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(B) The circuit consists of two $NAND$ gates whose outputs $X$ and $Y$ are fed into a $NOR$ gate to produce output $Z$.Inputs are $A, B, C$. The outpu

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$AND$

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(B) The circuit consists of two $NAND$ gates whose outputs $X$ and $Y$ are fed into a $NOR$ gate to produce output $Z$.Inputs are $A, B, C$. The outputs of the $NAND$ gates are $X = \overline{AB}$ and $Y = \overline{BC}$.The output of the $NOR$ gate is $Z = \overline{X + Y}$.Substituting $X$ and $Y$: $Z = \overline{\overline{AB} + \overline{BC}}$.Using De Morgan's Law: $Z = \overline{\overline{AB}} \cdot \overline{\overline{BC}} = AB \cdot BC = ABC$.Since the output $Z = ABC$ corresponds to a $3$-input $AND$ gate,the system is equivalent to an $AND$ gate.$A$$B$$C$$X = \overline{AB}$$Y = \overline{BC}$$Z = \overline{X+Y}$$0$$0$$0$$1$$1$$0$$1$$0$$0$$1$$1$$0$$0$$0$$1$$1$$1$$0$$1$$0$$1$$1$$1$$0$$0$$1$$0$$1$$1$$0$$1$$1$$0$$0$$1$$0$$0$$1$$1$$1$$0$$0$$1$$1$$1$$0$$0$$1$

The figure shows two $NAND$ gates followed by a $NOR$ gate. The system is equivalent to the following logic gate:

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