Gate-level equivalence using DeMorgan’s laws An AND gate with inversion “bubbles” on each input is functionally equivalent to which single gate?

Introduction / Context:Schematic “bubbles” indicate logical inversion. Recognizing bubble-pushing patterns and applying DeMorgan’s theorems enables quick translation between gate symbols and their equivalent logic functions, which is crucial in reading or redrawing schematics.

Given Data / Assumptions:

• Gate: AND with inverted inputs (bubbles on each input).
• No bubble on the output.
• Standard Boolean notation: complement shown with a prime (’).

Concept / Approach:The function of an AND gate with complemented inputs is Y = A’ · B’. DeMorgan’s theorem states that (A + B)’ = A’ · B’. Therefore, A’ · B’ equals (A + B)’. The latter is precisely the definition of a NOR gate.

Step-by-Step Solution:Write the AND-with-bubbles expression: Y = A’ · B’.Apply DeMorgan: A’ · B’ = (A + B)’.Recognize that Y = (A + B)’ is a 2-input NOR function.Hence, the composite symbol performs as a NOR gate.

Verification / Alternative check:Construct a truth table for A’ · B’ and compare with NOR(A, B); they match for all input combinations.

Why Other Options Are Wrong:NOT: a unary operation; our function depends on two inputs.OR: would be Y = A + B (no inversion).NAND: corresponds to (A · B)’ which is different from (A + B)’.

Common Pitfalls:Forgetting that bubbles invert signals and that DeMorgan converts between AND-of-complements and complement-of-OR.

Final Answer:NOR