Contrapositive of: $(x,y \in P) \implies xy \in P$, where $P$ is the set of real numbers

Question

One of the axioms of order for real numbers read:

$$(x,y \in P) \implies xy \in P$$

where P is the set of positive real numbers.

Then, the contrapositive of this statement is:

$$\sim ((x,y \in P) \implies xy \in P) = (xy \not\in P \implies (x \not\in P \; or \; y \not\in P))$$

This means if $xy$ is negative, $x \not\in P \; or \; y \not\in P$, meaning both $x$ and $y$ can be negative, which is obviously false. What am I missing?

Answer 1

Hang on, think about what you're saying. $x\not\in P$ OR $x\not\in P$ doesn't imply $x\not\in P$ AND $y\not\in P$. :-)

More detail: $xy\not\in P \Rightarrow x\not\in P \lor y\not\in P$ is true. Your concern is that $x\not\in P\land x\not\in P\Rightarrow xy\in P$ demonstrates a contradiction. But don't worry, because $x\not\in P\lor y\not\in P\not\Rightarrow x\not\in P\land y\not\in P$.