Difficulty: Easy
Correct Answer: AB + A'B'
Explanation:
Introduction / Context: Boolean simplification aims to reduce literal count and gate depth while preserving functionality. Some expressions are already minimal in a two-level implementation (sum-of-products or product-of-sums). Recognizing them avoids unnecessary algebra and helps in mapping to standard cells efficiently (e.g., XOR/XNOR implementations).Given Data / Assumptions:
Concept / Approach: Some identities: A + AB = A (absorption). A + A'B = A + B (consensus/absorption). AB + AB' = A(B + B') = A (complementation). However, AB + A'B' represents equivalence (XNOR) between A and B, which in two-level SOP uses two distinct product terms and cannot be reduced to a single literal or a single product term without introducing an XOR/XNOR gate primitive or changing the level of logic.Step-by-Step Solution:
Check option a: AB + AB' = A(B + B') = A → reducible.Check option b: A + AB = A → reducible by absorption.Check option d: A + A'B = (A + A')(A + B) = 1 * (A + B) = A + B → reducible.Option c: AB + A'B' corresponds to equivalence (A XNOR B) and is minimal as two product terms of two literals; not further reducible in two-level SOP.Verification / Alternative check:
Build a truth table: AB + A'B' is 1 when A = B, validating XNOR behavior and showing no single-literal collapse is possible.Why Other Options Are Wrong:
Each of a, b, and d simplifies to a single-literal or simpler expression using standard identities.Common Pitfalls:
Attempting to reduce XNOR to fewer product terms without allowing XOR/XNOR primitives.Confusing AB + A'B' with AB' + A'B (which is XOR).Final Answer:
AB + A'B'
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