The figure below shows a combination of NOT g | vedclass

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(D) The inputs to the $NOR$ gate are the inverted inputs $A$ and $B$. Let the inputs to the $NOR$ gate be $A' = \bar{A}$ and $B' = \bar{B}$.The output

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$AND$ gate

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(D) The inputs to the $NOR$ gate are the inverted inputs $A$ and $B$. Let the inputs to the $NOR$ gate be $A' = \bar{A}$ and $B' = \bar{B}$.The output $Y$ of the $NOR$ gate is given by $Y = \overline{A' + B'}$.Substituting the values of $A'$ and $B'$, we get $Y = \overline{\bar{A} + \bar{B}}$.According to De Morgan's theorem, $\overline{\bar{A} + \bar{B}} = \overline{\bar{A}} \cdot \overline{\bar{B}} = A \cdot B$.Therefore, the output $Y = A \cdot B$, which is the Boolean expression for an $AND$ gate.

Entry	$A$	$B$	$Y$
$1$	$0$	$1$	$0$
$2$	$1$	$0$	$0$
$3$	$1$	$1$	$1$
$4$	$0$	$0$	$1$

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

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Entry $A$ $B$ $Y$
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Four $NOR$ gates are connected as shown in the figure. The truth table for the given figure is:

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