Question regarding algebra of sets

[ssayani87]

Homework Statement

I'm working on a proof for Intro. to Algebra, and the problem deals with proving that the operation of Symmetric Difference imposes group structure on the power set 2^x (set of all subsets) of a given set X. There is a part of my proof that I'd like to get some advice on however...

I've worked out a proof, and I can see it relies heavily on this equation:

Let A, B be sets.

i. A - B = (AuB) - B

Homework Equations

The Attempt at a Solution

My proof goes like this: (=>)Let x in A - B. Then, x is in A, implying x is in AuB. Since x cannot be in B, then it also falls in AuB - B.
(<=): Let x in AuB - B. Then x cannot be in B, and must therefore be in AuB. x not in B but still in AuB - B implies that x must be in A and not in B, which means it also falls in A - B.

Did I miss anything/ any mistakes?

 

[Dick]

Homework Statement

I'm working on a proof for Intro. to Algebra, and the problem deals with proving that the operation of Symmetric Difference imposes group structure on the power set 2^x (set of all subsets) of a given set X. There is a part of my proof that I'd like to get some advice on however...

I've worked out a proof, and I can see it relies heavily on this equation:

Let A, B be sets.

i. A - B = (AuB) - B

Homework Equations

The Attempt at a Solution

My proof goes like this: (=>)Let x in A - B. Then, x is in A, implying x is in AuB. Since x cannot be in B, then it also falls in AuB - B.
(<=): Let x in AuB - B. Then x cannot be in B, and must therefore be in AuB. x not in B but still in AuB - B implies that x must be in A and not in B, which means it also falls in A - B.

Did I miss anything/ any mistakes?

Sounds fine to me.