Simplify $$$\overline{\overline{A \cdot B} + \left(\overline{D} \cdot A\right)}$$$
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Simplify the boolean expression $$$\overline{\overline{A \cdot B} + \left(\overline{D} \cdot A\right)}$$$.
Solution
Apply de Morgan's theorem $$$\overline{X + Y} = \overline{X} \cdot \overline{Y}$$$ with $$$X = \overline{A \cdot B}$$$ and $$$Y = \overline{D} \cdot A$$$:
$${\color{red}\left(\overline{\overline{A \cdot B} + \left(\overline{D} \cdot A\right)}\right)} = {\color{red}\left(\overline{\overline{A \cdot B}} \cdot \overline{\overline{D} \cdot A}\right)}$$Apply the double negation (involution) law $$$\overline{\overline{X}} = X$$$ with $$$X = A \cdot B$$$:
$${\color{red}\left(\overline{\overline{A \cdot B}}\right)} \cdot \overline{\overline{D} \cdot A} = {\color{red}\left(A \cdot B\right)} \cdot \overline{\overline{D} \cdot A}$$Apply de Morgan's theorem $$$\overline{X \cdot Y} = \overline{X} + \overline{Y}$$$ with $$$X = \overline{D}$$$ and $$$Y = A$$$:
$$A \cdot B \cdot {\color{red}\left(\overline{\overline{D} \cdot A}\right)} = A \cdot B \cdot {\color{red}\left(\overline{\overline{D}} + \overline{A}\right)}$$Apply the double negation (involution) law $$$\overline{\overline{X}} = X$$$ with $$$X = D$$$:
$$A \cdot B \cdot \left({\color{red}\left(\overline{\overline{D}}\right)} + \overline{A}\right) = A \cdot B \cdot \left({\color{red}\left(D\right)} + \overline{A}\right)$$Apply the commutative law:
$${\color{red}\left(A \cdot B \cdot \left(D + \overline{A}\right)\right)} = {\color{red}\left(A \cdot \left(D + \overline{A}\right) \cdot B\right)}$$Apply the commutative law:
$$A \cdot {\color{red}\left(D + \overline{A}\right)} \cdot B = A \cdot {\color{red}\left(\overline{A} + D\right)} \cdot B$$Apply the redundancy law $$$X \cdot \left(\overline{X} + Y\right) = X \cdot Y$$$ with $$$X = A$$$ and $$$Y = D$$$:
$${\color{red}\left(A \cdot \left(\overline{A} + D\right)\right)} \cdot B = {\color{red}\left(A \cdot D\right)} \cdot B$$Answer
$$$\overline{\overline{A \cdot B} + \left(\overline{D} \cdot A\right)} = A \cdot D \cdot B$$$