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(C) From the given logic circuit,the output of the $AND$ gate is $P = A \cdot B$ and the output of the $OR$ gate is $Q = A + B.$These are fed into a $

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$1, 1$

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(C) From the given logic circuit,the output of the $AND$ gate is $P = A \cdot B$ and the output of the $OR$ gate is $Q = A + B.$These are fed into a $NAND$ gate,so the final output is $Y = \overline{P \cdot Q} = \overline{(A \cdot B) \cdot (A + B)}.$Using Boolean algebra,$(A \cdot B) \cdot (A + B) = (A \cdot B \cdot A) + (A \cdot B \cdot B) = (A \cdot B) + (A \cdot B) = A \cdot B.$Therefore,$Y = \overline{A \cdot B}.$For the input sequence $(A, B) = (0, 0): Y = \overline{0 \cdot 0} = \overline{0} = 1.$For the input sequence $(A, B) = (0, 1): Y = \overline{0 \cdot 1} = \overline{0} = 1.$Thus,the output sequence is $1, 1.$

In the following logic circuit,the sequence of the inputs $A, B$ are $(0, 0), (0, 1).$ The output $Y$ for this sequence will be

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