Exclusive-NOR (XNOR) identity: Which Boolean expression correctly represents the two-input exclusive-NOR function?

Introduction / Context:
Exclusive-NOR (XNOR) outputs HIGH when its inputs are equal (both 0 or both 1) and LOW when they differ. Recognizing its Boolean identity is essential for algebraic manipulation, schematic implementation, and verification against truth tables or Karnaugh maps.

Given Data / Assumptions:

• Two-input function under positive logic.
• Goal is to match a standard Boolean expression to XNOR behavior.
• Complement notation: X′ denotes NOT X.

Concept / Approach:
The definition “equal → HIGH” translates directly to the sum of the two equal cases: A = 0, B = 0 and A = 1, B = 1. In minterm form, these are A′B′ and AB. Therefore, XNOR = A′B′ + AB. The complement of this (A′B + AB′) is the XOR function (different → HIGH).

Step-by-Step Solution:

Verification / Alternative check:
Karnaugh map placement of the 1s at (00) and (11) groups to yield A′B′ + AB; algebraically, note that XNOR = (A ⊙ B)′ = (A′B + AB′)′ which by De Morgan simplifies to A′B′ + AB.

Why Other Options Are Wrong:

• A′B + AB′: This is XOR (different → HIGH).
• (A + B) * (A′ + B′): Expands to A′B + AB′ (also XOR), not XNOR.
• A + B: Ordinary OR gate, 00 yields 0 not 1.
• AB′ + A′: Not symmetric and does not match the equality condition.

Common Pitfalls:
Mixing up XOR and XNOR; forgetting that XNOR is the complement of XOR and represents equality; misapplying De Morgan transformations.

Final Answer:
A′B′ + AB