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:


*

*commutativity

*associativity

*distributivity

*identity for addition ($0$)

*identity for multiplication ($1$)

*additive inverse

*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?
 A: It depends. What are your axioms? What is your definition of the notation $-m$?
If $-m$ is defined as the additive inverse of $m$, then no, you cannot factor $-m = -1\cdot m$ until you prove that this is true.
EDIT: For the particular axioms you have listed, your proof in fact may well be circular, since the most straightforward way of proving that $-1\cdot m = -m$ is to add $m$ to $-1\cdot m$ and show that this sum is zero. Here's a hint to get you started for your original problem:
$$m\cdot 0 = m\cdot (0+0)$$
by the additive identity axiom. Can you see how to take it from here?
A: 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. 
