1
$\begingroup$

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)?

$\endgroup$
  • 1
    $\begingroup$ "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)? $\endgroup$ – Apass.Jack May 17 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 May 17 at 13:55
1
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.