# Is this proved is correct?

Do I have Proving is true?

Question -> Prove:

(C-A)$\cup$(B-A)=(C$\cap$B)-A

My Proving:

(C-A)$\cup$(B-A)=((C$\cap$A')$\cup$ (B$\cap$A')

=(C$\cup$B)$\cap$A'

=(C$\cup$B)-A'

Do you think my Proving is correct?

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Did you mean $(C\setminus A)\cup(B\setminus A)=(C\cup B)\setminus A$, with a union on the righthand side? If you did, your argument is correct. – Brian M. Scott Mar 6 '13 at 16:02
Echoing Brian's comment, the right hand side of your first equation has $\cap$ instead of $\cup$. – copper.hat Mar 6 '13 at 16:04
I mean, this is exactly (C-A) ∪ (B-A) = (C ∩ B)-A – Software Mar 6 '13 at 16:06
No, it needs to be $(C-A)\cup (B-A)=(C\cup B)-A$ – Aang Mar 6 '13 at 16:07
Is the wrong question? – Software Mar 6 '13 at 16:09

In fact, $(C-A)\cup (B-A)=(C \cap A')\cup (B \cap A')$ which is $(C \cup B) \cap A'$ and it equals to $(C \cup B)-A$ not $(C \cup B)-A'$

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Yes exactly Thanks for your very complete Answer :) – Software Mar 6 '13 at 16:15
+1 Helpful...and I'll keep my look-out, per your request! – amWhy Mar 6 '13 at 16:55