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Let $S$ be a non-empty set. Consider the statement $$\forall A\subseteq S \quad \forall B\subseteq S\quad\exists D\subseteq S \ (D\neq \emptyset \ \wedge \ D\cap (A\cup B)=\emptyset) $$ The negation is $$\exists A \subseteq S \quad \exists B\subseteq S \quad \forall D \subseteq S \ (D=\emptyset \ \vee \ D\cap (A \cup B)\neq \emptyset). $$ I have to determine which statement is true.
I had a long think and thought that the second is true since I think the first is false.
Let $S = \{1,2,3\}$.
Then if I pick $A= \{1,2\}$ and $B=\{2,3\}$. If $D$ is empty then the statement is false. If $D$ is not empty, then $D$ must be a set which has common members with $A$ or $B$ and so the intersection is nonempty.
Is this correct or is there a better way to think about it?

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    $\begingroup$ Yes, this is correct. $\endgroup$ – Mees de Vries Nov 11 '17 at 11:20
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You are correct.

To see how this works for any $S$: Pick $A=B=S$. Then $A \subseteq S$, $B \subseteq S$, and $A \cup B =S$. Hence, there cannot be any non-empty $D \subseteq S$ that does not share any elements with $A \cup B$.

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