prove $(A \times B)^c = (A^c\times U)\cup(U\times B^c)$ got stuck proving that $(A \times B)^c = (A^c\times U)\cup(U\times B^c)$.
would appreciate your help, this is what I got so far:
 A: Let $(x,y)$ be an element of the left-hand side. Then $(x,y)\not\in A\times B$, and therefore $x\not \in A$, or $y\not\in B$  (or both).
Suppose $x\not\in A$. Then $x\in A^c$. It follows that $(x,y)\in A^c\times U$, and therefore $(x,y)$ is an element of the right-hand side.
A similar argument works if $y\not\in B$.
We leave it to you to show that every element of the right-hand side is an element of the left-hand side. 
A: Ok, so this is what I got. would appreciate your attention if I made any mistakes.. thanks again.

A: $\begin{align}(A\times B)^c & = ((A\times \mathscr U)\cap (\mathscr U\times B))^c
\\ & = (A\times \mathscr U)^c\cup (\mathscr U\times B)^c
\\ & = (A^c\times \mathscr U)\cup (\mathscr U\times B^c)
\end{align}$
The first step, $(A\times B)= (A\times \mathscr U)\cap (\mathscr U\times B)$ is equivalent to $$\forall x\,\forall y\;\Big(\big(x\in A\,\wedge\, y\in B\big) \,\leftrightarrow\,\big( x\in A\cap \mathscr U\,\wedge\, y\in \mathscr U\cap B\big)\Big)$$
Which is fairly incontrovertable.
The second step is just deMorgan's Law for Set Complements
The last step hinges on $(A\times \mathscr U)^c = (A^c\times \mathscr U)$, and the symmetric claim. 
This is equivalent to the claim that:  $\forall x\,\forall y\;\Big((x,y)\notin A\times \mathscr U \leftrightarrow (x,y)\in A^c\times \mathscr U\Big)$
Clearly this is the case, because is distills down to $\forall x\;(x\notin A \leftrightarrow x\in A^c)$ which is true by definition of set complement.
