For the circuit shown above,the equivalent te | vedclass

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(A) Let the inputs be $A$ and $B$. The circuit consists of a $	ext{NAND}$ gate and a $	ext{NOR}$ gate feeding into a $	ext{NAND}$ gate.$1$. The out

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$	ext{OR}$ gate

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(A) Let the inputs be $A$ and $B$. The circuit consists of a $	ext{NAND}$ gate and a $	ext{NOR}$ gate feeding into a $	ext{NAND}$ gate.$1$. The output of the top $	ext{NAND}$ gate is $Y_1 = \overline{A \cdot B}$.$2$. The output of the bottom $	ext{NOR}$ gate is $Y_2 = \overline{A + B}$.$3$. These are fed into the final $	ext{NAND}$ gate,so the final output is $Y = \overline{Y_1 \cdot Y_2} = \overline{(\overline{A \cdot B}) \cdot (\overline{A + B})}$.Using De Morgan's Law,$\overline{X \cdot Y} = \overline{X} + \overline{Y}$,we get $Y = \overline{(\overline{A \cdot B})} + \overline{(\overline{A + B})} = (A \cdot B) + (A + B)$.Since $(A \cdot B)$ is always a subset of $(A + B)$,the expression simplifies to $Y = A + B$,which is the Boolean expression for an $	ext{OR}$ gate.$A$$B$$Y_1 = \overline{A \cdot B}$$Y_2 = \overline{A + B}$$Y = \overline{Y_1 \cdot Y_2}$$0$$0$$1$$1$$0$$0$$1$$1$$0$$1$$1$$0$$1$$0$$1$$1$$1$$0$$0$$1$

For the circuit shown above,the equivalent $\text{GATE}$ is:

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