I am working through Apostol's Calculus I and just read about the order axioms. He presents a new undefined concept called positiveness, gives the axioms and then defines symbols $<, >, \leq, \geq$ in terms of this new concept of positiveness. Apostol then states
Thus we have x > 0 if and only if x is positive.
I attempted a proof of this before reading other proofs and quickly ran into something I know is incorrect (I read other proofs later) but haven't developed an intuition for, hoping someone can set me straight.
I'll provide the definitions and axioms from Apostol for reference, then show my attempt.
From Apostol
We shall assume that there exists a certain subset $\mathbb{R}^{+} \subset \mathbb{R}$ called the set of positive numbers, which satisfies the following three order axioms:
Axiom 7: $\text{If } x \text{ and } y \text{ are in } \mathbb{R}^{+}, \text{ so are } x + y \text{ and } xy$
Axiom 8: $\text{For every real } x \neq 0 \text{, either } x \in \mathbb{R}^{+} \text{ or } -x \in \mathbb{R}^{+}, \text{ but not both }$
Axiom 9: $0 \notin \mathbb{R}^{+}$
...classifies the symbols...
(1) $x < y$ means that $y-x$ is positive;
(2) $y > x$ means that $x < y$;
(3) $x \leq y$ means that either $x < y$ or $x = y$;
(4) $y \geq x$ means that $x \leq y$
Thus we have x > 0 if and only if x is positive.
My attempt $$\begin{align} x > 0 &\iff x \in \mathbb{R}^{+}\\ x > 0 &= x-0 \in \mathbb{R}^{+} &\text{(2) and (1)}\\ x - 0 &= x + (-0) &\text{theorem 1.3: b - a = b + (-a)} \\ x + (-0) &= x + (0) &\text{exercise in the previous section}\\ x + 0 &\notin \mathbb{R}^{+} &\text{ this is where I'm hung up } \end{align}$$
I arrive at the last line because we know $0 \notin \mathbb{R}^{+}$ from axiom 9. Then, from axiom 7, $(x + 0) \notin \mathbb{R}^{+}$.
Perhaps I can't conclude that because axiom 7 is in form $P \implies Q$ which allows for $x + y \in \mathbb{R}^{+}$ to be true even when the statement $(x \in \mathbb{R}^{+} \wedge y \in \mathbb{R}^{+})$ is false. Am I correct in my assessment of where I made the mistake?