$A\backslash (B\cap C) = (A\backslash B)\cup (A\backslash C)$; only one inclusion seems to work I encountered the following problem:
$$A\backslash (B \cap C) = (A\backslash B)\cup(A\backslash C).$$
So I need to prove two things:


*

*$A\backslash (B\cap C) \subseteq (A\backslash B) \cup (A\backslash C)$

*$(A\backslash B) \cup (A\backslash C) \subseteq A\backslash (B\cap C)$
The first one is fairly easy to me; what I don't understand is 2. My approach was to suppose $x \in A \land \neg(x \in B)$ or $x \in A \land \neg(x\in C)$ and try to derive $x \in A \land \neg(x \in B) \land \neg(x\in C)$, i.e.
$\forall x \ ((x\in A \land \neg(x\in B)) \lor (x\in A \land \neg (x\in C)) \to (x\in A \land \neg (x\in B) \land \neg(x\in C)))$
But this doesn't seem possible. For all I know there are two possible situations: 1). Both parts of the antecedent are true, 2). Either $((x\in A \land \neg(x \in B))$ or $((x \in A \land \neg(x\in C))$ is true. But if the latter is true, it would seem impossible to try and derive the consequent: I would have $x \in A$, but either I can't derive $\neg(x\in B)$ or $\neg(x \in C)$.
Could someone offer some advice please?
 A: You need De Morgan's Law and that intersection distributes over union, in other words $$(A\cap B)^c=A^c\cup B^c$$ and $$A\cap(B\cup C)=(A\cap B)\cup(A\cap C).$$ Also note that $$A\setminus B=A\cap B^c.$$  So using these three facts we have 
$$A\setminus(B\cap C)=A\cap(B\cap C)^c\overset{\text{De Morgan}}{=}$$
$$A\cap(B^c\cup C^c)\overset{\text{distribute}}=(A\cap B^c)\cup(A\cap C^c)$$
$$=(A\setminus B)\cup (A\setminus C)$$
A: To prove 2, you must suppose that $x\in A\setminus B$ or $x\in A\setminus C$. So far, so good.
But what you must prove is that $x$ is in $A$ but not in $B\cap C$. You are trying to prove something much stronger (and false, in general); namely, that $x$ is in $A$ but not in $B$ and not in $C$.
In formulas, you must derive this:
$$x\in A\wedge\neg(x\in B\wedge x\in C)$$
and this is what you are trying:
$$x\in A\wedge x\notin B\wedge x\notin C$$
A: 
Could someone offer some advice please? 

My advice is this: only use an element-chasing proof, which is what you are trying to do, when you must. Whenever possible, use set algebra to prove set identities. Why? Basically, your proof will be clearer and much shorter. Sometimes, using set algebra is rather difficult, and it turns out to be easier to construct an element-chasing proof, but this is not the case oftentimes. See if you can follow the proof below that uses set algebra (it is very similar to Gregory Grant's answer): 
\begin{align}
A\setminus (B\cap C)&= A\cap(B\cap C)^C\tag{by definition}\\[0.5em]
&= A\cap(B^C\cup C^C)\tag{De Morgan}\\[0.5em]
&= (A\cap B^C)\cup(A\cap C^C)\tag{distributivity}\\[0.5em]
&= (A\setminus B)\cup (A\setminus C).\tag{by definition}
\end{align}
Were you able to follow that proof? The linked to pdf and Gregory Grant's answer explicate how the definition, distributivity, and De Morgan's law work. 
Hopefully now you can see the advantages of using set algebra to establish set identities instead of relying on element-chasing proofs. The proof above is both shorter and clearer than any element-chasing proof could be. 
A: It is possible to see this without  Venn diagrams or de Morgan's laws.
Let $B$ be the set of people who have visited Belgium, and $C$ denote those people who have visited Canada. Think of $A$ as the set of Americans.
So in your question the  LHS, $A\setminus(B\cap C)$  asks us to  remove from the set of Americans 
those persons who have visited both Belgium and Canada.
Suppose we first removed all Americans who had visited Belgium. This would be an overkill, a mistake, we have removed more than what we were asked to. SO we have to put back some people. That means take the  union with a suitable set. 
What is that suitable set? Except those who visited Canada we have to put back all others. This means precisely that  we take union with $A\setminus $C$.
A: Suppose $x\in (A\setminus B)\cup (A\setminus C)$. Then $x\notin B$ and $x\notin C$, so $x \in A\setminus (B\cap C)$. Hence $ (A\setminus B)\cup (A\setminus C)\subseteq A\setminus (B\cap C)$.
