The combination of the gates shown in the fig | vedclass

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(B) The circuit consists of two $NOT$ gates (formed by $NAND$ gates with shorted inputs) followed by a $NAND$ gate.Let the inputs be $A$ and $B$.The o

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$OR$ gate

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

$A$	$B$	$Y$
$1$	$1$	$0$
$1$	$0$	$0$
$0$	$1$	$0$
$0$	$0$	$1$

The combination of the gates shown in the figure below produces:

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