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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 $\overline{A \cdot B}$.These two outputs are fed into th

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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 $\overline{A \cdot B}$.These two outputs are fed into the $AND$ gate $(G_3)$,so the final output $Y$ is:$Y = (A + B) \cdot (\overline{A \cdot B})$Using De Morgan's theorem,$\overline{A \cdot B} = \bar{A} + \bar{B}$.Substituting this into the equation:$Y = (A + B) \cdot (\bar{A} + \bar{B})$$Y = A \cdot \bar{A} + A \cdot \bar{B} + B \cdot \bar{A} + B \cdot \bar{B}$Since $A \cdot \bar{A} = 0$ and $B \cdot \bar{B} = 0$,the expression simplifies to:$Y = 0 + A \cdot \bar{B} + \bar{A} \cdot B + 0$$Y = A \cdot \bar{B} + \bar{A} \cdot B$This is the Boolean expression for an $XOR$ gate.

The following configuration of gate is equivalent to

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