Identify the logic operation carried out.

We need to identify the overall logic operation carried out by the given combinational digital circuit.

1. Analyze the Input Gates

Let's look at the first layer of the circuit connected to inputs $$A$$ and $$B$$:

• Top Gate: The input $$A$$ is split and fed into both terminals of a NAND gate. When the inputs of a NAND gate are shorted together, it functions as a NOT gate.

  The output of this gate is: $$\bar{A}$$

• Bottom Gate: Similarly, the input $$B$$ is split and fed into both terminals of another NAND gate, which also acts as a NOT gate.

  The output of this gate is: $$\bar{B}$$

2. Analyze the Output Gate

The outputs from the first stage ($$\bar{A}$$ and $$\bar{B}$$) serve as the inputs to the final gate, which is a NOR gate.

A standard NOR gate produces an output that is the inverted sum of its inputs. Therefore, the final output $$Y$$ is given by:

$$Y = \overline{\bar{A} + \bar{B}}$$

3. Simplify Using De Morgan's Laws

According to De Morgan's Theorem, the complement of a sum is equal to the product of the complements:

$$\overline{X + Y} = \bar{X} \cdot \bar{Y}$$

Applying this theorem to our output equation by treating $$\bar{A}$$ as $$X$$ and $$\bar{B}$$ as $$Y$$:

$$Y = \overline{(\bar{A})} \cdot \overline{(\bar{B})}$$

Since a double inversion cancels out ($$\overline{\bar{X}} = X$$), the expression simplifies to:

$$Y = A \cdot B$$

4. Conclusion

The simplified Boolean expression $$Y = A \cdot B$$ represents the standard operation of an AND gate.

Final Answer: Option B (AND)