You don't really need to prove anything. All you need to do is show that the null set fits the properties of 0. For example, let's say I define a group , where Z is the set of integers and is the regular addition but remove one from the sum, ie for all a,b in Z. In this case, 0 is not the identity for the group, because . The identity is 1, since . And that's all I need to do to show that 1 takes the place of 0 in the axiom.

So for you all you'd need to do is say that a+null set = a for any set a. Therefore, the null set can be used interchangeably with 0 when talking about sets.

I hope my explanation makes sense. It's these simple parts that are usually the toughest to explain well...