How to prove a set equality? Statement: Let $A$, $B$, and $C$ be sets. Then $(A \setminus B) \cap (A \setminus C) = A \setminus (B \cup C)$. 
Proof: Let $(x,y) \in ( A \setminus B) \cap (A \setminus C)$, where $ x \in A$ and $y \in (B \cap C$ by definition of intersection. So, $(x,y) \in (A \setminus B \cap C)$ which show that $(A \setminus C) \cap (A \setminus C) \subseteq A \setminus (A \cup C)$. \ Now let $(x,y) \in A \setminus(B \cup C)$, where $ x \in A $. So, $ (x,y) \in (A \setminus B) \cap (A \setminus C)$ by definition of intersection, which shows that $A \setminus (B \cup C) \subseteq (A \setminus B) \cap (A \setminus C)$. Therefore, by the Axiom Set of Equality we know that $(A \setminus B) \cap (A \setminus C) = A \setminus (B \cup C)$ . 
Am I on the right track to proving that this is a true statement or is there a better way to show this? 
 A: I turn this into words as
the items in A but not in B
and in A but not in C
are the items
that are in A
but not in either B or C.
I then rewrite this,
as Behrouz Maleki did,
 as
$(A \cap B^c) \cap(A \cap C^c)
=A \cap (B\cup C)^c
$.
Since
$\cap$
is associative,
commutative,
and idempotent,
the left side becomes
$A \cap B^c\cap C^c
=A \cap (B^c\cap C^c)
$
and we are done if
$B^c\cap C^c
=(B\cup C)^c
$.
For the last,
page De Morgan.
A: The def'n of $B\cup C$ is that for any $x$ we have $x\in B\cup C\iff (x\in B\lor x\in C).$ Therefore, for any $x$ we have $x\not \in B\cup C\iff$ $ \neg (x\in B \lor x\in C)\iff  (x\not \in B\land x\not \in C).$
So for any $x$ we have $x\in (A$ \ $B)\cap (A$ \ $C)\iff$ $$ (x\in A\land x\not \in B)\land (x\in A\land x\not \in C)\iff$$ $$ (x\in A)\land (x\in A)\land (x\not \in B)\land (x\not \in C)\iff$$ $ (x\in A)\land (x\not \in B\land x \not \in C)\iff$ $ (x\in A)\land (x\not \in B\cup C)\iff x\in A$ \ $(B\cup C).$
A: $$(A-B)\cap(A-C)=(A\cap B^c)\cap (A\cap C^c)=A\cap(B^c\cap C^c)=A\cap (B\cup C)^c=A-(B\cup C)$$
