Proof of $0 \cdot\vec{v}=\vec{0} $ for all $ \vec{v}$ in the vector space According to this webpage from this Lemma 1.17, which states:

In any vector space  V , for any  $\vec{v}\in V  \text{ and } 
r \in\mathbb{R} ,\text{ we have } $
$$0\cdot\vec{v}=\vec{0} $$

Their Proof:
For 1, note that  $\vec{v}=(1+0)\cdot\vec{v}=\vec{v}+(0\cdot\vec{v})$ . Add to both sides the additive inverse of  $\vec{v}$ , the vector  $\vec{w} $ such that  $\vec{w}+\vec{v}=\vec{0}$ .
\begin{array}{rl}
\vec{w}+\vec{v}
&=\vec{w}+\vec{v}+0\cdot\vec{v} \\
\vec{0}
&=\vec{0}+0\cdot\vec{v} \\
\vec{0}
&=0\cdot\vec{v}
\end{array}
My Question:
Intuitively, I understand this proof. However, my concern is with its rigor. How do they know $\vec{v}=(1+0)\cdot\vec{v}=\vec{v}+(0\cdot\vec{v})$ based on the weblinks 10 definitions on the vector space? What definition(s) are they using to justify this?
To me , it seems a bit circular because they say hey we are going to show  $0\cdot\vec{v}=\vec{0} $, but we are going to use $\vec{v}+(0\cdot\vec{v})=\vec{v}$, implying $(0\cdot\vec{v})$ is a zero vector.
 A: They actually don't use the fact that $0 \cdot \vec{v} = \vec{0}$ in the proof but I understand (firsthand) how easy it is to be confused by that.
What actually happens is they establish that $0 \cdot \vec{v} = \vec{0}$ by starting with this equation:
$$\vec{v} = (1 + 0) \cdot \vec{v}$$
This equation is completely fine because of course $1 + 0 = 1$, and of course $\vec{v} = 1 \cdot \vec{v}$ (this is condition #10 in Definition 1.1).
The next step is to use condition #7 in Definition 1.1.  This gives us:
$$(1 + 0) \cdot \vec{v} = 1 \cdot \vec{v} + 0 \cdot \vec{v}$$
But since $1 \cdot \vec{v} = \vec{v}$, then we can simplify this and say:
$$(1 + 0) \cdot \vec{v} = \vec{v} + 0 \cdot \vec{v}$$
Combining this with the first equation I wrote above, we have
$$\vec{v} = \vec{v} + 0 \cdot \vec{v}.$$
Next they add $\vec{w}$ to both sides, where $\vec{w}$ is the additive inverse of $\vec{v}$.  Such a $\vec{w}$ is guaranteed to exist by condition 5 in Definition 1.1.  This gives us:
$$\vec{v} + \vec{w} = \vec{v} + \vec{w} + 0 \cdot \vec{v}$$
Now, because $\vec{w}$ is the additive inverse of $\vec{v}$, then by definition of additive inverse we have $\vec{v} + \vec{w} = \vec{0}$.  So the equation simplifies to:
$$\vec{0} = \vec{0} + 0 \cdot \vec{v}$$
And by condition 4 in Definition 1.1, $\vec{0} + 0 \cdot \vec{v} = 0 \cdot \vec{v}$.  Therefore the equation simplifies to:
$$\vec{0} = 0 \cdot \vec{v}$$
Finally, we conclude that $0 \cdot \vec{v}$ must equal $\vec{0}$.
A: Personally, I prefer:
$0 = 0 + 0$ (field axiom)
$0v = (0 + 0)v$ (multiplying by the same scalar on both sides)
$0v = 0v + 0v$ (vector space axiom - distribution of scalar multiplcation over vector addition)
$0v - 0v = (0v + 0v) - 0v$ (subtracting $0v$ from both sides)
$0 = 0v + (0v - 0v)$ (associative law of vector addition - I probably should have added $-0v$ to both sides, but I get lazy)
$0 = 0v + 0$ (some identity something-something for vector spaces)
$0 = 0v$ (where we wanted to go)....
