Boolean Algebra Question Bank Set 5

MATH 250 - INTRODUCTION TO DISCRETE MATHEMATICS - Boolean Algebra Question Bank - Set 5 Liberty University Question 1 Question Let F = AB + A′ C + BC. Show that F = AB + A′ C using Boolean algebra laws and properties. Solution Step 1: Apply the absorption law A + AB = A. Step 2: Simplify the expression AB + A′ C + BC using the absorption law. F = AB + A′ C + BC = AB + (A′ C + BC) = AB + A′ C Therefore, F = AB + A′ C. Question 2 Question Let f (x, y, z) be a Boolean function defined as f (x, y, z) = xyz+xyz+xyz+xyz. Simplify the function f using Boolean Algebra laws. Solution To simplify the given Boolean function f (x, y, z), we will use various Boolean Algebra laws such as identity laws, complement laws, etc. Step 1: Apply the distributive law: xy + xy = x(y + y) = x Step 2: Simplify each term in the given function: Starting with xyz, we can see that this term will be the result because it does not have any terms to combine with. Step 3: Simplify xyz: xyz = xyz(y + y) = xyzy + xyzy = 0 + xyzy = xyz Step 4: Simplify xyz: xyz = xyz + xyz + xyzy = xy(z + z) + 0 = xy Step 5: Simplify xyz: xyz = xyz + xyz(y + y) = xyz + 0 = xyz Step 6: Combine all simplified terms together: f (x, y, z) = xyz + xyz + xyz + xyz = xyz + xyz + xy + xyz = xy + xyz Therefore, the simplified form of the Boolean function f (x, y, z) is xy + xyz. Question 3 Question Simplify the following Boolean expression: (A + B · C) · (A + B + C) Solution To simplify the given Boolean expression, we will use the basic rules of Boolean algebra: X + X = X, X · X = X, and De Morgan’s laws: A · B = A + B, A + B = A · B. Step 1: Apply distributive law. (A + B · C) · (A + B + C) =A·A+A·B+A·C +B·C ·A+B·C ·B+B·C ·C Step 2: Simplify using X · X = 0 and X + X = 1. (A · A) + (A · B) + (A · C) + (B · C · A) + (B · C · B) + (B · C · C) =0+A·B+A·C +0+0+0 =A·B+A·C Step 3: Final simplified form of the given Boolean expression is A·B +A·C. 2 Question 4 Question Let A, B, and C be Boolean variables. Simplify the following Boolean expression: (A + B · C) · (A · B + C) Solution To simplify the given Boolean expression, we will use basic Boolean algebra rules and laws. Step 1: Distribute the terms (A + B · C) · (A · B + C) = A · A · B + A · C + B · C · A · B + B · C · C Step 2: Apply the complement law A·A=0 Step 3: Simplify the expression 0·B+A·C +0+B·C ·0=A·C +B·0=A·C Step 4: Final answer Therefore, the simplified form of (A+B·C)·(A·B+C) is A · C. Question 5 Question Simplify the following Boolean expression using Boolean algebra laws: F = A′ B ′ C + AB ′ C ′ + ABC + AB ′ C Solution To simplify the given Boolean expression F = A′ B ′ C + AB ′ C ′ + ABC + AB ′ C, we will use the laws of Boolean algebra. Step 1: Apply absorption law AB ′ C ′ + AB ′ C = AB ′ C ′ Now we have: F = A′ B ′ C + AB ′ C ′ + ABC + AB ′ C ′ Step 2: Apply absorption law again ABC + AB ′ C ′ = AB(C + C ′ ) = AB 3 Now we have: F = A′ B ′ C + AB ′ C ′ + AB Step 3: Apply consensus theorem (or theorem of consensus) law A′ B ′ C + AB = A′ C + AB Now we have: F = A′ C + AB ′ C ′ Step 4: No further simplification is possible, hence the simplified expression is F = A′ C + AB ′ C ′ Question 6 Question Simplify the following Boolean expression: (A + B + C)(A + B + C)(A + B + C). Solution Let’s simplify the given Boolean expression step by step. Step 1: Use the distributive law to expand the expression. (A+B+C)(A+B+C)(A+B+C) = A(A+B+C)(A+B+C)+B(A+B+C)(A+B+C)+C(A+B+C)(A+B+C) Step 2: Further simplify each term by distributing. = AA + AB + AC + BA + AB + BB + BC + CA + BC + CC Since XX = X for any Boolean variable X, we simplify AB, AB, BC, and CC terms: = A + A + AC + AB + B + 0 + BC + CA + 0 + C Step 3: Combine like terms. = A + AC + AB + B + BC + CA + C Step 4: Use the consensus theorem, XY + XZ + Y Z = XY + XZ, to simplify AC + AB + BC. = A(C + B) + BC Step 5: Apply the distributive law once again to obtain the final simplified expression. = A + BC Therefore, (A + B + C)(A + B + C)(A + B + C) simplifies to A + BC. 4 Question 7 Question Simplify the following Boolean expression using algebraic manipulation: F = A′ B ′ C ′ + A′ BC ′ + AB ′ C + ABC Solution Step 1: Apply the distributive law to factor out C ′ : F = A′ B ′ C ′ (1 + A) + AB ′ C(1 + A) Step 2: Apply the distributive law again to factor out A: F = A′ C ′ (A′ + B ′ ) + AB ′ C(A + 1) Step 3: Use the complement laws (XX ′ = 0 and X + X ′ = 1) to simplify the terms: F = A′ C ′ + AB ′ C Step 4: Apply the distributive law to factor out C: F = C(A′ + AB ′ ) Step 5: Apply the absorption law (X + XY = X) to simplify the expression: F = C(A′ + B ′ ) So, the simplified Boolean expression is F = C(A′ + B ′ ). Question 8 Question Simplify the Boolean expression: (A+B +C)(A+D)(B +C +D)(A+B +C +D). Solution Step 1: Use the distributive law to expand the given expression. (A + B + C)(A + D)(B + C + D)(A + B + C + D) = (A2 + AD + AB + BD + AC + CD + BC + BD)(A + B + C + D) Step 2: Simplify the expression using the complement law (AA′ = 0 and A + A′ = 1). (A2 + AD + AB + BD + AC + CD + BC + BD)(A + B + C + D) = (AD + AB + BD + AC + CD + BC)(A + B + C + D) 5 Step 3: Expand the expression further. (AD + AB + BD + AC + CD + BC)(A + B + C + D) = ADA + ADB + ADC + ADB + ACD + ACC + BCD + BCB Step 4: Simplify using the idempotent law (AA = A). ADA + ADB + ADC + ADB + ACD + ACC + BCD + BCB = 0 + ADB + ADC + ADB + ACD + 0 + BCD + 0 Step 5: Simplify further by combining terms. 0 + ADB + ADC + ADB + ACD + 0 + BCD + 0 = ADB + ADB + ACD + BCD = AD(B + B + C) + CD(A + B) = AD + CD Therefore, the simplified form of the Boolean expression is AD + CD. Question 9 Question Simplify the following Boolean algebra expression: (A+B +C)(A′ +B +C)(A+ B ′ + C)(A + B + C ′ ). Solution To simplify the given Boolean algebra expression, we will use the properties of Boolean algebra including the distributive law, complement law, identity law, and simplification rules. Step 1: Apply the distributive law: (A + B + C)(A′ + B + C)(A + B ′ + C)(A + B + C ′ ) (A+B+C)(A′ +B+C)(A+B ′ +C)(A+B+C ′ ) = (A+B)(A+B ′ +C)(A+B+C ′ ) Step 2: Apply the distributive law again: (A + B)(A + B ′ + C)(A + B + C ′ ) (A + B)(A + B ′ + C)(A + B + C ′ ) = AA + AB ′ + AC + BBA + BBB ′ + BBC ′ (A + B)(A + B ′ + C)(A + B + C ′ ) = A + AB ′ + AC + 0 + B ′ + 0 Step 3: Simplify the expression: A + AB ′ + AC + B ′ A + AB ′ + AC + B ′ = A(1 + B ′ ) + B ′ (1 + A) A + AB ′ + AC + B ′ = A + B ′ Therefore, the simplified Boolean algebra expression is A + B ′ . 6 Question 10 Question Simplify the following Boolean expression using laws of Boolean Algebra: (A + B) · (A′ + B) · (A + B ′ ). Solution Step 1: Apply the distributive law X · (Y + Z) = X · Y + X · Z. = (A + B) · (A + B ′ ) + (A + B) · (A′ + B) Step 2: Apply the distributive law again. = AA + AB ′ + BA + BB ′ + AA′ + AB Step 3: Simplify by applying the idempotent law XX = X and the identity law X + X ′ = 1. = A + AB ′ + AB + B + A′ + AB Step 4: Apply the idempotent law and simplify further. = A + B + A′ Step 5: Apply the idempotent law one more time to get the final simplified expression. =A+B+A Step 6: Apply the idempotent law for the last time to obtain the simplest form of the expression. =A+B Therefore, (A + B) · (A′ + B) · (A + B ′ ) simplifies to A + B. Question 11 Question Simplify the Boolean expression F = A′ B + AC + BC ′ . Solution To simplify the Boolean expression F = A′ B + AC + BC ′ , we will use Boolean algebra laws and rules to simplify the expression step by step. Step 1: Apply the distributive law to factor out A. F = A′ B + AC + BC ′ = A(B ′ + C) + BC ′ Step 2: Apply the distributive law again to factor out B. F = A(B ′ + C) + BC ′ = AB ′ + AC + BC ′ Step 3: Apply the consensus theorem, AB + A′ C + BC = AB + A′ C, where AB is the consensus term. F = AB ′ + AC + BC ′ = AB ′ + A′ C 7 Therefore, the simplified form of the Boolean expression F = A′ B+AC+BC ′ is F = AB ′ + A′ C. Question 12 Question Simplify the Boolean expression (A + B) · (A + B · C) using Boolean algebra laws. Solution Step 1: Apply the Distributive Law: X · (Y + Z) = X · Y + X · Z (A + B) · (A + B · C) = A · A + A · B · C + B · A + B · B · C = A + AB · C + A · B + 0 = A + AB · C + A · B Step 2: Apply the Idempotent Law: X + X = X A + AB · C + A · B = A + AB · C Therefore, the simplified form of the Boolean expression (A + B) · (A + B · C) is A + AB · C. Question 13 Question Simplify the following Boolean expression: (A + B)(A + B + C) using Boolean algebra laws. Solution To simplify the expression (A + B)(A + B + C), we will use the distributive law and De Morgan’s law. Step 1: Apply distributive law: (A + B)(A + B + C) = A(A + B + C) + B(A + B + C) Step 2: Apply distributive law again: = AA + AB + AC + BA + BB + BC Step 3: Apply the complement law (AA = 0 and BB = 0): = 0 + AB + AC + BA + 0 + BC 8 Step 4: Rearrange the terms: = AC + BC + AB + BA Step 5: Apply the absorption law (AC + BC = C): = C + AB + BA Step 6: Apply the commutative law: = C + BA + AB Therefore, the simplified form of the given Boolean expression is C + BA + AB. Question 14 Question Simplify the Boolean expression (A + B + C)(A′ + C ′ )(A′ + B). Solution We will simplify the given Boolean expression step by step using the laws of Boolean Algebra. Step 1: Apply the Distributive Law: X(Y + Z) = XY + XZ. (A + B + C)(A′ + C ′ )(A′ + B) = (A + B + C)(A′ A′ + A′ C ′ + AB + BC ′ + AC ′ + BC) = (A + B + C)(0 + A′ C ′ + AB + BC ′ + AC ′ + BC) = A′ C ′ (A + B + C) + AB(A + B + C) + BC ′ (A + B + C) = A′ C ′ A + A′ C ′ B + A′ C ′ C + ABA + ABB + ABC + BCA + BCB + BCC = 0 + 0 + 0 + AB + 0 + 0 + 0 + 0 + 0 = AB Step 2: Apply the Idempotent Law: X + X = X. AB = AB Therefore, the simplified Boolean expression is AB. Question 15 Question Simplify the Boolean expression F = A′ B ′ C + A′ BC + AB ′ C ′ + ABC ′ using Boolean algebra laws and theorems. 9 Solution To simplify the given Boolean expression F = A′ B ′ C + A′ BC + AB ′ C ′ + ABC ′ , we will use the laws and theorems of Boolean algebra. Step 1: Apply the absorption law: XY + XZ = X(Y + Z). F = A′ B ′ C + A′ BC + AB ′ C ′ + ABC ′ = A′ B ′ C + A′ BC + A′ B(C ′ + C ′ ) + AB ′ C ′ ′ ′ ′ ′ ′ = A B C + A BC + A B + AB C (Applying absorption law) ′ = A′ B ′ C + A′ BC + A′ B + A′ B ′ C ′ (Reordering terms) Step 2: Apply the consensus theorem: XY + X ′ Z + Y Z = XY + X ′ Z. F = A′ B ′ C + A′ BC + A′ B + A′ B ′ C ′ = A′ B ′ C + A′ BC + A′ B + A′ B ′ C (Applying consensus theorem) Therefore, the simplified Boolean expression is F = A′ B ′ C + A′ BC + A′ B + A′ B ′ C. Question 16 Question Simplify the following expression using Boolean Algebra: F = A′ B ′ C + AB ′ C + ABC ′ + ABC Solution To simplify the expression F = A′ B ′ C + AB ′ C + ABC ′ + ABC, we will use the properties of Boolean Algebra. Step 1: Apply the absorption law XY + XY ′ = X. F = A′ B ′ C + AB ′ C + ABC ′ + ABC = A′ B ′ C + AB ′ C + A(B + BC ′ ) Step 2: Apply the absorption law XY + X ′ Z = XY + X ′ Z + Y Z twice. F = A′ B ′ C + AB ′ C + A(B + BC ′ ) = A′ B ′ C + AB ′ C + AB + ABC ′ + ABC Step 3: Apply the consensus theorem (X + Y )(X + Z)(Y + Z) = (X + Y )(Y + Z). F = A′ B ′ C + AB ′ C + AB + ABC ′ + ABC = A′ B ′ C + AB ′ C + AB + ABC 10 Step 4: Apply the absorption law XY + X ′ Z = XY + X ′ Z + Y Z. F = A′ B ′ C + AB ′ C + AB + ABC = A + AB + AC Step 5: Apply the consensus theorem (X + Y )(X + Z)(Y + Z) = (X + Y )(Y + Z). F = A + AB + AC = A(1 + B) + AC Step 6: Apply the identity law X + X ′ Y = X + Y . F = A + AC = A(1 + C) Thus, the simplified expression for F = A′ B ′ C + AB ′ C + ABC ′ + ABC is F = A(1 + C). Question 17 Question Simplify the following Boolean expression using Boolean algebra rules: F = A′ B + AB ′ + ABC ′ + BC Solution To simplify the Boolean expression F = A′ B + AB ′ + ABC ′ + BC, we will use the rules of Boolean algebra to manipulate the expression. Step 1: Apply the absorption law (X + XY = X) to A′ B + AB ′ . A′ B + AB ′ = A′ B + AB ′ + ABC ′ + BC Step 2: Apply the consensus theorem (X + X ′ Y = X + Y ) to A′ B + ABC ′ . A′ B + AB ′ + ABC ′ = A′ B + BC Step 3: Apply the consensus theorem to A′ B + BC. A′ B + BC = A′ B + A′ C + BC Step 4: Apply the absorption law to A′ B + A′ C. A′ B + A′ C + BC = A′ B + A′ C + BC + BC Step 5: Apply the idempotent law (X + X = X) to BC + BC. A′ B + A′ C + BC + BC = A′ B + A′ C + BC Step 6: The simplified expression is A′ B + A′ C + BC. 11 Question 18 Question Given the Boolean expression (A ∧ B) ∨ (¬A ∧ C) ∨ (B ∧ ¬C), simplify the expression using Boolean algebra laws. Solution To simplify the given Boolean expression, we will use the laws of Boolean algebra, including the commutative law, associative law, distributive law, identity law, complement law, etc. Step 1: Apply the distributive law to the given expression. (A ∧ B) ∨ (¬A ∧ C) ∨ (B ∧ ¬C) = (A ∨ ¬A) ∧ (A ∨ C) ∧ (B ∨ C) Step 2: Apply the complement law to simplify (A ∨ ¬A). (A ∨ ¬A) ∧ (A ∨ C) ∧ (B ∨ C) = 1 ∧ (A ∨ C) ∧ (B ∨ C) Step 3: Apply the identity law to simplify 1 ∧ (A ∨ C). 1 ∧ (A ∨ C) ∧ (B ∨ C) = A ∨ C ∧ (B ∨ C) Step 4: Apply the absorption law to simplify (A ∨ C) ∧ (B ∨ C). A ∨ C ∧ (B ∨ C) = A ∨ C Therefore, the simplified Boolean expression is A ∨ C. Question 19 Question Simplify the following Boolean expression: (A + B + C)(A + B + C)(A + B + C) Solution To simplify the given Boolean expression, we will use the distributive property and Boolean algebra laws. Step 1: Apply the distributive property to expand the expression. (A + B + C)(A + B + C)(A + B + C) = (A + B + C)(A + B + C)(A + B + C) = A(A+B +C)(A+B +C)+B(A+B +C)(A+B +C)+C(A+B +C)(A+B +C) Step 2: Apply the Absorption Law XY + XY = X. = A(A + B + C) + B(A + B + C) + C(A + B + C) =A+B+C Therefore, the simplified form of the given Boolean expression is A + B + C. 12 Question 20 Question Simplify the following Boolean expression: (A + B)(A + B)(A + B). Solution To simplify the given Boolean expression, we will use the Boolean algebra laws, such as the distributive law, the complement law, and the idempotent law. Step 1: Apply the distributive law to expand the expression. (A + B)(A + B)(A + B) = (AA + AB + BA + BB)(A + B) Step 2: Simplify the expression by applying the idempotent law (X + X = X) and the complement law (XX = 0). (0 + AB + BA + B)(A + B) = (AB + BA + B)(A + B) Step 3: Apply the distributive law to expand the expression further. (AB + BA + B)(A + B) = ABA + ABB + BA + BB Step 4: Simplify the expression by applying the idempotent law and the complement law. ABA + 0 + BA + 0 = AB + BA Step 5: Rearrange the terms using the commutative law (XY = Y X). AB + BA = AB + AB Step 6: Apply the idempotent law to simplify the expression. AB + AB = AB Therefore, (A + B)(A + B)(A + B) simplifies to AB. Question 21 Question Simplify the following Boolean expression using algebraic manipulation: F = A′ B + AB ′ + AB 13 Solution Step 1: Apply the commutative property of OR (+) to rearrange the terms. F = A′ B + AB ′ + AB = A′ B + AB + AB ′ = A′ B + AB + BA (Step 1) Step 2: Apply the idempotent law to simplify AB + BA. F = A′ B + AB + BA = A′ B + A(B + B) = A′ B + A(1) = A′ B + A (Step 2) Step 3: Apply the absorption law to simplify A′ B + A. F = A′ B + A = A + A′ B = A(1 + B) =A (Step 3) Therefore, the simplified expression is F = A. Question 22 Question Simplify the following Boolean expression using Boolean algebra laws: (A + B + C)(A′ + B ′ C) + A′ B ′ . Solution To simplify the given Boolean expression, we will use the laws of Boolean algebra to manipulate the expression step by step. Step 1: Apply the distributive law: (A + B + C)(A′ + B ′ C) + A′ B ′ . Step 2: Use the distributive property: AA′ + AB ′ C + BA′ + BB ′ C + CA′ + CB ′ C + A′ B ′ . Step 3: Simplify the terms involving complement pairs: 0 + AB ′ C + 0 + 0 + 0 + 0 + A′ B ′ . Step 4: Further simplify the expression: AB ′ C + A′ B ′ . Step 5: Apply the complement law: AB ′ C + A′ B ′ = AB ′ C + A′ B ′ (C + C ′ ). Step 6: Use the distributive property: AB ′ C + A′ B ′ C + A′ B ′ C ′ . Step 7: Combine terms with C using the consensus theorem: AB ′ C + ′ ′ A B C + A′ B ′ C ′ = AB ′ C + A′ B ′ C ′ . Therefore, the simplified Boolean expression is AB ′ C + A′ B ′ C ′ . 14 Question 23 Question Simplify the following Boolean expression: (A + B)(A + B)(A + B). Solution To simplify the given Boolean expression, we will use the distributive property and De Morgan’s laws. Step 1: Apply the distributive property to expand the expression. (A + B)(A + B)(A + B) = (A + B)((A · B) + (A · B) + (B · A) + (B · B)) = (A + B)(AB + AB + AB + 0) (Expanding) (Complement law: B · B = 0) = (A + B)(AB + AB) Step 2: Apply the distributive property again to simplify further. (A + B)(AB + AB) = A(AB + AB) + B(AB + AB) (Distributive property) = AAB + AAB + BAB + BAB (Expanding) = AB + 0B + AB + 0 (Complement law: A · A = 0 and B · A = 0) = AB + AB = AB Therefore, the simplified form of the given Boolean expression is AB. Question 24 Question Simplify the following Boolean expression: (A + B + C)(A + B + C)(A + B + C). Solution To simplify the given Boolean expression (A + B + C)(A + B + C)(A + B + C), we can expand the expression using the distributive property and then simplify it using the rules of Boolean Algebra. Step 1: Expand the expression by distributing: (A + B + C)(A + B + C)(A + B + C) = A(A+B +C)(A+B +C)+B(A+B +C)(A+B +C)+C(A+B +C)(A+B +C) 15 Step 2: Apply the distributive property to each term: = AAA + AAB + AAC + ABA + ABB + ABC + ACA + ACB + ACC+ +BAA + BAB + BAC + BBA + BBB + BBC + BCA + BCB + BCC+ +CAA + CAB + CAC + CBA + CBB + CBC + CCA + CCB + CCC Step 3: Simplify the expression using the rules of Boolean Algebra. Note that XX = X and XX = 0: = 0+0+0+0+0+0+0+0+0+0+0+0+0+0+0+0+0+0+0+0+0+0+0+0+0+0+0+0 =0 Therefore, the simplified Boolean expression is 0. Question 25 Question Simplify the Boolean expression (A + BC)(AB ′ C + A′ B). Solution To simplify the given Boolean expression, we will use the basic laws of Boolean algebra: commutative law, associative law, distributive law, identity law, complement law, and absorption law. Step 1: Apply the distributive law (A + BC)(AB ′ C + A′ B). = AAB ′ C + AA′ B + ABCB ′ C + ABCB = A′ B + A′ B + ABC + ABC = A′ B + ABC Therefore, (A + BC)(AB ′ C + A′ B) simplifies to A′ B + ABC. Question 26 Question Simplify the Boolean expression F = AB + AC + BC + A′ B ′ C ′ . 16 Solution To simplify the given Boolean expression, we will use the laws of Boolean algebra. Step 1: Apply the absorption law X + XY = X to simplify AB + AC. AB + AC = A(B + C) Step 2: Apply the absorption law again to simplify A(B + C). A(B + C) = A Step 3: Using the identity X ′ X = 0, simplify A′ B ′ C ′ . A′ B ′ C ′ = 0 Step 4: Rewrite the simplified expressions and combine them to get the final simplified expression for F . F = AB + AC + BC + A′ B ′ C ′ = A + BC + 0 = A + BC Therefore, the simplified Boolean expression is F = A + BC. Question 27 Question Simplify the following Boolean expression: (A + B)(A + C) + (A + B). Solution To simplify the given Boolean expression, we will use the properties of Boolean algebra. Step 1: Apply the distributive law: (A + B)(A + C) + (A + B) = A(A + C) + B(A + C) + A + B. Step 2: Apply the distributive law again: A(A + C) + B(A + C) + A + B = AA + AC + BA + BC + A + B. Step 3: Use the idempotent law (AA = A) and zero property (XA = 0 if X is a variable and not complemented): AA + AC + BA + BC + A + B = A + AC + BA + BC + A + B. Step 4: Apply the idempotent law once more: A + AC + BA + BC + A + B = A + BA + BC + B. Step 5: Apply the absorption law (X + XY = X): A + BA + BC + B = A + B(1 + C) = A + B. Therefore, (A + B)(A + C) + (A + B) simplifies to A + B. 17 Question 28 Question Simplify the Boolean expression (A + B)(A + B + C)(A + B + C). Solution Step 1: Apply the distributive property to expand the expression. (A+B)(A+B+C)(A+B+C) = AA+BA+BA+BB+BC+AB+BC+BB+AB+ABC Step 2: Simplify the terms involving complement pairs. 0 + 0 + 0 + 0 + 0 + AB + BC + 0 + AB + ABC Step 3: Simplify the remaining terms. AB + BC + AB + ABC = AB(1 + C) + AB(1 + C) Step 4: Use the distributive property to factor out a common term. AB(1 + C) + AB(1 + C) = AB + AB Step 5: Apply the absorption law X + X ′ Y = X + Y . AB + AB = A + B Therefore, the simplified Boolean expression is A + B. Question 29 Question Simplify the Boolean expression: AB + A′ B + A′ C + BC + B ′ C Solution To simplify the Boolean expression AB + A′ B + A′ C + BC + B ′ C, we can use Boolean algebra laws and properties to simplify the expression step by step. Step 1: Apply the absorption law A + AB = A. AB + A′ B + A′ C + BC + B ′ C = AB + A′ B + A′ C + BC + B ′ C + 0 = AB + A′ B + A′ C + ABC + B ′ C = AB(1 + C) + A′ C(1 + B) = AB + A′ C 18 Step 2: Apply the consensus theorem: AB + A′ C = AB + B ′ C + AC. AB + A′ C = AB + B ′ C + AC = (A + B ′ )(B ′ + C)(A + C) Therefore, the simplified form of the Boolean expression is (A + B ′ )(B ′ + C)(A + C). Question 30 Question Let A, B, and C be three Boolean variables. Simplify the following Boolean expression: (A + B) · (A · C + B · C) Solution To simplify the given Boolean expression, we will use the rules of Boolean algebra such as distribution, absorption, and identity properties. Step 1: Distribute the OR operation over the AND operation (A + B) · (A · C + B · C) = A · (A · C + B · C) + B · (A · C + B · C) Step 2: Apply the Distributive Law =A·A·C +A·B·C +B·A·C +B·B·C Step 3: Simplify the first and last terms using the Absorption Law =A·C +A·B·C +B·A·C +B·C Step 4: Simplify using the Identity Law =A·C +A·B·C +A·C +B·C Step 5: Combine like terms =A·C +A·C +A·B·C +B·C Step 6: Simplify further =A·C +A·B·C +B·C Step 7: Factor out common terms = A · (C + B · C) + B · C Step 8: Apply the Absorption Law =A·1+B·C Step 9: Simplify using the Identity Law =A+B·C Therefore, the simplified form of the given Boolean expression is A + B · C. 19 Question 31 Question Simplify the following Boolean expression: (A′ B + C)(A + B ′ C) + AB + AC Solution To simplify the given Boolean expression, we will first expand the terms using the distributive property, then simplify using Boolean algebra laws. Step 1: Expand the expression (A′ B + C)(A + B ′ C) + AB + AC = A′ B · A + A′ B · B ′ C + C · A + C · B ′ C + AB + AC = A′ B · A + A′ B · B ′ C + C · A + C · B ′ C + AB + AC Step 2: Apply the AND absorption law A′ B · A + A′ B · B ′ C + C · A + C · B ′ C + AB + AC = A′ B + A′ B · C + C · A + C · B ′ C + AB + AC = A′ B + C(A′ + B ′ C) + C · B ′ C + AB + AC Step 3: Apply the distributive law A′ B + C(A′ + B ′ C) + C · B ′ C + AB + AC = A′ B + CA′ + CB ′ C + AB + AC = A′ B + A′ C + B ′ C + AB + AC Step 4: Apply the consensus theorem A′ B + A′ C + B ′ C + AB + AC = A′ B + AB + A′ C + AC + B ′ C = B + A′ C + AC + B ′ C = B + A′ C + C(B + B ′ ) = B + A′ C + C Therefore, the simplified form of the given Boolean expression is B+A′ C +C. Question 32 Question Simplify the following Boolean expression: (A + B) · (A + B). Solution To simplify the given Boolean expression, we will use the distributive property of Boolean algebra. 20 A 0 0 1 1 B 0 1 0 1 A+B 1 0 1 1 (A + B) · (A + B) 0 0 1 1 (A + B) · (A + B) = A · (A + B) + B · (A + B) (Distributive Property) =A·A+A·B+B·A+B·B =A+A·B+B·A+0 (Distributive Property) (Complement Property) = A + A · B + B · A (Identity Property) = A + AB + AB = A + AB = A(1 + B) =A·1 =A (Commutative Property) (Idempotent Law) (Distributive Property) (Complement Property) (Identity Property) Therefore, (A + B) · (A + B) simplifies to A. Question 33 Question Simplify the following Boolean expression: (A + B)(A + B + C). Solution Step 1: Use the distributive property to expand the expression. (A + B)(A + B + C) = A(A + B + C) + B(A + B + C) Step 2: Apply the distributive property again to simplify each term. A(A + B + C) = AA + AB + AC = A + AB + AC B(A + B + C) = BA + BB + BC = AB + BC Step 3: Combine the simplified terms. A + AB + AC + AB + BC = A + AB + AB + AC + BC Step 4: Use the idempotent law (X + X = X) and the absorption law (X + XY = X) to simplify further. A + AB + AB + AC + BC = A + AB + AC 21 = A(1 + B) + AC = A + AC Step 5: Apply absorption once again to get the final simplified expression. A + AC = A(1 + C) = A Therefore, the simplified expression is A. Question 34 Question Simplify the following Boolean expression: (A + BC)(AB + AC). Solution To simplify the given Boolean expression (A + BC)(AB + AC), we will use the distributive property and the fact that X + X = X for any variable X in Boolean algebra. Step 1: Apply the distributive property to expand the expression. (A + BC)(AB + AC) = AAB + AAC + BCAB + BCAC Step 2: Simplify the expression by combining like terms. AAB + AAC + BCAB + BCAC = A + AC + AB(C + C) Step 3: Use the fact that X +X = X in Boolean algebra to simplify further. A + AC + AB(C + C) = A + AC + AB Step 4: Apply the distributive property in reverse to factor out a common term. A + AC + AB = A(1 + C) + AB = A + AB = A(1 + B) = A Therefore, the simplified form of the Boolean expression (A+BC)(AB+AC) is A. Question 35 Question Simplify the following Boolean expression: F = A′ B ′ C + AB ′ C + ABC + ABC ′ 22 Question 4 Question Let A, B, and C be Boolean variables. Simplify the following Boolean expression: (A + B · C) · (A · B + C) Solution To simplify the given Boolean expression, we will use basic Boolean algebra rules and laws. Step 1: Distribute the terms (A + B · C) · (A · B + C) = A · A · B + A · C + B · C · A · B + B · C · C Step 2: Apply the complement law A·A=0 Step 3: Simplify the expression 0·B+A·C +0+B·C ·0=A·C +B·0=A·C Step 4: Final answer Therefore, the simplified form of (A+B·C)·(A·B+C) is A · C. Question 5 Question Simplify the following Boolean expression using Boolean algebra laws: F = A′ B ′ C + AB ′ C ′ + ABC + AB ′ C Solution To simplify the given Boolean expression F = A′ B ′ C + AB ′ C ′ + ABC + AB ′ C, we will use the laws of Boolean algebra. Step 1: Apply absorption law AB ′ C ′ + AB ′ C = AB ′ C ′ Now we have: F = A′ B ′ C + AB ′ C ′ + ABC + AB ′ C ′ Step 2: Apply absorption law again ABC + AB ′ C ′ = AB(C + C ′ ) = AB 3 Now we have: F = A′ B ′ C + AB ′ C ′ + AB Step 3: Apply consensus theorem (or theorem of consensus) law A′ B ′ C + AB = A′ C + AB Now we have: F = A′ C + AB ′ C ′ Step 4: No further simplification is possible, hence the simplified expression is F = A′ C + AB ′ C ′ Question 6 Question Simplify the following Boolean expression: (A + B + C)(A + B + C)(A + B + C). Solution Let’s simplify the given Boolean expression step by step. Step 1: Use the distributive law to expand the expression. (A+B+C)(A+B+C)(A+B+C) = A(A+B+C)(A+B+C)+B(A+B+C)(A+B+C)+C(A+B+C)(A+B+C) Step 2: Further simplify each term by distributing. = AA + AB + AC + BA + AB + BB + BC + CA + BC + CC Since XX = X for any Boolean variable X, we simplify AB, AB, BC, and CC terms: = A + A + AC + AB + B + 0 + BC + CA + 0 + C Step 3: Combine like terms. = A + AC + AB + B + BC + CA + C Step 4: Use the consensus theorem, XY + XZ + Y Z = XY + XZ, to simplify AC + AB + BC. = A(C + B) + BC Step 5: Apply the distributive law once again to obtain the final simplified expression. = A + BC Therefore, (A + B + C)(A + B + C)(A + B + C) simplifies to A + BC. 4 Question 7 Question Simplify the following Boolean expression using algebraic manipulation: F = A′ B ′ C ′ + A′ BC ′ + AB ′ C + ABC Solution Step 1: Apply the distributive law to factor out C ′ : F = A′ B ′ C ′ (1 + A) + AB ′ C(1 + A) Step 2: Apply the distributive law again to factor out A: F = A′ C ′ (A′ + B ′ ) + AB ′ C(A + 1) Step 3: Use the complement laws (XX ′ = 0 and X + X ′ = 1) to simplify the terms: F = A′ C ′ + AB ′ C Step 4: Apply the distributive law to factor out C: F = C(A′ + AB ′ ) Step 5: Apply the absorption law (X + XY = X) to simplify the expression: F = C(A′ + B ′ ) So, the simplified Boolean expression is F = C(A′ + B ′ ). Question 8 Question Simplify the Boolean expression: (A+B +C)(A+D)(B +C +D)(A+B +C +D). Solution Step 1: Use the distributive law to expand the given expression. (A + B + C)(A + D)(B + C + D)(A + B + C + D) = (A2 + AD + AB + BD + AC + CD + BC + BD)(A + B + C + D) Step 2: Simplify the expression using the complement law (AA′ = 0 and A + A′ = 1). (A2 + AD + AB + BD + AC + CD + BC + BD)(A + B + C + D) = (AD + AB + BD + AC + CD + BC)(A + B + C + D) 5 Step 3: Expand the expression further. (AD + AB + BD + AC + CD + BC)(A + B + C + D) = ADA + ADB + ADC + ADB + ACD + ACC + BCD + BCB Step 4: Simplify using the idempotent law (AA = A). ADA + ADB + ADC + ADB + ACD + ACC + BCD + BCB = 0 + ADB + ADC + ADB + ACD + 0 + BCD + 0 Step 5: Simplify further by combining terms. 0 + ADB + ADC + ADB + ACD + 0 + BCD + 0 = ADB + ADB + ACD + BCD = AD(B + B + C) + CD(A + B) = AD + CD Therefore, the simplified form of the Boolean expression is AD + CD. Question 9 Question Simplify the following Boolean algebra expression: (A+B +C)(A′ +B +C)(A+ B ′ + C)(A + B + C ′ ). Solution To simplify the given Boolean algebra expression, we will use the properties of Boolean algebra including the distributive law, complement law, identity law, and simplification rules. Step 1: Apply the distributive law: (A + B + C)(A′ + B + C)(A + B ′ + C)(A + B + C ′ ) (A+B+C)(A′ +B+C)(A+B ′ +C)(A+B+C ′ ) = (A+B)(A+B ′ +C)(A+B+C ′ ) Step 2: Apply the distributive law again: (A + B)(A + B ′ + C)(A + B + C ′ ) (A + B)(A + B ′ + C)(A + B + C ′ ) = AA + AB ′ + AC + BBA + BBB ′ + BBC ′ (A + B)(A + B ′ + C)(A + B + C ′ ) = A + AB ′ + AC + 0 + B ′ + 0 Step 3: Simplify the expression: A + AB ′ + AC + B ′ A + AB ′ + AC + B ′ = A(1 + B ′ ) + B ′ (1 + A) A + AB ′ + AC + B ′ = A + B ′ Therefore, the simplified Boolean algebra expression is A + B ′ . 6 Question 10 Question Simplify the following Boolean expression using laws of Boolean Algebra: (A + B) · (A′ + B) · (A + B ′ ). Solution Step 1: Apply the distributive law X · (Y + Z) = X · Y + X · Z. = (A + B) · (A + B ′ ) + (A + B) · (A′ + B) Step 2: Apply the distributive law again. = AA + AB ′ + BA + BB ′ + AA′ + AB Step 3: Simplify by applying the idempotent law XX = X and the identity law X + X ′ = 1. = A + AB ′ + AB + B + A′ + AB Step 4: Apply the idempotent law and simplify further. = A + B + A′ Step 5: Apply the idempotent law one more time to get the final simplified expression. =A+B+A Step 6: Apply the idempotent law for the last time to obtain the simplest form of the expression. =A+B Therefore, (A + B) · (A′ + B) · (A + B ′ ) simplifies to A + B. Question 11 Question Simplify the Boolean expression F = A′ B + AC + BC ′ . Solution To simplify the Boolean expression F = A′ B + AC + BC ′ , we will use Boolean algebra laws and rules to simplify the expression step by step. Step 1: Apply the distributive law to factor out A. F = A′ B + AC + BC ′ = A(B ′ + C) + BC ′ Step 2: Apply the distributive law again to factor out B. F = A(B ′ + C) + BC ′ = AB ′ + AC + BC ′ Step 3: Apply the consensus theorem, AB + A′ C + BC = AB + A′ C, where AB is the consensus term. F = AB ′ + AC + BC ′ = AB ′ + A′ C 7 Therefore, the simplified form of the Boolean expression F = A′ B+AC+BC ′ is F = AB ′ + A′ C. Question 12 Question Simplify the Boolean expression (A + B) · (A + B · C) using Boolean algebra laws. Solution Step 1: Apply the Distributive Law: X · (Y + Z) = X · Y + X · Z (A + B) · (A + B · C) = A · A + A · B · C + B · A + B · B · C = A + AB · C + A · B + 0 = A + AB · C + A · B Step 2: Apply the Idempotent Law: X + X = X A + AB · C + A · B = A + AB · C Therefore, the simplified form of the Boolean expression (A + B) · (A + B · C) is A + AB · C. Question 13 Question Simplify the following Boolean expression: (A + B)(A + B + C) using Boolean algebra laws. Solution To simplify the expression (A + B)(A + B + C), we will use the distributive law and De Morgan’s law. Step 1: Apply distributive law: (A + B)(A + B + C) = A(A + B + C) + B(A + B + C) Step 2: Apply distributive law again: = AA + AB + AC + BA + BB + BC Step 3: Apply the complement law (AA = 0 and BB = 0): = 0 + AB + AC + BA + 0 + BC 8 Step 4: Rearrange the terms: = AC + BC + AB + BA Step 5: Apply the absorption law (AC + BC = C): = C + AB + BA Step 6: Apply the commutative law: = C + BA + AB Therefore, the simplified form of the given Boolean expression is C + BA + AB. Question 14 Question Simplify the Boolean expression (A + B + C)(A′ + C ′ )(A′ + B). Solution We will simplify the given Boolean expression step by step using the laws of Boolean Algebra. Step 1: Apply the Distributive Law: X(Y + Z) = XY + XZ. (A + B + C)(A′ + C ′ )(A′ + B) = (A + B + C)(A′ A′ + A′ C ′ + AB + BC ′ + AC ′ + BC) = (A + B + C)(0 + A′ C ′ + AB + BC ′ + AC ′ + BC) = A′ C ′ (A + B + C) + AB(A + B + C) + BC ′ (A + B + C) = A′ C ′ A + A′ C ′ B + A′ C ′ C + ABA + ABB + ABC + BCA + BCB + BCC = 0 + 0 + 0 + AB + 0 + 0 + 0 + 0 + 0 = AB Step 2: Apply the Idempotent Law: X + X = X. AB = AB Therefore, the simplified Boolean expression is AB. Question 15 Question Simplify the Boolean expression F = A′ B ′ C + A′ BC + AB ′ C ′ + ABC ′ using Boolean algebra laws and theorems. 9 Solution To simplify the given Boolean expression F = A′ B ′ C + A′ BC + AB ′ C ′ + ABC ′ , we will use the laws and theorems of Boolean algebra. Step 1: Apply the absorption law: XY + XZ = X(Y + Z). F = A′ B ′ C + A′ BC + AB ′ C ′ + ABC ′ = A′ B ′ C + A′ BC + A′ B(C ′ + C ′ ) + AB ′ C ′ ′ ′ ′ ′ ′ = A B C + A BC + A B + AB C (Applying absorption law) ′ = A′ B ′ C + A′ BC + A′ B + A′ B ′ C ′ (Reordering terms) Step 2: Apply the consensus theorem: XY + X ′ Z + Y Z = XY + X ′ Z. F = A′ B ′ C + A′ BC + A′ B + A′ B ′ C ′ = A′ B ′ C + A′ BC + A′ B + A′ B ′ C (Applying consensus theorem) Therefore, the simplified Boolean expression is F = A′ B ′ C + A′ BC + A′ B + A′ B ′ C. Question 16 Question Simplify the following expression using Boolean Algebra: F = A′ B ′ C + AB ′ C + ABC ′ + ABC Solution To simplify the expression F = A′ B ′ C + AB ′ C + ABC ′ + ABC, we will use the properties of Boolean Algebra. Step 1: Apply the absorption law XY + XY ′ = X. F = A′ B ′ C + AB ′ C + ABC ′ + ABC = A′ B ′ C + AB ′ C + A(B + BC ′ ) Step 2: Apply the absorption law XY + X ′ Z = XY + X ′ Z + Y Z twice. F = A′ B ′ C + AB ′ C + A(B + BC ′ ) = A′ B ′ C + AB ′ C + AB + ABC ′ + ABC Step 3: Apply the consensus theorem (X + Y )(X + Z)(Y + Z) = (X + Y )(Y + Z). F = A′ B ′ C + AB ′ C + AB + ABC ′ + ABC = A′ B ′ C + AB ′ C + AB + ABC 10 Step 4: Apply the absorption law XY + X ′ Z = XY + X ′ Z + Y Z. F = A′ B ′ C + AB ′ C + AB + ABC = A + AB + AC Step 5: Apply the consensus theorem (X + Y )(X + Z)(Y + Z) = (X + Y )(Y + Z). F = A + AB + AC = A(1 + B) + AC Step 6: Apply the identity law X + X ′ Y = X + Y . F = A + AC = A(1 + C) Thus, the simplified expression for F = A′ B ′ C + AB ′ C + ABC ′ + ABC is F = A(1 + C). Question 17 Question Simplify the following Boolean expression using Boolean algebra rules: F = A′ B + AB ′ + ABC ′ + BC Solution To simplify the Boolean expression F = A′ B + AB ′ + ABC ′ + BC, we will use the rules of Boolean algebra to manipulate the expression. Step 1: Apply the absorption law (X + XY = X) to A′ B + AB ′ . A′ B + AB ′ = A′ B + AB ′ + ABC ′ + BC Step 2: Apply the consensus theorem (X + X ′ Y = X + Y ) to A′ B + ABC ′ . A′ B + AB ′ + ABC ′ = A′ B + BC Step 3: Apply the consensus theorem to A′ B + BC. A′ B + BC = A′ B + A′ C + BC Step 4: Apply the absorption law to A′ B + A′ C. A′ B + A′ C + BC = A′ B + A′ C + BC + BC Step 5: Apply the idempotent law (X + X = X) to BC + BC. A′ B + A′ C + BC + BC = A′ B + A′ C + BC Step 6: The simplified expression is A′ B + A′ C + BC. 11 Question 18 Question Given the Boolean expression (A ∧ B) ∨ (¬A ∧ C) ∨ (B ∧ ¬C), simplify the expression using Boolean algebra laws. Solution To simplify the given Boolean expression, we will use the laws of Boolean algebra, including the commutative law, associative law, distributive law, identity law, complement law, etc. Step 1: Apply the distributive law to the given expression. (A ∧ B) ∨ (¬A ∧ C) ∨ (B ∧ ¬C) = (A ∨ ¬A) ∧ (A ∨ C) ∧ (B ∨ C) Step 2: Apply the complement law to simplify (A ∨ ¬A). (A ∨ ¬A) ∧ (A ∨ C) ∧ (B ∨ C) = 1 ∧ (A ∨ C) ∧ (B ∨ C) Step 3: Apply the identity law to simplify 1 ∧ (A ∨ C). 1 ∧ (A ∨ C) ∧ (B ∨ C) = A ∨ C ∧ (B ∨ C) Step 4: Apply the absorption law to simplify (A ∨ C) ∧ (B ∨ C). A ∨ C ∧ (B ∨ C) = A ∨ C Therefore, the simplified Boolean expression is A ∨ C. Question 19 Question Simplify the following Boolean expression: (A + B + C)(A + B + C)(A + B + C) Solution To simplify the given Boolean expression, we will use the distributive property and Boolean algebra laws. Step 1: Apply the distributive property to expand the expression. (A + B + C)(A + B + C)(A + B + C) = (A + B + C)(A + B + C)(A + B + C) = A(A+B +C)(A+B +C)+B(A+B +C)(A+B +C)+C(A+B +C)(A+B +C) Step 2: Apply the Absorption Law XY + XY = X. = A(A + B + C) + B(A + B + C) + C(A + B + C) =A+B+C Therefore, the simplified form of the given Boolean expression is A + B + C. 12 Question 20 Question Simplify the following Boolean expression: (A + B)(A + B)(A + B). Solution To simplify the given Boolean expression, we will use the Boolean algebra laws, such as the distributive law, the complement law, and the idempotent law. Step 1: Apply the distributive law to expand the expression. (A + B)(A + B)(A + B) = (AA + AB + BA + BB)(A + B) Step 2: Simplify the expression by applying the idempotent law (X + X = X) and the complement law (XX = 0). (0 + AB + BA + B)(A + B) = (AB + BA + B)(A + B) Step 3: Apply the distributive law to expand the expression further. (AB + BA + B)(A + B) = ABA + ABB + BA + BB Step 4: Simplify the expression by applying the idempotent law and the complement law. ABA + 0 + BA + 0 = AB + BA Step 5: Rearrange the terms using the commutative law (XY = Y X). AB + BA = AB + AB Step 6: Apply the idempotent law to simplify the expression. AB + AB = AB Therefore, (A + B)(A + B)(A + B) simplifies to AB. Question 21 Question Simplify the following Boolean expression using algebraic manipulation: F = A′ B + AB ′ + AB 13 Solution Step 1: Apply the commutative property of OR (+) to rearrange the terms. F = A′ B + AB ′ + AB = A′ B + AB + AB ′ = A′ B + AB + BA (Step 1) Step 2: Apply the idempotent law to simplify AB + BA. F = A′ B + AB + BA = A′ B + A(B + B) = A′ B + A(1) = A′ B + A (Step 2) Step 3: Apply the absorption law to simplify A′ B + A. F = A′ B + A = A + A′ B = A(1 + B) =A (Step 3) Therefore, the simplified expression is F = A. Question 22 Question Simplify the following Boolean expression using Boolean algebra laws: (A + B + C)(A′ + B ′ C) + A′ B ′ . Solution To simplify the given Boolean expression, we will use the laws of Boolean algebra to manipulate the expression step by step. Step 1: Apply the distributive law: (A + B + C)(A′ + B ′ C) + A′ B ′ . Step 2: Use the distributive property: AA′ + AB ′ C + BA′ + BB ′ C + CA′ + CB ′ C + A′ B ′ . Step 3: Simplify the terms involving complement pairs: 0 + AB ′ C + 0 + 0 + 0 + 0 + A′ B ′ . Step 4: Further simplify the expression: AB ′ C + A′ B ′ . Step 5: Apply the complement law: AB ′ C + A′ B ′ = AB ′ C + A′ B ′ (C + C ′ ). Step 6: Use the distributive property: AB ′ C + A′ B ′ C + A′ B ′ C ′ . Step 7: Combine terms with C using the consensus theorem: AB ′ C + ′ ′ A B C + A′ B ′ C ′ = AB ′ C + A′ B ′ C ′ . Therefore, the simplified Boolean expression is AB ′ C + A′ B ′ C ′ . 14 Question 23 Question Simplify the following Boolean expression: (A + B)(A + B)(A + B). Solution To simplify the given Boolean expression, we will use the distributive property and De Morgan’s laws. Step 1: Apply the distributive property to expand the expression. (A + B)(A + B)(A + B) = (A + B)((A · B) + (A · B) + (B · A) + (B · B)) = (A + B)(AB + AB + AB + 0) (Expanding) (Complement law: B · B = 0) = (A + B)(AB + AB) Step 2: Apply the distributive property again to simplify further. (A + B)(AB + AB) = A(AB + AB) + B(AB + AB) (Distributive property) = AAB + AAB + BAB + BAB (Expanding) = AB + 0B + AB + 0 (Complement law: A · A = 0 and B · A = 0) = AB + AB = AB Therefore, the simplified form of the given Boolean expression is AB. Question 24 Question Simplify the following Boolean expression: (A + B + C)(A + B + C)(A + B + C). Solution To simplify the given Boolean expression (A + B + C)(A + B + C)(A + B + C), we can expand the expression using the distributive property and then simplify it using the rules of Boolean Algebra. Step 1: Expand the expression by distributing: (A + B + C)(A + B + C)(A + B + C) = A(A+B +C)(A+B +C)+B(A+B +C)(A+B +C)+C(A+B +C)(A+B +C) 15 Step 2: Apply the distributive property to each term: = AAA + AAB + AAC + ABA + ABB + ABC + ACA + ACB + ACC+ +BAA + BAB + BAC + BBA + BBB + BBC + BCA + BCB + BCC+ +CAA + CAB + CAC + CBA + CBB + CBC + CCA + CCB + CCC Step 3: Simplify the expression using the rules of Boolean Algebra. Note that XX = X and XX = 0: = 0+0+0+0+0+0+0+0+0+0+0+0+0+0+0+0+0+0+0+0+0+0+0+0+0+0+0+0 =0 Therefore, the simplified Boolean expression is 0. Question 25 Question Simplify the Boolean expression (A + BC)(AB ′ C + A′ B). Solution To simplify the given Boolean expression, we will use the basic laws of Boolean algebra: commutative law, associative law, distributive law, identity law, complement law, and absorption law. Step 1: Apply the distributive law (A + BC)(AB ′ C + A′ B). = AAB ′ C + AA′ B + ABCB ′ C + ABCB = A′ B + A′ B + ABC + ABC = A′ B + ABC Therefore, (A + BC)(AB ′ C + A′ B) simplifies to A′ B + ABC. Question 26 Question Simplify the Boolean expression F = AB + AC + BC + A′ B ′ C ′ . 16 Solution To simplify the given Boolean expression, we will use the laws of Boolean algebra. Step 1: Apply the absorption law X + XY = X to simplify AB + AC. AB + AC = A(B + C) Step 2: Apply the absorption law again to simplify A(B + C). A(B + C) = A Step 3: Using the identity X ′ X = 0, simplify A′ B ′ C ′ . A′ B ′ C ′ = 0 Step 4: Rewrite the simplified expressions and combine them to get the final simplified expression for F . F = AB + AC + BC + A′ B ′ C ′ = A + BC + 0 = A + BC Therefore, the simplified Boolean expression is F = A + BC. Question 27 Question Simplify the following Boolean expression: (A + B)(A + C) + (A + B). Solution To simplify the given Boolean expression, we will use the properties of Boolean algebra. Step 1: Apply the distributive law: (A + B)(A + C) + (A + B) = A(A + C) + B(A + C) + A + B. Step 2: Apply the distributive law again: A(A + C) + B(A + C) + A + B = AA + AC + BA + BC + A + B. Step 3: Use the idempotent law (AA = A) and zero property (XA = 0 if X is a variable and not complemented): AA + AC + BA + BC + A + B = A + AC + BA + BC + A + B. Step 4: Apply the idempotent law once more: A + AC + BA + BC + A + B = A + BA + BC + B. Step 5: Apply the absorption law (X + XY = X): A + BA + BC + B = A + B(1 + C) = A + B. Therefore, (A + B)(A + C) + (A + B) simplifies to A + B. 17 Question 28 Question Simplify the Boolean expression (A + B)(A + B + C)(A + B + C). Solution Step 1: Apply the distributive property to expand the expression. (A+B)(A+B+C)(A+B+C) = AA+BA+BA+BB+BC+AB+BC+BB+AB+ABC Step 2: Simplify the terms involving complement pairs. 0 + 0 + 0 + 0 + 0 + AB + BC + 0 + AB + ABC Step 3: Simplify the remaining terms. AB + BC + AB + ABC = AB(1 + C) + AB(1 + C) Step 4: Use the distributive property to factor out a common term. AB(1 + C) + AB(1 + C) = AB + AB Step 5: Apply the absorption law X + X ′ Y = X + Y . AB + AB = A + B Therefore, the simplified Boolean expression is A + B. Question 29 Question Simplify the Boolean expression: AB + A′ B + A′ C + BC + B ′ C Solution To simplify the Boolean expression AB + A′ B + A′ C + BC + B ′ C, we can use Boolean algebra laws and properties to simplify the expression step by step. Step 1: Apply the absorption law A + AB = A. AB + A′ B + A′ C + BC + B ′ C = AB + A′ B + A′ C + BC + B ′ C + 0 = AB + A′ B + A′ C + ABC + B ′ C = AB(1 + C) + A′ C(1 + B) = AB + A′ C 18 Step 2: Apply the consensus theorem: AB + A′ C = AB + B ′ C + AC. AB + A′ C = AB + B ′ C + AC = (A + B ′ )(B ′ + C)(A + C) Therefore, the simplified form of the Boolean expression is (A + B ′ )(B ′ + C)(A + C). Question 30 Question Let A, B, and C be three Boolean variables. Simplify the following Boolean expression: (A + B) · (A · C + B · C) Solution To simplify the given Boolean expression, we will use the rules of Boolean algebra such as distribution, absorption, and identity properties. Step 1: Distribute the OR operation over the AND operation (A + B) · (A · C + B · C) = A · (A · C + B · C) + B · (A · C + B · C) Step 2: Apply the Distributive Law =A·A·C +A·B·C +B·A·C +B·B·C Step 3: Simplify the first and last terms using the Absorption Law =A·C +A·B·C +B·A·C +B·C Step 4: Simplify using the Identity Law =A·C +A·B·C +A·C +B·C Step 5: Combine like terms =A·C +A·C +A·B·C +B·C Step 6: Simplify further =A·C +A·B·C +B·C Step 7: Factor out common terms = A · (C + B · C) + B · C Step 8: Apply the Absorption Law =A·1+B·C Step 9: Simplify using the Identity Law =A+B·C Therefore, the simplified form of the given Boolean expression is A + B · C. 19 Question 31 Question Simplify the following Boolean expression: (A′ B + C)(A + B ′ C) + AB + AC Solution To simplify the given Boolean expression, we will first expand the terms using the distributive property, then simplify using Boolean algebra laws. Step 1: Expand the expression (A′ B + C)(A + B ′ C) + AB + AC = A′ B · A + A′ B · B ′ C + C · A + C · B ′ C + AB + AC = A′ B · A + A′ B · B ′ C + C · A + C · B ′ C + AB + AC Step 2: Apply the AND absorption law A′ B · A + A′ B · B ′ C + C · A + C · B ′ C + AB + AC = A′ B + A′ B · C + C · A + C · B ′ C + AB + AC = A′ B + C(A′ + B ′ C) + C · B ′ C + AB + AC Step 3: Apply the distributive law A′ B + C(A′ + B ′ C) + C · B ′ C + AB + AC = A′ B + CA′ + CB ′ C + AB + AC = A′ B + A′ C + B ′ C + AB + AC Step 4: Apply the consensus theorem A′ B + A′ C + B ′ C + AB + AC = A′ B + AB + A′ C + AC + B ′ C = B + A′ C + AC + B ′ C = B + A′ C + C(B + B ′ ) = B + A′ C + C Therefore, the simplified form of the given Boolean expression is B+A′ C +C. Question 32 Question Simplify the following Boolean expression: (A + B) · (A + B). Solution To simplify the given Boolean expression, we will use the distributive property of Boolean algebra. 20 A 0 0 1 1 B 0 1 0 1 A+B 1 0 1 1 (A + B) · (A + B) 0 0 1 1 (A + B) · (A + B) = A · (A + B) + B · (A + B) (Distributive Property) =A·A+A·B+B·A+B·B =A+A·B+B·A+0 (Distributive Property) (Complement Property) = A + A · B + B · A (Identity Property) = A + AB + AB = A + AB = A(1 + B) =A·1 =A (Commutative Property) (Idempotent Law) (Distributive Property) (Complement Property) (Identity Property) Therefore, (A + B) · (A + B) simplifies to A. Question 33 Question Simplify the following Boolean expression: (A + B)(A + B + C). Solution Step 1: Use the distributive property to expand the expression. (A + B)(A + B + C) = A(A + B + C) + B(A + B + C) Step 2: Apply the distributive property again to simplify each term. A(A + B + C) = AA + AB + AC = A + AB + AC B(A + B + C) = BA + BB + BC = AB + BC Step 3: Combine the simplified terms. A + AB + AC + AB + BC = A + AB + AB + AC + BC Step 4: Use the idempotent law (X + X = X) and the absorption law (X + XY = X) to simplify further. A + AB + AB + AC + BC = A + AB + AC 21 = A(1 + B) + AC = A + AC Step 5: Apply absorption once again to get the final simplified expression. A + AC = A(1 + C) = A Therefore, the simplified expression is A. Question 34 Question Simplify the following Boolean expression: (A + BC)(AB + AC). Solution To simplify the given Boolean expression (A + BC)(AB + AC), we will use the distributive property and the fact that X + X = X for any variable X in Boolean algebra. Step 1: Apply the distributive property to expand the expression. (A + BC)(AB + AC) = AAB + AAC + BCAB + BCAC Step 2: Simplify the expression by combining like terms. AAB + AAC + BCAB + BCAC = A + AC + AB(C + C) Step 3: Use the fact that X +X = X in Boolean algebra to simplify further. A + AC + AB(C + C) = A + AC + AB Step 4: Apply the distributive property in reverse to factor out a common term. A + AC + AB = A(1 + C) + AB = A + AB = A(1 + B) = A Therefore, the simplified form of the Boolean expression (A+BC)(AB+AC) is A. Question 35 Question Simplify the following Boolean expression: F = A′ B ′ C + AB ′ C + ABC + ABC ′ 22 Question 4 Question Let A, B, and C be Boolean variables. Simplify the following Boolean expression: (A + B · C) · (A · B + C) Solution To simplify the given Boolean expression, we will use basic Boolean algebra rules and laws. Step 1: Distribute the terms (A + B · C) · (A · B + C) = A · A · B + A · C + B · C · A · B + B · C · C Step 2: Apply the complement law A·A=0 Step 3: Simplify the expression 0·B+A·C +0+B·C ·0=A·C +B·0=A·C Step 4: Final answer Therefore, the simplified form of (A+B·C)·(A·B+C) is A · C. Question 5 Question Simplify the following Boolean expression using Boolean algebra laws: F = A′ B ′ C + AB ′ C ′ + ABC + AB ′ C Solution To simplify the given Boolean expression F = A′ B ′ C + AB ′ C ′ + ABC + AB ′ C, we will use the laws of Boolean algebra. Step 1: Apply absorption law AB ′ C ′ + AB ′ C = AB ′ C ′ Now we have: F = A′ B ′ C + AB ′ C ′ + ABC + AB ′ C ′ Step 2: Apply absorption law again ABC + AB ′ C ′ = AB(C + C ′ ) = AB 3 Now we have: F = A′ B ′ C + AB ′ C ′ + AB Step 3: Apply consensus theorem (or theorem of consensus) law A′ B ′ C + AB = A′ C + AB Now we have: F = A′ C + AB ′ C ′ Step 4: No further simplification is possible, hence the simplified expression is F = A′ C + AB ′ C ′ Question 6 Question Simplify the following Boolean expression: (A + B + C)(A + B + C)(A + B + C). Solution Let’s simplify the given Boolean expression step by step. Step 1: Use the distributive law to expand the expression. (A+B+C)(A+B+C)(A+B+C) = A(A+B+C)(A+B+C)+B(A+B+C)(A+B+C)+C(A+B+C)(A+B+C) Step 2: Further simplify each term by distributing. = AA + AB + AC + BA + AB + BB + BC + CA + BC + CC Since XX = X for any Boolean variable X, we simplify AB, AB, BC, and CC terms: = A + A + AC + AB + B + 0 + BC + CA + 0 + C Step 3: Combine like terms. = A + AC + AB + B + BC + CA + C Step 4: Use the consensus theorem, XY + XZ + Y Z = XY + XZ, to simplify AC + AB + BC. = A(C + B) + BC Step 5: Apply the distributive law once again to obtain the final simplified expression. = A + BC Therefore, (A + B + C)(A + B + C)(A + B + C) simplifies to A + BC. 4 Question 7 Question Simplify the following Boolean expression using algebraic manipulation: F = A′ B ′ C ′ + A′ BC ′ + AB ′ C + ABC Solution Step 1: Apply the distributive law to factor out C ′ : F = A′ B ′ C ′ (1 + A) + AB ′ C(1 + A) Step 2: Apply the distributive law again to factor out A: F = A′ C ′ (A′ + B ′ ) + AB ′ C(A + 1) Step 3: Use the complement laws (XX ′ = 0 and X + X ′ = 1) to simplify the terms: F = A′ C ′ + AB ′ C Step 4: Apply the distributive law to factor out C: F = C(A′ + AB ′ ) Step 5: Apply the absorption law (X + XY = X) to simplify the expression: F = C(A′ + B ′ ) So, the simplified Boolean expression is F = C(A′ + B ′ ). Question 8 Question Simplify the Boolean expression: (A+B +C)(A+D)(B +C +D)(A+B +C +D). Solution Step 1: Use the distributive law to expand the given expression. (A + B + C)(A + D)(B + C + D)(A + B + C + D) = (A2 + AD + AB + BD + AC + CD + BC + BD)(A + B + C + D) Step 2: Simplify the expression using the complement law (AA′ = 0 and A + A′ = 1). (A2 + AD + AB + BD + AC + CD + BC + BD)(A + B + C + D) = (AD + AB + BD + AC + CD + BC)(A + B + C + D) 5 Step 3: Expand the expression further. (AD + AB + BD + AC + CD + BC)(A + B + C + D) = ADA + ADB + ADC + ADB + ACD + ACC + BCD + BCB Step 4: Simplify using the idempotent law (AA = A). ADA + ADB + ADC + ADB + ACD + ACC + BCD + BCB = 0 + ADB + ADC + ADB + ACD + 0 + BCD + 0 Step 5: Simplify further by combining terms. 0 + ADB + ADC + ADB + ACD + 0 + BCD + 0 = ADB + ADB + ACD + BCD = AD(B + B + C) + CD(A + B) = AD + CD Therefore, the simplified form of the Boolean expression is AD + CD. Question 9 Question Simplify the following Boolean algebra expression: (A+B +C)(A′ +B +C)(A+ B ′ + C)(A + B + C ′ ). Solution To simplify the given Boolean algebra expression, we will use the properties of Boolean algebra including the distributive law, complement law, identity law, and simplification rules. Step 1: Apply the distributive law: (A + B + C)(A′ + B + C)(A + B ′ + C)(A + B + C ′ ) (A+B+C)(A′ +B+C)(A+B ′ +C)(A+B+C ′ ) = (A+B)(A+B ′ +C)(A+B+C ′ ) Step 2: Apply the distributive law again: (A + B)(A + B ′ + C)(A + B + C ′ ) (A + B)(A + B ′ + C)(A + B + C ′ ) = AA + AB ′ + AC + BBA + BBB ′ + BBC ′ (A + B)(A + B ′ + C)(A + B + C ′ ) = A + AB ′ + AC + 0 + B ′ + 0 Step 3: Simplify the expression: A + AB ′ + AC + B ′ A + AB ′ + AC + B ′ = A(1 + B ′ ) + B ′ (1 + A) A + AB ′ + AC + B ′ = A + B ′ Therefore, the simplified Boolean algebra expression is A + B ′ . 6 Question 10 Question Simplify the following Boolean expression using laws of Boolean Algebra: (A + B) · (A′ + B) · (A + B ′ ). Solution Step 1: Apply the distributive law X · (Y + Z) = X · Y + X · Z. = (A + B) · (A + B ′ ) + (A + B) · (A′ + B) Step 2: Apply the distributive law again. = AA + AB ′ + BA + BB ′ + AA′ + AB Step 3: Simplify by applying the idempotent law XX = X and the identity law X + X ′ = 1. = A + AB ′ + AB + B + A′ + AB Step 4: Apply the idempotent law and simplify further. = A + B + A′ Step 5: Apply the idempotent law one more time to get the final simplified expression. =A+B+A Step 6: Apply the idempotent law for the last time to obtain the simplest form of the expression. =A+B Therefore, (A + B) · (A′ + B) · (A + B ′ ) simplifies to A + B. Question 11 Question Simplify the Boolean expression F = A′ B + AC + BC ′ . Solution To simplify the Boolean expression F = A′ B + AC + BC ′ , we will use Boolean algebra laws and rules to simplify the expression step by step. Step 1: Apply the distributive law to factor out A. F = A′ B + AC + BC ′ = A(B ′ + C) + BC ′ Step 2: Apply the distributive law again to factor out B. F = A(B ′ + C) + BC ′ = AB ′ + AC + BC ′ Step 3: Apply the consensus theorem, AB + A′ C + BC = AB + A′ C, where AB is the consensus term. F = AB ′ + AC + BC ′ = AB ′ + A′ C 7 Therefore, the simplified form of the Boolean expression F = A′ B+AC+BC ′ is F = AB ′ + A′ C. Question 12 Question Simplify the Boolean expression (A + B) · (A + B · C) using Boolean algebra laws. Solution Step 1: Apply the Distributive Law: X · (Y + Z) = X · Y + X · Z (A + B) · (A + B · C) = A · A + A · B · C + B · A + B · B · C = A + AB · C + A · B + 0 = A + AB · C + A · B Step 2: Apply the Idempotent Law: X + X = X A + AB · C + A · B = A + AB · C Therefore, the simplified form of the Boolean expression (A + B) · (A + B · C) is A + AB · C. Question 13 Question Simplify the following Boolean expression: (A + B)(A + B + C) using Boolean algebra laws. Solution To simplify the expression (A + B)(A + B + C), we will use the distributive law and De Morgan’s law. Step 1: Apply distributive law: (A + B)(A + B + C) = A(A + B + C) + B(A + B + C) Step 2: Apply distributive law again: = AA + AB + AC + BA + BB + BC Step 3: Apply the complement law (AA = 0 and BB = 0): = 0 + AB + AC + BA + 0 + BC 8 Step 4: Rearrange the terms: = AC + BC + AB + BA Step 5: Apply the absorption law (AC + BC = C): = C + AB + BA Step 6: Apply the commutative law: = C + BA + AB Therefore, the simplified form of the given Boolean expression is C + BA + AB. Question 14 Question Simplify the Boolean expression (A + B + C)(A′ + C ′ )(A′ + B). Solution We will simplify the given Boolean expression step by step using the laws of Boolean Algebra. Step 1: Apply the Distributive Law: X(Y + Z) = XY + XZ. (A + B + C)(A′ + C ′ )(A′ + B) = (A + B + C)(A′ A′ + A′ C ′ + AB + BC ′ + AC ′ + BC) = (A + B + C)(0 + A′ C ′ + AB + BC ′ + AC ′ + BC) = A′ C ′ (A + B + C) + AB(A + B + C) + BC ′ (A + B + C) = A′ C ′ A + A′ C ′ B + A′ C ′ C + ABA + ABB + ABC + BCA + BCB + BCC = 0 + 0 + 0 + AB + 0 + 0 + 0 + 0 + 0 = AB Step 2: Apply the Idempotent Law: X + X = X. AB = AB Therefore, the simplified Boolean expression is AB. Question 15 Question Simplify the Boolean expression F = A′ B ′ C + A′ BC + AB ′ C ′ + ABC ′ using Boolean algebra laws and theorems. 9 Solution To simplify the given Boolean expression F = A′ B ′ C + A′ BC + AB ′ C ′ + ABC ′ , we will use the laws and theorems of Boolean algebra. Step 1: Apply the absorption law: XY + XZ = X(Y + Z). F = A′ B ′ C + A′ BC + AB ′ C ′ + ABC ′ = A′ B ′ C + A′ BC + A′ B(C ′ + C ′ ) + AB ′ C ′ ′ ′ ′ ′ ′ = A B C + A BC + A B + AB C (Applying absorption law) ′ = A′ B ′ C + A′ BC + A′ B + A′ B ′ C ′ (Reordering terms) Step 2: Apply the consensus theorem: XY + X ′ Z + Y Z = XY + X ′ Z. F = A′ B ′ C + A′ BC + A′ B + A′ B ′ C ′ = A′ B ′ C + A′ BC + A′ B + A′ B ′ C (Applying consensus theorem) Therefore, the simplified Boolean expression is F = A′ B ′ C + A′ BC + A′ B + A′ B ′ C. Question 16 Question Simplify the following expression using Boolean Algebra: F = A′ B ′ C + AB ′ C + ABC ′ + ABC Solution To simplify the expression F = A′ B ′ C + AB ′ C + ABC ′ + ABC, we will use the properties of Boolean Algebra. Step 1: Apply the absorption law XY + XY ′ = X. F = A′ B ′ C + AB ′ C + ABC ′ + ABC = A′ B ′ C + AB ′ C + A(B + BC ′ ) Step 2: Apply the absorption law XY + X ′ Z = XY + X ′ Z + Y Z twice. F = A′ B ′ C + AB ′ C + A(B + BC ′ ) = A′ B ′ C + AB ′ C + AB + ABC ′ + ABC Step 3: Apply the consensus theorem (X + Y )(X + Z)(Y + Z) = (X + Y )(Y + Z). F = A′ B ′ C + AB ′ C + AB + ABC ′ + ABC = A′ B ′ C + AB ′ C + AB + ABC 10 Step 4: Apply the absorption law XY + X ′ Z = XY + X ′ Z + Y Z. F = A′ B ′ C + AB ′ C + AB + ABC = A + AB + AC Step 5: Apply the consensus theorem (X + Y )(X + Z)(Y + Z) = (X + Y )(Y + Z). F = A + AB + AC = A(1 + B) + AC Step 6: Apply the identity law X + X ′ Y = X + Y . F = A + AC = A(1 + C) Thus, the simplified expression for F = A′ B ′ C + AB ′ C + ABC ′ + ABC is F = A(1 + C). Question 17 Question Simplify the following Boolean expression using Boolean algebra rules: F = A′ B + AB ′ + ABC ′ + BC Solution To simplify the Boolean expression F = A′ B + AB ′ + ABC ′ + BC, we will use the rules of Boolean algebra to manipulate the expression. Step 1: Apply the absorption law (X + XY = X) to A′ B + AB ′ . A′ B + AB ′ = A′ B + AB ′ + ABC ′ + BC Step 2: Apply the consensus theorem (X + X ′ Y = X + Y ) to A′ B + ABC ′ . A′ B + AB ′ + ABC ′ = A′ B + BC Step 3: Apply the consensus theorem to A′ B + BC. A′ B + BC = A′ B + A′ C + BC Step 4: Apply the absorption law to A′ B + A′ C. A′ B + A′ C + BC = A′ B + A′ C + BC + BC Step 5: Apply the idempotent law (X + X = X) to BC + BC. A′ B + A′ C + BC + BC = A′ B + A′ C + BC Step 6: The simplified expression is A′ B + A′ C + BC. 11 Question 18 Question Given the Boolean expression (A ∧ B) ∨ (¬A ∧ C) ∨ (B ∧ ¬C), simplify the expression using Boolean algebra laws. Solution To simplify the given Boolean expression, we will use the laws of Boolean algebra, including the commutative law, associative law, distributive law, identity law, complement law, etc. Step 1: Apply the distributive law to the given expression. (A ∧ B) ∨ (¬A ∧ C) ∨ (B ∧ ¬C) = (A ∨ ¬A) ∧ (A ∨ C) ∧ (B ∨ C) Step 2: Apply the complement law to simplify (A ∨ ¬A). (A ∨ ¬A) ∧ (A ∨ C) ∧ (B ∨ C) = 1 ∧ (A ∨ C) ∧ (B ∨ C) Step 3: Apply the identity law to simplify 1 ∧ (A ∨ C). 1 ∧ (A ∨ C) ∧ (B ∨ C) = A ∨ C ∧ (B ∨ C) Step 4: Apply the absorption law to simplify (A ∨ C) ∧ (B ∨ C). A ∨ C ∧ (B ∨ C) = A ∨ C Therefore, the simplified Boolean expression is A ∨ C. Question 19 Question Simplify the following Boolean expression: (A + B + C)(A + B + C)(A + B + C) Solution To simplify the given Boolean expression, we will use the distributive property and Boolean algebra laws. Step 1: Apply the distributive property to expand the expression. (A + B + C)(A + B + C)(A + B + C) = (A + B + C)(A + B + C)(A + B + C) = A(A+B +C)(A+B +C)+B(A+B +C)(A+B +C)+C(A+B +C)(A+B +C) Step 2: Apply the Absorption Law XY + XY = X. = A(A + B + C) + B(A + B + C) + C(A + B + C) =A+B+C Therefore, the simplified form of the given Boolean expression is A + B + C. 12 Question 20 Question Simplify the following Boolean expression: (A + B)(A + B)(A + B). Solution To simplify the given Boolean expression, we will use the Boolean algebra laws, such as the distributive law, the complement law, and the idempotent law. Step 1: Apply the distributive law to expand the expression. (A + B)(A + B)(A + B) = (AA + AB + BA + BB)(A + B) Step 2: Simplify the expression by applying the idempotent law (X + X = X) and the complement law (XX = 0). (0 + AB + BA + B)(A + B) = (AB + BA + B)(A + B) Step 3: Apply the distributive law to expand the expression further. (AB + BA + B)(A + B) = ABA + ABB + BA + BB Step 4: Simplify the expression by applying the idempotent law and the complement law. ABA + 0 + BA + 0 = AB + BA Step 5: Rearrange the terms using the commutative law (XY = Y X). AB + BA = AB + AB Step 6: Apply the idempotent law to simplify the expression. AB + AB = AB Therefore, (A + B)(A + B)(A + B) simplifies to AB. Question 21 Question Simplify the following Boolean expression using algebraic manipulation: F = A′ B + AB ′ + AB 13 Solution Step 1: Apply the commutative property of OR (+) to rearrange the terms. F = A′ B + AB ′ + AB = A′ B + AB + AB ′ = A′ B + AB + BA (Step 1) Step 2: Apply the idempotent law to simplify AB + BA. F = A′ B + AB + BA = A′ B + A(B + B) = A′ B + A(1) = A′ B + A (Step 2) Step 3: Apply the absorption law to simplify A′ B + A. F = A′ B + A = A + A′ B = A(1 + B) =A (Step 3) Therefore, the simplified expression is F = A. Question 22 Question Simplify the following Boolean expression using Boolean algebra laws: (A + B + C)(A′ + B ′ C) + A′ B ′ . Solution To simplify the given Boolean expression, we will use the laws of Boolean algebra to manipulate the expression step by step. Step 1: Apply the distributive law: (A + B + C)(A′ + B ′ C) + A′ B ′ . Step 2: Use the distributive property: AA′ + AB ′ C + BA′ + BB ′ C + CA′ + CB ′ C + A′ B ′ . Step 3: Simplify the terms involving complement pairs: 0 + AB ′ C + 0 + 0 + 0 + 0 + A′ B ′ . Step 4: Further simplify the expression: AB ′ C + A′ B ′ . Step 5: Apply the complement law: AB ′ C + A′ B ′ = AB ′ C + A′ B ′ (C + C ′ ). Step 6: Use the distributive property: AB ′ C + A′ B ′ C + A′ B ′ C ′ . Step 7: Combine terms with C using the consensus theorem: AB ′ C + ′ ′ A B C + A′ B ′ C ′ = AB ′ C + A′ B ′ C ′ . Therefore, the simplified Boolean expression is AB ′ C + A′ B ′ C ′ . 14 Question 23 Question Simplify the following Boolean expression: (A + B)(A + B)(A + B). Solution To simplify the given Boolean expression, we will use the distributive property and De Morgan’s laws. Step 1: Apply the distributive property to expand the expression. (A + B)(A + B)(A + B) = (A + B)((A · B) + (A · B) + (B · A) + (B · B)) = (A + B)(AB + AB + AB + 0) (Expanding) (Complement law: B · B = 0) = (A + B)(AB + AB) Step 2: Apply the distributive property again to simplify further. (A + B)(AB + AB) = A(AB + AB) + B(AB + AB) (Distributive property) = AAB + AAB + BAB + BAB (Expanding) = AB + 0B + AB + 0 (Complement law: A · A = 0 and B · A = 0) = AB + AB = AB Therefore, the simplified form of the given Boolean expression is AB. Question 24 Question Simplify the following Boolean expression: (A + B + C)(A + B + C)(A + B + C). Solution To simplify the given Boolean expression (A + B + C)(A + B + C)(A + B + C), we can expand the expression using the distributive property and then simplify it using the rules of Boolean Algebra. Step 1: Expand the expression by distributing: (A + B + C)(A + B + C)(A + B + C) = A(A+B +C)(A+B +C)+B(A+B +C)(A+B +C)+C(A+B +C)(A+B +C) 15 Step 2: Apply the distributive property to each term: = AAA + AAB + AAC + ABA + ABB + ABC + ACA + ACB + ACC+ +BAA + BAB + BAC + BBA + BBB + BBC + BCA + BCB + BCC+ +CAA + CAB + CAC + CBA + CBB + CBC + CCA + CCB + CCC Step 3: Simplify the expression using the rules of Boolean Algebra. Note that XX = X and XX = 0: = 0+0+0+0+0+0+0+0+0+0+0+0+0+0+0+0+0+0+0+0+0+0+0+0+0+0+0+0 =0 Therefore, the simplified Boolean expression is 0. Question 25 Question Simplify the Boolean expression (A + BC)(AB ′ C + A′ B). Solution To simplify the given Boolean expression, we will use the basic laws of Boolean algebra: commutative law, associative law, distributive law, identity law, complement law, and absorption law. Step 1: Apply the distributive law (A + BC)(AB ′ C + A′ B). = AAB ′ C + AA′ B + ABCB ′ C + ABCB = A′ B + A′ B + ABC + ABC = A′ B + ABC Therefore, (A + BC)(AB ′ C + A′ B) simplifies to A′ B + ABC. Question 26 Question Simplify the Boolean expression F = AB + AC + BC + A′ B ′ C ′ . 16 Solution To simplify the given Boolean expression, we will use the laws of Boolean algebra. Step 1: Apply the absorption law X + XY = X to simplify AB + AC. AB + AC = A(B + C) Step 2: Apply the absorption law again to simplify A(B + C). A(B + C) = A Step 3: Using the identity X ′ X = 0, simplify A′ B ′ C ′ . A′ B ′ C ′ = 0 Step 4: Rewrite the simplified expressions and combine them to get the final simplified expression for F . F = AB + AC + BC + A′ B ′ C ′ = A + BC + 0 = A + BC Therefore, the simplified Boolean expression is F = A + BC. Question 27 Question Simplify the following Boolean expression: (A + B)(A + C) + (A + B). Solution To simplify the given Boolean expression, we will use the properties of Boolean algebra. Step 1: Apply the distributive law: (A + B)(A + C) + (A + B) = A(A + C) + B(A + C) + A + B. Step 2: Apply the distributive law again: A(A + C) + B(A + C) + A + B = AA + AC + BA + BC + A + B. Step 3: Use the idempotent law (AA = A) and zero property (XA = 0 if X is a variable and not complemented): AA + AC + BA + BC + A + B = A + AC + BA + BC + A + B. Step 4: Apply the idempotent law once more: A + AC + BA + BC + A + B = A + BA + BC + B. Step 5: Apply the absorption law (X + XY = X): A + BA + BC + B = A + B(1 + C) = A + B. Therefore, (A + B)(A + C) + (A + B) simplifies to A + B. 17 Question 28 Question Simplify the Boolean expression (A + B)(A + B + C)(A + B + C). Solution Step 1: Apply the distributive property to expand the expression. (A+B)(A+B+C)(A+B+C) = AA+BA+BA+BB+BC+AB+BC+BB+AB+ABC Step 2: Simplify the terms involving complement pairs. 0 + 0 + 0 + 0 + 0 + AB + BC + 0 + AB + ABC Step 3: Simplify the remaining terms. AB + BC + AB + ABC = AB(1 + C) + AB(1 + C) Step 4: Use the distributive property to factor out a common term. AB(1 + C) + AB(1 + C) = AB + AB Step 5: Apply the absorption law X + X ′ Y = X + Y . AB + AB = A + B Therefore, the simplified Boolean expression is A + B. Question 29 Question Simplify the Boolean expression: AB + A′ B + A′ C + BC + B ′ C Solution To simplify the Boolean expression AB + A′ B + A′ C + BC + B ′ C, we can use Boolean algebra laws and properties to simplify the expression step by step. Step 1: Apply the absorption law A + AB = A. AB + A′ B + A′ C + BC + B ′ C = AB + A′ B + A′ C + BC + B ′ C + 0 = AB + A′ B + A′ C + ABC + B ′ C = AB(1 + C) + A′ C(1 + B) = AB + A′ C 18 Step 2: Apply the consensus theorem: AB + A′ C = AB + B ′ C + AC. AB + A′ C = AB + B ′ C + AC = (A + B ′ )(B ′ + C)(A + C) Therefore, the simplified form of the Boolean expression is (A + B ′ )(B ′ + C)(A + C). Question 30 Question Let A, B, and C be three Boolean variables. Simplify the following Boolean expression: (A + B) · (A · C + B · C) Solution To simplify the given Boolean expression, we will use the rules of Boolean algebra such as distribution, absorption, and identity properties. Step 1: Distribute the OR operation over the AND operation (A + B) · (A · C + B · C) = A · (A · C + B · C) + B · (A · C + B · C) Step 2: Apply the Distributive Law =A·A·C +A·B·C +B·A·C +B·B·C Step 3: Simplify the first and last terms using the Absorption Law =A·C +A·B·C +B·A·C +B·C Step 4: Simplify using the Identity Law =A·C +A·B·C +A·C +B·C Step 5: Combine like terms =A·C +A·C +A·B·C +B·C Step 6: Simplify further =A·C +A·B·C +B·C Step 7: Factor out common terms = A · (C + B · C) + B · C Step 8: Apply the Absorption Law =A·1+B·C Step 9: Simplify using the Identity Law =A+B·C Therefore, the simplified form of the given Boolean expression is A + B · C. 19 Question 31 Question Simplify the following Boolean expression: (A′ B + C)(A + B ′ C) + AB + AC Solution To simplify the given Boolean expression, we will first expand the terms using the distributive property, then simplify using Boolean algebra laws. Step 1: Expand the expression (A′ B + C)(A + B ′ C) + AB + AC = A′ B · A + A′ B · B ′ C + C · A + C · B ′ C + AB + AC = A′ B · A + A′ B · B ′ C + C · A + C · B ′ C + AB + AC Step 2: Apply the AND absorption law A′ B · A + A′ B · B ′ C + C · A + C · B ′ C + AB + AC = A′ B + A′ B · C + C · A + C · B ′ C + AB + AC = A′ B + C(A′ + B ′ C) + C · B ′ C + AB + AC Step 3: Apply the distributive law A′ B + C(A′ + B ′ C) + C · B ′ C + AB + AC = A′ B + CA′ + CB ′ C + AB + AC = A′ B + A′ C + B ′ C + AB + AC Step 4: Apply the consensus theorem A′ B + A′ C + B ′ C + AB + AC = A′ B + AB + A′ C + AC + B ′ C = B + A′ C + AC + B ′ C = B + A′ C + C(B + B ′ ) = B + A′ C + C Therefore, the simplified form of the given Boolean expression is B+A′ C +C. Question 32 Question Simplify the following Boolean expression: (A + B) · (A + B). Solution To simplify the given Boolean expression, we will use the distributive property of Boolean algebra. 20 A 0 0 1 1 B 0 1 0 1 A+B 1 0 1 1 (A + B) · (A + B) 0 0 1 1 (A + B) · (A + B) = A · (A + B) + B · (A + B) (Distributive Property) =A·A+A·B+B·A+B·B =A+A·B+B·A+0 (Distributive Property) (Complement Property) = A + A · B + B · A (Identity Property) = A + AB + AB = A + AB = A(1 + B) =A·1 =A (Commutative Property) (Idempotent Law) (Distributive Property) (Complement Property) (Identity Property) Therefore, (A + B) · (A + B) simplifies to A. Question 33 Question Simplify the following Boolean expression: (A + B)(A + B + C). Solution Step 1: Use the distributive property to expand the expression. (A + B)(A + B + C) = A(A + B + C) + B(A + B + C) Step 2: Apply the distributive property again to simplify each term. A(A + B + C) = AA + AB + AC = A + AB + AC B(A + B + C) = BA + BB + BC = AB + BC Step 3: Combine the simplified terms. A + AB + AC + AB + BC = A + AB + AB + AC + BC Step 4: Use the idempotent law (X + X = X) and the absorption law (X + XY = X) to simplify further. A + AB + AB + AC + BC = A + AB + AC 21 = A(1 + B) + AC = A + AC Step 5: Apply absorption once again to get the final simplified expression. A + AC = A(1 + C) = A Therefore, the simplified expression is A. Question 34 Question Simplify the following Boolean expression: (A + BC)(AB + AC). Solution To simplify the given Boolean expression (A + BC)(AB + AC), we will use the distributive property and the fact that X + X = X for any variable X in Boolean algebra. Step 1: Apply the distributive property to expand the expression. (A + BC)(AB + AC) = AAB + AAC + BCAB + BCAC Step 2: Simplify the expression by combining like terms. AAB + AAC + BCAB + BCAC = A + AC + AB(C + C) Step 3: Use the fact that X +X = X in Boolean algebra to simplify further. A + AC + AB(C + C) = A + AC + AB Step 4: Apply the distributive property in reverse to factor out a common term. A + AC + AB = A(1 + C) + AB = A + AB = A(1 + B) = A Therefore, the simplified form of the Boolean expression (A+BC)(AB+AC) is A. Question 35 Question Simplify the following Boolean expression: F = A′ B ′ C + AB ′ C + ABC + ABC ′ 22 Solution To simplify the given Boolean expression F = A′ B ′ C + AB ′ C + ABC + ABC ′ , we will use Boolean algebra laws and rules. 1. Use the absorption law to simplify the expression: F = A′ B ′ C + AB ′ C + ABC + ABC ′ = B ′ C(A′ + A) + BC(A + A′ ) = B ′ C + BC = C(B ′ + B) =C Therefore, the simplified form of the Boolean expression is F = C. 23