Negation distribution

Question

I just have a quick question on how negation distributes to universal quantifiers and predicted in first order logic. As show below:

$$ \neg(\neg\forall x \neg p(x)$$

Does this become:

$$ \forall x \neg p(x)$$

or

$$ \forall x p(x)$$

Answer 1

$\neg\neg\varphi$ is always equivalent to $\varphi$, so the first of your two alternatives is correct. To get rid of the $\neg$ following the $\forall$, you'll need to turn the quantifier into an $\exists$, i.e. you'd have $\neg\exists xp(x)$ then.

Answer 2

$$\neg(\neg\forall x (\neg p(x)))\equiv \lnot\lnot\Big(\forall x (\lnot p(x))\Big) \equiv \forall x (\lnot p(x))$$

Note that the transformation above has little to do with the presence of a quantifier; it's simply an application of double negation to a quantified statement.

Note also that what we ended with can equivalently be expressed as follows: $$\forall x (\lnot p(x)) \equiv \lnot \exists x(p(x))$$

This is due to the nature of quantifier negation ("moving negation inward," "moving negation outward"):

$$\lnot \forall x (p(x)) \equiv \exists x (\lnot p(x))$$

$$\lnot \exists x (p(x)) \equiv \forall x (\lnot p(x))$$