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?
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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 $3$.
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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.
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 ...