The output ($Y$) of the given logic implementation is similar to the output of an/a _________ gate.

To find the gate equivalent to the given logic circuit, we need to determine the Boolean expression for the output $Y$ in terms of the inputs $A$ and $B$.Step 1: Analyze Individual GatesThe circuit consists of three logic gates. Let's find the output of each:Top Gate (NOR): Takes inputs $A$ and $B$.Output 1 = $\overline{A+B}$ Bottom Gate (NAND): Takes inputs $B$ and $A$.Output 2 = $\overline{A\cdot B}$ Final Gate (AND): Takes Output 1 and Output 2 as its inputs.$Y=(\text{Output 1})\cdot (\text{Output 2})$$Y=(\overline{A+B})\cdot (\overline{A\cdot B})$ Step 2: Simplify the Boolean ExpressionWe can simplify $Y$ using De Morgan's Laws ($\overline{X+Y}=\bar{X}\cdot \bar{Y}$ and $\overline{X\cdot Y}=\bar{X}+\bar{Y}$):From the first part: $\overline{A+B}=\bar{A}\cdot \bar{B}$ From the second part: $\overline{A\cdot B}=\bar{A}+\bar{B}$ Substitute these back into the equation for $Y$:
$$Y=(\bar{A}\cdot \bar{B})\cdot (\bar{A}+\bar{B})$$
Distribute $(\bar{A}\cdot \bar{B})$ over the terms in the parentheses:
$$Y=(\bar{A}\cdot \bar{B}\cdot \bar{A})+(\bar{A}\cdot \bar{B}\cdot \bar{B})$$
Using the Idempotent Law ($X\cdot X=X$):$\bar{A}\cdot \bar{A}=\bar{A}$$\bar{B}\cdot \bar{B}=\bar{B}$ The expression becomes:
$$Y=(\bar{A}\cdot \bar{B})+(\bar{A}\cdot \bar{B})$$
$$Y=\bar{A}\cdot \bar{B}$$
Applying De Morgan’s Law in reverse ($\bar{A}\cdot \bar{B}=\overline{A+B}$):
$$Y=\overline{A+B}$$
Step 3: ConclusionThe final expression $Y=\overline{A+B}$ is the standard Boolean expression for a NOR gate.Verification via Truth Table



The output column for $Y$ matches exactly with a NOR gate.Correct Option: (4) NOR