Using the definition of a vector space to prove that $0x = 0$ and $(-1)x = -x$ Let $x$ be an element of a vector space. I am trying to use the definition of a vector space to prove that $0x = 0$ and $(-1)x = -x$.
Let $w + x = x$  for all $x \in E$.
$$\begin{align} 0x &= 0(w + x) \\ &= 0w + 0x \\ &= 0w + 0(-w) \\ &= 0w - 0w \\ &= 0 \end{align}$$
Now, I think this is correct for proving that $0x = 0$, but I don't think it actually proves that $(-1)x = -x$, but rather just assumes it to be true. Am I correct in my assessment of this? If so, I would greatly appreciate it if people would please take the time to explain how we are supposed to prove both of these at the same time.
EDIT
The definition of vector space is given as follows:

By a vector space we mean a nonempty set $E$ with two operations:
$(x, y) \mapsto x + y$ from $E \times E$ into $E$ called addition.
$(\lambda, x) \mapsto \lambda x$ from $\mathbb{F} \times E$ into $E$ multiplication by scalars,
such that the following conditions are satisfied for all $x, y, z \in E$ and $\alpha, \beta \in \mathbb{F}$:
(a) $x + y = y + x$;
(b) $(x + y) + z = x + (y + z)$
(c) For every $x, y \in E$ there exists a $z \in E$ such that $x + z = y$;
(d) $\alpha (\beta x) = (\alpha \beta) x$;
(e) $(\alpha + \beta)x = \alpha x + \beta x$;
(f) $\alpha(x + y) = \alpha x + \alpha y$
(g) $1x = x$
Elements of $E$ are called vectors. If $\mathbb{F} = \mathbb{R}$, then $E$ is called a real vector space, and if $\mathbb{F} = \mathbb{C}$, $E$ is called a complex vector space.

From chapter 1 of Introduction to Hilbert Spaces with Applications, third edition, by Debnath and Mikusinski.
I found this (see 3.(c) on page 5) solution:

Since $0x = (0 + 0)x = 0x + 0x$, it follows that $0x = 0$. Then
$$\begin{align} x + (-1)x &= 1x + (-1)x \\ &= [1 + (-1)]x \\ &= 0x \\ &= 0 \end{align}$$
Therefore $(-1)x = 0 - x = -x$.

How is the first line of this solution valid reasoning?

Since $0x = (0 + 0)x = 0x + 0x$, it follows that $0x = 0$.

In order to deduce from $0x = 0x + 0x$ that $0x = 0$, doesn't one already have to assume that $0x = 0$?
 A: I'll use the vector space axioms given here. Let $V$ be a vector space over a field $K$ and let $a \in K$, $\mathbf{v} \in V$. By the definition of $0$, $$0\mathbf{v}=(a+(-a))\mathbf{v}=a\mathbf{v} - a\mathbf{v}=a(\mathbf{v}-\mathbf{v})=a\mathbf{0}=\mathbf{0}.$$
All these statements follow directly from the axioms given on Wikipedia.
A: Indeed, the axioms as you give them don't even explicitly state that a zero vector exists, and therefore that fact first has to be proven.
What axiom (c) guarantees is that for every vector $x$ there exists a vector $0_x$ such that $x + 0_x = x$. Note that at this point we don't know yet that $0_x$ is unique, nor that $0_x = 0_y$ for $x\ne y$.
Let us first prove uniqueness of $0_x$ for given $x$. For this, let us first note that axiom (c) also guarantees us a vector $n_x$ so that $x+n_x=0_x$.
Now let's assume there is a second vector $0'_x$ such that $x+0'_x=x$. Again, axiom (c) guarantees us an element $n'_x$ such that $x+n'_x = 0'_x$.
Thus we have
$$0_x = x + n_x = (0'_x + x) + n_x = 0'_x + (x+n_x) = 0'_x + 0_x = (n'_x + x) + 0_x = n'_x + (x + 0_x) = n'_x + x = 0'_x$$
That is, the element $0_x$ is indeed unique for each $x$. We also can easily verify that $0x = 0_x$:
We have $x = 1x = (1+0)x = 1x + 0x = x+0x$. But we have seen above that the solution $x+y=x$ has the unique solution $0_x$, therefore $0x=0_x$ (and I'll use that notation exclusively from now on).
What remains to be shown is that $0x = 0y$ for $x\ne y$.
Be $z$ the vector such that $0x+z=0y$; this vector is guaranteed to exist by axiom (c). Adding $0x+0y$ to both sides gives us $0x+0y+z=0x+0y$, therefore $z=0(0x+0y) = 0x+0y$. Inserting into the original equation gives us $0x+0x+0y=0y$, that is, $0x+0y=0y$. By exchanging the roles of $0x$ and $0y$ we also get $0x+0y=0x$, and therefore $0x=0y$. Since $x$ and $y$ were arbitrary, we have now established that all products of the form $0x$ result in the same vector, which by convention we call $0$.
Now that we have showed the existence and uniqueness of $0$, we can also show the uniqueness of $n_x$ for given $x$: Let's assume there are $n_x$ and $n'_x$ such that $x+n_x = x+n'_x = 0$. Then we have $n_x = n_x + 0 = n_x + x + n'_x = 0 + n'_x = n'_x$.
Having shown the uniqueness of $n_x$, we can now also give it the more conventional name $-x$.
With this in place, it is now easy to show that $(-1)x=-x$, using the definition that $-x$ is the element that fulfils $x+(-x)=0$:
$$x+(-1)x = 1x+(-1)x = (1+(-1))x = 0x = 0$$
The uniqueness of $-x$ then guarantees that $(-1)x=-x$.
