Please read math.stackexchange.com/q/30127. It is not a proof by contradiction.

The antecedent is true for every set $P$ of natural numbers except for the sets $P = \{\ \}$ and $P = \{1\}$. The consequent is true only when $P$ is an infinite set. There exist many, many, many, many sets of natural numbers that are not infinite.

However, in saying that N is the set of natural numbers, I meant that N is the set of all natural numbers. Hence, I use "the" as in "the set of natural numbers" rather than "a".

Besides this, P stands for the set of prime numbers, which your comment didn't specify.

While (3) happens to be true because the consequent is true, as a general step in a proof it is nonsense: replace $P$ for example by $\{1,2\}$. Then $\exists X\exists Y(X\in\mathbb{N}\wedge Y\in P\wedge Y>X)$ is true but $\forall X\exists Y(X\in\mathbb{N}\wedge Y\in P\wedge Y>X)$ is blatantly false. So how do you plan to justify (3) in the case $P=$primes without first proving the infinitude of primes anyways?

@NoahSchweber, but is (3) a tautology though because of its logical form?

"(3) is a tautology." How on earth can you think that? The statement is saying if there exist an $x$ and $y$ so that there is a $y \in P$ (and $P$ can be ANY set) where $y > x$ then somehow for every $x$ there will always be a larger $y$ in $P$. That should be obviously absurd. Let $P=$ all integers less than $100$ the as $23 < 59$ you are going to conclude that if $x=678$ there must be a $y$ so that $678 < y < 100$???? Note, you proof never did anything about primeness so if is true for primes it must be true for everything. Obvious not everything is infinite.

"but is (3) a tautology though because of its logical form? " what form is that? Are you claimin $\exists x\exists y (Q(x,y)) \to \forall x \exists y (Q(x,y))$ is a tautology? It most certainly is not.

@fleablood, it should have looked like this: ∃X∃Y((X∈N)∧(Y∈P)∧(Y>X))→∀X∃Y((X∈N)→(Y∈P)∧(Y>X))

@fleablood, in generalized form it is this ∃X∃Y(N(X)∧P(Y)∧G(Y))→∀X∃Y(N(X)→P(Y)∧G(Y))

I repeat, why would you think that is a tautology? It obviously is not.

@fleablood, did you use a logic calculator to see whether or not what I said is a tautology. I think it is a tautology because I used the following logic calculator to figure it out: somerby.net/mack/logic/en/index.html

A counter example would be $P = \{7\}$ and $x = 3$ and $y=7$. Then the antecedent is true. As for the the counter example let $x = 8$. There is is no $y$ so that $y \in P$ and $y > 8$.... Basically you are claiming if a single example exists, then something must be true for all natural numbers. That's obviously false in general.

I didn't need to because it is obviously false. But I tried plugging it into your calculator and got "Error: Expected predication or parenthesized expression, found "∃X∃Y(N(X)∧P(Y)∧G(Y))→∀X∃Y(N(X)→P(Y)∧G(Y))"" I'm not going to waste time trying to figure out the syntax to make this work.

@Fleablood, before you judge my proof, may you look at this: umsu.de/trees/…

@NoahSchweber, please look at this. For my proof is a form of this: umsu.de/trees/…

Here we go! Run this instead: ∃X∃Y(N(X)∧P(Y)∧G(X,Y))→∀X∃Y(N(X)→P(Y)∧G(X,Y)). You will get that that is INVALID. Here $G(X,Y)$ must somehow relate $x$ and $y$. In your case you were just claiming $G(y)$ that only pertains to $y$ and has nothing to do with $X$.

The thing that really should have raised alarms is your symbolic proof was completely divorced from any meaning of the natural numbers, primeness and the relative size of numbers. If such a meaning-divorced proof is ever true it must always be true about everyhing. As it stand you statement states "For any $x,y$ where $x\in \mathbb N$ and $y$ is prime and $G(y)$ then for any $x$ there exists a $y$ where $x\in \mathbb N\implies y$ is prime and $G(y)$". That is very true because the $y$ in the consequence can be the exact same $y$ as the $y$ in the antecedent.