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(D) The inputs to the two $OR$ gates are $A$ and $B$. Therefore,the outputs of the two $OR$ gates are $Y_1 = A + B$ and $Y_2 = A + B$.These outputs $Y

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$NOR$

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(D) The inputs to the two $OR$ gates are $A$ and $B$. Therefore,the outputs of the two $OR$ gates are $Y_1 = A + B$ and $Y_2 = A + B$.These outputs $Y_1$ and $Y_2$ are fed into a $NAND$ gate. The output $Y$ of the $NAND$ gate is given by:$Y = \overline{Y_1 \cdot Y_2}$Substituting the values of $Y_1$ and $Y_2$:$Y = \overline{(A + B) \cdot (A + B)}$Using the Boolean identity $X \cdot X = X$,we have $(A + B) \cdot (A + B) = A + B$.Therefore,$Y = \overline{A + B}$.The expression $\overline{A + B}$ represents the output of a $NOR$ gate.Thus,the equivalent logic gate is a $NOR$ gate.

For the combination of logic gates shown in the figure,the equivalent logic gate is

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