de morgan law $A\setminus (B \cap C) = (A\setminus B) \cup (A\setminus C) $ First part :
I want to prove the following De Morgan's law :  ref.(dfeuer)
$A\setminus (B \cap C) = (A\setminus B) \cup (A\setminus C) $

Second part:
Prove that $(A\setminus B) \cup (A\setminus C) = A\setminus (B \cap C) $

Proof:
Let $y\in (A\setminus B) \cup (A\setminus C)$

$(A\setminus B) \cup (A\setminus C) = (y \in A\;  \land y \not\in
 B\;) \vee (y \in A\; \land y \not\in C\;)$
$\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;= y \in A\ \land (
 y \not\in B\; \vee \; y \not\in C ) $
$\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;= y \in A \land
 (\lnot ( y\in B) \lor \lnot( y\in C) )$

According to De-Morgan's theorem :
$\lnot( B \land C) \Longleftrightarrow (\lnot B \lor \lnot C)$ thus

$y \in A \land (\lnot ( y\in B) \lor \lnot( y\in C) ) = y \in A \land y \not\in (B \land C)$

We can conclude that  $(A\setminus B) \cup (A\setminus C) = A\setminus (B \cap C)$
 A: I'll guide you through one direction. You will need to figure the other out yourself.
Let $x\in A \setminus (B\cap C)$.
Then by the definition of $\setminus$, $x\in A$ and $x\notin B\cap C$. The latter statement can be translated into $\lnot (x \in B \cap C)$.
By the definition of $B\cap C$, $x\in B\cap C \iff (x \in B) \land (x\in C)$.
Thus we conclude that $\lnot ((x \in B) \land (x\in C))$.
By De Morgan's laws for logic, we can rewrite this as
$\lnot(x\in B)\lor\lnot(x\in C)$.
Suppose that $\lnot (x\in B)$. Then since $x\in A$, $x\in A\setminus B$.
Suppose instead that $\lnot (x \in C)$. Then since $x\in A$, $x\in A \setminus C$.
As one of these must be true, $(x\in A\setminus B)\lor (x\in A \setminus C)$.
By the definition of union, $x\in (A\setminus B)\cup(A \setminus C)$.
A: You write,

Can we say that:
$x∈A$ and $x∉B$ and $x∈C$ OR
$x∈A$ and $x∉C$ and $x∈B$

In short, the answer is no, we cannot say that, because you've not told us what $x$ is. You've got to tell us what this $x$ is before you start talking about it.
To show that sets $S$ and $T$ are equal, a typical strategy is to show $S \subseteq T$ and $T\subseteq S$. In order to do this, you're going to let $x$ be some arbitrary member of $S$ and show that this implies that it is in $T$, and then let $y$ be some arbitrary member of $T$ and show that this implies that it is in $S$. Once that's done, you've won.
In your example, $A\setminus (B\cap C)$ is your $S$. So begin by explicitly writing:
"Let $x\in A\setminus (B\cap C)$."
Now what do we know about $x$? Well, if $x\in A\setminus (B\cap C)$ then $x\in A$ but $x\notin (B\cap C)$. Unpacking the second part of that last sentence, if $x\notin (B\cap C)$ then either $x\notin B$ or $x\notin C$—otherwise $x$ would be in their intersection. So this gives us two cases: either $x\notin B$ or $x\notin B$. Suppose that $x\notin B$. We know that $x\in A$ so this means that $x\in A\setminus B$, and so $x\in (A\setminus B)\cup (A\setminus C)$. In the second case, we know that $x\in A$ and $x\notin C$, which means $x\in A\setminus C$, which further implies that $x\in (A\setminus B)\cup (A\setminus C)$. Since these are the only two possible cases, we conclude that $x\in (A\setminus B)\cup (A\setminus C)$.
To go to the other direction, again, begin by explicitly writing
"Let $y\in (A\setminus B) \cup (A\setminus C)$"
And from there, determine what you can say about $y$. I'll let you finish the other direction.
A: If $x \in A \setminus (B \cap C ) $. then this occur
$$ \iff x \notin B\cup C \iff x \notin B \; \; \text{or} \; \; x \notin C \iff$$
$$ \iff x  \in A \setminus B \; \; \text{or} \; \; x \in A \setminus C \iff x \in(A \setminus B ) \cup (A \setminus C )$$
$$ \therefore A \setminus (B \cap C )  = (A \setminus B ) \cup (A \setminus C ) $$
