How should I interpret $y-x$ in this context? I asked a related question here. I am meticulously working my way trough Mathematical Analysis by Apostol. I am reading about axioms, specifically axiom 4:


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*Given any 2 real numbers $x$ and $y$, there exists a real number $z$ such that $x + z = y$. This $z$ is denoted by $y - x$; the number $x
   - x$ is denoted by $0$. (It can be proved that $0$ is independent of $x$.)  We write $-x$ for $0 - x$ and call $-x$ the negative of $x$.


In the related question, it was concluded that $y-x$ should not be interpreted as $y + (-x)$; it can be proven to be such, but it does not follow directly from the axiom. 
The problem is that the only operators we have explicitly assumed in the book are addition and multiplication. 
So my question is, I'm supposed to treat the $"-"$ like a meaningless symbol for now (just a part of the symbol representing $z$), or am I supposed to take it as subtraction but the author just didn't bother to explicitly assume subtraction? Am I allowed to break it down and write $y + y - x = 2y - x$? (I think I would probably need to prove that $x + x = 2x$ before anything though.) Or is there another way I should think about it?
The axioms that were given before this were the associative, commutative, and distributive laws.

You should interpret $y-x$ as just a whole symbol when representing $z$ in the axiom 4. The first sentence does not define $-x$, the last sentence does instead.
What immediately follows from the first sentence is that there exists a real number $z$ such that $x+z=x$. We symbolically denote this $z$ by $x-x$, or alternatively $0(x)$. Notice that I didn't omit "$(x)$" for $0$. What is special about zero is that $0(x)$ is independent of $x$, that is, $0(x)=0(y)$ for $\forall x,y\in \mathbb{R}$. This follows from commutative laws and associative laws of addition, and uniquness of $0(x+y)$ (which is guaranteed in the first sentence of the axiom 4). Given this, we can safely denote $0(x)$ simply by $0$, and it immediately follows from the first sentence that there exists a real number $w$ such that $x+w=0$. While a 'straightforward' notation of $w$ is $0-x$, we denote it by $-x$ in practice. Thus the negative of $x$ is finally defined, and we can easily check that $y-x$ defined in the first sentence coincides with $y+(-x)$.
What the discussion above tells us is that $x+(-x)=0$ is trivial; it is the definition of $-x$! Note that we don't have to prove the uniquness of $-x$, because it is guaranteed by the first sentence of the axiom 4. Also, you can show that $y+(y-x)=(y+y)-x$ without knowledge of $y-x=y+(-x)$ (if you have any difficulty in proving this, please let me know).
Appendix: Proof of the uniqueness of zero
Let $x,y\in\mathbb{R}$. Associative laws of addition and definition of $0(x)$ indicate $$(x+y)+0(y)=x+(y+0(y))=x+y.$$ Commutative laws of addition is additionally required to show that $$(x+y)+0(x)=x+(y+0(x))=x+(0(x)+y)=(x+0(x))+y=x+y.$$
After all, we obtain $(x+y)+0(x)=x+y$ and $(x+y)+0(y)=x+y$, and uniqueness of $0(x+y)$ requires $0(x)=0(y)$.
