The figure shows a combination of two NOT gat | vedclass

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(C) The inputs to the $NOR$ gate are $\bar{A}$ and $\bar{B}$ because they pass through $NOT$ gates. The output $Y$ of the $NOR$ gate is given by the B

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$AND$ gate

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(C) The inputs to the $NOR$ gate are $\bar{A}$ and $\bar{B}$ because they pass through $NOT$ gates. The output $Y$ of the $NOR$ gate is given by the Boolean expression: $Y = \overline{\bar{A} + \bar{B}}$.According to De Morgan's theorem,$\overline{\bar{A} + \bar{B}} = \overline{\bar{A}} \cdot \overline{\bar{B}} = A \cdot B$.This is the Boolean expression for an $AND$ gate.Truth table:$A$$B$$\bar{A}$$\bar{B}$$\bar{A} + \bar{B}$$Y = \overline{\bar{A} + \bar{B}}$$0$$0$$1$$1$$1$$0$$0$$1$$1$$0$$1$$0$$1$$0$$0$$1$$1$$0$$1$$1$$0$$0$$0$$1$Thus,the combination is equivalent to an $AND$ gate.

The figure shows a combination of two $NOT$ gates and a $NOR$ gate. The combination is equivalent to a

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