May be this is a stupid question but I was thinking we know that suppose $D = \varnothing$ then $\forall\, x \in D \,P(x)$ is true vacuously and $\exists y\in D\, P(y)$ is false. What is you mix the two like $\forall x \in D \,\,\exists \,y \in D \,Q(x, y)$. Then is $Q$ true or false or neither?
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2Can't say anything about $Q$, but the entire thing is vacuously true because of the outermost $\forall x \in D$. – Daniel Fischer Jun 26 '13 at 12:33
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So if order is switched, then it's false? – mathnoob Jun 26 '13 at 12:34
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Yes. setting $P(x)=\exists y\in D:Q(x,y)$, we know $\forall x\in D: P(x)$ is vacuously true. Switching them makes it false. – Thomas Andrews Jun 26 '13 at 12:36
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True. (True that it's false if you put the $\exists y \in D$ outermost.) – Daniel Fischer Jun 26 '13 at 12:36
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Ok thanks for the clarification. – mathnoob Jun 26 '13 at 12:37
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@DanielFischer Probably good to promote these comments to an answer to prevent this from entering the Unanswered queue. – Lord_Farin Jun 26 '13 at 12:38
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This sort of thing is one reason why domains are usually taken to be nonempty by convention. – Carl Mummert Jun 26 '13 at 12:47
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$\bigl(\forall x \in D\bigr) \bigl(\exists y \in D\bigr) Q(x,y)$ has the form $\bigl(\forall x \in D\bigr) P(x)$, where $P(x)$ expands to $\bigl(\exists y \in D\bigr) Q(x,y)$. So it is vacuously true.
If you switch the order of the quantors, it becomes vacuously false, since you get a proposition of the form $\bigl(\exists y \in D\bigr) R(y)$ (with $R(y) = \bigl(\forall x \in D\bigr) Q(x,y)$).
When you have an empty domain $D$, only the outermost quantor matters, any outermost $\bigl(\forall x \in D\bigr)$ makes it vacuously true, $\bigl(\exists y \in D\bigr)$ vacuously false.
Daniel Fischer
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