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(A) The circuit consists of an $OR$ gate followed by an $AND$ gate.$1$. The inputs to the $OR$ gate are $A$ and $B$. Therefore,the output of the $OR$

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(A) The circuit consists of an $OR$ gate followed by an $AND$ gate.$1$. The inputs to the $OR$ gate are $A$ and $B$. Therefore,the output of the $OR$ gate is $(A + B)$.$2$. The inputs to the $AND$ gate are $A$ and the output of the $OR$ gate $(A + B)$.$3$. The final output $Y$ of the $AND$ gate is given by the Boolean expression: $Y = A \cdot (A + B)$.$4$. Using the distributive law of Boolean algebra: $Y = A \cdot A + A \cdot B$.$5$. Since $A \cdot A = A$,we have $Y = A + A \cdot B$.$6$. Factoring out $A$,we get $Y = A \cdot (1 + B)$.$7$. Since $(1 + B) = 1$,the final expression is $Y = A \cdot 1 = A$.

What is the output of the logic gate combination shown below?

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