Identity check — distributive equivalence used in simplification Evaluate whether the Boolean expressions A + (B · C) and (A + B) · (A + C) are equivalent for all binary inputs.

Introduction / Context:
Recognizing standard identities enables fast simplification and structured design using NAND/NOR implementations. One key identity is the distributive relationship connecting sum with product terms.

Given Data / Assumptions:

• Variables A, B, C are Boolean (0 or 1).
• We compare A + (B · C) with (A + B) · (A + C).

Concept / Approach:
In Boolean algebra, addition (OR) distributes over multiplication (AND) and vice versa in specific forms. The identity A + (B · C) = (A + B) · (A + C) holds for all input combinations and is frequently used to realize logic using only NAND or only NOR gates.

Step-by-Step Solution:
If A = 1, left side: 1 + (B · C) = 1; right side: (1 + B)(1 + C) = 1 · 1 = 1.If A = 0, left side: 0 + (B · C) = B · C; right side: (0 + B)(0 + C) = B · C.Therefore equality holds for both A=1 and A=0, which covers all cases.

Verification / Alternative check:
Construct a short truth-table fragment for A=0 and A=1 and see both sides match for all B, C. Karnaugh-map verification also confirms overlap of minterms.

Why Other Options Are Wrong:
“Not equivalent” contradicts the identity. Conditional equivalence options limit the identity to special cases, which is incorrect because it holds unconditionally.

Common Pitfalls:
Confusing this with DeMorgan’s theorems, or misapplying distribution in the reverse form without parentheses, leading to mistakes.

Final Answer:
Equivalent (Correct)