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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!!

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You could accept any answers that have been useful by clicking on the check mark next to them. One acceptance per question. – Ross Millikan Apr 20 '12 at 16:21
Okie dokie i will do – Xabi Apr 20 '12 at 16:27
up vote 0 down vote accepted

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 $X\cap X'=\emptyset$.

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Thanks for explaining that clearly :) – Xabi Apr 20 '12 at 17:39

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