I wanted to check if my proof of this question is sufficient. This is the question:
Prove that for all $a \in \mathbb{R}$, $a \cdot 0 = 0$.
And the proof:
Lets assume $a=a$ which implies $a-a=0$. We must show that $a(0) = 0$. By (A7), $a(1) = a$ and by (A9) $a(0+1) = a(1) = a$. By (A3) $a + a(0) = a$ and therefore by subtracting $a$ from each side, $a(0) = 0$.
The axioms are here: field axioms
On
Let $x=a\cdot 0$. Then $x + x = a\cdot 0 + a \cdot 0 = a\cdot(0+0)= a \cdot 0 = x$ where we use A9 first and then A3 to see that $0+0=0$.
Now use the existence of $-x$ (A4): $x+x = x$ so $(x + x) + (-x)=x + (-x)$ (adding $-x$ to both sides of an equality).
But $x + -x = 0$ by A4, while A2 + A4 + A3 tells us that $(x+x) + -x = x + (x + -x)=x +0=x$. So $x=0$ as we wanted to show.
This looks pretty good! A few comments:
What is the purpose of your first sentence “Let's assume $a=a$ which implies $a-a=0$.”? There's no need to explicitly assume that $a=a$; it goes without saying. Furthermore, the statement $a-a=0$ isn't needed or used anywhere else in the proof.
When you're writing your first proofs with these axioms, it helps to put each equation on its own line, citing the axiom used on that line. For instance, you write “By (A9), $a(0+1) = a(1) = a$.” But only the first $=$ is due to (A9); the second is due to (A7).
Henno's answer reminds me that “subtracting $a$ from both sides” is a bit too quick when you're going step by step with the field axioms. So do as he suggests, and instead add $(-a)$ to both sides, regroup, and cancel.