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Please help me

Do I have Proving is true?

Question -> Prove:


My Proving:

(C-A)$\cup$(B-A)=((C$\cap$A')$\cup$ (B$\cap$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
up vote 2 down vote accepted

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
very elegant answer – Adi Dani Mar 6 '13 at 19:02
By the way: good mornin', Babak! – amWhy Mar 7 '13 at 6:17
We had snow yesterday. Quite a bit! Kind of a slow night tonight again. But that's okay. You had a nice day again yesterday! – amWhy Mar 7 '13 at 6:27


(C−A)∪(B−A)=(C∩A ′ )∪(B∩A ′ )

(C∪B)∩A ′ = (C∪B)−A not (C∪B)−A ′

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