Mixing logical and mathematical notation when negating statements

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Working through Hammack's Book of Proof I have come at an impasse. Exercise 3 of Section 2.10 requires that you negate proposition (1). The answer that naturally follows and is confirmed by the book's solutions is proposition (2).

(1) For every prime number $p$, there is another prime number $q$ with $q > p$.
(2) There is a prime number $p$ such that for every prime number $q$, $q \leq p$.

This makes sense given the notation given by the book up to now. Assuming a set of all the prime numbers, $P$, the negation of $$\forall p, p \in P, \exists q, q \in P, q > p$$ is simply $$\exists q, q \in P, \forall p, p \in P, q \leq p$$ As if the negation proceeded as follows:

$$\neg (\forall p, p \in P, \exists q, q \in P, q > p)$$

$$\exists p, p \in P, \neg(\exists q, q \in P, q > p)$$

$$\exists q, q \in P, \forall p, p \in P, \neg (q > p)$$

$$\exists q, q \in P, \forall p, p \in P, q \leq p$$

I say "as if" because here comes my question: I also want to express these two propositions using nothing but logical operators. However I cannot seem to do it in a satisfactory way.

Take the above sentences, using the glossary:

$P(x)$: $x$ is a prime number
$L(x, y)$: $x$ is larger than $y$

The two sentences look like they should have the similar translations, as apart from the negated quantifiers and the negated last predicate, they both share the same shape. However I cannot find a good translation that suits this constraint.

For example, if (1) is taken to mean

$$\forall p \exists q ((P(p) \land P(q)) \rightarrow L(p, q))$$

then (2) should be

$$\exists p \forall q ((P(p) \land P(q)) \rightarrow \neg L(p, q))$$

but this cannot be as the negation of (1):

$$\exists p \forall q ((P(p) \land P(q)) \land \neg L(p, q))$$

is not the same as (2).

If (1) is taken to mean

$$\forall p \exists q (P(p) \land P(q) \land L(p, q))$$

then (2) should be

$$\exists p \forall q (P(p) \land P(q) \land \neg L(p, q))$$

which is not the same as the negation of (1) which is

$$\exists p \forall q (\neg P(p) \lor \neg P(q) \lor \neg L(p, q))$$

and so on.

What am I missing?