Let a & b be sets:

A u (B-A)U(C-A)=(B u C)-A

How do I solve?

The other ones I have are:

(A n B) E A

A E (A U B)

A-B E A

A n (B - A) 0

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- Jul 25th 2009, 04:47 AMorendaclVin diagrams? is there any other way..
Let a & b be sets:

A u (B-A)U(C-A)=(B u C)-A

How do I solve?

The other ones I have are:

(A n B) E A

A E (A U B)

A-B E A

A n (B - A) 0 - Jul 25th 2009, 02:10 PMGammabe careful
your first one is not even true.

A=(0,1)

B=(1,2)

C=(2,3)

Your LHS yields $\displaystyle (0,1)\cup (1,2) \cup (2,3)$

but the RHS is $\displaystyle (1,2) \cup (2,3)$

you should learn a few TeX codes so set theory makes sense.

\cup is union

\cap is intersection

\subset is subset

\supset goes the other way

\in is element of

\emptyset is the nullset

I have no idea what your E is.

Maybe type your question again correctly so we can help? - Jul 25th 2009, 02:31 PMPlato
[tex] \cup [/tex] gives $\displaystyle \cup $ is union

[tex] \cap [/tex] gives $\displaystyle \cap $ is intersection

[tex] \subseteq [/tex] gives $\displaystyle \subseteq $ is subset

[tex] \in [/tex] gives $\displaystyle \in $ is element

[tex] \emptyset [/tex] gives $\displaystyle \emptyset $ is emptyset - Jul 25th 2009, 03:13 PMAlephZero
Unfortunately I have learned to speak some ASCII... E is apparently often equivalent to $\displaystyle \in;$ although, rarely, it is sometimes seen outside of its usual habitat as being equivalent to $\displaystyle \subset$ or $\displaystyle \subseteq,$ which I suspect is what's meant in this case.

Sorry for the mixed metaphor, there. - Jul 25th 2009, 03:34 PMGamma

Okay so I will show you how to do these without vinn diagrams. I will take Alephzero's suggestion and take E as $\displaystyle \subset$

1)Let $\displaystyle x\in A \cap B$ then x is by definition in both A and B, therefore x is in A, so $\displaystyle x\in A \Rightarrow A \cap B \subset A$ as desired

2) Let $\displaystyle x\in A$, then x is certainly in A or B since it is in A. Thus $\displaystyle x\in A \cup B \Rightarrow A \subset (A\cup B)$ as desired

3) Let $\displaystyle x\in A-B$. Then x is in A, but not in B. But this means $\displaystyle x\in A$ so $\displaystyle A-B \subset A$ as desired.

4) I think this is supposed to say $\displaystyle A \cap (B-A)=\emptyset$

So you show both containments.

Clearly $\displaystyle A \cap (B-A)\supset \emptyset$.

So let $\displaystyle x\in A \cap (B-A)$. Then $\displaystyle x\in A$ and $\displaystyle x\in B-A$. But this means x is in A, and x is in B, but not in A. So clearly no x can both be in A and not be in A, so this is precisely the empty set, so we have the desired $\displaystyle A \cap (B-A)=\emptyset$

QED.

It might be worth at least noting what happens if you cannot chose an x in the set, like for instance if you have the case where like $\displaystyle A-B=\emptyset$ or something, then you just have the containment being trivially true.

As for the first one, you are gonna have to get the correct statement before I can show you how to prove it because as it stands it is not true as my counterexample demonstrates.