# Request to vet the logical rep. for the negative of $p\mid p^l$, for $p$ a prime, $l \in \mathbb{Z+}$

Intend to state the logical representation of the negation of the above proposition.

For the given proposition, the logical equivalent is: $\forall l \in \mathbb{Z+}, \exists n \in \mathbb{Z}, np = p^l.$
The logical equivalent reads: for all positive integer values of $l$, there exists some integer $n$ s.t. the equality $np = p^l$ holds true.

So, the negation of the given proposition will be :$\lnot (\forall l \in \mathbb{Z+}, \exists n \in \mathbb{Z}, np = p^l) = \exists l \in \mathbb{Z+}, \forall n \in \mathbb{Z}, np \ne p^l.$
The logical equivalent for the negated proposition reads:
for some positive integer value(s) of $l$, for all integers $n$, the equality $np = p^l$ never holds true.

By the above elaboration, it seems that the variable $n$ doesn't hold significance, and consequently can state the negated proposition's logical equivalent as:
$\lnot (\forall l \in \mathbb{Z+}, \exists n \in \mathbb{Z}, np = p^l) = \exists l \in \mathbb{Z+}, p \nmid p^l.$

• Not clear... It is correct that $\lnot (\forall l \ \exists n \ \varphi)$ is equiv to $(\exists l \ \forall n \ \lnot \varphi)$. – Mauro ALLEGRANZA Feb 27 '18 at 16:15
• But this does not mean that you can omit $n$. – Mauro ALLEGRANZA Feb 27 '18 at 16:15
• To say that $p \nmid p^l$ means exactly that for no $n$ we have : $np=p^l$. – Mauro ALLEGRANZA Feb 27 '18 at 16:17
• @MauroALLEGRANZA Then the existential quantifier for $l$ is insignificant for all practical purposes in the negated proposition. In fact, even though $l$ has the existential quantifier as : $\exists$, it serves as if the $\forall$ is in force actually. By that I mean that $\forall l \in \mathbb{Z}, \forall n \in \mathbb{Z}, np\ne p^l$ If so, then does the correct way to interpret the significance of the existential quantifiers depends on context? – jitender Feb 27 '18 at 16:24
• Not exactly... If you write $p∤p^l$, this already includes the quantification on $n$, because it is simply: $\forall n (p^l \ne pn)$ i.e. $\lnot \exists n (p^l=pn)$. But you want to assert it for some $l$ or for all $l$, and this needs the appropriate quantification on $l$. – Mauro ALLEGRANZA Feb 27 '18 at 16:47

The negation of the formula is correct: $¬(∀l \ ∃n \ \varphi)$ is equivalent to $(∃l \ ∀n \ ¬ \varphi)$.

We have the true fact that $p∣p^l$, for every $l \in \mathbb Z_{+}$, $p$ divides every power of $p$.

What is its negation ? There is a power of $p$ such that $p$ does not divide it, i.e.

$∃l \ (p^l \text { is not a multiple of } p)$,

and this is $p∤p^l$, for some $l \in \mathbb Z_{+}$.

If we unwind it we get:

$∃l \ \forall n \ (p^l \ne pn)$,

and this is exactly what you have written.