The combination of the gates shown in the fig | vedclass

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(B) The circuit consists of two $NAND$ gates used as $NOT$ gates,followed by a $NAND$ gate.$1$. The first two $NAND$ gates have their inputs shorted,a

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$OR$ gate

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(B) The circuit consists of two $NAND$ gates used as $NOT$ gates,followed by a $NAND$ gate.$1$. The first two $NAND$ gates have their inputs shorted,acting as $NOT$ gates. Thus,the outputs are $\bar{A}$ and $\bar{B}$.$2$. These outputs are fed into a third $NAND$ gate.$3$. The final output $Y$ is given by: $Y = \overline{\bar{A} \cdot \bar{B}}$.$4$. Using De Morgan's theorem,$\overline{\bar{A} \cdot \bar{B}} = \overline{\bar{A}} + \overline{\bar{B}} = A + B$.$5$. The expression $Y = A + B$ represents the Boolean operation of an $OR$ gate.

The combination of the gates shown in the figure below produces which gate?

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