Set algebra Question I'm studying for my finals, and I came across the following set/Boolean algebra question:

Use the laws of set algebra to show that:
  $$\varnothing = \Bigl( (X\cup Y)\cap (X\cup Y^c)\Bigr)\cap\Bigl( (X^c\cup Y)\cap (X^c\cup Y^c)\Bigr).$$

my answer:
Distributive law: $X\cup (Y\cap Y^c)\cap X\cup(Y\cap Y^c)$
Complement laws: $X\cup\varnothing \cap  X\cup\varnothing$
Complement laws: $X\cap X^c=\varnothing$
Am I doing this right?
I would attach the exact rules I have to use to make it more clearer, by I don't no how to.
Thanks in advance!!
 A: It is not clear to how you obtained your first expression. Moreover, your expressions are ambiguous; they need parentheses somewhere.  Your second expression $X\cup \emptyset\cap X\cup\emptyset$ seems incorrect. How do you get $X\cap X'$ out of this (if that's what you intended to say)?
Let's hack things out; starting with
$$\tag{1}
\bigl(\color{maroon}{(X\cup Y) \cap (X\cup Y') }\bigr) 
\cap \bigl(\color{darkgreen}{(X'\cup Y) \cap (X'\cup Y')}\bigr),
$$let's first simplify the term $\color{maroon}{(X\cup Y) \cap (X\cup Y')} $: 
We will do this rather methodically, using the distributive laws
$$\eqalign{
(X\cup Y) \cap (X\cup Y') 
&= \bigl( \color{darkblue}{ (X\cup Y)\cap X \bigr) }\cup  
\bigl(\color{teal}{(X\cup Y)\cap Y'}\bigr)\cr
&=    \color{darkblue}{X}\cup \bigl(\color{teal}{ (X\cap Y') \cup (\color{maroon}{Y\cap Y'})}\bigr)\cr
&=    \color{darkblue}{X}\cup \bigl(\color{teal}{ (X\cap Y') \cup ( \color{maroon}{\emptyset})})\cr
&= X\cup( X\cap Y')\cr
&=  (\color{darkblue}{X\cup X}) \cap (X\cup Y') \cr
&=  \color{darkblue}{X}\cap (X\cup Y') \cr
&=X.
}$$
Similarly (or just appealing to the previous computation substituting "$X$" with "$X'$"), you can show that 
$\color{darkgreen}{(X'\cup Y) \cap (X'\cup Y')} =X'$
So $(1)$  reduces to $X\cap X'=\emptyset$.
