Interpreting DeMorgan’s theorems: One statement of DeMorgan’s theorems is (A • B)' = A' + B'. Simply stated, which pair of gates are logically equivalent under this interpretation?

Introduction / Context:
DeMorgan’s theorems connect AND/OR operations with inversion in Boolean algebra and gate-level design. They enable trading bubbles (inversions) between inputs and outputs, greatly simplifying gate implementation with universal NAND/NOR families.

Given Data / Assumptions:

• DeMorgan form: (A • B)' = A' + B' and (A + B)' = A' • B'.
• Bubble notation: a bubble denotes inversion on a pin.
• Equivalence concerns logical function, not physical symbol shape.

Concept / Approach:
(A • B)' represents a NAND function. The right-hand side A' + B' is an OR gate whose inputs are inverted (bubbles on inputs). Therefore, a NAND gate is functionally identical to an OR gate with inverted inputs. Similarly, a NOR gate equals an AND gate with inverted inputs by (A + B)' = A' • B'.

Step-by-Step Solution:
Start with NAND definition: Y = (A • B)'.Apply DeMorgan: Y = A' + B'.Gate mapping: an OR gate whose two inputs each have bubbles implements A' + B'.Therefore: NAND ≡ OR-with-bubbled-inputs.

Verification / Alternative check:
Truth tables for NAND and for OR with both inputs inverted match exactly for all input combinations, confirming equivalence.

Why Other Options Are Wrong:
NOR ≡ AND with bubbled inputs, not AND with bubbled output.NOR vs NAND with bubbled output: unrelated equivalence.NAND vs OR with bubbled output: bubbled output OR is NOR, not NAND.

Common Pitfalls:
Confusing input bubbles with output bubbles. DeMorgan moves inversions across a gate only if the gate type flips AND↔OR and all pins crossing the inversion are inverted.

Final Answer:
a NAND gate and an OR gate with bubbled inputs