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(D) $NOR$ gate can be constructed using $NAND$ gates by following these steps:$1$. Connect the inputs of two $NAND$ gates together to act as $NOT$ gat

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$4$

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(D) $NOR$ gate can be constructed using $NAND$ gates by following these steps:$1$. Connect the inputs of two $NAND$ gates together to act as $NOT$ gates (inverters).$2$. Feed the inputs $A$ and $B$ into these two $NOT$ gates to get $\bar{A}$ and $\bar{B}$.$3$. Connect $\bar{A}$ and $\bar{B}$ to the inputs of a third $NAND$ gate.$4$. The output of this third $NAND$ gate will be $\overline{\bar{A} \cdot \bar{B}}$.$5$. By De Morgan's Law,$\overline{\bar{A} \cdot \bar{B}} = A + B$,which is the output of an $OR$ gate.$6$. To get a $NOR$ gate,we need to invert this output,requiring a fourth $NAND$ gate used as a $NOT$ gate.$7$. However,a more efficient configuration exists using $4$ $NAND$ gates to realize a $NOR$ gate. Thus,the minimum number of $NAND$ gates required is $4$.

To obtain a $NOR$ gate,what is the minimum number of $NAND$ gates required?

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