What is the relationship regarding a condtional worded as "not a necessary condition"

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I came across a problem the other day in my discrete mathematics class, and while I understand the relationships of necessary and sufficient, I could not get passed the wording of this one. "Being not odd is not a necessary condition for an integer to be not prime."

If p means an integer is composite (not prime), and q means an integer is even (not odd), then "not odd is a necessary condition for an integer to be not prime" would be represented as $p\rightarrow q$ (please correct me if I'm wrong there as well). But what would it be when the wording is "...NOT a necessary condition for..." Would it simply be the negation of the conditional?

This wording came after a universal quantifier.

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"Being not odd is not a necessary condition for an integer to be not prime."

This is the negation of: "Being not odd is a necessary condition for an integer to be not prime.", so let's first think about this positive statement.

Since in general, "$P$ is a necessary condition for $Q$ translates as $Q \to P$, the positive statement is of the form:

$\neg Prime(x) \to \neg Odd(x)$

though that should of course be universally quantified: "For every number $x$: not being odd is a necessary condition for it to be not prime", and so we get:

$\forall x (\neg Prime(x) \to \neg Odd(x))$

OK, so that's the positive statement. But the original statement is the negation of that, so that works out to:

$\neg \forall x (\neg Prime(x) \to \neg Odd(x))$

which is equivalent to:

$\exists x (\neg Prime(x) \land Odd(x))$

And that all makes sense:

To say that "Being not odd is not a necessary condition for an integer to be not prime." is to say that "Being even is not necessary to not be a prime". And that is a true statement, because there are all kinds of odd numbers that are not prime ... which is the very translation of $\exists x (\neg Prime(x) \land Odd(x))$