Looked up a solution and the result I found looked nothing like what I had, and so I was wondering if a solution along these lines would work?
I tried to solve the following problem as follows:
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$0 \le x < y \implies [x \in P \land (y/x > 1)] \lor x = 0$
Case 1: $x = 0,x^n = 0$ for any nonnegative n or 0, and since 0 < y it follows $x^n = 0 < y^n$.
Case 2: $x \neq 0 \land x < y.$ If x < y yet again it implies y/x > 1
$x^n < y^n$, we divide both sides by $x^n (x \neq 0)$ and get $1 < (y/x)^n$. Since we know y/x > 1 it follows $(y/x)^n$ to any nonnegative power (except 0, which we don't consider) will still be bigger than 1.
It feels like it should work.
This is very well suited for mathematical induction... The induction base is trivial, as $0\leq x < y \Rightarrow 0 \leq x < y$. We just need to show that when $0 \leq x < y$ and $x^n < y^n$, we also have that $0 \leq x^{n+1} < y^{n+1}$. This is immediate, since $$ x^{n+1} = x x^n < x y^n < y y^n = y^{n+1} $$
In the previous inequalities I first used the induction hypothesis and then the given fact that $x < y$.