When I can reverse the logical operators? I heard say that is logically equivalent to say it:
$$\neg (p \vee q) = p \land q$$
So every time you have a negation operator in front can make a "distributive" altering the operator from within? And if $\neg(p \vee \neg q)$ then result is $p \land \neg q$? Can anyone give some examples?
 A: No, that does not work. What does work is De Morgan's law:
$$ \neg(p\lor q) \iff \neg p \land \neg q $$
Your formulation is missing the negations on the right-hand side.
If you want, you can think of this as "negation distributes over $\land$ and $\lor$, except that it changes each of them to the other". I'm not entirely sure that's a helpful way of thinking, but whatever works for you ...
A: If I can add something:
The De Morgan's Laws holds exclusively for conjunction and disjunction, thus having something to do with their duality. It is a pair of laws, expressed informally as follows:

The negation of a conjunction is the disjunction of the negations.
The negation of a disjunction is the conjunction of the negations.

Symbolically:
$$¬(p\land q) \Leftrightarrow ¬p\vee ¬q$$
$$¬(p∨q)\Leftrightarrow ¬p∧¬q$$
But as Makholm observed, your formulation neglects the inner negations in the RHS.
A: Example, as requested...
Suppose I want to know what is the domain of the function
$$
\frac{1}{x^2-4x+3}
$$
That is, I want to find all the $x$ where that denominator is not zero.
So I factor the denominator.  $x^2-4x+3=(x-1)(x-3)$.
When is it zero?  Either $x=1$ or $x=3$.  So when is it nonzero?
$$
\neg\;\big(x=1\;\lor\;x=3\big)
$$
That is, (by the principle explained in Henning's answer)
$$
x\neq 1\;\land \;x\neq 3
$$
In order to belong to the domain, $x$ must be different from $1$ and different from $3$.
