de Morgan's laws on generalised union and intersection.

Hi, I need to prove de Morgan's laws for generalised union and intersections, I came up with a proof but I don't know if it's works with what I know.
Here is what I know:
Given A, a set and B such that $B \subseteq A$ then $B^{c}=A-B$ .
$\mathcal{A}$ is a collection of sets.
$\bigcup_{A \in \mathcal{A}}A=\{ x| \exits A \in \mathcal {A} \text{ such that } x \in A \}$ .
$\bigcap_{A \in \mathcal{A}}A= \{x| x \in A, \forall A \in \mathcal A\}$ .

proof:
$\text{1. }(\bigcup_{A \in \mathcal{A}}A)^{c}=\{x| x \notin A, \forall A \in \mathcal{A}\}=\{x|x \in A^{c}, \forall A \in \mathcal{A}\}=\bigcap_{A \in \mathcal{A}}A^{c}$ .
$\text{2. }(\bigcap_{A \in \mathcal{A}}A)^{c}=\{x| \exits A \in \mathcal{A} \text{ such that } x \notin A\}= \{ x| \exists A \in \mathcal{A} \text{ such that } x \in A^{c}\}=\bigcup_{A \in \mathcal{A}}A^c$ .

Thanks in advance!.

Re: de Morgan's laws on generalised union and intersection.

I think the proof is OK.

I wanted to make a remark about the lack of a quantifier over A in the definition of union, but then I saw that it is a LaTeX error. Since you put existential quantifier in the beginning, I would do the same with the universal quantifier. I would also add an extra equality $\left(\bigcup_{A \in \mathcal{A}}A\right)^{c}=\{x\mid\neg\exists A\in\mathcal{A}.\,x\in A\}$ for clarity, and similarly for intersection.

Hint: use \mid instead of | in the set-builder notation. It creates correct spaces around it.

Re: de Morgan's laws on generalised union and intersection.

Thank you.