# Math Help - discrete math.proof.set theory.

1. ## discrete math.proof.set theory.

Theorem: A\(BUC)=(A\B)intersection(A\C)
How to proof ?Any ideas would be useful.

Thanks,
Sasha

2. Originally Posted by sasha1111
Theorem: A\(BUC)=(A\B)intersection(A\C)
How to proof ?Any ideas would be useful.

Thanks,
Sasha
Don't you dare look until you've tried it yourself. Ok?

Spoiler:

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

3. 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))

4. Originally Posted by sasha1111
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))
It's not a proof per se. but drawing a diagram will let you see what is going on better:

(diagram is "borrowed" from wikipedia.)

5. Originally Posted by sasha1111
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))
Originally Posted by Swlabr
It's not a proof per se. but drawing a diagram will let you see what is going on better:

(diagram is "borrowed" from wikipedia.)
Really, the God-honest simple way is to do what I showed you in the first post. Use the algebra of sets.

6. Originally Posted by Drexel28
Really, the God-honest simple way is to do what I showed you in the first post. Use the algebra of sets.
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.

7. Originally Posted by Swlabr
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.
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!

8. Originally Posted by sasha1111
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))
sasha,

Your method is called proof of implication. Drexel did it by set algebra, and Swlabr is by Venn Diagram. I use all of them, but I think if you can do it by set algebra, you can also do logic. The set algebra can make you very smart with logic.

9. Doing it by implication is just long-winded.

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

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

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

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

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

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

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

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

I accidentally made some redundant steps. But, you get the picture.

10. Thank you.Smooth way!

11. Originally Posted by Drexel28
Doing it by implication is just long-winded.

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

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

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

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

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

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

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

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

I accidentally made some redundant steps. But, you get the picture.
Drexel,

I am so glad you showed the above. The method is called prove by definition. I may say we have all the methods laid out for our friend, sasha.

12. Well done Drexel.Looking forward for cooperation.

Greetings from Lithuania