Boolean algebra duality principle (repair applied): Which statement correctly expresses the duality theorem for Boolean expressions?

Introduction / Context:
The original stem was incomplete. Applying the Recovery-First Policy, we restate the concept as a complete, teachable item. The duality theorem in Boolean algebra is a cornerstone identity used to derive valid expressions and implement alternative but equivalent logic forms. Mastering duality improves your ability to transform AND–OR circuits into OR–AND counterparts and to check identities succinctly.

Given Data / Assumptions:

• Standard Boolean algebra with operators + (OR) and · (AND).
• Constants 1 (TRUE) and 0 (FALSE).
• Variables and their complements (A, A’, etc.) remain as written unless explicitly complemented.

Concept / Approach:
Duality states that if an algebraic identity is valid, its dual is also valid. To form the dual of any Boolean expression: interchange + and · everywhere, and interchange 1 and 0 everywhere. You do not alter variables or complements. This rule mirrors dual operator roles in the Boolean lattice and ensures that basic laws (e.g., A + 0 = A) have valid duals (A · 1 = A).

Step-by-Step Solution:

Verification / Alternative check:
Check a known law and its dual. Identity: A + 0 = A. Dual via swaps ⇒ A · 1 = A. Both hold universally, verifying the method.

Why Other Options Are Wrong:
Swapping variables with complements changes the function, not its dual. De Morgan conversions are related transformations but not the definition of duality. Duality is general and applies to any Boolean expression, not only minterms.

Common Pitfalls:
Accidentally complementing variables when forming the dual; forgetting to swap constants; mixing duality with De Morgan’s complement rules (which involve negation), whereas duality does not introduce complements.

Final Answer:
Obtain the dual by interchanging + with · and 1 with 0, leaving variables and complements unchanged