Prove that $(A-B) \cap (A-C) = A \cap (B \cup C)^c$ for any three sets A, B, C. I was given a question that says Prove that $(A-B) \cap (A-C) = A \cap (B \cup C)^c$ for any three sets A, B, C.
I'm completely lost with this question. In a previous question that says $A \cap C \subseteq A- (B-C)$. I used this proof
Let $x \in A \cap C$. Then $x \in A$ and $x \in C$. Also note that $x \notin B-C$. As $x \in A$ and $x \notin B-C$, we see that $x \in A - (B-C)$. Therefore $A \cap C \subseteq A- (B-C)$. 
With the question I just did i tried to apply that method with the question i struggled with but I couldn't see how it would work.
 A: The set $(A-B)\cap(A-C)$ is "everything in $A$ that is not in $B$ and not in $C$", and the set $A\cap(B\cup C)^{\mathsf{c}}$ is "everything that is in $A$ and not in ($B$ or $C$)". 
Can you see why these are the same thing?
A: Suppose $x\in (A-B)\cap (A-C)$. Then $x\in A$ and $x$ is not in $B$ and not in $C$, which means $x\in (B\cup C)^c$. Therefore, $(A-B)\cap (A-C)\subseteq A\cap(B\cup C)^c$.
Conversely, suppose that $x\in A\cap (B\cup C)^c$. This means that $x\in A$ and $x\in (B\cup C)^c$. This means that $x$ is in $A$, but $x$ is not in $B\cup C$, which means that $x$ is not in $B$ and $x$ is not in $C$. Therefore, $x\in (A-B)\cap (A-C)$, so $A\cap (B\cup C)^c\subseteq (A-B)\cap (A-C)$.
Combining these two yields the result.
A: Usually, one defines $X \setminus S = X \cap S^c$.  If you're used to the more English definition of "everything in X that is not in S", you should think about why these say the same thing.
Regardless, we now get that $(A\setminus B) \cap (A \setminus C)
\\ = (A \cap B^c) \cap (A \cap C^c)
\\ = A \cap B^c \cap A \cap C^c
\\ = A \cap B^c \cap C^c$,
after we removed the redundant $A$ factor.  Now, applying D'morgan's laws, we have $B^c \cap C^c = (B \cup C)^c$ and we're done.
A: We redefine $\cap$ as $\cdot$ and $\cup$ as $+$, and $A-B=A\cdot (A-B)$
so $(A-B)(A-C)=A(A-B)A(A-C)=(A-AB)(A-AC)=(A-AB)-(AC-ABC)=A((E-B)-(C-BC))=A(E-B-C)=A(E-B)(E-C)$ 
so the we can also get $(A-B)\cap (A-C)=A\cap B^{c}\cap C^{c}=A\cap (B\cup C)^{c}$ in this equation. It's Boole Algebra.
A: Here is an extremely easy way to prove your set identity without recourse to an element-chasing proof (that is, this proof simply relies on set algebra):
\begin{align}
(A-B)\cap(A-C)&= (A\cap B^c)\cap(A\cap C^c)\tag{by definition}\\[0.5em]
&= (A\cap A)\cap(B^c\cap C^c)\tag{associativity of $\cap$}\\[0.5em]
&= A\cap(B\cup C)^c.\tag{simplify & DeMorgan}
\end{align}
I imagine the goal of the exercise is for you to realize how you can effectively exploit the associative property of $\cap$. As you can see above, if you are clued into associativity, then the proof is a piece of cake. 
