Theorem 2.6: (Distributivity of Unions across Intersections) Let A, B, and C be three sets. then

Proof: Let

Then

or both

If

then both

by Theorem 1.2.

so

by Definition 1.4. If on the other hand,

then it is in both B and C, so it is in

and

again by Theorem 1.2. Hence,

by Definition 1.4. Conversely suppose that

This says that

and

There are two cases. Either

in which case

by Theorem 1.2 or if x is not in A, then since it must be in B.

We also have that

and if it is not in A then it must be in C. So

in either case.

Since we have containment both ways we conclude that

by Definition 1.2.