Set Theory: Simplify ((A∩(B∪C))∩(A−B))∩(B∪Cc) I am attempting to simplify this expression:
$((A\cap(B\cup C))\cap(A−B))\cap(B\cup C^c)$
With venn diagrams I am able to figure out that I should be able to simplify it down to an empty set. But I am getting stuck trying to write a proper proof for it. I have gotten this so far:
$((A\cap(B\cup C))\cap(A−B))\cap(B\cup C^c)$
$((A\cap(B\cup C))\cap(A\cap B^c))\cap(B\cup C^c)$ Set Difference Law
$(((A\cup B)\cap(A\cup C))\cap(A\cap B^c))\cap(B\cup C^c)$ Distributive law
what are the next steps?
 A: You've made a small, but very significant error in your application of the distributive law:
You've written $A \cap (B \cup C) = (A \cup B) \cap (A \cup C)$, but the distributive law says $A \cap (B \cup C) = (A \cap B) \cup (A \cap C)$.
So all together you now have:
$$[(A \cap B) \cup (A \cap C)] \cap (A \cap B^C) \cap (A \cap C^C)$$ 
Edit to explain the empty set:
We can now apply the distributive property again to get:
$$\left[(A \cap B) \cap (A \cap B^C) \cap (A \cap C^C)\right] \cup \left[(A \cap C) \cap (A \cap B^C) \cap (A \cap C^C)\right]$$
Note that if $x \in (A \cap B) \cap (A \cap B^C) \cap (A \cap C^C)$, then $x \in A \cap B$ and $x \in A \cap B^C$, so in particular $x \in B$ and $x \in B^C$. No such $x$ exists! Thus $(A \cap B) \cap (A \cap B^C) \cap (A \cap C^C) = \emptyset$. Similarly, $(A \cap C) \cap (A \cap B^C) \cap (A \cap C^C) = \emptyset$.
A: 
I have gotten this so far:
$((A\cap(B\cup C))\cap(A−B))\cap(B\cup C^c)$
$((A\cap(B\cup C))\cap(A\cap B^c))\cap(B\cup C^c)$ Set Difference Law
$\require{cancel}\color{red}{\cancel{\color{black}{(((A\cup B)\cap(A\cup C))\cap(A\cap B^c))\cap(B\cup C^c)}}}$ Distributive law

$\quad(((A\cap B)\cup(A\cap C))\cap(A\cap B^\mathsf c))\cap(B\cup C^\mathsf c)$ Distributive law
But instead of that, I would suggest distributing out the common conjunct.
$\quad (A\cap ((B\cup C)\cap B^\mathsf c))\cap(B\cup C^\mathsf c)$ Distributive law
$\quad A\cap (C\cap B^\mathsf c)\cap(B\cup C^\mathsf c)$ Absortion
The way forward should be clear from here.
