De Morgan's Law Operation order

Question

I have the following boolean logic: $$ \overline {\overline {\overline {B+C+D} + \overline {DA}} + \overline {\overline {\overline {A+E} + \overline { B}} + \overline {E}}} $$

I am trying to simplify the logic, what confuses me is why I cannot apply De Morgan's Law $\overline {A + B} = \overline {A} \cdot \overline {B}$.

In this case: $$ \overline {\overline {\overline {B+C+D} + \overline {DA}}} \cdot \overline {\overline {\overline {\overline {A+E} + \overline { B}} + \overline {E}}} $$

You are correct. Remember $\overline{\overline{X}} = X$. DeMorgans can be applied to the whole expression or terms as required. As stated, your formatting was a bit of a mess to read. Please check that I have edited it correctly. — StainlessSteelRat, Jun 20 '16 at 14:21
I don't understand why you claim you cannot use De Morgan's law(s). You can and you must. — , Jun 20 '16 at 16:56

score 0 · Answer 1 · edited Jul 25 '17 at 22:42

You are correct.

In general, De Morgan's Law can be applied as needed to simplify the expression.

In this case, it is best to start with the full expression as you have done.
$$\begin{align} \overline {\overline {\overline {B+C+D} + \overline {DA}} + \overline {\overline {\overline {A+E} + \overline { B}} + \overline {E}}} \\[2ex] \left(\overline {\overline {\overline {B+C+D} + \overline {DA}}}\right) \cdot \left(\overline {\overline {\overline {\overline {A+E} + \overline { B}} + \overline {E}}}\right) \\[2ex] \left(\overline {B+C+D} + \overline {DA}\right) \cdot \left(\overline {\overline {A+E} + \overline { B}} + \overline {E}\right) \end{align}$$

Then apply De Morgan's Law to terms to get the sum of products and simplify.

De Morgan's Law Operation order

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