Propositional logic De-morgans theorem question

Question

the theorem states that $(A\wedge B) = \neg (\neg A\vee \neg B)$, where $A$ and $B$ are propositional formulas.
Can't I turn $\neg (\neg A\vee \neg B)$ to $(\neg \neg A\vee \neg \neg B)$ then cancel the double negations so its $(A\vee B)$, because that seems to be allowed if $A$ and $B$ were propositions.

Answer 1

By DeMorgan's, $\lnot(\lnot A\lor\lnot B)$ would be $(\lnot\lnot A \land \lnot\lnot B)$, which is what you started with. You didn't flip the $\lor$ to a $\land$ in the second case.

Answer 2

You're not allowed to just move the ¬ inside the ( ) and distribute it over each term. As RandomUser says, you have to flip the $\vee$ to a $\wedge$ at the same time.

It's true that in arithmetic $-(a+b)=-a+(-b)$, but note that $-(a \times b)$ doesn't equal $(-a) \times (-b)$. Each operation, $\vee$, $\wedge$, $+$, and $-$, has its specific properties, and what works for one doesn't automatically work for another.

Answer 3

$$\neg(\neg A \vee \neg B) \iff \neg\neg A \wedge \neg\neg B \iff A \wedge B $$