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'
  • $\begingroup$ 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. $\endgroup$ Feb 14, 2016 at 15:13

1 Answer 1


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

  • $\begingroup$ Explain how please. $\endgroup$
    – rert588
    Feb 14, 2016 at 15:11
  • $\begingroup$ That is not the expression $\endgroup$
    – rert588
    Feb 14, 2016 at 15:11
  • $\begingroup$ Read the original expression again. $\endgroup$
    – rert588
    Feb 14, 2016 at 15:12
  • $\begingroup$ @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. $\endgroup$ Feb 14, 2016 at 15:12
  • $\begingroup$ @rert588 Edited in more explanation. $\endgroup$ Feb 14, 2016 at 15:17

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