In the logic-gate network shown below, the overall Boolean function realised at output \(Y\) is equivalent to which basic logic gate?
NAND gate
Step 1 — Simplify the NOR output using De Morgan.
\[M \;=\; \overline{\bar{A}+\bar{B}} \;=\; \bar{\bar{A}}\cdot\bar{\bar{B}} \;=\; A\cdot B.\]
Step 2 — Apply the final NAND with B.
\[Y \;=\; \overline{M\cdot B} \;=\; \overline{(A\cdot B)\cdot B} \;=\; \overline{A\cdot(B\cdot B)} \;=\; \overline{A\cdot B}.\]
(Used the idempotent law \(B\cdot B = B\).)
Step 3 — Verify with a truth table.
\[\begin{array}{cc|cc|c|c|c}A & B & \bar A & \bar B & M=\overline{\bar A+\bar B} & M\cdot B & Y=\overline{M\cdot B}\\ \hline 0 & 0 & 1 & 1 & 0 & 0 & 1\\ 0 & 1 & 1 & 0 & 0 & 0 & 1\\ 1 & 0 & 0 & 1 & 0 & 0 & 1\\ 1 & 1 & 0 & 0 & 1 & 1 & 0\end{array}\]
The output column \(Y = 1,1,1,0\) is exactly the NAND truth table of \((A,B)\).
∴ The network implements a NAND gate, matching option (D).
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