Converting W(X + YZ) to standard SOP form By applying which Boolean algebra law can the expression W(X + YZ) be expanded into a sum-of-products (SOP) form?
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Aassociative law
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Bcommutative law
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Cdistributive law
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Dnone of the above
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Eidempotent law
Answer
Correct Answer: distributive law
Explanation
Introduction / Context:In digital logic design, we frequently convert mixed expressions into a standard form for implementation with basic gates or programmable logic. The two most common canonical forms are sum-of-products (SOP) and product-of-sums (POS). This question asks which Boolean law lets us expand W(X + YZ) into SOP, where each term is a product (AND) of literals and the overall expression is a sum (OR) of those products.
Given Data / Assumptions:
- The expression is W(X + YZ).
- Variables W, X, Y, Z are Boolean (take values 0 or 1).
- Goal: express as SOP (sum of product terms).
Concept / Approach:
The key law is Boolean distribution, which lets a factor multiply (AND) into a sum (OR). The distributive law appears in two useful forms:
- A(B + C) = AB + AC
- A + BC = (A + B)(A + C) (the dual, for POS work)
Here, we need the first form to push W across the parentheses so that the result is a sum of product terms suitable for SOP implementation.
Step-by-Step Solution:
Start with: F = W(X + YZ)Apply distributive law: W(X + YZ) = WX + W(YZ)Associate the second product (implicit): W(YZ) = (WY)Z = WYZTherefore, F = WX + WYZ, which is clearly SOP (sum of the products WX and WYZ).Verification / Alternative check:
You can construct a quick truth table for W, X, Y, Z and verify that F = 1 exactly when either WX = 1 or WYZ = 1. A Karnaugh map grouping would also show the same on-cells, confirming that distribution produced an equivalent SOP expression without changing functionality.
Why Other Options Are Wrong:
- Associative law changes grouping only, such as (A + B) + C = A + (B + C); it does not expand across a sum.
- Commutative law swaps order, such as AB = BA or A + B = B + A; it does not expand products over sums.
- Idempotent law (A + A = A, AA = A) simplifies duplicate literals, not distribute factors.
- “None of the above” is incorrect because distribution is exactly what is used.
Common Pitfalls:
- Stopping at WX + W(YZ) and forgetting that W(YZ) is already a valid product term WYZ; no further expansion is needed for SOP.
- Accidentally turning W + (YZ) into (W + Y)(W + Z), which is a POS transformation, not SOP.
Final Answer:
distributive law