Cartesian Product in Element Method

I'm trying to do proofs using element method, but a few problems popped up with Cartesian products in them. How do these fit in?

For example, I know I can break

(A $\cap$ B) into $\ [x \in A] \wedge [x \in B]$

How do I work Cartesian products into this?

One example problem I'm working on is:

$$(A\times B)-(A\times C) \subset A\times(B-C)$$

Can someone help step me through this? Thank you!

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You know that the elements of any Cartesian Products is of form $(x,y)$. Now let $$(x,y)\in(A\times B)-(A\times C)$$ so as $(A-B)=A\cap B'$ so $$(x,y)\in(A\times B),~~(x,y)\notin(A\times C)$$ so $$(x,y)\in(A\times B)\to x\in A,~y\in B$$ and $$(x,y)\notin(A\times C)\to x\in A,~y\notin C$$ These two latter ones mean that $x\in A, y\in B, y\notin C$.

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Ah okay, so the whole Cartesian product just adds almost like a new variable into everything. Thanks, this helped clear it up. –  K. Barresi Feb 8 '13 at 6:18
Nice exposition! +1 –  amWhy Feb 9 '13 at 0:05

$x\in (A\times B)-(A\times C)$ means that $x\in A\times B$ and $x\notin A\times C$. Now, $x\in A\times B$ means $x=(a,b)$ with $a\in A, b\in B$. Since $x=(a,b)\notin A\times C$ it follows from $a\in A$ that $b\notin C$. Therefore, $x=(a,b)\in A\times (B-C)$.

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Thanks Ittay, this helps clear things up a bit. –  K. Barresi Feb 8 '13 at 6:18