Question

According to De Morgan's first theorem a NOR Gate is equivalent to a __________ Gate.

• A. Bubbled NOR
• B. XNOR
• C. XAND
• D. Bubbled AND

Hint:

Rewrite NOT(A+B) as (NOT A)(NOT B), an AND with inverted inputs.

Answer 1

The Correct Option is D

Start from the definition of NOR and push the inversion inside using De Morgan.

\[\overline{A + B} = \overline{A}\,\overline{B}\]

The transformed form is an AND of the two inputs, but each input is inverted first. In gate-symbol language, an inversion drawn at an input is shown as a small bubble, so an AND gate with bubbles on both inputs is a $bubbled\ AND$.

Hence a NOR gate and a bubbled AND gate produce identical outputs for every input combination.

\[\boxed{\text{Bubbled AND (D)}}\]