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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)?

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    $\begingroup$ "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)? $\endgroup$
    – John L.
    Commented May 17, 2019 at 13:49
  • $\begingroup$ 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 $P$. $\endgroup$
    – chi
    Commented May 17, 2019 at 13:55

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The semantics of negation are very simple:

$\lnot \alpha$ 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.

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