Interpreting De Morgan’s theorem Given the theorem (A * B)' = A' + B', which pair of gate descriptions are logically equivalent according to this statement?

Introduction / Context:De Morgan’s theorems are fundamental identities used to transform logic expressions and convert between gate types. They allow designers to swap between NAND/NOR implementations and AND/OR structures with input or output inversions, which is crucial for technology mapping and minimizing parts in practical circuits.

Given Data / Assumptions:

• Use positive logic (1 = HIGH, 0 = LOW).
• De Morgan’s theorem provided: (A * B)' = A' + B'.
• Symbol * denotes logical AND; + denotes logical OR; ' denotes inversion.

Concept / Approach:

The given identity states that the complement of an AND equals the OR of the complements. In gate terms, an AND followed by an inverter (i.e., a NAND) is equivalent to an OR gate whose inputs are both inverted. This is the classic translation used to realize functions with only NAND gates, or to interpret bubble-logic schematics correctly.

Step-by-Step Solution:

Verification / Alternative check:

Build a 2-input truth table for both expressions. For all four input combinations of A and B, (A * B)' and A' + B' match exactly. A quick logic simulator test or simple gate-level substitution confirms the equivalence in practical circuits.

Why Other Options Are Wrong:

• NOR equals an AND with inverted inputs per the other De Morgan identity (A + B)' = A' * B', not the one given here.
• “AND and NOR with inverted inputs” mismatches the provided form.
• “NOR and NAND with inverted inputs” is not a direct De Morgan pair for this identity.
• “NAND and AND with inverted inputs” confuses complement placement; an AND with inverted inputs corresponds to a NOR under the other identity.

Common Pitfalls:

• Confusing which De Morgan form you are using. Remember: complement of AND becomes OR of complements; complement of OR becomes AND of complements.
• Misreading bubble logic: bubbles on the inputs of an OR create a NAND-equivalent; bubbles on the inputs of an AND create a NOR-equivalent.

Final Answer:

a NAND and an OR gate with inverted inputs