Expression conversion using de Morgan's laws I'm sorry strongly, because it's a very dummy question... I have an example in the algebra of logic. I need to convert an expression using the rules of de Morgan - replace by the conjunction of disjunctions, and disjunction - on the conjunction. 
I have a solution, but teacher said that it isn't correct. Please help! 
 A: Your middle term would have to be inverted.  Your first term is correct, but the rest are wrong. 
DeMorgan's is used to get all terms the same AND/OR to allow the expression to be simplified.  
$$\overline {x1\ x4} + x2\ \overline {x3}\ x4 + \overline{x1 + \overline {x1} \ x2 \ x3\ x4}$$
Take deMorgan's to eliminate NORs.
$$\overline {x1}+\overline {x4} + x2\ \overline {x3}\ x4 + \overline{x1} \cdot \overline{\overline {x1} \ x2 \ x3\ x4}$$
Take deMorgan's to eliminate NAND in last term.
$$\overline {x1}+\overline {x4} + x2\ \overline {x3}\ x4 + \overline{x1} \cdot ({x1}+\overline{x2}+\overline{x3}+\overline{x4})$$
$$\overline {x1}+\overline {x4} + x2\ \overline {x3}\ x4 + \overline{x1}{x1}+\overline{x1}\ \overline{x2}+\overline{x1}\ \overline{x3}+\overline{x1}\ \overline{x4}$$
Sum of Products, so it can be simplified. 
Apply complement: $\overline X \cdot X = 0$ and identity $X + 0 = X$.
$$\overline {x1}+\overline {x4} + x2\ \overline {x3}\ x4 + 0 +\overline{x1}\ \overline{x2}+\overline{x1}\ \overline{x3}+\overline{x1}\ \overline{x4}$$
Look for common factors (Distributive) and apply annulment $X + 1 = 1$.
$$\overline {x1} (1+\overline{x2} + \overline{x3} + \overline{x4}) + \overline {x4} + x2\ \overline {x3}\ x4$$
$$\overline {x1} + \overline {x4} + x2\ \overline {x3}\ x4$$
The $x4$ term is not needed (redundant) because of the opposite term $\overline {x4}$.
So $ F = \overline {x1} + \overline {x4} + x2\ \overline {x3}$.
