What does the following combination of gates | vedclass

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(A) The given circuit consists of four gates: $G_1$ and $G_2$ are $NOT$ gates, $G_3$ is a $NOR$ gate, and $G_4$ is a $NOT$ gate.Let the inputs be $A$

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$NAND$ gate

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(A) The given circuit consists of four gates: $G_1$ and $G_2$ are $NOT$ gates, $G_3$ is a $NOR$ gate, and $G_4$ is a $NOT$ gate.Let the inputs be $A$ and $B$.The outputs of $G_1$ and $G_2$ are $\bar{A}$ and $\bar{B}$ respectively.These are fed into the $NOR$ gate $G_3$. The output of $G_3$ is $\overline{\bar{A} + \bar{B}}$.By De Morgan's Law, $\overline{\bar{A} + \bar{B}} = \overline{\bar{A}} \cdot \overline{\bar{B}} = A \cdot B$.This output is then passed through the $NOT$ gate $G_4$.The final output $Y$ is $\overline{A \cdot B}$, which is the Boolean expression for a $NAND$ gate.

What does the following combination of gates produce?

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