# Set Proof

1. Nov 17, 2012

### 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.

Last edited: Nov 17, 2012
2. Nov 17, 2012

### 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.

3. Nov 17, 2012

### 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.

4. Nov 17, 2012

### gopher_p

What does the associative law say for intersections?

5. Nov 17, 2012

### nicnicman

Associative law for intersections:

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

6. Nov 17, 2012

### 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?

7. Nov 17, 2012

### nicnicman

Okay, yeah I see it now. Thanks!

And, the other one is pretty straight forward.

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