# Simplification of a set theory expression with complements

I have the set theory expression that I need to simplify.

I have used De Morgan's Laws as well as other identities to simplify this expression but I always end up expanding it and then coming back to the original expression.

The expression is: (X intersect Y)' intersect Y'.


Could someone help me with this?

Thank you.

One of my attempts is :
(X intersect Y)' intersect Y
= (Y union (X intersect Y))'
= ((Y union X) intersect (Y union Y))'
= ((Y union X) intersect Y)'
= (Y union X)' union Y'

• To show $A\cap B,$ use ´A\cap B´ between dollar signs; for $A\cup B,$ use ´A\cup B´. Could you share one of your attempts with us? It seems like you're probably very close to the solution. Feb 14, 2016 at 15:13

The expression is $(X^C \cup Y^C) \cap Y^C$, which simplifies to $Y^C$.

Indeed, $z \in (X \cap Y)^C$ if and only if $z \not \in X \cap Y$, if and only if it is either not in $X$ or not in $Y$. That is, if and only if it is in $X^C$ or is in $Y^C$; that is, iff it is in $X^C \cup Y^C$.

Therefore, $X^C \cup Y^C = (X \cap Y)^C$.

$X^C \cup Y^C \supseteq Y^C$, so $(X^C \cup Y^C) \cap Y^C = Y^C$.

• Explain how please. Feb 14, 2016 at 15:11
• That is not the expression Feb 14, 2016 at 15:11
• Read the original expression again. Feb 14, 2016 at 15:12
• @rert588 I don't take well to such blunt imperatives when I give my time for free. Maybe I'll come back to this in a bit. Feb 14, 2016 at 15:12
• @rert588 Edited in more explanation. Feb 14, 2016 at 15:17