About the use of Set algebra

Question

I'm stocked with the following exercise:
Show that $A\cap(B\setminus C) =(A\cap B)\setminus (A\cap C) $.

Proof (Attempt): $A\cap (B\setminus C)= A\cap(B\cap C^c) = (A\cap B)\cap C^c $ since intersection is associative. But then... Iv'e got no idea on a following step, or even if this is the right way to get the proof.
So, any help would be grateful.

Answer 1

Another possible solution (not the shortest, probably, but it does the job): now start working with the right-hand side, to have "meet in the middle": $$(A\cap B)\setminus(A\cap C)=(A\cap B)\cap(A\cap C)^c=(A\cap B)\cap(A^c\cup C^c)=(A\cap B\cap A^c)\cup(A\cap B\cap C^c)=\varnothing\cup(A\cap B\cap C^c),$$ as desired.

Answer 2

If $x \in A \cap (B -C)$, then $x$ is in $A$ and $B$, but not in $C$. Then certainly $x$ is not in $A \cap C$, so $x \in A \cap B-(A \cap C)$. Conversely, if $x \in A \cap B-(A \cap C)$, then $x$ is in both $A$ and $B$, but not in $C$. Hence $x$ is in $A$ and $x$ is in $B-C$.

Answer 3

$A\cap (B\setminus C)= A\cap(B\cap C^c) = (A\cap B)\cap C^c=(A\cap B)\setminus(C)=(A\cap B)\setminus(C\cap A)$, since you're removing some elements from $A\cap B$ (the last step), so we take care of elements of $C$ in $A$.