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

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)