Verified answer

(A-C) = {x in A and x not in C}

(B-C) = {x in B and x not in C}

(A-C) union (B-C) = {x in A and x not in C} or {x in B and x not in C}

Both those sets include x not in C, so we can "factor" that out:

(A-C) union (B-C) = {x not in C} and ( {x in A} or {x in B} )

And the right hand side of that is just union of A and B, so we have:

(A-C) union (B-C) = {x not in C} and (A union B)

which we can rewrite by swapping the terms as:

(A union B) - C.

Hope that all makes sense - it's not easy to use proper set notation on Yahoo Answers, but hopefully you'll understand the proof.

B union C

It's good to plow through the statement that way in order to get a basic understanding of what these things mean, and as an aid in doing that, you should use a Venn diagram.

Inside a rectangle representing the universal set, draw 3 circles that mutually intersect, so that all 8 of the individual regions appear. Identify the regions that each side of your equation picks out, and show that they are the same.

But you should also be able to use the definition of set difference:

A-B == A∩B'

along with the set-theory versions of DeMorgan's Laws:

A∪(B∩C) = (A∪B) ∩ (A∪C)

A∩(B∪C) = (A∩B) ∪ (A∩C)

in order to work the problem symbolically. And even then, it helps to use a Venn diagram.

(A∩C') ∪ (B∩C')

= (A∪(B∩C')) ∩ (C'∪(B∩C'))

= (A∪B) ∩ (A∪C') ∩ (C'∪B) ∩ (C'∪C')

= (A∪B) ∩ (A∪C') ∩ (C'∪B) ∩ C'

= (A∪B) ∩ (A∪C') ∩ C'

= (A∪B) ∩ C'

= (A∪B) - C

EDIT:

Matthew has a quicker way, that translates like this

(the 1st of the 2 steps uses the 2nd DeMorgan's Law "in reverse"):

(A∩C') ∪ (B∩C')

= (A∪B) ∩ C'

= (A∪B) - C

What you have done is correct, but if it would make you feel better, argue as follows: Since x is in A, it follows that x is in AUB, so x is in (AUB)-C.