Theorem: A\(BUC)=(A\B)intersection(A\C)

How to proof ?Any ideas would be useful.

Thanks,

Sasha

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- Jan 11th 2010, 12:28 PMsasha1111discrete math.proof.set theory.
Theorem: A\(BUC)=(A\B)intersection(A\C)

How to proof ?Any ideas would be useful.

Thanks,

Sasha - Jan 11th 2010, 01:50 PMDrexel28
- Jan 12th 2010, 01:07 AMsasha1111
A\(BUC) = (A\B) ∩ (A\C)

PROOF:

x are A\(BUC) => (x aren’t (A)) and (x are (BUC)) => (x aren’t (AUA)) and (x are (BUC)) => (x aren’t (A)) or (x are (B∩C)) => ((x aren’t A) or (x are B)) and ((x aren’t A) or (x are C)) => (x are (A\B)) and (x are (A\C)) => x are ((A\B) ∩ (A\C)) - Jan 12th 2010, 07:27 AMSwlabr
It's not a proof per se. but drawing a diagram will let you see what is going on better:

http://upload.wikimedia.org/wikipedi...m_cmyk.svg.png

(diagram is "borrowed" from wikipedia.) - Jan 12th 2010, 07:29 AMDrexel28
- Jan 12th 2010, 07:40 AMSwlabr
I hate to disagree, but I do!

What you did required some sort of knowledge of what was actually going on, or at least someone telling you some rules. The other two ways don't.

I've never been able to derive proofs just using what I know about sets...I prefer to either look at an element or draw a diagram. - Jan 12th 2010, 07:43 AMDrexel28
I wasn't directing that comment at you by the way. My point was, using the algebra of sets is just as fundamental, and even more intuitive, than showing the inclusion/reverse-inclusion method. I quoted your picture so that although each step I made was valid, she can verify logically with your picture!

- Jan 12th 2010, 11:23 AMnovice
- Jan 12th 2010, 11:43 AMDrexel28
Doing it by implication is just long-winded.

Let

$\displaystyle x\in A-\left(B\cup C\right)$

$\displaystyle \Longleftrightarrow x\in A\text{ and }x\notin \left(B\cup C\right)$

$\displaystyle \Longleftrightarrow x\in A\text{ and }\left(x\notin B\text{ and }x\notin C\right)$

$\displaystyle \Longleftrightarrow \left(x\in A\text{ and }x\in A\right)\text{ and }\left(x\in B'\text{ and }x\in C'\right)$

$\displaystyle \Longleftrightarrow \left(x\in A\text{ and }x\in B'\right)\text{ and }\left(x\in A\text{ and }x\in C'\right)$

$\displaystyle \Longleftrightarrow \left(x\in A\text{ and }x\notin B\right)\text{ and }\left(x\in A\text{ and }x\notin C\right)$

$\displaystyle \Longleftrightarrow x\in\left(A-B\right)\text{ and }x\in\left(A-C\right)$

$\displaystyle \Longleftrightarrow x\in\left(\left(A-B\right)\cap\left(A-C\right)\right)$

I accidentally made some redundant steps. But, you get the picture. - Jan 12th 2010, 12:07 PMsasha1111
Thank you.Smooth way!

- Jan 12th 2010, 01:59 PMnovice
- Jan 13th 2010, 12:53 AMsasha1111
Well done Drexel.Looking forward for cooperation.

Greetings from Lithuania