# Proving a set equality

Hi I am learning group theory and encountered this: $$(B\cap (A\cap B)')\cup (B'\cap (A\cap B)) = B\cap (A'\cup B').$$

I don't understand how this is true, could someone please show me proof?

Thanks

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Are $A$ and $B$ subgroups? What does $A'$ mean here? Derived subgroup? –  lhf May 25 '11 at 14:33
What are $A$, $B$, etc.? Where is the union and intersection taking place? –  Jiangwei Xue May 25 '11 at 14:33
A better word for "equals statement" is "equality". –  Stefan Walter May 25 '11 at 14:43
Have you tried a truth table? It would be huge but I think it would be a good exercise in learning equivalence. –  scrappedcola May 25 '11 at 16:09

Assuming the prime denotes complementation: The second term on the left, $B'\cap(A\cap B)$, is empty, since $B$ and $B'$ are disjoint. That leaves $B\cap(A\cap B)'$. You can use De Morgan's law $(A\cap B)'=A'\cup B'$ to transform this into the right-hand side.

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@Jason, if this answers your question, please clarify it and retag it as (elementary-set-theory). –  lhf May 25 '11 at 14:40
@Lhf: I have done it anyway. Jason, can put it back if needed. –  Aryabhata May 25 '11 at 15:18
@joriki: Aw so me thanks I was missing the disjoint definition :) btw how do you print the math formulas so nicely? –  Jason May 25 '11 at 17:41
@Jason: Seems you've figured that one out in the meantime? :-) –  joriki May 25 '11 at 18:42
I guys I have he he :) –  Jason May 25 '11 at 22:06