find a condition on B,C so set difference is associative (A\B)\C=A\(B\C) Im trying to find a some conditions on the sets $B,C$ so the following will be correct:
for all $A$, $(A\setminus B)\setminus C= A\setminus (B\setminus C)$ if and only if    ** condition **
(set theory)
 A: Break things into logical statements.  Note that
$$
x \in (A \setminus B) \setminus C \iff (x \in A \text{ and } x \notin B) \text{ and } x \notin C\\
x \in A \setminus (B \setminus C) \iff x \in A \text{ and } x \notin B \setminus C\\
\iff x \in A \text{ and not}(x \in B \text{ and } x \notin C)\\
\iff x \in A \text{ and } (x \notin B \text{ or }x \in C)
$$
In other words, we want conditions that will guarantee that 
$$
x \in A \text{ and } x \notin B \text{ and } x \notin C \iff x \in A \text{ and } (x \notin B \text{ or }x \in C).
$$
So, the sets coincide iff given that $x \in A$, it must be the case that $x$ satisfies
$$
x \notin B \text{ and } x \notin C \iff x \notin B \text{ or }x \in C.
$$
Note that the statement on the left implies the statement on the right.  So really, what we need is that for all $x \in A$,
$$
x \notin B \text{ or }x \in C \implies x \notin B \text{ and } x \notin C.
$$
In other words, the sets coincide if the possibilities 
$$
x \notin B \text{ and } x \in C, \quad x\in B \text{ and } x\in C
$$
never occur when $x$ is in $A$.  In other words, the sets coincide iff $x \notin C$ for all $x \in A$.  In other words, the sets coincide iff
$$
A \cap C = \emptyset.
$$
This will only be true for every $A$ if $C = \emptyset$.
A: Another way:
Since $(A\setminus B)\setminus C = A \setminus (B\cup C)$, the associativity equation can equivalently be written as
$$A\setminus (B\cup  C) = A\setminus (B\setminus C)$$
Now obviously that is true for all $A$ exactly if
$$B\cup C = B\setminus C$$
But those two sets differ exactly by the elements of $C$, thus they are equal if and only if $C=\emptyset$.
A: First observe that: $$\left(A\setminus B\right)\setminus C=A\cap B^{\complement}\cap C^{\complement}$$
and: $$A\setminus\left(B\setminus C\right)=A\cap\left(B\cap C^{\complement}\right)^{\complement}=A\cap\left(B^{\complement}\cup C\right)=\left(A\cap B^{\complement}\right)\cup\left(A\cap C\right)$$
So we have: $$\left(A\setminus B\right)\setminus C=A\cap B^{\complement}\cap C^{\complement}\subseteq A\cap B^{\complement}\subseteq\left(A\cap B^{\complement}\right)\cup\left(A\cap C\right)=A\setminus\left(B\setminus C\right)\tag1$$
And what we need is: $$A\cap B^{\complement}\cap C^{\complement}=A\cap B^{\complement}=\left(A\cap B^{\complement}\right)\cup\left(A\cap C\right)\tag2$$
For the second equality it is necessary and sufficient that $A\cap C=\varnothing$
If that is satisfied then $A\cap C^{\complement}=A$ so that also
$A\cap B^{\complement}\cap C^{\complement}=A\cap B^{\complement}$.
So the associativity is accomplished if and only if $A\cap C=\varnothing$.
If this must be the case for every set $A$ then it is unescapable that $C=\varnothing$.
A: If for all A , (A \ B)\ C=A \ (B \ C) , then take A=C , it gives (C \ B)\C=C \ (B\C) , thus $\emptyset=C$ is the condition you are searching for
