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Proof of uniqueness of identity element of addition of vector space
This proof is solely based on vector space axioms. Axiom names are italicised. They are defined in Wikipedia (vector space).
Let $V$ be a vector space. We prove the uniqueness of an identity element of addition (IEOA). By Identity element of addition, there exists an IEOA. Let this element be denoted by $0$. For the sake of contradiction, we assume that the IEOA is not unique. That is, there exists an IEOA $0'$ such that $0' \ne 0$. Obviously, $V \ne \emptyset$. Let $v \in V$. By Identity element of addition, $v + 0 = v$ and $v + 0' = v$. Hence, $$v + 0 = v + 0'.$$ By Commutativity of addition, $$0 + v = 0' + v.$$ By Inverse elements of addition, there exists an additive inverse of $v$. Let $-v$ denote the additive inverse of $v$. Due to the foregoing equality, $$(0 + v) + (-v) = (0' + v) + (-v).$$ By Associativity of addition, $$0 + (v + (-v)) = 0' + (v + (-v)).$$ By Inverse elements of addition, $$0 + 0 = 0' + 0.$$ By Identity element of addition, $$0 = 0'.$$ The foregoing equality contradicts our assumption. Thus, our assumption is false and its negation is true. That is, the IEOA is unique. QED
P.S.: I wrote another version of the proof, which is less roundabout.