The output $(X)$ of the logic circuit shown in figure will be
The output $(X)$ of the logic circuit shown in figure will be
$X=\overline{\bar{A}} \cdot \overline{\bar{B}}$
$X=\overline{A \cdot B}$
$X=A \cdot B$
$X=\overline{A+B}$
Solution
$X=\overline{\overline{A B}}=A \cdot B$ (i.e., AND gate)
If the output $X$ of NAND gate is connected to the input of NOT gate (made from NAND gate by joining two inputs) from the given figure then we get back an AND gate.
