Proving the Distributive Laws for Sets Using Commutative and Complement Laws

[nicnicman]

If A, B, and C are sets prove that (A-C) - (B-C) = (A-B) - C

Note: n = intersection, u = union, and ' = complement.

(A-C)-(B-C)
= (AnC') n (BnC')' by definition of complement, intersection, and subtraction
= (AnC') n (B'uC'') by DeMorgan's laws
= (AnC') n (B'uC) by Complementation law
= A n (C' n (B'uC)) by Commutative laws
= A n ((C'nB') u (C'nC)) by Distributive laws
= A n ((C'nB') u empty set) by Complement laws
= A n (C'nB') by Identity laws
= A n (B'nC') by Commutative laws
= (AnB') n C' by Commutative laws
= (A-B) - C by definition of complement, intersection, and subtraction

How is this? Did I use the Commutative laws correctly?
Thanks for any suggestions.

 

[gopher_p]

$\left(A\cap C^c\right)\cap\left(B^c\cup C\right)=A\cap \left(C^c\cap\left(B^c\cup C\right)\right)$ and $A\cap\left(B^c\cap C^c\right)=\left(A\cap B^c\right)\cap C^c$ look more like associative laws than commutative laws to me.

Other than that and the typo (which I'm sure you'll find on your own) it looks good.

 

[nicnicman]

Yeah, those are the parts I was not sure of. I saw similar problems elsewhere and followed their logic, but I'm unsure of how those two steps are using the Associative laws.
Maybe someone could shed some light.

Oh, and I'll fix the typo.

Thanks.

 

[gopher_p]

What does the associative law say for intersections?

 

[nicnicman]

Associative law for intersections:

A n (B n C) = (A n B) n C

 

[gopher_p]

That's correct. Do you see immediately how that applies to $A\cap\left(B^c\cap C^c\right)=\left(A\cap B^c\right)\cap C^c$?

For $(A\cap C^c)\cap(B^c\cup C)=A\cap(C^c\cap(B^c\cup C))$, rewrite that equality with $D$ written in place of $(B^c\cup C)$. Now do you see how the associative law applies here?

 

[nicnicman]

Okay, yeah I see it now. Thanks!

And, the other one is pretty straight forward.