Need to simplify $(A ∩ \varnothing)' \cap (A \cup B)'$

Question

I have the following expression I need to simplify:

$$(A \cap \varnothing)' \cap (A \cup B)'$$

So, far my solution is to use DeMorgan's Law to simplify it as follows:

$$(A' \cup \varnothing') \cap (A' \cap B')$$

But I'm not sure where to go from here. I was perhaps thinking of using Communicative Law to swap the $A'$ and $\varnothing'$ in the middle:

$$A' \cup \varnothing' \cap A \cap B'$$

So, it becomes:

$$(A' \cup A') \cap (\varnothing' \cap B)'$$

And then go from there. But I'm not sure if that's allowed.

Is there another way to simplify this?

Answer 1

Better: $A \cap \emptyset = \emptyset.$ Let $X$ be the universal set. Then

$$ X \cap (A \cup B)' = X \cap A' \cap B'$$

Now as A and B are in the universal set, so are their complements. This best simplifies as $A' \cap B'$.

Answer 2

You need parentheses in the second line. Union and intersection do not associate however you want. As $(A \cap \emptyset)'=A' \cup \emptyset'=A' \cup U=U'$ where $U$ is the universe the second line becomes $(A' \cup U)\cap (A' \cap B')=U \cap (A' \cap B')=A' \cap B$