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(C) Let the inputs of the $OR$ gate be $A$ and $B$. The output of the $OR$ gate is $X = A + B$.This output $X$ is connected to both inputs of a $NAND$

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$NOR$ gate

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(C) Let the inputs of the $OR$ gate be $A$ and $B$. The output of the $OR$ gate is $X = A + B$.This output $X$ is connected to both inputs of a $NAND$ gate. Let the inputs of the $NAND$ gate be $I_1 = X$ and $I_2 = X$.The output $Y$ of the $NAND$ gate is given by $Y = \overline{I_1 \cdot I_2} = \overline{X \cdot X} = \overline{X}$.Substituting $X = A + B$,we get $Y = \overline{A + B}$.The expression $\overline{A + B}$ represents the Boolean operation of a $NOR$ gate.Therefore,the combination acts as a $NOR$ gate.

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

The output of an $OR$ gate is connected to both the inputs of a $NAND$ gate. The combination will serve as a

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$A$ $B$ $Y$
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