I am trying to translate the following statements into predicate logic using the following predicates: $E(x) = x$ is even, $P(x) = x$ is prime, $L(x,y) = x < y$
(i): Some Primes are Odd.
(ii): Every even number is greater than 1.
(iii): There are infinitely many primes.
(iv): The only even prime is 2.
My attempts were
(i): $\neg (\forall x P(x) \implies E(x))$
My reasoning behind this was that saying $\exists x(P(x) \land \neg E(x))$ just says there is at least 1, rather than some odd primes.
(ii): $\forall x E(x) \implies L(1,x)$
(iii): $\forall x \exists y (P(x) \land P(y) \land L(x,y)$
My reasoning for this one was that this means that for every prime, there is a prime that is greater than it, which means there are infinitely many primes.
I am not really sure how to go about (iv)
Any help is greatly appreciated!!
For $ (i) $, both your attempt and the proposition you argue against are actually equivalent, as $$\neg (\forall x, P(x) \rightarrow E(x)) \iff \exists x, \neg(P(x) \rightarrow E(x))$$ $$\iff \exists x, \neg( \neg P(x) \vee E(x))$$ $$\iff \exists x, P(x) \wedge \neg E(x)$$
Your attempt for $ (ii) $ is correct.
Your attempt for $ (iii) $ is incorrect, as that is stating that for all $x$, there exists $y$ such that both $x$ and $y$ are prime and that $x$ is less than $y$. But of course, not every natural number is prime, so the statement must be false. What you want is something like
$$ \forall x, \exists y, L(x, y) \wedge P(y) .$$
This loosely says that primes can be arbitrarily large.
$ (iv) $ corresponds to $$ (P(x) \wedge E(x)) \rightarrow (L(1, x) \wedge L(x, 3))$$
if we're given that $ x \in \mathbb{N} $ and only have access to predicates $L, E, P$.