Simplify Set Math Problem: A\cup (B\cup C - B\cap C) - A\cap (B\cup C - B\cap C)

[Treadstone 71]

I have no formal training in set theory, and I need to simplify the following:

$$(A\cup B - A\cap B)\cup C - (A\cup B - A\cap B)\cap C$$

Preferably, it should somehow end up as:

$$A\cup (B\cup C - B\cap C) - A\cap (B\cup C - B\cap C)$$

 

[StatusX]

The union of A and B is the set of points in A or B or both, so subtracting off the part in both A and B leaves the part in either A or B. Call this set D. Then you do the same thing with D and C. So you want the points that are either in C or D (not both), which means they are either in A, or B, or C, but not more than one of these. Does that help at all?

 

[AKG]

I don't see how the second thing is a simplification of the first, but the two are the same. You could use:

(A + B + C) - (AB + BC + AC)

where + is union, and juxtaposition is intersection.

 

[CarlB]

This is basically the XOR relation. That is,

(A xor B) xor C = A xor (B xor C) = A xor B xor C,

well familiar to those of us who design integrated circuits.

Carl

 

[Treadstone 71]

What are the basic properties of unions and intersections? For example, what is

$$(A-B)\cap C$$

I know how unions and intersections interact with one another, but what about the above case?

 

[StatusX]

We can pretend we are given a base space X containing all the sets, and then we can define a set X-B, the complement of B in X, and then this becomes:

$$(A-B) \cap C = (A \cap (X-B))\cap C = A \cap (X-B)\cap C$$

Since intersection is associative, and so you can do what you want with this:

$$(A-B) \cap C = (C-B) \cap A = (A\cap C) -B$$