The correct truth table for the given logic circuit is

To find the correct truth table for the given logic circuit, we analyze the output of each gate step-by-step.

1. Analyzing the Circuit Structure:
The circuit has two inputs: A (top input) and B (bottom input).
- The top gate has both of its inputs tied together to input A. A NAND/NOR gate with tied inputs acts as a NOT gate. Therefore, the output of the top gate, which we will call $X_1$, is:

$X_1=\overline{A}$

- The bottom-left gate is a NAND gate with inputs A and B. Its output, which we will call $X_2$, is:

$X_2=\overline{A·B}$

- The bottom-right gate is another NOT gate (a NAND gate with tied inputs) receiving input from $X_2$. Its output, which we will call $X_3$, is:

$X_3=\overline{X_2}=A·B$

- The final output gate acts as a NOR gate, taking inputs $X_1$ and $X_3$. The final output Y is:

$Y=\overline{X_1+X_3}=\overline{\overline{A}+(A·B)}$

2. Simplifying the Boolean Expression:
Using De Morgan's Law $\overline{P+Q}=\overline{P}·\overline{Q}$:

$Y=\overline{\overline{A}}·\overline{A·B}$

Since the double negation $\overline{\overline{A}}=A$, and applying De Morgan's Law again to the second term:

$Y=A·(\overline{A}+\overline{B})$

Distributing A:

$Y=(A·\overline{A})+(A·\overline{B})$

Since $A·\overline{A}=0$, the expression simplifies to:

$Y=A·\overline{B}$

This indicates that the output Y is 1 only when A is 1 and B is 0.

3. Constructing the Truth Table:
Let us evaluate all input combinations using the simplified Boolean expression $Y=A·\overline{B}$: - For A = 0, B = 0: $Y=0·\overline{0}=0·1=0$ - For A = 0, B = 1: $Y=0·\overline{1}=0·0=0$ - For A = 1, B = 0: $Y=1·\overline{0}=1·1=1$ - For A = 1, B = 1: $Y=1·\overline{1}=1·0=0$

Thus, the correct truth table is:
A   B   Y
0   0   0
0   1   0
1   0   1
1   1   0