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?

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:

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:

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)