Proving that the axioms for addition hold in $R$ -- Associativity. Principles of Mathematical Analysis by Walter Rudin. I'm studying the proof for theorem 1.19 from Principles of Mathematical Analysis by Walter Rudin (public online copy here).

There exists an ordered field $R$ which has the least-upper-bound property.
Moreover, $R$ contains $Q$ as a subfield.

The lengthy proof for this theorem is contained in the appendix to chapter 1 on page 17.
I am stuck on Step 4:

If $\alpha \in R$ and $\beta \in R$ we define $\alpha + \beta$ to be the set of all sums $r + s$, where $r \in \alpha$ and $s \in \beta$.
We define $0^*$ to be the set of all negative rational numbers. It is clear that $0^*$ is a cut. We verify that the axioms for addition (see Definition 1.12) hold in $R$, with $0^*$ playing the role of $0$.

Specifically, the author does not prove associativity (A3):

(A3) As above, this follows from the associative law in $Q$.

I don't really understand what the author is saying here; nor am I sure how to demonstrate this myself (I've tried).
Definition 1.12:

(A3) Addition is associative: $(x + y) + z = x + (y + z)$ for all $x, y, z \in F$.

The entire reason students such as myself buy textbooks is to have these main concepts explained to them, which Rudin has, for whatever reason, decided not to do. I would greatly appreciate it if people could please take the time to fill this gap by proving the associative property.
 A: For $x, y, z \in R$, which are 3 cuts of $Q$, you have to prove that:
$$(x+y)+z = z+(y+z).$$
What is an element of the cut $(x+y)+z$? It is the sum of an element $p + s$ where $p \in x+y$ and $s \in z$. And the element $p$ can be written (by definition of the cut $x+y$) as $q+r$ where $q \in x$ and $r \in y$.
Therefore an element of $(x+y)+z$ is having the form $(q+r) + s$ where $(q,r,s) \in (x,y,z)$. Now you can use the associative property of the addition in $Q$ to say that $(q+r)+s = q+(r+s) \in x+(y+z)$. We have proven that:
$$(x+y)+z \subseteq x+(y+z)$$.
You can prove in a similar way the converse inclusion which brings you to the conclusion.
A: As discussed in the comments to mathcounterexamples.net's answer, there seem to be errors? 
I think a correct proof is as follows:
We want to show that $(x + y) + z = x + (y + z)$.
Let $p \in x + y$. Then $p = r + s$, where $r \in x, s \in y$.
Let $c \in z$.
$\therefore (x + y) + z = (r + s) + c$
$(r + s) + c = r + (s + c) \in x + (y + z)$
$\therefore (x + y) + z \subset x + (y + z)$
