To get the truth table shown from the followi | vedclass

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(C) The given circuit consists of an $OR$ gate where one input is $A$ and the other input is the output of gate $G$,let's call it $C$. The output of t

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$NAND$

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(C) The given circuit consists of an $OR$ gate where one input is $A$ and the other input is the output of gate $G$,let's call it $C$. The output of the $OR$ gate is $Y = A + C$.From the truth table,we observe that $Y$ is always $1$ for all combinations of inputs $A$ and $B$.If $G$ is a $NAND$ gate,its output $C = \overline{A \cdot B}$.The final output is $Y = A + \overline{A \cdot B}$.Let's verify this for all inputs:$1$. If $A=0, B=0$: $C = \overline{0 \cdot 0} = 1$. Then $Y = 0 + 1 = 1$.$2$. If $A=0, B=1$: $C = \overline{0 \cdot 1} = 1$. Then $Y = 0 + 1 = 1$.$3$. If $A=1, B=0$: $C = \overline{1 \cdot 0} = 1$. Then $Y = 1 + 1 = 1$.$4$. If $A=1, B=1$: $C = \overline{1 \cdot 1} = 0$. Then $Y = 1 + 0 = 1$.Since the output $Y$ is $1$ in all cases,the gate $G$ must be a $NAND$ gate.

To get the truth table shown from the following logic circuit,the logic gate $G$ should be

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