Probability set theory: $\Omega \backslash (A \cap B) = (\Omega \backslash A) \cup (\Omega \backslash B)$? My probability theory book (Grimmett & Welsh) contains the following formula:

$\Omega \backslash (A \cap B) = (\Omega \backslash A) \cup (\Omega$
$\backslash B)$

But drawing Venn diagrams, this equality does not seem to hold up?
$\Omega \backslash (A \cap B)$ gives me a Venn diagram where only $(A \cap B)$ is missing, so $A \backslash B$ and $B \backslash A$ are not missing. 
$(\Omega \backslash A) \cup (\Omega \backslash B)$ gives me a Venn diagram where both $A$ and $B$ are completely missing.
What am I doing wrong?
 A: In the Venn diagram below, let $A$ ($B$) denote the content of the circle on the left (right), so $A\cap B$ is the yellow part and $\Omega\backslash (A\cap B)$ is the union of the red, blue and white parts (the white part being outside both circles). Meanwhile, $\Omega\backslash A$ is anything that's blue or white, $\Omega\backslash B$ is anything that's red or white, and $(\Omega\backslash A)\cup(\Omega\backslash B)$ is anything that's red, blue or white, as expected. I suggest you look over your own Venn diagram again to see which of these facts you misunderstood. I suspect you accidentally considered $(\Omega\backslash A)\cap(\Omega\backslash B)$, which is just the white part.

A: I believe you would also appreciate a proof that uses the logical definition of the meaning of equality $\Omega \backslash (A \cap B) = (\Omega \backslash A) \cup (\Omega \backslash B)$ without appealing to a drawing.
Note that each of the following statements are equivalent:


*

*$\Omega \backslash (A \cap B) = (\Omega \backslash A) \cup (\Omega \backslash B)$

*$\Omega \backslash (A \cap B) \subseteq (\Omega \backslash A) \cup (\Omega \backslash B)$ and  $(\Omega \backslash A) \cup (\Omega \backslash B)\subseteq \Omega \backslash (A \cap B)$

*$x\in \Omega \backslash (A \cap B) $ implies $x\in (\Omega \backslash A) \cup (\Omega \backslash B)$ and  $x\in (\Omega \backslash A) \cup (\Omega \backslash B)  $ implies $x\in \Omega \backslash (A \cap B) $
First we will prove the implication: $x\in \Omega \backslash (A \cap B) $ implies $x\in (\Omega \backslash A) \cup (\Omega \backslash B)$. Note that each of the statements below results in the next statement.


*

*$x\in \Omega \backslash (A \cap B) $,

*$x\in \Omega$ and  $x\notin(A \cap B) $,

*$x\in \Omega$ and  $x\notin A$ and $x\notin B $,

*$x\in \Omega$ and  $x\notin A$ and  $x\in \Omega$  and $x\notin B $

*$x\in \Omega\backslash A$ and $x\in \Omega\backslash B$

*$x\in (\Omega\backslash A)\cap (\Omega\backslash B)$
Now we will prove the implication $x\in (\Omega \backslash A) \cup (\Omega \backslash B)  $ implies $x\in \Omega \backslash (A \cap B) $. Notice again that each of the statements below results in the next statement.


*

*$x\in (\Omega\backslash A)\cap (\Omega\backslash B)$,

*$x\in \Omega\backslash A$ and $x\in \Omega\backslash B$,

*$x\in \Omega$ and  $x\notin A$ and  $x\in \Omega$  and $x\notin B $

*$x\in \Omega$ and  $x\notin A$ and $x\notin B $,

*$x\in \Omega$ and  $x\notin(A \cap B) $,

