Question

Consider a Boolean gate (D) where the output $Y$ is related to the inputs $A$ and $B$ as \[ Y = A + \overline{B}, \] where $+$ denotes logical OR. The Boolean inputs ‘0’ and ‘1’ are also separately available. Using only D gates and inputs ‘0’ and ‘1’, ________________ (select the correct option(s)).

• A. NAND logic can be implemented
• B. OR logic cannot be implemented
• C. NOR logic can be implemented
• D. AND logic cannot be implemented

Hint:

Once NOT is available, OR + NOT is functionally complete and can emulate NAND and NOR.

Answer 1

The Correct Option is A, C

The D gate implements: \[ Y = A + \overline{B}. \] Step 1: Implement NOT. Set $A=0$: \[ Y = 0 + \overline{B} = \overline{B}. \] Thus NOT is implementable. Step 2: Implement NOR. NOR is: \[ \overline{A + B}. \] Using NOT (from Step 1), invert inputs so that D’s OR + NOT operation can emulate NOR. Thus NOR is implementable. → Option (C) is correct. Step 3: Implement NAND. NAND is: \[ \overline{AB}. \] Using De Morgan: \[ \overline{AB} = \overline{A} + \overline{B}. \] Since D can produce $\overline{A}$ and $\overline{B}$ and an OR output → NAND is implementable. → Option (A) is correct. Step 4: AND logic cannot be implemented using only OR + NOT with fixed polarity. Thus (D) is not correct. Final Answer: (A), (C)