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

but the RHS is

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 is union

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

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

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

[tex] \emptyset [/tex] gives is emptyset - Jul 25th 2009, 03:13 PMAlephZero
- 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

1)Let then x is by definition in both A and B, therefore x is in A, so as desired

2) Let , then x is certainly in A or B since it is in A. Thus as desired

3) Let . Then x is in A, but not in B. But this means so as desired.

4) I think this is supposed to say

So you show both containments.

Clearly .

So let . Then and . 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

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 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.