To get the truth table shown from the followi | vedclass

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(C) The given circuit consists of an $OR$ gate with two inputs. One input is $A$ and the other input is the output of gate $G$, which takes $A$ and $B

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$AND$ gate.

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(C) The given circuit consists of an $OR$ gate with two inputs. One input is $A$ and the other input is the output of gate $G$, which takes $A$ and $B$ as inputs. Let the output of gate $G$ be $X$. Then the final output $Y = A + X$.From the truth table:For $A=0, B=0$, $Y=0$. Since $Y = A + X$, $0 = 0 + X \implies X=0$.For $A=0, B=1$, $Y=0$. Since $Y = A + X$, $0 = 0 + X \implies X=0$.For $A=1, B=0$, $Y=1$. Since $Y = A + X$, $1 = 1 + X \implies X$ can be $0$ or $1$.For $A=1, B=1$, $Y=1$. Since $Y = A + X$, $1 = 1 + X \implies X$ can be $0$ or $1$.Looking at the values of $X$ derived from the inputs $A$ and $B$: when $A=0, B=0 \implies X=0$; when $A=0, B=1 \implies X=0$; when $A=1, B=0 \implies X=0$ (consistent with $Y=1$); when $A=1, B=1 \implies X=1$ (consistent with $Y=1$).This truth table for $X$ (inputs $A, B$ and output $X$) corresponds to the $AND$ gate, where $X = A \cdot B$.

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

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