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(D) The given circuit consists of two $NOT$ gates ($G_1$ and $G_2$) and one $NOR$ gate $(G_3)$.$1$. The input $A$ passes through the $NOT$ gate $G_1$,

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$AND, A \cdot B$

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(D) The given circuit consists of two $NOT$ gates ($G_1$ and $G_2$) and one $NOR$ gate $(G_3)$.$1$. The input $A$ passes through the $NOT$ gate $G_1$,so the output at $C$ is $C = \overline{A}$.$2$. The input $B$ passes through the $NOT$ gate $G_2$,so the output at $D$ is $D = \overline{B}$.$3$. These outputs $C$ and $D$ are fed into the $NOR$ gate $G_3$. The output $Y$ of a $NOR$ gate is the complement of the $OR$ of its inputs.$4$. Therefore,$Y = \overline{C + D}$.$5$. Substituting the values of $C$ and $D$,we get $Y = \overline{\overline{A} + \overline{B}}$.$6$. According to De Morgan's theorem,$\overline{\overline{A} + \overline{B}} = \overline{\overline{A}} \cdot \overline{\overline{B}} = A \cdot B$.$7$. The Boolean expression $Y = A \cdot B$ represents an $AND$ gate.Thus,the resultant gate is an $AND$ gate with the expression $A \cdot B$.

The resultant gate and its Boolean expression for the given circuit is

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In Boolean algebra,$Y = A + B$ represents . . . . . .

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