The combination of logic gates shown below be | vedclass

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(D) The given circuit consists of two $NAND$ gates used as $NOT$ gates (since their inputs are shorted) followed by a $NAND$ gate.Let the inputs be $A

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$OR$ gate

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(D) The given circuit consists of two $NAND$ gates used as $NOT$ gates (since their inputs are shorted) followed by a $NAND$ gate.Let the inputs be $A$ and $B$.The output of the first $NAND$ gate (acting as $NOT$) is $\bar{A}$.The output of the second $NAND$ gate (acting as $NOT$) is $\bar{B}$.These two outputs are fed into a third $NAND$ gate.The final output $Y$ is given by $Y = \overline{(\bar{A} \cdot \bar{B})}$.Using De Morgan's theorem,$\overline{(\bar{A} \cdot \bar{B})} = \overline{\bar{A}} + \overline{\bar{B}} = A + B$.This is the Boolean expression for an $OR$ gate.Therefore,the combination acts as an $OR$ gate.

The combination of logic gates shown below becomes:

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