Question 1

Karnaugh map (K-map) fundamentals: Which statement best describes what a Karnaugh map is and how it relates to the truth table?

Accepted Answer

Introduction / Context: A Karnaugh map (K-map) is a visual method for simplifying Boolean expressions and designing logic with minimal gates. It retains the exact information of a truth table but organizes it spatially so that adjacent cells differ by only one variable (Gray code ordering). This makes it easy to spot groups of 1s (or 0s) and derive simplified sum-of-products or product-of-sums forms. Given Data / Assumptions: We are dealing with combinational logic functions of typically up to 4–6 variables for practical hand simplification. Grouping on a K-map follows powers of two (1, 2, 4, 8, ...). Gray code labeling ensures adjacency reflects a single-bit change. Concept / Approach: The K-map is nothing more than a truth table rearranged in two dimensions to expose adjacency. Every cell corresponds to one minterm (or maxterm) of the original function. By grouping adjacent 1s into largest possible rectangles, we factor out variables that do not change within the group, yielding a minimized Boolean expression that implements with fewer gates or literals. Step-by-Step Solution: Start from the truth table and place each output value into the K-map cell addressed by its input combination (Gray-coded axes).Identify adjacent 1s and form groups of size 2^n (wrap-around adjacency is allowed).Translate each group into a product term (for SOP) by including only variables that remain constant over the group.Combine all product terms to get the simplified expression. Verification / Alternative check: Cross-verify by expanding the simplified expression back into minterms and confirming it yields the same truth table as the original function. Karnaugh and algebraic minimization give identical results when done correctly. Why Other Options Are Wrong: Eliminates the need for NAND/NOR: K-maps do not dictate gate families; they only simplify logic expressions. Variable complements can be eliminated: Complements may remain; K-maps do not inherently remove inversion. Replace Boolean rules: K-maps complement algebra; they do not replace the underlying Boolean laws. Common Pitfalls: Forgetting Gray code ordering, missing wrap-around adjacency, or making non-power-of-two groups. Also, confusing don't-care conditions with zeros can lead to suboptimal or incorrect simplifications. Final Answer: It is simply a rearranged truth table.

Question 2

Digital logic design: The systematic reduction and simplification of logic circuits to equivalent but simpler forms is primarily accomplished by which method?

Accepted Answer

Introduction / Context: In digital electronics, designers routinely simplify logic expressions and gate networks to reduce chip count, power, and propagation delay. The recognized theoretical toolset for this task is Boolean algebra, which provides axioms and theorems to transform logic expressions without changing their functionality. Given Data / Assumptions: We want a general, systematic method for simplification, not a hardware family or a measurement tool. Equivalence of logic functions must be preserved after simplification. Applies to both gate-level networks and algebraic expressions. Concept / Approach: Boolean algebra includes identities such as commutative, associative, distributive laws; DeMorgan’s theorems; idempotent, absorption, and consensus theorems. Using these, complex sum-of-products (SOP) or product-of-sums (POS) expressions can be reduced to minimal or near-minimal forms. Karnaugh maps and Quine–McCluskey are visual/algorithmic companions, but the underlying correctness relies on Boolean algebraic equivalence. Step-by-Step Solution: Identify the target function in algebraic form (e.g., Y = A B + A B̄ C).Apply Boolean identities strategically (e.g., absorption: X + X Y = X).Consolidate terms to remove redundancies and reduce literal count.Map the simplified expression back to gates with fewer inputs or stages. Verification / Alternative check: Construct a truth table for original and simplified expressions; row-by-row equality confirms correctness. Alternatively, test-vector simulation or logic-equivalence checking tools validate the transformation in practice. Why Other Options Are Wrong: Symbolic reduction: vague label; not a formal methodology by itself.TTL logic: a hardware family, not a simplification method.Using a truth table: verifies equivalence but does not directly provide systematic reduction steps. Common Pitfalls: Over-reliance on intuitive gate cancellation without formal rules can introduce errors. Always back reductions with Boolean identities and, when possible, confirm with a truth table. Final Answer: using Boolean algebra

Question 3

Boolean identities: Which option correctly expresses the commutative law of multiplication for logical variables?

Accepted Answer

Introduction / Context: Commutativity is a foundational property in Boolean algebra and ordinary algebra. It states that the order of operands in a binary operation does not affect the result. Recognizing commutative properties helps reorder and simplify expressions without changing their logical function. Given Data / Assumptions: Boolean variables A, B, C represent logic 0/1. Operations: + is OR, • is AND. We seek the law for multiplication (AND). Concept / Approach: The commutative law for AND says A • B = B • A. Similarly, for OR: A + B = B + A. These laws allow rearrangement of terms, which often enables grouping or factoring that simplifies the expression further. Step-by-Step Solution: Identify operation: multiplication corresponds to logical AND.Apply commutativity: order of A and B is interchangeable for AND.Confirm by truth table: evaluate both sides for all combinations of A and B; outputs match.Use in simplification: reordering can expose common factors or enable distributive reductions. Verification / Alternative check: Karnaugh maps are unaffected by permutation of variable order; this reflects underlying commutativity. Algebraic proofs also follow directly from Boolean postulates. Why Other Options Are Wrong: A + B = B + A: commutative law of addition (OR), not multiplication.A • B = B + A: mixes AND and OR; not an identity.A • (B • C) = (A • B) • C: this is the associative law, not the commutative law. Common Pitfalls: Confusing commutative with associative properties; the former swaps operands, the latter changes grouping. Keep symbols consistent to avoid misapplication. Final Answer: A • B = B • A

Question 4

Identifying canonical forms: Which Boolean output expression is explicitly in product-of-sums (POS) form, indicating an implementation as an AND of OR terms?

Accepted Answer

Introduction / Context: Digital designers often express logic functions in canonical or standard forms to map them to gate-level structures. Two widely used forms are sum-of-products (SOP) and product-of-sums (POS). Recognizing POS is key when you plan to realize a function using OR gates feeding an AND gate (or NAND–NAND with DeMorgan equivalents). Given Data / Assumptions: SOP means OR of AND terms (e.g., AB + ĀC). POS means AND of OR terms (e.g., (A + B)(Ā + C)). Variables may appear complemented (with a bar) or uncomplemented. Concept / Approach: To classify an expression, look at the outermost operator. If the outermost operator is AND (a product) and its factors are sums (OR groupings inside parentheses), the expression is in POS form. Conversely, if the outermost operator is OR and the terms inside are products (concatenations), the expression is SOP. Step-by-Step Solution: Inspect option a: Y = (A + B)(Ā + C). Outermost operator is multiplication; each factor is an OR term → POS.Inspect option b: Y = AB + ĀC. Outermost operator is addition; terms are products → SOP, not POS.Inspect option c: Y = (A B)(C + D) mixes a product term (A B) with a sum; not canonical POS.Inspect option d: Y = A + B + C is a simple sum, not a product of sums. Verification / Alternative check: Gate mapping: POS maps to OR gates feeding a final AND. Option a naturally synthesizes as two OR gates whose outputs are ANDed, confirming the POS structure. Why Other Options Are Wrong: AB + ĀC: clear SOP.(A B)(C + D): first factor is not a sum; not POS.A + B + C: no AND of sums. Common Pitfalls: Judging by appearance rather than the outermost operator hierarchy. Always check whether sums are being multiplied (POS) or products are being added (SOP). Final Answer: Y = (A + B)(Ā + C)

Question 5

Form classification in logic design: Which of the following Boolean expressions is in sum-of-products (SOP) form, meaning an OR of AND terms?

Accepted Answer

Introduction / Context: Recognizing SOP form is useful because it directly maps to AND gates feeding an OR gate, or to NAND–NAND structures via DeMorgan's transformations. Many minimization techniques such as Karnaugh maps naturally produce SOP results for straightforward implementation. Given Data / Assumptions: SOP = sum (OR) of product (AND) terms. POS = product (AND) of sum (OR) terms. Concatenation implies AND, + implies OR. Concept / Approach: To decide whether an expression is SOP, check that the outermost operator is addition (OR) and that each term being added is a product of literals (possibly with complements). If so, the expression is SOP. If the outermost operator is multiplication of sums, it is POS instead. Step-by-Step Solution: Option a: (A + B)(C + D) is an AND of two sums → POS, not SOP.Option b: AB + CD is an OR of two product terms → SOP.Option c: (A + B + C) is just one sum term → neither SOP nor POS canonical form.Option d: A(B + C) is mixed and not expanded; in expanded form it becomes AB + AC (which would be SOP), but as written it is not in SOP form. Verification / Alternative check: Expanding option d gives AB + AC, confirming that SOP indeed looks like an OR of products, as in option b. Why Other Options Are Wrong: (A + B)(C + D): POS structure.(A + B + C): only a sum, no product terms.A(B + C): not in finalized SOP until distributed. Common Pitfalls: Confusing unexpanded factored forms with SOP. Always ensure each added term is a product, not a sum or mixed factor, to qualify as SOP. Final Answer: Y = AB + CD

Question 6

Interpreting DeMorgan’s theorems: One statement of DeMorgan’s theorems is (A • B)' = A' + B'. Simply stated, which pair of gates are logically equivalent under this interpretation?

Accepted Answer

Introduction / Context: DeMorgan’s theorems connect AND/OR operations with inversion in Boolean algebra and gate-level design. They enable trading bubbles (inversions) between inputs and outputs, greatly simplifying gate implementation with universal NAND/NOR families. Given Data / Assumptions: DeMorgan form: (A • B)' = A' + B' and (A + B)' = A' • B'. Bubble notation: a bubble denotes inversion on a pin. Equivalence concerns logical function, not physical symbol shape. Concept / Approach: (A • B)' represents a NAND function. The right-hand side A' + B' is an OR gate whose inputs are inverted (bubbles on inputs). Therefore, a NAND gate is functionally identical to an OR gate with inverted inputs. Similarly, a NOR gate equals an AND gate with inverted inputs by (A + B)' = A' • B'. Step-by-Step Solution: Start with NAND definition: Y = (A • B)'.Apply DeMorgan: Y = A' + B'.Gate mapping: an OR gate whose two inputs each have bubbles implements A' + B'.Therefore: NAND ≡ OR-with-bubbled-inputs. Verification / Alternative check: Truth tables for NAND and for OR with both inputs inverted match exactly for all input combinations, confirming equivalence. Why Other Options Are Wrong: NOR ≡ AND with bubbled inputs, not AND with bubbled output.NOR vs NAND with bubbled output: unrelated equivalence.NAND vs OR with bubbled output: bubbled output OR is NOR, not NAND. Common Pitfalls: Confusing input bubbles with output bubbles. DeMorgan moves inversions across a gate only if the gate type flips AND↔OR and all pins crossing the inversion are inverted. Final Answer: a NAND gate and an OR gate with bubbled inputs

Question 7

Boolean distributive law: Which example correctly demonstrates distribution of AND over OR in Boolean algebra?

Accepted Answer

Introduction / Context: The distributive law allows factoring and expansion of Boolean expressions, just as in ordinary algebra but with logical operations. It is essential for moving between factored forms and canonical SOP forms in synthesis and optimization. Given Data / Assumptions: • denotes AND, + denotes OR. Variables A, B, C are Boolean. We seek distribution of AND over OR. Concept / Approach: In Boolean algebra, there are two distributive laws: AND over OR, and OR over AND. The one asked here is: A • (B + C) = (A • B) + (A • C). This identity justifies expanding a mixed expression into SOP form or, conversely, factoring SOP terms into a compact factored form. Step-by-Step Solution: Consider the left side: A is 1 and (B + C) is 1 if either B or C is 1.Right side: (A • B) + (A • C) is 1 if A and B are 1, or A and C are 1.Enumerate all eight input combinations; both sides agree for each case.Therefore, the equality holds and demonstrates proper distribution. Verification / Alternative check: A Karnaugh map with A as a common factor shows that minterms covered by A(B + C) equal the union of minterms for AB and AC, confirming the distributive relationship. Why Other Options Are Wrong: A • (B • C) = (A • B) + C: mixes operations incorrectly; not a valid identity.A + (B + C) = (A • B) + (A • C): attempts to distribute OR into AND; incorrect form.(A + B) + C = A + (B + C): this is the associative law of OR, not distributive. Common Pitfalls: Swapping AND/OR in distribution or forgetting that Boolean algebra also supports OR distributing over AND: A + BC = (A + B)(A + C). Final Answer: A • (B + C) = (A • B) + (A • C)

Question 8

Gate equivalence by bubble movement: The observation that a bubbled-input OR gate performs the same logic function as a bubbled-output AND gate is known as what?

Accepted Answer

Introduction / Context: Bubble notation captures logical inversion on gate inputs and outputs. DeMorgan’s theorems formalize how inversions move across AND/OR gates while swapping the gate type, a powerful technique for designing with NAND/NOR-only libraries and for reading schematics quickly. Given Data / Assumptions: Bubbled input: inverted input variable to a gate. Bubbled output: inverted output of a gate. Goal: name the principle equating bubbled-input OR to bubbled-output AND. Concept / Approach: DeMorgan’s second theorem states (A + B)' = A' • B'. Interpreted in gate form, an OR with both inputs inverted (bubbled-input OR) yields the same function as an AND with an inverted output (bubbled-output AND), i.e., NOR ≡ AND-with-inverted-inputs and equally NAND ≡ OR-with-inverted-inputs depending on the specific form used. Step-by-Step Solution: Start with Y = (A + B)'.Apply DeMorgan: Y = A' • B'.Map to gates: left side is an OR with output bubble (NOR) or equivalently an OR with bubbled inputs feeding an AND.Thus, a bubbled-input OR and a bubbled-output AND implement the same logic. Verification / Alternative check: Truth tables or logic-simulation confirm functional identity. Schematic re-drawing with bubble-pushing gives identical simplified forms. Why Other Options Are Wrong: Karnaugh map: a minimization tool, not a law.Commutative / associative laws: reorder or regroup operations; they do not involve inversion movement. Common Pitfalls: Confusing input versus output bubbles. DeMorgan requires flipping the gate type while inverting all lines that cross the inversion boundary. Final Answer: DeMorgan's second theorem

Question 9

Logical equivalence of gates: Which implementation produces the same Boolean function as a NOR gate?

Accepted Answer

Introduction / Context: Universal gate design often relies on recognizing when combinations of simple gates emulate others. Understanding these equivalences allows flexible implementation with limited gate libraries and aids schematic interpretation. Given Data / Assumptions: NOR is defined as Y = (A + B)'. INVERTER provides logical complementation of its input. We compare cascaded gates to a single NOR gate. Concept / Approach: By definition, a NOR gate outputs the inversion of an OR function. Therefore, an OR gate whose output is fed into an inverter produces exactly the same truth table as a NOR gate. This follows directly from the composition Y = NOT(OR(A, B)). Step-by-Step Solution: Start with OR output: Z = A + B.Invert Z: Y = Z' = (A + B)'.Recognize equality: Y equals the NOR of A and B.Hence, OR followed by INVERTER is functionally identical to NOR. Verification / Alternative check: Truth table enumeration shows identical outputs for all four input combinations. Logic simulators will show gate-level equivalence after optimization. Why Other Options Are Wrong: NAND then INVERTER: equals AND (since NOT(NAND) = AND), not NOR.AND then INVERTER: equals NAND, not NOR.NOR then INVERTER: equals OR, the opposite of NOR. Common Pitfalls: Confusing the order of gates or assuming inverter placement does not matter; inversion placement is critical for function. Final Answer: OR gate immediately followed by an INVERTER

Question 10

Meaning of the commutative laws: What do the commutative laws of addition (OR) and multiplication (AND) indicate about the order of operands in Boolean expressions?

Accepted Answer

Introduction / Context: Commutative laws are core algebraic properties that carry over to Boolean algebra. They empower designers to rearrange terms without changing the logic function, enabling cleaner factoring, easier Karnaugh mapping, and standardized gate ordering in schematics. Given Data / Assumptions: Commutative law for OR: A + B = B + A. Commutative law for AND: A • B = B • A. Variables represent binary logic values. Concept / Approach: Order independence simplifies manipulation: whether A precedes B or vice versa, the value of the expression is unchanged. This is distinct from associative laws (which change grouping) and distributive laws (which change structure via factoring or expansion). Step-by-Step Solution: Identify operation type (OR or AND) and the operands.Swap operand order and observe unchanged truth table outputs.Apply this property to reorganize terms to reveal common factors or to line up literals for minimization.Use in practice during algebraic reductions and when labeling gate inputs; hardware is indifferent to the labeling order. Verification / Alternative check: Truth tables for A op B versus B op A (op ∈ {+, •}) are identical for all four input pairs. Tool-based equivalence checkers confirm commutativity automatically during logic synthesis. Why Other Options Are Wrong: Grouping any way we want: that is associativity, not commutativity.Term-by-term multiplication: describes distribution/expansion, not commutativity.Factoring requirement statement: unrelated and misleading. Common Pitfalls: Conflating commutative and associative properties; reordering and regrouping are different operations with separate laws. Final Answer: the way we OR or AND two variables is unimportant because the result is the same

Question 11

Associative law for logical OR — evaluate the statement: “A + (B + C) = (A + B) + C” for Boolean variables A, B, C.

Accepted Answer

Introduction / Context: Boolean algebra mirrors several familiar algebraic laws but with logical operators. The associative law establishes that grouping (parentheses) does not change the outcome for repeated OR or repeated AND operations. This property is fundamental when simplifying logic expressions and designing gate networks. Given Data / Assumptions: A, B, C are Boolean variables (0 or 1). Operator “+” denotes logical OR. No timing hazards or non-ideal hardware effects are considered—pure logic algebra. Concept / Approach: Associativity of OR states that A + (B + C) = (A + B) + C = A + B + C. Because OR is associative, the order of evaluation (grouping) does not alter the truth table. This enables regrouping terms freely to match gate fan-in or to combine like terms during simplification without affecting functionality. Step-by-Step Solution: Construct a truth table for A, B, C covering 8 combinations.Compute B + C, then A + (B + C).Compute A + B, then (A + B) + C.Verify equality holds for all rows → expressions are identical. Verification / Alternative check: Use set theory analogy: OR corresponds to union; union is associative: A ∪ (B ∪ C) = (A ∪ B) ∪ C. Alternatively, apply Boolean algebra axioms where associativity is a postulate. Why Other Options Are Wrong: “Incorrect”: contradicts the associative axiom of OR.“Only true when A = 1” or “mutually exclusive”: associativity is unconditional for Boolean variables.“Arithmetic only”: Boolean OR is also associative by definition. Common Pitfalls: Confusing associativity with commutativity or distributivity; all three are valid but serve different rearrangements. Another pitfall is mixing OR with XOR; XOR is also associative but has different algebraic properties. Final Answer: Correct

Question 12

Recovery (clarified stem): Absorption law equivalence — are the Boolean expressions “A + A'B” and “A + B” logically equivalent for all possible values of A and B?

Accepted Answer

Introduction / Context: Incomplete original wording has been repaired using the Recovery-First Policy. A classic identity in Boolean algebra is the absorption law, which enables significant simplification of logic by removing redundant terms. One common form is A + A'B = A + B, which this question asks you to validate. Given Data / Assumptions: A and B are Boolean variables taking values 0 or 1. Operator “+” denotes OR; juxtaposition or “·” denotes AND; prime (′) denotes NOT. We test equivalence across all input combinations. Concept / Approach: Using Boolean algebra: A + A'B = (A + A') (A + B) by distributive expansion. Since A + A' = 1 (complementarity), the expression reduces to 1 · (A + B) = A + B, proving equivalence. Intuitively, if A is 1, both sides are 1 regardless of B; if A is 0, the left side reduces to A'B = B, and the right side is 0 + B = B — again identical. Step-by-Step Solution: Start with A + A'B.Apply distributive law: X + YZ = (X + Y)(X + Z) with X=A, Y=A', Z=B.Compute (A + A') = 1 via complementarity.Reduce: 1 * (A + B) = A + B. Equivalence shown. Verification / Alternative check: Create a 4-row truth table for (A, B) ∈ {(0,0), (0,1), (1,0), (1,1)}. Evaluate both expressions; results match in all rows, confirming equivalence. Why Other Options Are Wrong: “Not equivalent”: contradicts algebraic proof and truth table.Conditional statements (only when A=0, B=1, or A= B′): unnecessary constraints; the identity holds universally. Common Pitfalls: Misapplying distributive law or misreading A'B as AB; forgetting that A + A' = 1 and A · A' = 0 are fundamental complementarity relations. Final Answer: Equivalent (Correct)

Question 13

Pushing negations using DeMorgan — applying DeMorgan’s theorems to expressions with overbars (inversions) converts double and grouped inversions into equivalent forms with only single negations on single variables. Is this statement accurate?

Accepted Answer

Introduction / Context: DeMorgan’s theorems are cornerstone identities in Boolean algebra used to manipulate logic expressions, especially when moving between NAND/NOR and AND/OR implementations. They “push” inversions inward so that negations apply directly to individual variables rather than to grouped sums or products. Given Data / Assumptions: Operators: + for OR, · for AND, overbar/prime for NOT. Typical patterns: (X + Y)′ = X′ · Y′ and (X · Y)′ = X′ + Y′. Double inversions can be removed since (Z)′′ = Z. Concept / Approach: To eliminate group-level overbars, apply DeMorgan to replace a complemented sum with a product of complements or a complemented product with a sum of complements. After iterative application, any expression with complex nested overbars is transformed into an equivalent form where only variables carry the negation operator. This is especially useful when designing with NAND or NOR gates, since they naturally implement complemented sums/products. Step-by-Step Solution: Start from an expression like F = (A + BC)′.Apply DeMorgan: F = A′ · (BC)′.Apply again: (BC)′ = B′ + C′.Result: F = A′ · (B′ + C′) — only single-variable complements remain. Verification / Alternative check: Build truth tables before and after transformation to confirm equality for all input combinations. Alternatively, implement both forms with gates and simulate; outputs match identically. Why Other Options Are Wrong: “Inaccurate”: contradicts theorems themselves.“Only for ≥3 variables” or “only XOR”: DeMorgan applies to sums/products of any number of variables; XOR is unrelated to DeMorgan’s specific forms.“Arithmetic only”: DeMorgan is a Boolean-logic identity, not arithmetic. Common Pitfalls: Forgetting to flip the operator (OR↔AND) when moving the bar; neglecting to complement every term inside the group; missing that double negation cancels. Final Answer: Accurate (Correct)

Question 14

Commutative law for OR and AND — according to the commutative property, the order of variables does not affect the result of A + B or A · B. Is this universally valid for Boolean variables?

Accepted Answer

Introduction / Context: Commutativity is a fundamental property in Boolean algebra that simplifies expression manipulation and gate realization. It states that swapping operands in OR or AND does not change the outcome, allowing flexible wiring and algebraic rearrangement without altering logic functionality. Given Data / Assumptions: A and B are Boolean variables (0 or 1). Operators: + for OR, · for AND. We assume ideal Boolean behavior independent of timing. Concept / Approach: For all combinations of A and B, the truth tables of OR and AND are symmetric under operand interchange. That is, A + B equals B + A and A · B equals B · A in every input case. This symmetry enables hardware optimization since gate inputs are interchangeable, and it supports algebraic simplifications in logic design and Karnaugh mapping. Step-by-Step Solution: List the four input cases: (0,0), (0,1), (1,0), (1,1).Evaluate A + B and B + A for each case; values match identically.Evaluate A · B and B · A likewise; values match identically.Therefore, the commutative property holds universally for OR and AND. Verification / Alternative check: Use set theory analogy: union and intersection of sets are commutative (A ∪ B = B ∪ A; A ∩ B = B ∩ A). This mirrors Boolean OR/AND. Why Other Options Are Wrong: “Not valid” or conditional variants contradict the truth tables.Limiting to OR only ignores that AND is equally commutative.“Arithmetic only” confuses Boolean and numeric addition; both enjoy commutativity, but the statement here is about logic. Common Pitfalls: Confusing commutativity with associativity or distributivity; all are valid but represent different transformation freedoms. Final Answer: Universally valid (Correct)

Question 15

Boolean forms refresher (SOP vs POS): In digital logic simplification, a sum-of-products (SOP) Boolean expression is formed by ORing together two or more product terms (each product term is an AND of literals). The statement that SOP “describes the …

Accepted Answer

Introduction / Context: Sum-of-products (SOP) and product-of-sums (POS) are the two canonical ways to write Boolean expressions used for logic design and simplification. Many exam questions test whether you can distinguish SOP from POS because mixing them up leads to wrong gate-level implementations (AND–OR vs OR–AND structures). The prompt claims SOP is “the ANDing of two or more OR functions.” We will evaluate this definition precisely. Given Data / Assumptions: SOP means “sum” (OR) of “products” (ANDed literals). A product term is an AND of one or more literals (e.g., A·B·C’). POS means “product” (AND) of “sums” (ORed literals). Literals may be complemented or uncomplemented without changing the form classification. Concept / Approach: By definition, SOP is built by ORing several product terms: Y = P1 + P2 + P3, where each Pi is an AND of literals. Conversely, POS is built by ANDing several sum terms: Y = S1 · S2 · S3, where each Si is an OR of literals. Thus the statement in the stem swaps these: it describes POS, not SOP. Correct identification ensures you choose the right two-level gate structure (AND–OR for SOP, OR–AND for POS). Step-by-Step Solution: State the definition: SOP = OR of product terms, POS = AND of sum terms.Map the stem: “ANDing of OR functions” ⇒ product (AND) of sums (OR) ⇒ POS.Conclude the stem mislabels POS as SOP ⇢ the claim is incorrect. Verification / Alternative check: Write a small example. SOP example: Y = A·B + A’·C. POS example: Y = (A + B)·(A’ + C). Implementations correspond to AND–OR (SOP) vs OR–AND (POS) at the gate level—another confirmation of the definitions. Why Other Options Are Wrong: “Correct” directly contradicts the standard definition. The other choices introduce conditions (single variable, De Morgan, literal polarity) that do not change how SOP/POS are defined. Common Pitfalls: Confusing the spoken order: “sum of products” means OR of ANDs, while “product of sums” means AND of ORs. Also, believing complements change the form—they do not. Final Answer: Incorrect

Question 16

Arithmetic property check: Is subtraction a commutative operation in arithmetic and digital design (i.e., does A − B always equal B − A)?

Accepted Answer

Introduction / Context: Commutativity is a core algebraic property. Addition and multiplication are commutative over integers (A + B = B + A; A * B = B * A). However, subtraction and division generally are not. In digital systems, misunderstanding this can cause incorrect ALU design, bit-slice microarchitecture errors, or wrong simplifications in fixed-point pipelines. The prompt asks whether subtraction is commutative, which we will evaluate precisely with examples and reasoning. Given Data / Assumptions: Standard integer arithmetic (the argument extends to reals and fixed-point). No special modulo constraints unless explicitly declared. Binary subtractors follow the same algebraic property as decimal subtraction. Concept / Approach: Operation op is commutative if A op B = B op A for all A, B in a set. For subtraction, test with a counterexample: 7 − 3 = 4, but 3 − 7 = −4, which are not equal. Therefore, subtraction fails the commutativity test. In digital logic, two’s-complement subtraction A − B = A + (two’s-complement of B) preserves noncommutativity: flipping operands changes the result (and the sign) except in special cases. Step-by-Step Solution: Define commutativity formally: A − B must equal B − A for all A, B to be commutative.Provide a counterexample: pick A = 7, B = 3 ⇒ 7 − 3 = 4, 3 − 7 = −4.Since 4 ≠ −4, subtraction is not commutative.Conclude that the claim is incorrect. Verification / Alternative check: Consider symbolic relation: A − B = −(B − A). Equality holds only when A = B (both sides 0). This reinforces the general noncommutativity claim. Why Other Options Are Wrong: “Correct” contradicts basic arithmetic. The other conditions (nonnegative, modulo-2) do not fix noncommutativity; in modulo-2 arithmetic subtraction equals addition (XOR), but then the operator is effectively addition, not subtraction. Common Pitfalls: Conflating subtraction with adding a negative; forgetting that operand order matters in difference operations; misapplying properties of addition to subtraction. Final Answer: Incorrect

Question 17

Boolean algebra duality principle (repair applied): Which statement correctly expresses the duality theorem for Boolean expressions?

Accepted Answer

Introduction / Context: The original stem was incomplete. Applying the Recovery-First Policy, we restate the concept as a complete, teachable item. The duality theorem in Boolean algebra is a cornerstone identity used to derive valid expressions and implement alternative but equivalent logic forms. Mastering duality improves your ability to transform AND–OR circuits into OR–AND counterparts and to check identities succinctly. Given Data / Assumptions: Standard Boolean algebra with operators + (OR) and · (AND). Constants 1 (TRUE) and 0 (FALSE). Variables and their complements (A, A’, etc.) remain as written unless explicitly complemented. Concept / Approach: Duality states that if an algebraic identity is valid, its dual is also valid. To form the dual of any Boolean expression: interchange + and · everywhere, and interchange 1 and 0 everywhere. You do not alter variables or complements. This rule mirrors dual operator roles in the Boolean lattice and ensures that basic laws (e.g., A + 0 = A) have valid duals (A · 1 = A). Step-by-Step Solution: Start with an expression, e.g., X = A + (B · 1).Apply duality: replace + with · and 1 with 0 ⇒ X_d = A · (B + 0).Simplify using identities: B + 0 = B ⇒ X_d = A · B.Confirm both original and dual are valid Boolean expressions linked by the theorem. Verification / Alternative check: Check a known law and its dual. Identity: A + 0 = A. Dual via swaps ⇒ A · 1 = A. Both hold universally, verifying the method. Why Other Options Are Wrong: Swapping variables with complements changes the function, not its dual. De Morgan conversions are related transformations but not the definition of duality. Duality is general and applies to any Boolean expression, not only minterms. Common Pitfalls: Accidentally complementing variables when forming the dual; forgetting to swap constants; mixing duality with De Morgan’s complement rules (which involve negation), whereas duality does not introduce complements. Final Answer: Obtain the dual by interchanging + with · and 1 with 0, leaving variables and complements unchanged

Question 18

Involution (double inversion) rule in Boolean algebra: Does inverting a Boolean variable twice return it to its original value (i.e., (A’’) = A)?

Accepted Answer

Introduction / Context: The involution law (double inversion) is one of the simplest and most used identities in Boolean algebra and logic circuit design. It states that if you complement a Boolean variable twice, you get back the original variable. This is fundamental for simplifying expressions and for understanding how active-low signals propagate through cascaded inverters or NAND/NOR structures used as inverters. Given Data / Assumptions: Variables take values in {0, 1}. Complement operator (’ or overbar) flips 0 to 1 and 1 to 0. Ideal, time-independent Boolean evaluation (no timing hazards). Concept / Approach: If A ∈ {0, 1}, then A’ = 1 − A, and applying complement again yields (A’)’ = 1 − (1 − A) = A. Hence (A’’) = A. At the gate level, two inverters in series reproduce the original logic level, a useful trick for buffering or delay but not for changing logic function. This law is frequently used in algebraic reductions and in transforming complements through expressions using De Morgan’s theorems. Step-by-Step Solution: Start with A ∈ {0, 1}.Compute first inversion: A’ = 1 − A.Compute second inversion: (A’)’ = 1 − (1 − A) = A.Therefore, double inversion returns the original variable. Verification / Alternative check: Truth table check: if A = 0 ⇒ A’ = 1 ⇒ (A’)’ = 0; if A = 1 ⇒ A’ = 0 ⇒ (A’)’ = 1. In both cases, (A’’) = A. Why Other Options Are Wrong: It does not depend on canonical forms, input being 1, or propagation delay (the identity is algebraic; delays are implementation concerns). Common Pitfalls: Confusing the double negation identity with removing pairs of inversions incorrectly when parentheses or operator precedence matter; always track complements carefully across grouped expressions. Final Answer: Correct

Question 19

Operator symbols in Boolean algebra: Is Boolean multiplication denoted by the OR symbol “+” (as in A + B), or is that statement incorrect?

Accepted Answer

Introduction / Context: In Boolean algebra, precise operator symbols prevent design mistakes. By convention, “+” denotes OR, the dot “·” (or adjacency) denotes AND (multiplication), and the prime or overbar denotes NOT. Mislabeling operators leads to wrong logic equations and incorrect gate-level implementations. The prompt asks whether the OR symbol “+” represents Boolean multiplication, which we will evaluate. Given Data / Assumptions: Standard Boolean operator notation: + for OR, · for AND, ’ for NOT. Some texts omit “·” and write AB to mean A AND B. We are not using arithmetic addition; Boolean OR is distinct. Concept / Approach: Boolean multiplication corresponds to the AND operation. Its outputs mirror intersection logic: the result is 1 only when all inputs are 1. By contrast, the OR operation (“+”) yields 1 when any input is 1. Therefore, equating “+” with multiplication is incorrect. Correct mapping is essential when translating algebra to gates: AND gates implement multiplication (·), OR gates implement +, and inverters implement complements. Step-by-Step Solution: State symbol map: AND ⇒ ·, OR ⇒ +, NOT ⇒ ’.Check the claim: “+” as multiplication would imply OR equals AND, which is false.Conclude that Boolean multiplication is not “+” but “·” (or adjacency). Verification / Alternative check: Truth tables distinguish AND and OR: AND outputs 1 only for (1,1); OR outputs 1 for (1,0), (0,1), and (1,1). Symbols reflect these distinct behaviors. Why Other Options Are Wrong: Operator meaning does not depend on minterms, K-map layout, or complementing variables. The symbols are conventions used uniformly across Boolean algebra. Common Pitfalls: Mixing arithmetic “+” with Boolean “+” and writing ambiguous expressions without dots or parentheses; always clarify operator precedence and use parentheses liberally. Final Answer: Incorrect

Question 20

Three-input AND function check: The statement claims the Boolean expression for a three-input AND gate is Y = A · B + C. Is that correct for a gate whose output is 1 only when A, B, and C are all 1?

Accepted Answer

Introduction / Context: Recognizing canonical expressions for basic gates is essential. A three-input AND gate outputs 1 only when all three inputs are 1. The stem proposes Y = A · B + C, which mixes an AND with an OR. We must determine whether this equals the strict three-input AND function or something else. Misidentification leads to incorrect truth tables and wrong hardware implementations. Given Data / Assumptions: Desired function: Y = 1 only when A = B = C = 1. Proposed expression: Y = A · B + C. Standard operator meanings: · is AND, + is OR. Concept / Approach: The true three-input AND is Y = A · B · C. The proposed Y = A · B + C equals 1 when either (A AND B) is 1 or when C is 1. That expression produces many minterms beyond A = B = C = 1, for example A = 0, B = 0, C = 1 gives Y = 1. Therefore, the proposed expression does not implement the strict three-input AND function and the claim is incorrect. Step-by-Step Solution: Write target truth: only the minterm ABC should assert Y.Evaluate proposed expression for A=0,B=0,C=1 ⇒ Y = 0 + 1 = 1 (incorrect behavior).Confirm correct expression: Y = A · B · C.Conclude the stem statement is false. Verification / Alternative check: Construct K-maps: the three-input AND corresponds to a single cell (A=1,B=1,C=1), whereas A · B + C covers an entire row/column group, clearly not the same region. Why Other Options Are Wrong: “Correct” contradicts direct evaluation. Conditional statements about De Morgan or active-low inputs do not transform A · B + C into A · B · C without additional complements and gates. Common Pitfalls: Assuming that adding inputs with “+” always tightens a condition—it actually loosens it (more chances to be 1). For strict AND across three signals, use a triple product term. Final Answer: Incorrect

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