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The output ($Y$) of the given logic implementation (NOR gate with inputs $A, B$ feeding an AND gate alongside a NAND gate with inputs $B, A$) is similar to the output of an/a _______ gate.
A
AND
B
NAND
C
OR
D
NOR
Explanation
Evaluating the truth table: $Y = \overline{A+B} \cdot \overline{A \cdot B} = \overline{A+B}$, which is identical to a NOR gate.
Detailed Solution
Let the top gate be a NOR gate: $C = \overline{A + B}$. Let the bottom gate be a NAND gate: $D = \overline{A \cdot B}$. The output gate is an AND gate: $Y = C \cdot D = (\overline{A + B}) \cdot (\overline{A \cdot B})$. By Boolean algebra, $\overline{A + B} = \bar{A} \bar{B}$. When $\bar{A} \bar{B} = 1$, $A=0, B=0$, so $\overline{A \cdot B} = 1$. Thus $(\overline{A + B}) \cdot (\overline{A \cdot B}) = \overline{A + B}$. For input combinations $(0,0) \to Y=1$; $(0,1) \to Y=0$; $(1,0) \to Y=0$; $(1,1) \to Y=0$. This is precisely the truth table of a NOR gate.