What will be the input of A and B for the Boo | vedclass

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(A) The given Boolean expression is $Y = (\overline{A+B}) \cdot (\overline{A \cdot B})$.Using De Morgan's laws,we know that $\overline{A+B} = \bar{A}

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(A) The given Boolean expression is $Y = (\overline{A+B}) \cdot (\overline{A \cdot B})$.Using De Morgan's laws,we know that $\overline{A+B} = \bar{A} \cdot \bar{B}$ and $\overline{A \cdot B} = \bar{A} + \bar{B}$.Substituting these into the expression:$Y = (\bar{A} \cdot \bar{B}) \cdot (\bar{A} + \bar{B})$$Y = (\bar{A} \cdot \bar{B} \cdot \bar{A}) + (\bar{A} \cdot \bar{B} \cdot \bar{B})$Since $\bar{A} \cdot \bar{A} = \bar{A}$ and $\bar{B} \cdot \bar{B} = \bar{B}$,we get:$Y = (\bar{A} \cdot \bar{B}) + (\bar{A} \cdot \bar{B}) = \bar{A} \cdot \bar{B}$.For $Y = 1$,both $\bar{A}$ and $\bar{B}$ must be $1$,which means $A = 0$ and $B = 0$.$A, B$$Y$$0, 0$$1$$0, 1$$0$$1, 0$$0$$1, 1$$0$

What will be the input of $A$ and $B$ for the Boolean expression $(\overline{A + B}) \cdot (\overline{A \cdot B}) = 1$?

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