The following configuration of gates is equiv | vedclass

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(B) The output of the $OR$ gate $(G_1)$ is $(A+B)$.The output of the $NAND$ gate $(G_2)$ is $\overline{A \cdot B}$.The final output $Y$ from the $AND$

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$XOR$

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(B) The output of the $OR$ gate $(G_1)$ is $(A+B)$.The output of the $NAND$ gate $(G_2)$ is $\overline{A \cdot B}$.The final output $Y$ from the $AND$ gate $(G_3)$ is the product of these two inputs:$Y = (A+B) \cdot (\overline{A \cdot B})$Using De Morgan's theorem,$\overline{A \cdot B} = \overline{A} + \overline{B}$.So,$Y = (A+B) \cdot (\overline{A} + \overline{B})$$Y = A \cdot \overline{A} + A \cdot \overline{B} + B \cdot \overline{A} + B \cdot \overline{B}$Since $A \cdot \overline{A} = 0$ and $B \cdot \overline{B} = 0$,we get:$Y = 0 + A \cdot \overline{B} + \overline{A} \cdot B + 0$$Y = A \cdot \overline{B} + \overline{A} \cdot B$This is the Boolean expression for an $XOR$ gate.

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

The following configuration of gates is equivalent to:

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