Proof of $(-a)b=(-b)a$ starting from axioms

Question

Prove from axioms that $ \forall_{a,b\in\mathbb{R}}:(-a)b=(-b)a$ , but $a0=0$ is allowed.

My attempt: $LHS=(-a)b=(-a)b+0=(-a)b+(ba+(-ba))=((-a)b+ba)+(-ba)=b(-a+a)+(-ba)=b0+(-ba)=0+(-ba)=-ba$

So, now I need show that $-ba=(-b)a$

Answer 1

You're almost there, by the looks of it. You just need to prove that $ba + (-b)a = 0$, so that $(-b)a = -(ba)$ by uniqueness of additive inverses. But you can do this using distributivity, commutativity and the rule $a0=0$: $$ba+(-b)a = (b-b)a = 0a = a0 = 0$$ [If you haven't proved uniqueness of additive inverses, then you should do that first.]