The resultant logic gate from the combination | vedclass

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Question

(A) The given circuit consists of two $NOT$ gates followed by a $NAND$ gate.Let the inputs be $A$ and $B$.The outputs of the two $NOT$ gates are $\bar

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$OR$

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(A) The given circuit consists of two $NOT$ gates followed by a $NAND$ gate.Let the inputs be $A$ and $B$.The outputs of the two $NOT$ gates are $\bar{A}$ and $\bar{B}$.These outputs are fed into a $NAND$ gate.The output $Y$ of the $NAND$ gate is given by $Y = \overline{(\bar{A} \cdot \bar{B})}$.Using De Morgan's theorem,$\overline{(\bar{A} \cdot \bar{B})} = \overline{\bar{A}} + \overline{\bar{B}} = A + B$.Since the final output $Y = A + B$,the combination acts as an $OR$ gate.

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

The resultant logic gate from the combination of the following gates is:

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$A$ $B$ $C$ $Y$
$0$ $0$ $0$ $1$
$0$ $0$ $1$ $1$
$0$ $1$ $0$ $0$
$0$ $1$ $1$ $0$
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What will be the input of $A$ and $B$ for the Boolean expression $\overline{(A+B) \cdot(A \cdot B)}=1$?

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

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