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