Is my proof wrong? "Prove that the additive inverse is unique" "Prove that for each x $\in V$, where $V$ is a vector space, the additive inverse $-x$ is unique."
My proof is as follows;
Let $V$ be a vector space, let $x \in V$. For contradiction, assume the additive inverse $-x$ is not unique. Then, there exists $-x'$ such that
\begin{align*}
x + (-x') = 0 \\
x + (-x) = 0 \\
\end{align*}
We can rearrange this to see
\begin{align*}
x = -(-x') \\
x = -(-x) \\
\Rightarrow -(-x) = -(-x') \\
-x = -x' \\
\end{align*}
Therefore, the additive inverse $-x$ is unique.
In the book and other places I've seen other approaches to this problem, and I'm wondering if this proof is wrong? Other proofs referred to the axioms of vector spaces but I didn't find it necessary. I guess rate my proof would be the point of this question as I'm still uncertain if my proofs are solid.
edit: The axioms referenced are the 8 axioms on Vector Space operations found in "A course in linear algebra" by David Damiano.
 A: If we have a Vector Space $(V,+)$ with binary composition +, please note that there really is only that, addition. Subtraction is defined as addition with the inverse, i.e. $x - y := x + (-y)$. Hence, this definition and way of writing only makes sense if you know already that the inverse is unique.
Hence, the proof of uniquess of the inverse should in my opinion avoid the minus sign altogether, just to optimally have no circularity (i hope that's the correct word). The proof follows just from left cancellation, since V is in particular a Monoid with left cancellation.
Let y $\in$ V. Suppose we have $x, x' \in V$ that satisfy
$$\begin{align*}
y + x = 0\\
y + x' = 0\\
\end{align*}$$
Since $x$ for example is also a left inverse of $y$, we can add it from the left
to get the following:
$$\begin{align*}
x + (y + x) = x\\
x + (y + x') = x\\
\end{align*}$$
Since $V$ is a Monoid, the associative law holds, so we have:
$$\begin{align*}
(x + y) + x = x\\
(x + y) + x' = x\\
\end{align*}$$
Since $x+ y = 0$, the second line yields
$$\begin{align*}
x' = x\\
\end{align*}$$
Just as we wanted. Notice that i actually used the associative law. I believe it is absolutely required for uniqueness of inverses. It is not required for uniqueness of identity elements, but i think for the inverses it is, feel free to correct me though.  I know this is a long answer, but i wanted to be as detailed as possible, after all that's what we mathematicians are taught to be, to avoid mistakes. This is my first ever post too, i hope it does not look too bad.
