Showing that a Boolean algebra is a Boolean ring

I've proved that a Boolean ring is a Boolean algebra but I am having trouble with the converse. The operation for + is defined as the symmetric difference for elements $a$ and $b$ from the Boolean algebra and multiplication is standard juxtaposition for $\lor$, $\land$ respectively. I know I must show that under "addition" the Boolean algebra is an Abelian group and then show associativity for multiplication and then the distributive laws. I am stuck with how to show closure of "addition", is it even necessary? I just need a few hints to get on with the proof. Thanks.

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Closure just says: is the symmetric difference of two elements in the Boolean algebra also in the Boolean algebra? This follows from the closure of the various Boolean algebra operations. Though you do generally need to know that operations satisfy closure, it's not much discussed because it's a little too obvious. Without closure, you don't even have a well-defined function! – Paul VanKoughnett Mar 11 '13 at 4:43
Thanks for the response. So the fact that the symmetric difference was defined as addition implies that this Boolean algebra is closed under addition? If that is the case then all I need to show are the other properties of an Abelian group (commutativity, existence of inverse, existence of identity,...) followed by the other ring properties? – Philip.E Mar 11 '13 at 4:59
@Philip: No, you've got it the wrong way around. Symmetric difference wasn't defined as addition, addition was defined as symmetric difference, and symmetric difference in turn is defined in terms of the Boolean algebra operations, whose properties, including closure, are part of the assumption that you have a Boolean algebra; closure of the prospective Boolean ring under addition follows directly from the fact that addition is defined in terms of the Boolean algebra operations for which closure holds. – joriki Mar 11 '13 at 5:56