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! 
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}$.
