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Simplify $$$\left(\left(1 \cdot 0\right) + \overline{0}\right) \cdot \left(\overline{1} + 0 + 1\right)$$$
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Your Input
Simplify the boolean expression $$$\left(\left(1 \cdot 0\right) + \overline{0}\right) \cdot \left(\overline{1} + 0 + 1\right)$$$.
Solution
Apply the negation law $$$\overline{0} = 1$$$:
$$\left(\left(1 \cdot 0\right) + {\color{red}\left(\overline{0}\right)}\right) \cdot \left(\overline{1} + 0 + 1\right) = \left(\left(1 \cdot 0\right) + {\color{red}\left(1\right)}\right) \cdot \left(\overline{1} + 0 + 1\right)$$Apply the negation law $$$\overline{1} = 0$$$:
$$\left(\left(1 \cdot 0\right) + 1\right) \cdot \left({\color{red}\left(\overline{1}\right)} + 0 + 1\right) = \left(\left(1 \cdot 0\right) + 1\right) \cdot \left({\color{red}\left(0\right)} + 0 + 1\right)$$Apply the dominant (null, annulment) law $$$x + 1 = 1$$$ with $$$x = 1 \cdot 0$$$:
$${\color{red}\left(\left(1 \cdot 0\right) + 1\right)} \cdot \left(0 + 0 + 1\right) = {\color{red}\left(1\right)} \cdot \left(0 + 0 + 1\right)$$Apply the dominant (null, annulment) law $$$x + 1 = 1$$$ with $$$x = 0$$$:
$$1 \cdot \left(0 + {\color{red}\left(0 + 1\right)}\right) = 1 \cdot \left(0 + {\color{red}\left(1\right)}\right)$$Apply the dominant (null, annulment) law $$$x + 1 = 1$$$ with $$$x = 0$$$:
$$1 \cdot {\color{red}\left(0 + 1\right)} = 1 \cdot {\color{red}\left(1\right)}$$Apply the identity law $$$x \cdot 1 = x$$$ with $$$x = 1$$$:
$${\color{red}\left(1 \cdot 1\right)} = {\color{red}\left(1\right)}$$Answer
$$$\left(\left(1 \cdot 0\right) + \overline{0}\right) \cdot \left(\overline{1} + 0 + 1\right) = 1$$$
The DNF is $$$\text{True}$$$.
The CNF is $$$\text{True}$$$.
The NNF is $$$\text{True}$$$.