elementary number theory - My proof of $m cdot 0 = 0 = 0 cdot m$ for all $m in mathbb{Z}$

I have the following proposition to prove:

For all $m \in\mathbb\ Z$, $m \cdot 0 = 0 = 0 \cdot m$

I can use the following axioms:

1. commutativity


2. associativity


3. distributivity


4. identity for addition ($0$)


5. identity for multiplication ($1$)


6. additive inverse


7. cancellation: Let $m,n,p$ be integers. If $m \cdot n = m \cdot p$ and $m \ne 0$, then $n = p$.

Here is my proof:

\begin{align*}
m \cdot 0 &= m \cdot (m + (-m))\\
m \cdot 0 &= (m \cdot m) + (m \cdot (-m))\\
m \cdot 0 &= (m \cdot m) +(m \cdot -1 \cdot m) \\
m \cdot 0 &= (m \cdot m) +-1 \cdot (m \cdot m) \\
m \cdot 0 &= (m \cdot m) - (m \cdot m) \\
m \cdot 0 &= 0
\end{align*}

However, I am not sure, given a simple set of axioms, that this solution is correct. More specifically, is factoring $-m$ as $-1 \cdot m$ acceptable? Or is another proposition that I should prove beforehand?

Answer

Assume that m is an integer. By the commutative property we know that m.0 = 0.m.

Now, we only need to prove only that m.0 = 0. We use m = m, then

m.1 = m.1 because 1 is the identity under multiplication.

m.(1+0) = m.1 because 0 is the identity under addition.

Using the distributive property,

(m.1)+(m.0) = (m.1)

m +(m.0) = m

-m + m +(m.0) = -m + m (-m is the inverse of m under addition.)

(-m + m) +(m.0) = (-m + m), associative property.

0 +(m.0) = 0 because the definition of the identity under addition.

m.0 = 0 Q.E.D.