# Predicate Logics - double negation HELP me understand

Sorry for maybe a silly question but i need to understand how ¬(¬∀x ¬A(x)) equals ∀x ¬A(x)

In my mind, the negation before the parenthesis will be applied to both ¬∀x and ¬A(x). So it would look like this:

¬(¬∀x ¬A(x)) = ¬¬∀x ¬¬A(x)

A double negation would become positive and would then let ¬(¬∀x ¬A(x)) equal to ∀x A(x).

So, why would ¬(¬∀x ¬A(x)) equal ∀x ¬A(x)? Why wouldnt ¬(¬∀x ¬A(x)) equal ∀x A(x)?

• "In my mind, the negation before the parenthesis will be applied to both ¬∀x and ¬A(x)." Consider ¬(∀y B(y)). Is it correct to apply the negation to both ∀y and B(y)? – John L. May 17 '19 at 13:49
• Your formula $\lnot(\lnot\forall x\ \lnot A(x))$ is of the form $\lnot(\lnot P)$ where $P$ is the formula $\forall x\ \lnot A(x)$. So after simplification we get $P$. – chi May 17 '19 at 13:55

$$\lnot \alpha$$ is True iff $$\alpha$$ is false.
In your example, $$\lnot (\lnot \forall x \lnot A(x))$$ is True iff $$\lnot \forall x \lnot A(x)$$ is False iff $$\forall x \lnot A(x)$$ is True iff for all $$x$$, $$\lnot A(x)$$ is True iff for all $$x$$, $$A(x)$$ is false.
Perhaps the issue would become simpler if we fully parenthesize your expression: $$\lnot \stackrel 1( \lnot \stackrel2( \forall x \stackrel3( \lnot \stackrel4( A\stackrel5( x \stackrel 5) \stackrel 4) \stackrel 3) \stackrel 2) \stackrel 1).$$ The superscripts count the nesting depth.