Prove the set identity using the laws of set theory 
$A\cap(B\cup A')\cap B'=\emptyset$

Using the distributive law I got:
$(A\cap B)\cup (A \cap A')\cap B'=\emptyset$
But I don't see any rules to simplify $(A \cap A')$
Any tips in helping me to progress through this problem are appreciated!
Edit:
Staring at the problem more I see I can also organize it this way:
$(A\cap B)\cup (A'\cap B')=\emptyset$
Just not sure how of to use the laws to simplify this into equally the empty set.  
 A: Assuming you're using $A^\prime$ to denote the compliment of $A$, note the following.
Simply use the distributivity laws to note:
$A \cap (B \cup A^\prime) \cap B^\prime = ((A \cap B) \cup (A \cap A^\prime)) \cap B^\prime.$
Now, we know $A\cap A^\prime = \emptyset$, so:
$A \cap (B \cup A^\prime) \cap B^\prime = ((A \cap B) \cup \emptyset) \cap B^\prime.$
Note that $(A \cap B) \cup \emptyset$ is the set containing all of the elements in $A \cap B$ and all of the elements in $\emptyset$. However, there are no elements in $\emptyset$, so $(A \cap B) \cup \emptyset$ simply contains all of the elements in $A$. Therefore, $(A \cap B) \cup \emptyset = A \cap B$. Using this, we have:
$A \cap (B \cup A^\prime) \cap B^\prime = (A \cap B) \cap B^\prime.$
Now, set intersection is distributive, so we see
$A \cap (B \cup A^\prime) \cap B^\prime = A \cap (B \cap B^\prime).$
Similar to before, we know $B \cap B^\prime = \emptyset$, so
$A \cap (B \cup A^\prime) \cap B^\prime = A \cap \emptyset = \emptyset.$
