Law: AA = 0, AB = AB, AC = AC, BA = AB, BB = 0, BC = BC, CA = AC, CB = BC, and CC = 0. Step 4: Combine the simplified terms: AB + AC + AB + BC + AC = AB + AC + BC . Question 13 Question Simplify the following Boolean expression: (A + B + C)(A + B + C)(A + B + C)(A + B + C). Solution Step 1: Use the distributive property to expand the expression. (A + B + C)(A + B + C)(A + B + C)(A + B + C) = A(A + B + C)(A + B + C)(A + B + C) + B(A + B + C)(A + B + C)(A + B + C) + C(A + B + C)(A + B + C)(A + B + C) Step 2: Apply the distributive property again. = A(AA + AB + AC + AB + AB + BC + ABC + BB + BC) + B(AA + AB + AC + AB + AB + BC + ABC + BB + BC) + C(AA + AB + AC + AB + AB + BC + ABC + BB + BC) Step 3: Simplify the terms. 8 = A(AC + AB + BC + ABC + BB + BC) + B(AC + AB + BC + ABC + BB + BC) + C(AC + AB + BC + ABC + BB + BC) = A(AC + AB + BB + BC) + B(AC + AB + BB + BC) + C(AC + AB + BB + BC) = A(C + B) + B(C + A) + C(B + A) = AC + AB + BC Therefore, the simplified expression is AC + AB + BC. Question 14 Question Simplify the following Boolean expression using algebraic manipulation techniques: (A + B + C)(A + B + C)(A + B + C)(A + B + C). Solution 1. We start by expanding the given expression using the distributive law: (A + B + C)(A + B + C)(A + B + C)(A + B + C) = (A + B + C)(A(B + C) + B(B + C) + C(A + B)) = (A + B + C)(AB + AC + BB + BC + CA + CB) = (A + B + C)(AB + AC + BC + CA + CB) = ABA + ABC + AAC + ACB + ACC + ABC + BA + BC + CA + CB + CC = 0 + ABC + A + 0 + 0 + ABC + BA + 0 + 0 + C. 9 2. Next, simplify the expression further by combining terms: ABC + A + ABC + BA + C = A(BC + 1 + BC) + BA + C = A(1 + B + C) + BA + C = A + AB + AC + BA + C
= A 1 + B + AC + C = A + AC + C = A(1 + C) + C = A + C. Therefore, the simplified form of the given Boolean expression is A + C. Question 15 Question Simplify the following Boolean expression using algebraic manipulation: F = A′ B ′ C ′ D′ +A′ B ′ C ′ D+A′ BC ′ D+A′ BCD+AB ′ C ′ D′ +AB ′ C ′ D+AB ′ CD+ABCD Solution To simplify the given Boolean expression, we will use the laws of Boolean algebra including absorption, identity, complement, and distributive laws. Step 1: Apply the absorption law. F = A′ B ′ (C ′ D′ +C ′ D+CD)+AB ′ (C ′ D′ +C ′ D+CD)+A(B ′ C ′ D′ +B ′ C ′ D+BCD) Step 2: Apply the idempotent law XX = X. F = A′ B ′ (D′ + D) + AB ′ (D′ + D) + A(C ′ D′ + C ′ D + CD) Step 3: Apply the complement law X + X ′ Y = X + Y . F = A′ B ′ + AB ′ + AC + AD Step 4: Apply the consensus law XY + X ′ Z + Y Z = XY + X ′ Z. F = A(B ′ + B) + AC + AD Step 5: Apply the idempotent law XX = X. F = A + AC + AD 10 Step 6: Apply the absorption law. F = A(1 + C) Step 7: Apply the identity law X1 = X. F =A Therefore, the simplified Boolean expression is F = A. Question 16 Question Simplify the Boolean expression (A + B) · (A′ · B ′ ) + A · B. Solution We will simplify the given Boolean expression using the laws of Boolean algebra. Step 1: Apply the Distributive Law to expand the expression. (A + B) · (A′ · B ′ ) + A · B = A · A′ · B ′ + B · A′ · B ′ + A · B Step 2: Apply the Complement Law to simplify the expression A · A′ . A · A′ · B ′ + B · A′ · B ′ + A · B = 0 · B ′ + B · A′ · B ′ + A · B Step 3: Apply the Annulment Law to simplify the term 0 · B ′ . 0 · B ′ + B · A′ · B ′ + A · B = 0 + B · A′ · B ′ + A · B Step 4: Apply the Identity Law to remove the term 0. B · A′ · B ′ + A · B Step 5: Apply the Absorption Law to the expression B · A′ · B ′ . A·B Therefore, the simplified form of the expression (A + B) · (A′ · B ′ ) + A · B is A · B. Question 17 Question Simplify the following Boolean expression using algebraic manipulation: (A′ B + AB ′ )(A + B) + AB 11 Solution Let’s simplify the given Boolean expression step by step: Step 1: Apply the distributive law: ′ ′ (A B + AB )(A + B) + AB = A′ B(A + B) + AB ′ (A + B) + AB Step 2: Apply the distributive law again: = A′ BA + A′ BB + AB ′ A + AB ′ B + AB Step 3: Apply the idempotent law: = AB + 0 + 0 + 0 + AB Step 4: Apply the identity law: = AB + AB Step 5: Apply the idempotent law again: = AB Therefore, the simplified form of the Boolean expression is AB . Question 18 Question Simplify the following Boolean expression: (A + B)(A + C)(A + D). Solution To simplify the Boolean expression (A + B)(A + C)(A + D), we will use the following Boolean algebra identities: ˆ Complement Law: A + A = 1 ˆ Absorption Law: A + AB = A Step 1: Apply the Complement Law to A + D: (A + B)(A + C)(A + D) = (A + B)(A + C)(1) Step 2: Simplify the expression by using the Absorption Law: (A + B)(A + C)(1) = (A + B)(A + C) Step 3: Apply the Distributive Law to expand the expression: (A + B)(A + C) = AA + AC + BA + BC 12 = A + AC + BA + BC Step 4: Use the Absorption Law to simplify the expression: A + AC + BA + BC = A + AC Step 5: Apply the Distributive Law to factor out A: A + AC = A(1 + C) Step 6: Apply the Complement Law to simplify the expression: A(1 + C) = A Therefore, the simplified form of (A + B)(A + C)(A + D) is A. Question 19 Question Simplify the Boolean expression (A + B ′ )(A′ B + AB ′ ). Solution Step 1: Use the distributive property to expand the expression. (A + B ′ )(A′ B + AB ′ ) = A · A′ B + A · AB ′ + B ′ · A′ B + B ′ · AB ′ = AA′ B + AAB ′ + B ′ A′ B + B ′ AB ′ = AAB ′ + AAB ′ + AB ′ B + AB ′ B =A+A+0+0 =A Step 2: Simplify further to obtain the final simplified expression. (A + B ′ )(A′ B + AB ′ ) = A Question 20 Question Simplify the following Boolean expression: F = A′ B ′ C ′ + AB ′ C + ABC + A′ BC ′ 13 Solution To simplify the Boolean expression F = A′ B ′ C ′ + AB ′ C + ABC + A′ BC ′ , we will use Boolean algebra laws and rules. Step 1: Apply the absorption law (A + AB = A) to eliminate redundant terms. F = A′ B ′ C ′ + AB ′ C + ABC + A′ BC ′ = A′ B ′ C ′ + AB ′ C + A(BC + B ′ C ′ ) ′ ′ ′ ′ ′ = A B C + AB C + A(C(B + B )) = A′ B ′ C ′ + AB ′ C + AC (Grouping terms) (Distributive law) (Complement law: B + B ′ = 1) Step 2: Apply the consensus theorem (AB + AC + BC = AB + AC) to simplify the expression further. F = A′ B ′ C ′ + AB ′ C + AC = A′ B ′ C ′ + AB ′ C + A′ C (Consensus theorem) Step 3: Finally, we apply the consensus theorem once more to obtain the simplified expression. F = A′ B ′ C ′ + AB ′ C + A′ C = A′ C(B ′ + B) + A′ B ′ C ′ ′ ′ ′ =AC +AB C ′ ′ ′ (Complement law: B ′ + B = 1) = C(A + B ) + A B ′ C ′ ′ ′ =C +AB C ′ ′ (Distributive law) (Distributive law) (Complement law: A′ + B ′ = 1) Therefore, the simplified Boolean expression is F = C + A′ B ′ C ′ . Question 21 Question Simplify the Boolean expression (A + B + C) · (A + B) + A · B · C. Solution Step 1: Use the distributive law to expand the expression. (A + B + C) · (A + B) + A · B · C = (A · A + A · B + B · A + B · B + C · A + C · B) + A · B · C = (0 + A · B + A · B + 0 + C · A + C · B) + A · B · C = (A · B + A · B + C · A + C · B) + A · B · C 14 Step 2: Use the consensus theorem to simplify the expression. (A · B + A · B + C · A + C · B) + A · B · C = (A · B + A · B + C · A + C · B) = (A · B + A · B + C · (A + B)) = (A · B + A · B + C) Therefore, the simplified Boolean expression is A · B + A · B + C. Question 22 Question Simplify the following Boolean expression: (A + B + C)(A + B + C)(A + B + C). Solution Step 1: Use the distributive property to expand the expression. (A + B + C)(A + B + C)(A + B + C) = (A(A + B + C) + B(A + B + C) + C(A + B + C))(A + B + C) = (A + AB + AC + BA + B 2 + BC + CA + BC + CC)(A + B + C) = (A + AB + AC + A + B + BC + CA + BC + 0)(A + B + C) = (A + B + AC + BC + CA + BC)(A + B + C) Step 2: Continue simplifying by removing redundant terms. (A + B + AC + BC + CA + BC)(A + B + C) = (A(1 + C) + B(1 + C) + AC + BC)(A + B + C) = (A + B + AC + BC)(A + B + C) = (A(1 + C) + B(1 + C))(A + B + C) = (A + B)(A + B + C) = A + AB + BA + B + BC + BC = A + B + BA + BC + BC = A + B + BC + BC = A + (B + BC) + BC = A + B + BC = A + BC Therefore, the simplified expression is A + BC. 15 Question 23 Question Simplify the Boolean expression F = A′ B ′ C ′ + AB ′ C + ABC + A′ B ′ C Solution To simplify the given Boolean expression, we will use the laws of Boolean algebra. Step 1: Apply the idempotent law: X + X = X F = A′ B ′ C ′ + AB ′ C + ABC + A′ B ′ C Step 2: Group the terms which contain the same variables: F = A′ B ′ C ′ + A′ B ′ C + AB ′ C + ABC Step 3: Apply the idempotent law: X + X = X F = A′ B ′ C ′ + AB ′ C + ABC Step 4: Apply the absorption law: X + XY = X F = A′ B ′ (C ′ + C) + ABC Step 5: Apply the complement law: X + X ′ Y = X + Y F = A′ B ′ + ABC Step 6: Apply the absorption law: X + XY = X F = B(A′ + AC) Thus, the simplified form of the given Boolean expression is F = B(A′ +AC). Question 24 Question Simplify the following Boolean expression: (A + B)(A′ C + BC). Solution To simplify the given Boolean expression (A + B)(A′ C + BC), we will use the distributive property, complement property, and absorption property of Boolean algebra. Step 1: Apply the distributive property to expand the expression: (A + B)(A′ C + BC) = A · A′ C + A · BC + B · A′ C + B · BC = AA′ C + ABC + BA′ C + BBC Step 2: Simplify using the complement property AA′ = 0: AA′ C + ABC + BA′ C + BBC = 0C + ABC + 0B + B0 = 0 + AB + 0 + 0 = AB Step 3: Final answer: The simplified expression is AB. 16 Question 25 Question Simplify the following Boolean expression using algebraic manipulation techniques: F (A, B, C, D) = (A + B + C) · (A+!B + C) + (!A + B + C) · (A + B + C) Solution Step 1: Apply the Distributive Property to expand the expression. F (A, B, C, D) = (A + B + C) · (A+!B + C) + (!A + B + C) · (A + B + C) = A · A + A·!B + A · C + B · A + B·!B + B · C + C · A + C·!B + C · C + (!A) · A + (!A) · B + (!A) · C + B · A + B · B + B · C + C · A + C · B + C · C Step 2: Simplify the expanded expression by removing redundant terms and applying the Idempotent Law (X · X = X). F (A, B, C, D) = A + A·!B + A · C + B + 0 + B · C + C + 0 + 0 + 0 + 0 + 0 + A · B + B + 0 + C + 0 + 0 = A + A·!B + A · C + B + B · C + C + A · B + C =A+A·C +B+B·C +C +A·B Step 3: Apply the Absorption Law (X + X · Y = X) and the Complement Law (X+!X = 1) to simplify the expression further. F (A, B, C, D) = A + A · C + B + B · C + C + A · B =A+B+C Therefore, the simplified Boolean expression is F (A, B, C, D) = A + B + C. Question 26 Question Simplify the following Boolean expression: (A + B + C)(A + B + C)(A + B + C)(A + B + C) Solution To simplify the given Boolean expression, we will use the laws of Boolean algebra, including the distributive law, the absorption law, and the complement law. Step 1: Apply the distributive law: (A + B + C)(A + B + C)(A + B + C)(A + B + C) ⇒ (A + B)(A + B + C)(A + B + C)(A + B + C) 17 Step 2: Apply the distributive law again: (A + B)(A + B + C)(A + B + C)(A + B + C) ⇒ (A + B)(A + B + C)(A + B + C) Step 3: Apply the absorption law: (A + B)(A + B + C)(A + B + C) ⇒ (A + B)(A + B + C) Step 4: Apply the distributive law once more: (A + B)(A + B + C) ⇒ AA + AB + AC + BB + BBC Step 5: Apply the complement law: BB = 0 ⇒ AB + AC Therefore, the simplified Boolean expression is AB + AC. Question 27 Question Simplify the Boolean expression (A + B) · (A + B) · (A + B + C) using Boolean algebra laws. Solution Step 1: Apply the Distributive Law to the first two terms: (A+B)·(A+B) = A·A+A·B +B ·A+B ·B = A+AB +AB +0 = A+AB +AB Step 2: Apply the Associative Law to the result of Step 1 with the third term: (A+AB +AB)·(A+B +C) = A·(A+B +C)+AB ·(A+B +C)+AB ·(A+B +C) = A + AB + AC + AB + ABB + ABC + AB + ABB + ABC = A + AB + AC + AB + 0 + ABC + AB + 0 + ABC = A + AB + AC + AB + ABC + AB + ABC Step 3: Apply the Absorption Law to the result of Step 2: A + AB + AC + AB + ABC + AB + ABC = A(1 + B + C) + AB + ABC = A + AB + ABC = A(1 + B + BC) Therefore, the simplified expression is A(1 + B + BC). Question 28 Question Simplify the following Boolean expression: F = AB ′ + A′ B + (A + B ′ )′ 18 Solution Step 1: Apply De Morgan’s Theorem to simplify the third term. (A + B ′ )′ = A′ B ′′ Step 2: Simplify the second term using the complement law. A′ B = A · B Step 3: Substitute the simplified expressions back into the original expression. F = AB ′ + A · B + A′ B ′′ = AB ′ + AB + A′ B Step 4: Use the absorption law to simplify the expression. F = AB ′ + AB + A′ B = AB ′ + A′ B Therefore, the simplified form of the given Boolean expression is AB ′ + A′ B. Question 29 Question Simplify the Boolean expression (A + B)(A + C)(B + C) + ABC. Solution To simplify the given Boolean expression, we will first expand it using the distributive property, then use the Boolean algebra laws to simplify the resulting expression. Step 1: Expand the given expression (A + B)(A + C)(B + C) + ABC = (A + B)(A + C)B + (A + B)(A + C)C + ABC = AB + BB + AC + BC + AC + CC + ABC = AB + B + AC + BC + AC + C + ABC = AB + B + AC + BC + AC + C + AC + BC = AB + B + AC + BC + C + BC = AB + AC + BC + B + C Step 2: Apply Boolean algebra laws to simplify the expression AB + AC + BC + B + C = B(A + C) + C(A + B) + B + C =B+C +B+C =C +B 19 Therefore, the simplified form of the Boolean expression (A+B)(A+C)(B + C) + ABC is C + B. Question 30 Question Simplify the Boolean expression (A + B) · (A + B) using Boolean algebra rules. Solution 1. Let’s first expand the expression using the distributive law: (A + B) · (A + B) = A · A + A · B + B · A + B · B 2. Use the idempotent law X · X = X to simplify A · A: A·A+A·B+B·A+B·B =A+A·B+B·A+B·B 3. Use the absorption law X + X · Y = X to simplify B · B: A+A·B+B·A+B·B =A+A·B+B·A 4. Use the commutative law XY = Y X to rearrange A · B and B · A: A+A·B+B·A=A+A·B+A·B 5. Use the distribution law X + XY = X to simplify A · B + A · B: A + A · B + A · B = A + A · (B + B) 6. Use the complement law X + X ′ Y = X + Y to simplify B + B: A + A · (B + B) = A + A · 1 = A Therefore, (A + B) · (A + B) simplifies to A . Question 31 Question Simplify the Boolean expression F = A′ B + AB ′ + AC + BC ′ . 20 Solution To simplify the Boolean expression F = A′ B + AB ′ + AC + BC ′ , we will use Boolean algebra rules to simplify term by term. Step 1: Apply the absorption law: X + XY = X. We have: F = A′ B + AB ′ + AC + BC ′ = B(A′ + A) + AC + BC ′ = B + AC + BC ′ Step 2: Apply the consensus theorem: X + X ′ Z + Y Z = X + Y Z. We can rewrite the expression as: F = B + AC + BC ′ = B(1 + C ′ ) + AC = B + AC Step 3: Apply the consensus theorem again: X + X ′ Y + Z = X + Y + Z. Simplifying further, we get: F = B + AC = B(1 + C) + A(C + C ′ ) =B+A Therefore, the simplified form of the Boolean expression F = A′ B + AB ′ + AC + BC ′ is F = A + B. Question 32 Question Simplify the following Boolean expression: F = (A + B ′ )(A′ + B)(A′ + B ′ ) Solution To simplify the given Boolean expression, we will use the laws of Boolean algebra including the distributive law, complement law, and identity law. Step 1: Apply the distributive law to the expression: F = (A + B ′ )(A′ + B)(A′ + B ′ ) F = (A + B ′ )(A′ A′ + A′ B + B ′ A′ + B ′ B ′ ) F = (A + B ′ )(0 + A′ B + 0 + 0) 21 F = (A + B ′ )A′ B Step 2: Use the absorption law to simplify the expression: F = A′ B + A′ B ′ B F = A′ B Therefore, the simplified expression is F = A′ B. Question 33 Question Simplify the Boolean expression (A + B)(A + C)(B + C). Solution To simplify the given Boolean expression, we will use the following Boolean algebra identities: 1. Commutative Property: A + B = B + A and A · B = B · A 2. Associative Property: (A + B) + C = A + (B + C) and (A · B) · C = A · (B · C) 3. Distributive Property: A · (B + C) = A · B + A · C and A + (B · C) = (A + B) · (A + C) 4. Absorption Law: A + (A · B) = A and A · (A + B) = A 5. Identity Law: A + 0 = A and A · 1 = A 6. Zero Law: A · 0 = 0 and A + 1 = 1 Step 1: Apply the Distributive Property: (A + B)(A + C)(B + C) = ((A + B)A + (A + B)C)(B + C) Step 2: Apply the Distributive Property again: = ((A · A + B · A) + (A · C + B · C))(B + C) Step 3: Simplify using the Absorption Law and the Identity Law: = (A + B) + (A + C)(B + C) Step 4: Apply the Distributive Property once more: = (A + B) + AB + AC + BC Step 5: Apply the Commutative Property and the Associative Property: = A + AB + AC + BC + B Step 6: Apply the Absorption Law: = A + B + AC Therefore, the simplified Boolean expression (A + B)(A + C)(B + C) is equivalent to A + B + AC. 22 Question 34 Question Simplify the Boolean expression (A′ + B) · (A + B ′ ) + (A + B ′ ) · (A′ + B). Solution We simplify the given Boolean expression step by step using Boolean algebra laws. Step 1: Apply the Distributive Law: X · (Y + Z) = X · Y + X · Z. (A′ +B)·(A+B ′ )+(A+B ′ )·(A′ +B) = (A′ ·A+A′ ·B ′ +B·A+B·B ′ )+(A·A′ +A·B+B ′ ·A′ +B ′ ·B) Step 2: Apply the Complement Law: X · X ′ = 0. = (0 + A′ · B ′ + B · A + 0) + (0 + A · B + B ′ · A′ + 0) Step 3: Apply the Identity Law: X + 0 = X. = A′ · B ′ + B · A + A · B + B ′ · A′ Step 4: Apply the Commutative Law: X · Y = Y · X and X + Y = Y + X. = A · B + B · A + A′ · B ′ + B ′ · A′ Step 5: Apply the Idempotent Law: X + X = X and X · X = X. = A · B + A′ · B ′ Therefore, (A′ + B) · (A + B ′ ) + (A + B ′ ) · (A′ + B) simplifies to A · B + A′ · B ′ . Question 35 Question Simplify the following Boolean expression: (A + B + C)(A + B + C). Solution Step 1: Apply the distributive property to expand the expression. (A + B + C)(A + B + C) = A(A + B + C) + B(A + B + C) + C(A + B + C) Step 2: Apply the absorption law X + X = 1 to simplify the terms. = A + B(A + B + C) + C(A + B + C) = A + B + C(A + B + C) 23 Step 5: Absorption Law Finally, we can simplify the expression using the absorption law: A + A · (B + C) + B · C = A + B · C Therefore, the simplified form of (A + B) · (A + C) is A + B · C. Question 2 Question Simplify the following Boolean expression: (A + B + C)(A + B + C)(A + B + C). Solution Step 1: Use the distributive law to multiply the first two terms together. (A + B + C)(A + B + C) = (A · A + A · B + A · C + B · A + B · B + B · C + C · A + C · B + C · C) = (A + AB + AC + BA + 0 + BC + AC + 0 + 0) = (A + AC + AB + AB + BC + AC) = (A + A(C + C) + AB + BC) = (A + A + AB + BC) = (A + AB + BC) Step 2: Now, multiply the result from step 1 with the third term and simplify. (A + AB + BC)(A + B + C) = (AA + AB + ABC + AAB + ABB + ABBC + AAC + ABC + ABBC) = (0 + AB + ABC + 0 + 0 + 0 + 0 + 0 + 0) = (AB + ABC) Therefore, the simplified Boolean expression is AB + ABC. Question 3 Question Let F (A, B, C) = (A + B)(A + B + C)(A + B + C) where A, B, and C are Boolean variables. Simplify the Boolean expression F to its simplest form using Boolean Algebra laws. 2 Solution To simplify the Boolean expression F (A, B, C) = (A+B)(A+B+C)(A+B+C), we will apply various Boolean Algebra laws. Step 1: Apply the distributive law: XY + XZ = X(Y + Z). F (A, B, C) = (A + B)(A + B + C)(A + B + C) = (A + B)(A + B + C + A + B + C) = (A + B)(A + A + B + B + C + C) = (A + B)(1 + B + C) = (A + B) Therefore, the simplified form of the Boolean expression F (A, B, C) = (A + B). Question 4 Question Simplify the following Boolean expression using Boolean Algebra: (A + B)(A + C)(B + C). Solution Step 1: Apply the Distributive Law to expand the expression. (A + B)(A + C)(B + C) = A(A + C)(B + C) + B(A + C)(B + C) = AA + AC + AB + BC + BC = A + AC + AB + BC Step 2: Apply the Idempotent Law (AA = A). A + AC + AB + BC = A + A(C + B) + BC = A + (A)(C + B) + BC = A + AC + AB + BC Step 3: Apply the Absorption Law (A + AB = A). A + AC + AB + BC = A + AC + A(C + B) = A + AC + A = A(1 + C) + A =A+A =A Therefore, the simplified expression is A. 3 Question 5 Question Let F (A, B, C) = A′ B ′ C ′ + A′ BC + AB ′ C ′ + ABC. Simplify F (A, B, C) using Boolean algebra laws. Solution To simplify the given Boolean expression F (A, B, C) = A′ B ′ C ′ + A′ BC + AB ′ C ′ + ABC, we will apply various Boolean algebra laws. Step 1: Apply the absorption law X + XY = X. F (A, B, C) = A′ B ′ C ′ + A′ B(C + C ′ ) + AB ′ C ′ + AB(C + C ′ ) = A′ B ′ C ′ + A′ B + AB ′ C ′ + AB = A′ B(1 + 1) + AB(1 + 1) = A′ B + AB Step 2: Apply the consensus theorem X + X ′ Y = X + Y . F (A, B, C) = A′ B + AB = B(A′ + A) = B(1) =B Therefore, the simplified form of F (A, B, C) is B. Question 6 Question Simplify the Boolean expression (A + B + C)(A + B)(A + C) using Boolean algebra laws and theorems. Solution We can simplify the given Boolean expression step by step using Boolean algebra laws and theorems. (A + B + C)(A + B)(A + C) = (A + B + C)(AA + AC + BA + BC) = (A + B + C)(0 + AC + BA + BC) Exp App = AC(A + B + C) + BA(A + B + C) + BC(A + B + C) = ACA + ACB + ACC + BAA + BAB + BAC + BCA + BCB + BCC = 0 + 0 + AC + 0 + 0 + BC + 0 + 0 + 0 = AC + BC 4 Appl Therefore, the simplified Boolean expression is AC + BC. Question 7 Question Simplify the following Boolean expression: (A + B)(A + B) + A. Solution To simplify the given Boolean expression, we will first expand the terms and then apply Boolean algebra rules to simplify it step by step. Step 1: Expand the terms using the distributive law. (A + B)(A + B) = AA + AB + BA + BB Step 2: Apply the idempotent law (AA = A) and null law (AB + BB = A + B). AA + AB + BA + BB = A + A + B Step 3: Simplify the expression. A+A+B =A+B Thus, the simplified form of the Boolean expression (A + B)(A + B) + A is A + B. Question 8 Question Simplify the Boolean expression (A + B + C)(A′ + B + C ′ )(A + B ′ ). Solution To simplify the Boolean expression, we will use the properties of Boolean algebra, such as idempotent law, identity law, complementary law, etc. Step 1: Use the distributive law to expand the expression. (A+B+C)(A′ +B+C ′ )(A+B ′ ) = (AA′ +AB+AC+BA′ +BB+BC ′ +CA+CB+CC ′ )(A+B ′ ) Step 2: Simplify the expression by removing terms AA′ , BB, CC ′ . (AB + AC + BA′ + BC ′ + CA + CB)(A + B ′ ) Step 3: Use the absorption law to further simplify the expression. AB + AC + BA′ + BC ′ + CA + CB 5 Step 4: Use the consensus theorem to simplify the expression. AB + AC + CA Step 5: Use the consensus theorem again to simplify the expression. AB + AC Therefore, the simplified form of (A+B+C)(A′ +B+C ′ )(A+B ′ ) is AB+AC. Question 9 Question Let F (A, B, C) = A′ BC ′ + AB ′ C + A′ B ′ C ′ . Simplify the expression using Boolean algebra properties. Solution To simplify the given expression F (A, B, C) = A′ BC ′ + AB ′ C + A′ B ′ C ′ , we will apply Boolean algebra properties such as the distributive law, complement law, idempotent law, etc. Step 1: Apply the distributive law (A(B + C) = AB + AC) to simplify the expression: F (A, B, C) = A′ BC ′ + AB ′ C + A′ B ′ C ′ = A′ C ′ (B + B ′ ) + A′ B(C + C ′ ) Step 2: Use the complement law (X + X ′ Y = X + Y ) to simplify the expression further: F (A, B, C) = A′ C ′ + A′ B Step 3: Apply the distributive law in reverse (X + XY = X) to simplify the expression: F (A, B, C) = A′ C ′ + A′ B = A′ (C ′ + B) Therefore, the simplified expression for F (A, B, C) is A′ (C ′ + B). Question 10 Question Simplify the following Boolean expression: (A + B + C)(A + B + C)(A + B + C)(A + B + C). 6 Solution Step 1: Expand the given expression using the distributive property. (A + B + C)(A + B + C)(A + B + C)(A + B + C) = A(A+B+C)(A+B+C)(A+B+C)+B(A+B+C)(A+B+C)(A+B+C)+C(A+B+C)(A+B+C)(A+B+C) Step 2: Apply the distributive property again to expand each term. (A+AB+AC)(A+B+C)(A+B+C)+(BA+B+BC)(A+B+C)(A+B+C)+(CA+CB+CC)(A+B+C)(A+B+C Step 3: Continue expanding and simplifying each term. (A + AB + AC)(A + B + C)(A + B + C) = (AA + AB + AC + AB + ABB + ABC + AAC + AC + ACC) = (A + AB + AC + AB + ABC + AC + AC) = A(1 + B + C + B + BC + C + C) = A(1) =A Step 4: Similarly simplify the other two expanded terms to get the final simplified expression. Final expression: A + B + C Therefore, (A + B + C)(A + B + C)(A + B + C)(A + B + C) simplifies to A + B + C. Question 11 Question Simplify the boolean expression (A+B)(A+C)(B+C) using the laws of Boolean Algebra. Solution We will simplify the given expression step by step using the laws of Boolean Algebra. (A + B)(A + C)(B + C) = (A + B)(AB + AC + BC + BC) Expand (A + C)(B + C) using the distributive = ABB + ABC + ABC + ACC + BCC + BCC = AB + ABC + ABC + AC + BC + BC = AB + AC + BC Simplify using XX = X and X + X = X Simplify using XX = X and X + X = X Therefore, (A + B)(A + C)(B + C) simplifies to AB + AC + BC. 7 Distribute (A + B) over (AB + AC Question 12 Question Simplify the following Boolean expression using Boolean algebra rules: (A + B + C)(A + B + C). Solution We will simplify the given Boolean expression step by step using Boolean algebra rules. Step 1: Apply the Distributive Law: (A + B + C)(A + B + C) = A(A + B + C) + B(A + B + C) + C(A + B + C). Step 2: Use the Absorption Law: AA = 0 and A+AB = A+B. Apply these laws to simplify each term: AA + AB + AC + BA + BB + BC + CA + CB + CC. Step 3: Further simplify using the Complement Law: AA = 0, AB = AB, AC = AC, BA = AB, BB = 0, BC = BC, CA = AC, CB = BC, and CC = 0. Step 4: Combine the simplified terms: AB + AC + AB + BC + AC = AB + AC + BC . Question 13 Question Simplify the following Boolean expression: (A + B + C)(A + B + C)(A + B + C)(A + B + C). Solution Step 1: Use the distributive property to expand the expression. (A + B + C)(A + B + C)(A + B + C)(A + B + C) = A(A + B + C)(A + B + C)(A + B + C) + B(A + B + C)(A + B + C)(A + B + C) + C(A + B + C)(A + B + C)(A + B + C) Step 2: Apply the distributive property again. = A(AA + AB + AC + AB + AB + BC + ABC + BB + BC) + B(AA + AB + AC + AB + AB + BC + ABC + BB + BC) + C(AA + AB + AC + AB + AB + BC + ABC + BB + BC) Step 3: Simplify the terms. 8 = A(AC + AB + BC + ABC + BB + BC) + B(AC + AB + BC + ABC + BB + BC) + C(AC + AB + BC + ABC + BB + BC) = A(AC + AB + BB + BC) + B(AC + AB + BB + BC) + C(AC + AB + BB + BC) = A(C + B) + B(C + A) + C(B + A) = AC + AB + BC Therefore, the simplified expression is AC + AB + BC. Question 14 Question Simplify the following Boolean expression using algebraic manipulation techniques: (A + B + C)(A + B + C)(A + B + C)(A + B + C). Solution 1. We start by expanding the given expression using the distributive law: (A + B + C)(A + B + C)(A + B + C)(A + B + C) = (A + B + C)(A(B + C) + B(B + C) + C(A + B)) = (A + B + C)(AB + AC + BB + BC + CA + CB) = (A + B + C)(AB + AC + BC + CA + CB) = ABA + ABC + AAC + ACB + ACC + ABC + BA + BC + CA + CB + CC = 0 + ABC + A + 0 + 0 + ABC + BA + 0 + 0 + C. 9 2. Next, simplify the expression further by combining terms: ABC + A + ABC + BA + C = A(BC + 1 + BC) + BA + C = A(1 + B + C) + BA + C = A + AB + AC + BA + C
= A 1 + B + AC + C = A + AC + C = A(1 + C) + C = A + C. Therefore, the simplified form of the given Boolean expression is A + C. Question 15 Question Simplify the following Boolean expression using algebraic manipulation: F = A′ B ′ C ′ D′ +A′ B ′ C ′ D+A′ BC ′ D+A′ BCD+AB ′ C ′ D′ +AB ′ C ′ D+AB ′ CD+ABCD Solution To simplify the given Boolean expression, we will use the laws of Boolean algebra including absorption, identity, complement, and distributive laws. Step 1: Apply the absorption law. F = A′ B ′ (C ′ D′ +C ′ D+CD)+AB ′ (C ′ D′ +C ′ D+CD)+A(B ′ C ′ D′ +B ′ C ′ D+BCD) Step 2: Apply the idempotent law XX = X. F = A′ B ′ (D′ + D) + AB ′ (D′ + D) + A(C ′ D′ + C ′ D + CD) Step 3: Apply the complement law X + X ′ Y = X + Y . F = A′ B ′ + AB ′ + AC + AD Step 4: Apply the consensus law XY + X ′ Z + Y Z = XY + X ′ Z. F = A(B ′ + B) + AC + AD Step 5: Apply the idempotent law XX = X. F = A + AC + AD 10 Step 6: Apply the absorption law. F = A(1 + C) Step 7: Apply the identity law X1 = X. F =A Therefore, the simplified Boolean expression is F = A. Question 16 Question Simplify the Boolean expression (A + B) · (A′ · B ′ ) + A · B. Solution We will simplify the given Boolean expression using the laws of Boolean algebra. Step 1: Apply the Distributive Law to expand the expression. (A + B) · (A′ · B ′ ) + A · B = A · A′ · B ′ + B · A′ · B ′ + A · B Step 2: Apply the Complement Law to simplify the expression A · A′ . A · A′ · B ′ + B · A′ · B ′ + A · B = 0 · B ′ + B · A′ · B ′ + A · B Step 3: Apply the Annulment Law to simplify the term 0 · B ′ . 0 · B ′ + B · A′ · B ′ + A · B = 0 + B · A′ · B ′ + A · B Step 4: Apply the Identity Law to remove the term 0. B · A′ · B ′ + A · B Step 5: Apply the Absorption Law to the expression B · A′ · B ′ . A·B Therefore, the simplified form of the expression (A + B) · (A′ · B ′ ) + A · B is A · B. Question 17 Question Simplify the following Boolean expression using algebraic manipulation: (A′ B + AB ′ )(A + B) + AB 11 Solution Let’s simplify the given Boolean expression step by step: Step 1: Apply the distributive law: ′ ′ (A B + AB )(A + B) + AB = A′ B(A + B) + AB ′ (A + B) + AB Step 2: Apply the distributive law again: = A′ BA + A′ BB + AB ′ A + AB ′ B + AB Step 3: Apply the idempotent law: = AB + 0 + 0 + 0 + AB Step 4: Apply the identity law: = AB + AB Step 5: Apply the idempotent law again: = AB Therefore, the simplified form of the Boolean expression is AB . Question 18 Question Simplify the following Boolean expression: (A + B)(A + C)(A + D). Solution To simplify the Boolean expression (A + B)(A + C)(A + D), we will use the following Boolean algebra identities: ˆ Complement Law: A + A = 1 ˆ Absorption Law: A + AB = A Step 1: Apply the Complement Law to A + D: (A + B)(A + C)(A + D) = (A + B)(A + C)(1) Step 2: Simplify the expression by using the Absorption Law: (A + B)(A + C)(1) = (A + B)(A + C) Step 3: Apply the Distributive Law to expand the expression: (A + B)(A + C) = AA + AC + BA + BC 12 = A + AC + BA + BC Step 4: Use the Absorption Law to simplify the expression: A + AC + BA + BC = A + AC Step 5: Apply the Distributive Law to factor out A: A + AC = A(1 + C) Step 6: Apply the Complement Law to simplify the expression: A(1 + C) = A Therefore, the simplified form of (A + B)(A + C)(A + D) is A. Question 19 Question Simplify the Boolean expression (A + B ′ )(A′ B + AB ′ ). Solution Step 1: Use the distributive property to expand the expression. (A + B ′ )(A′ B + AB ′ ) = A · A′ B + A · AB ′ + B ′ · A′ B + B ′ · AB ′ = AA′ B + AAB ′ + B ′ A′ B + B ′ AB ′ = AAB ′ + AAB ′ + AB ′ B + AB ′ B =A+A+0+0 =A Step 2: Simplify further to obtain the final simplified expression. (A + B ′ )(A′ B + AB ′ ) = A Question 20 Question Simplify the following Boolean expression: F = A′ B ′ C ′ + AB ′ C + ABC + A′ BC ′ 13 Solution To simplify the Boolean expression F = A′ B ′ C ′ + AB ′ C + ABC + A′ BC ′ , we will use Boolean algebra laws and rules. Step 1: Apply the absorption law (A + AB = A) to eliminate redundant terms. F = A′ B ′ C ′ + AB ′ C + ABC + A′ BC ′ = A′ B ′ C ′ + AB ′ C + A(BC + B ′ C ′ ) ′ ′ ′ ′ ′ = A B C + AB C + A(C(B + B )) = A′ B ′ C ′ + AB ′ C + AC (Grouping terms) (Distributive law) (Complement law: B + B ′ = 1) Step 2: Apply the consensus theorem (AB + AC + BC = AB + AC) to simplify the expression further. F = A′ B ′ C ′ + AB ′ C + AC = A′ B ′ C ′ + AB ′ C + A′ C (Consensus theorem) Step 3: Finally, we apply the consensus theorem once more to obtain the simplified expression. F = A′ B ′ C ′ + AB ′ C + A′ C = A′ C(B ′ + B) + A′ B ′ C ′ ′ ′ ′ =AC +AB C ′ ′ ′ (Complement law: B ′ + B = 1) = C(A + B ) + A B ′ C ′ ′ ′ =C +AB C ′ ′ (Distributive law) (Distributive law) (Complement law: A′ + B ′ = 1) Therefore, the simplified Boolean expression is F = C + A′ B ′ C ′ . Question 21 Question Simplify the Boolean expression (A + B + C) · (A + B) + A · B · C. Solution Step 1: Use the distributive law to expand the expression. (A + B + C) · (A + B) + A · B · C = (A · A + A · B + B · A + B · B + C · A + C · B) + A · B · C = (0 + A · B + A · B + 0 + C · A + C · B) + A · B · C = (A · B + A · B + C · A + C · B) + A · B · C 14 Step 2: Use the consensus theorem to simplify the expression. (A · B + A · B + C · A + C · B) + A · B · C = (A · B + A · B + C · A + C · B) = (A · B + A · B + C · (A + B)) = (A · B + A · B + C) Therefore, the simplified Boolean expression is A · B + A · B + C. Question 22 Question Simplify the following Boolean expression: (A + B + C)(A + B + C)(A + B + C). Solution Step 1: Use the distributive property to expand the expression. (A + B + C)(A + B + C)(A + B + C) = (A(A + B + C) + B(A + B + C) + C(A + B + C))(A + B + C) = (A + AB + AC + BA + B 2 + BC + CA + BC + CC)(A + B + C) = (A + AB + AC + A + B + BC + CA + BC + 0)(A + B + C) = (A + B + AC + BC + CA + BC)(A + B + C) Step 2: Continue simplifying by removing redundant terms. (A + B + AC + BC + CA + BC)(A + B + C) = (A(1 + C) + B(1 + C) + AC + BC)(A + B + C) = (A + B + AC + BC)(A + B + C) = (A(1 + C) + B(1 + C))(A + B + C) = (A + B)(A + B + C) = A + AB + BA + B + BC + BC = A + B + BA + BC + BC = A + B + BC + BC = A + (B + BC) + BC = A + B + BC = A + BC Therefore, the simplified expression is A + BC. 15 Question 23 Question Simplify the Boolean expression F = A′ B ′ C ′ + AB ′ C + ABC + A′ B ′ C Solution To simplify the given Boolean expression, we will use the laws of Boolean algebra. Step 1: Apply the idempotent law: X + X = X F = A′ B ′ C ′ + AB ′ C + ABC + A′ B ′ C Step 2: Group the terms which contain the same variables: F = A′ B ′ C ′ + A′ B ′ C + AB ′ C + ABC Step 3: Apply the idempotent law: X + X = X F = A′ B ′ C ′ + AB ′ C + ABC Step 4: Apply the absorption law: X + XY = X F = A′ B ′ (C ′ + C) + ABC Step 5: Apply the complement law: X + X ′ Y = X + Y F = A′ B ′ + ABC Step 6: Apply the absorption law: X + XY = X F = B(A′ + AC) Thus, the simplified form of the given Boolean expression is F = B(A′ +AC). Question 24 Question Simplify the following Boolean expression: (A + B)(A′ C + BC). Solution To simplify the given Boolean expression (A + B)(A′ C + BC), we will use the distributive property, complement property, and absorption property of Boolean algebra. Step 1: Apply the distributive property to expand the expression: (A + B)(A′ C + BC) = A · A′ C + A · BC + B · A′ C + B · BC = AA′ C + ABC + BA′ C + BBC Step 2: Simplify using the complement property AA′ = 0: AA′ C + ABC + BA′ C + BBC = 0C + ABC + 0B + B0 = 0 + AB + 0 + 0 = AB Step 3: Final answer: The simplified expression is AB. 16 Question 25 Question Simplify the following Boolean expression using algebraic manipulation techniques: F (A, B, C, D) = (A + B + C) · (A+!B + C) + (!A + B + C) · (A + B + C) Solution Step 1: Apply the Distributive Property to expand the expression. F (A, B, C, D) = (A + B + C) · (A+!B + C) + (!A + B + C) · (A + B + C) = A · A + A·!B + A · C + B · A + B·!B + B · C + C · A + C·!B + C · C + (!A) · A + (!A) · B + (!A) · C + B · A + B · B + B · C + C · A + C · B + C · C Step 2: Simplify the expanded expression by removing redundant terms and applying the Idempotent Law (X · X = X). F (A, B, C, D) = A + A·!B + A · C + B + 0 + B · C + C + 0 + 0 + 0 + 0 + 0 + A · B + B + 0 + C + 0 + 0 = A + A·!B + A · C + B + B · C + C + A · B + C =A+A·C +B+B·C +C +A·B Step 3: Apply the Absorption Law (X + X · Y = X) and the Complement Law (X+!X = 1) to simplify the expression further. F (A, B, C, D) = A + A · C + B + B · C + C + A · B =A+B+C Therefore, the simplified Boolean expression is F (A, B, C, D) = A + B + C. Question 26 Question Simplify the following Boolean expression: (A + B + C)(A + B + C)(A + B + C)(A + B + C) Solution To simplify the given Boolean expression, we will use the laws of Boolean algebra, including the distributive law, the absorption law, and the complement law. Step 1: Apply the distributive law: (A + B + C)(A + B + C)(A + B + C)(A + B + C) ⇒ (A + B)(A + B + C)(A + B + C)(A + B + C) 17 Step 2: Apply the distributive law again: (A + B)(A + B + C)(A + B + C)(A + B + C) ⇒ (A + B)(A + B + C)(A + B + C) Step 3: Apply the absorption law: (A + B)(A + B + C)(A + B + C) ⇒ (A + B)(A + B + C) Step 4: Apply the distributive law once more: (A + B)(A + B + C) ⇒ AA + AB + AC + BB + BBC Step 5: Apply the complement law: BB = 0 ⇒ AB + AC Therefore, the simplified Boolean expression is AB + AC. Question 27 Question Simplify the Boolean expression (A + B) · (A + B) · (A + B + C) using Boolean algebra laws. Solution Step 1: Apply the Distributive Law to the first two terms: (A+B)·(A+B) = A·A+A·B +B ·A+B ·B = A+AB +AB +0 = A+AB +AB Step 2: Apply the Associative Law to the result of Step 1 with the third term: (A+AB +AB)·(A+B +C) = A·(A+B +C)+AB ·(A+B +C)+AB ·(A+B +C) = A + AB + AC + AB + ABB + ABC + AB + ABB + ABC = A + AB + AC + AB + 0 + ABC + AB + 0 + ABC = A + AB + AC + AB + ABC + AB + ABC Step 3: Apply the Absorption Law to the result of Step 2: A + AB + AC + AB + ABC + AB + ABC = A(1 + B + C) + AB + ABC = A + AB + ABC = A(1 + B + BC) Therefore, the simplified expression is A(1 + B + BC). Question 28 Question Simplify the following Boolean expression: F = AB ′ + A′ B + (A + B ′ )′ 18 Solution Step 1: Apply De Morgan’s Theorem to simplify the third term. (A + B ′ )′ = A′ B ′′ Step 2: Simplify the second term using the complement law. A′ B = A · B Step 3: Substitute the simplified expressions back into the original expression. F = AB ′ + A · B + A′ B ′′ = AB ′ + AB + A′ B Step 4: Use the absorption law to simplify the expression. F = AB ′ + AB + A′ B = AB ′ + A′ B Therefore, the simplified form of the given Boolean expression is AB ′ + A′ B. Question 29 Question Simplify the Boolean expression (A + B)(A + C)(B + C) + ABC. Solution To simplify the given Boolean expression, we will first expand it using the distributive property, then use the Boolean algebra laws to simplify the resulting expression. Step 1: Expand the given expression (A + B)(A + C)(B + C) + ABC = (A + B)(A + C)B + (A + B)(A + C)C + ABC = AB + BB + AC + BC + AC + CC + ABC = AB + B + AC + BC + AC + C + ABC = AB + B + AC + BC + AC + C + AC + BC = AB + B + AC + BC + C + BC = AB + AC + BC + B + C Step 2: Apply Boolean algebra laws to simplify the expression AB + AC + BC + B + C = B(A + C) + C(A + B) + B + C =B+C +B+C =C +B 19 Therefore, the simplified form of the Boolean expression (A+B)(A+C)(B + C) + ABC is C + B. Question 30 Question Simplify the Boolean expression (A + B) · (A + B) using Boolean algebra rules. Solution 1. Let’s first expand the expression using the distributive law: (A + B) · (A + B) = A · A + A · B + B · A + B · B 2. Use the idempotent law X · X = X to simplify A · A: A·A+A·B+B·A+B·B =A+A·B+B·A+B·B 3. Use the absorption law X + X · Y = X to simplify B · B: A+A·B+B·A+B·B =A+A·B+B·A 4. Use the commutative law XY = Y X to rearrange A · B and B · A: A+A·B+B·A=A+A·B+A·B 5. Use the distribution law X + XY = X to simplify A · B + A · B: A + A · B + A · B = A + A · (B + B) 6. Use the complement law X + X ′ Y = X + Y to simplify B + B: A + A · (B + B) = A + A · 1 = A Therefore, (A + B) · (A + B) simplifies to A . Question 31 Question Simplify the Boolean expression F = A′ B + AB ′ + AC + BC ′ . 20 Solution To simplify the Boolean expression F = A′ B + AB ′ + AC + BC ′ , we will use Boolean algebra rules to simplify term by term. Step 1: Apply the absorption law: X + XY = X. We have: F = A′ B + AB ′ + AC + BC ′ = B(A′ + A) + AC + BC ′ = B + AC + BC ′ Step 2: Apply the consensus theorem: X + X ′ Z + Y Z = X + Y Z. We can rewrite the expression as: F = B + AC + BC ′ = B(1 + C ′ ) + AC = B + AC Step 3: Apply the consensus theorem again: X + X ′ Y + Z = X + Y + Z. Simplifying further, we get: F = B + AC = B(1 + C) + A(C + C ′ ) =B+A Therefore, the simplified form of the Boolean expression F = A′ B + AB ′ + AC + BC ′ is F = A + B. Question 32 Question Simplify the following Boolean expression: F = (A + B ′ )(A′ + B)(A′ + B ′ ) Solution To simplify the given Boolean expression, we will use the laws of Boolean algebra including the distributive law, complement law, and identity law. Step 1: Apply the distributive law to the expression: F = (A + B ′ )(A′ + B)(A′ + B ′ ) F = (A + B ′ )(A′ A′ + A′ B + B ′ A′ + B ′ B ′ ) F = (A + B ′ )(0 + A′ B + 0 + 0) 21 F = (A + B ′ )A′ B Step 2: Use the absorption law to simplify the expression: F = A′ B + A′ B ′ B F = A′ B Therefore, the simplified expression is F = A′ B. Question 33 Question Simplify the Boolean expression (A + B)(A + C)(B + C). Solution To simplify the given Boolean expression, we will use the following Boolean algebra identities: 1. Commutative Property: A + B = B + A and A · B = B · A 2. Associative Property: (A + B) + C = A + (B + C) and (A · B) · C = A · (B · C) 3. Distributive Property: A · (B + C) = A · B + A · C and A + (B · C) = (A + B) · (A + C) 4. Absorption Law: A + (A · B) = A and A · (A + B) = A 5. Identity Law: A + 0 = A and A · 1 = A 6. Zero Law: A · 0 = 0 and A + 1 = 1 Step 1: Apply the Distributive Property: (A + B)(A + C)(B + C) = ((A + B)A + (A + B)C)(B + C) Step 2: Apply the Distributive Property again: = ((A · A + B · A) + (A · C + B · C))(B + C) Step 3: Simplify using the Absorption Law and the Identity Law: = (A + B) + (A + C)(B + C) Step 4: Apply the Distributive Property once more: = (A + B) + AB + AC + BC Step 5: Apply the Commutative Property and the Associative Property: = A + AB + AC + BC + B Step 6: Apply the Absorption Law: = A + B + AC Therefore, the simplified Boolean expression (A + B)(A + C)(B + C) is equivalent to A + B + AC. 22 Question 34 Question Simplify the Boolean expression (A′ + B) · (A + B ′ ) + (A + B ′ ) · (A′ + B). Solution We simplify the given Boolean expression step by step using Boolean algebra laws. Step 1: Apply the Distributive Law: X · (Y + Z) = X · Y + X · Z. (A′ +B)·(A+B ′ )+(A+B ′ )·(A′ +B) = (A′ ·A+A′ ·B ′ +B·A+B·B ′ )+(A·A′ +A·B+B ′ ·A′ +B ′ ·B) Step 2: Apply the Complement Law: X · X ′ = 0. = (0 + A′ · B ′ + B · A + 0) + (0 + A · B + B ′ · A′ + 0) Step 3: Apply the Identity Law: X + 0 = X. = A′ · B ′ + B · A + A · B + B ′ · A′ Step 4: Apply the Commutative Law: X · Y = Y · X and X + Y = Y + X. = A · B + B · A + A′ · B ′ + B ′ · A′ Step 5: Apply the Idempotent Law: X + X = X and X · X = X. = A · B + A′ · B ′ Therefore, (A′ + B) · (A + B ′ ) + (A + B ′ ) · (A′ + B) simplifies to A · B + A′ · B ′ . Question 35 Question Simplify the following Boolean expression: (A + B + C)(A + B + C). Solution Step 1: Apply the distributive property to expand the expression. (A + B + C)(A + B + C) = A(A + B + C) + B(A + B + C) + C(A + B + C) Step 2: Apply the absorption law X + X = 1 to simplify the terms. = A + B(A + B + C) + C(A + B + C) = A + B + C(A + B + C) 23 Step 3: Distribute C over (A + B + C). = A + B + CA + CB + CC = A + B + CA + CB Step 4: Use the idempotent law X + X = X to simplify CC. A + B + CA + CB Step 5: Apply the complement law XX = 0 to simplify CA and CB. A+B Therefore, the simplified Boolean expression is A + B. 24