Can you combine axioms for a vector space? This is what I wrote...initially I wanted to write that it is false because an axiom is a basic property and wouldn't be so basic if you start combining them.
In the axioms for a vector space, can Axioms (2)  and (3) be replaced by a single axiom that states $(u + v) + w = v + (u + w)$
Recall that axioms (2) and (3) for a vector space are:


(2) For all $u$ and $v \in V$,   $u + v = v + u$.
             (Commutativity of vector addition)
(3) For any three vectors $u,v,$ and $w$, $(u + v) + w = u + (v + w)$.
            (Associativity of vector addition)


Yes, we can combine the axioms of commutativity and associativity into one single axiom since in order to get from the left hand side of the equation to the right hand side of the equation the idea of commutativity and associativity are displayed as follows:
$(u + v) + w =
    u + (v + w) =$.................................( associativity axiom)
$(u + v) + w =$.................................(associativy axiom again)
$(v+u) + w =$....................................(commutitivity axiom)
$v + (u + w)$ 
Therefore the above axiom has the axioms (2) and (3) combined into one axiom.
Thanks.
 A: Perhaps I am misunderstanding the question. But.


*

*Commutativity: set $w=0$, this implies $$u+v = (u+v)+0 = v+(u+0) = v+u,\forall u,v\in V,$$
where the first and thirth equality hold because $0$ is neutral for addition, the second holds by the axiom.

*Associativity:
$$ (u+v)+w = (v+u)+w = u+(v+w),$$
the first equality is commutativity, the second one is a direct application of the axiom.

A: Yes, you can combine them. A binary operation where the following three axiom hold:


(1) there is an element $0 \in V$ that for any $u \in V$, $u + 0 = 0 + u = u$.
(2) For all $u$ and $v \in V$,   $u + v = v + u$.
             (Commutativity of vector addition)
(3) For any three vectors $u,v,$ and $w$, $(u + v) + w = u + (v + w)$.
            (Associativity of vector addition)


are equivalent to 


(1) there is an element $0 \in V$ that for any $u \in V$, $u + 0 = 0 + u = u$.
(4) For any three vectors $u,v,$ and $w$, $(u + v) + w = v + (u + w)$


You showed that(4) can be derived from (2),(3) and @Myself showed  that (1),(2),(3) can be derived from (1),(4). 
a vectorspace satisfies (1) (2) (3) so axioms (2)(3) can be replaced by (4)
On the other hand if you omit axiom (1) the statements (2),(3) are not equivalent to (4). You showed that (4) follows from (2)(3) even if (1) does not hold. But (2)(3) does not follow from (4) if (1) does not hold. The binary operation $+$ defined by


$u + v := v$


satisfies (4) (and (3)) but not (2).
