Another proof of uniqueness of identity element of addition of vector space

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The following proof is solely based on vector space axioms.
Axiom names are italicised.
They are defined in Wikipedia (vector space).

Another proof of uniqueness of identity element of addition of vector space

We prove the uniqueness of an identity element of addition of a vector space.

Let $V$ be a vector space.
It remains to prove that an identity element of addition of $V$ is unique.
We define a predicate $P$ over $V$ as follows.
For every $v \in V$,
$$P(v) \quad := \quad \big(\, u + v = u \; \text{ for all } u \in V \,\big).$$

By Identity element of addition,
there exists an element of $V$ for which $P$ holds.
Let $0 \in V$ such that $P(0 )$ holds;
let $0′ \in V$ such that $P(0′)$ holds.
It remains to prove that $0 = 0’$.
\begin{align*}
0 & = 0 + 0′ && \text{since } P(0′) \text{ holds.} \\
& = 0′ + 0 && \text{by }\textit{Commutativity of addition.}\ \\
& = 0′ && \text{since } P(0) \text{ holds.}
\end{align*}
QED

P.S.: This is intended to improve my previous, roundabout proof.
Of course, I could only incorporate a small subset of the helpful advice given there.

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