Q.13) The output $(\mathrm{Y})$ of the given logic implementation is similar to the output of an/a $\qquad$gate.
A) NOR
B) AND
C) NAND
D) OR
Solution:
Top gate: $A$ OR $B=A+B$
Bottom gate: A OR $\mathrm{A}=\mathrm{A}$
Final gate: $A N D$ of $(A+B)$ and $A=(A+B) \cdot A=A \cdot A+B \cdot A=A+A B=A$
But this passes through a NOT gate, so $Y=\bar{A}$, only when $B=1$.
Truth table shows the behavior matches that of a NAND gate.
Hence, the answer is option (3) NAND.