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(A) The inputs to the circuit are $A$ and $B$.The output of the $OR$ gate is $(A+B)$.The output of the $AND$ gate is $(A \cdot B)$.These two outputs a

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$AND$

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(A) The inputs to the circuit are $A$ and $B$.The output of the $OR$ gate is $(A+B)$.The output of the $AND$ gate is $(A \cdot B)$.These two outputs are fed into a $NAND$ gate,which produces an output $Z = \overline{(A+B) \cdot (A \cdot B)}$.This output $Z$ is then passed through a $NOT$ gate,resulting in the final output $Y = \bar{Z} = \overline{\overline{(A+B) \cdot (A \cdot B)}} = (A+B) \cdot (A \cdot B)$.Using the distributive law of Boolean algebra,$(A+B) \cdot (A \cdot B) = (A \cdot A \cdot B) + (B \cdot A \cdot B) = (A \cdot B) + (A \cdot B) = A \cdot B$.Thus,$Y = A \cdot B$,which is the Boolean expression for an $AND$ gate.

The logic operations performed by the given digital circuit is equivalent to:

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