Does the order of "unbounded" quantifiers matter?

Question

Is $(\exists x)(\forall y)P(x,y) \Leftrightarrow (\forall y)(\exists x)P(x,y)$ ?

Please understand that I am neither asking about $(\exists x)(\forall y)P(x,y) \Leftrightarrow (\forall x)(\exists y)P(x,y)$ nor $(\exists x)(\forall y)P(x,y) \Leftrightarrow (\exists y)(\forall x)P(x,y)$ nor about "bounded" quantifications like $\forall x \in X, \exists y \in Y$.

Answer 1

No. For example, let $P(x,y)$ mean $x\ne y$, and assume there are at least two things in the universe we quantify over.

Then $\exists x \forall y (x\ne y)$ is false, but $\forall y \exists x (x \ne y)$ is true.

Answer 2

Inspired by Henning's answer...

Let a and b be two distinct objects in the domain of quantification $U$ (the "universe" set). We have:

$$a\in U$$ $$b\in U$$ $$a\neq b$$

Yes, I know the OP said "unbounded quantifiers," but I think you really need to make the "universe" set explicit to understand the principle at work here.

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Part 1

Required to prove: $\neg \exists x\in U: \forall y\in U:x\neq y$

Suppose to the contrary. Let $c$ be such that $c\in U$ and $\forall y\in U: c\neq y$. This leads to the obvious contradiction $c\neq c$.

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Part 2

Required to prove: $\forall y\in U: \exists x\in U: x\neq y$

Let $c$ be such that $c\in U$. Consider two cases.

Case 1: Suppose $a=c$.

Then $b\neq c$ by substitution and symmetry. Since $b\in U$, we then have $\exists x\in U: x\neq c$.

Case 2: Suppose $a\neq c$.

Since $a\in U$, we also have $\exists x\in U: x\neq c$.

In both cases, we have $\exists x\in U: x\neq c$. Generalizing on $c$, we have, as required:

$\forall y\in U: \exists x\in U: x\neq y$

Answer 3

Take for example the following two propositions paraphrased from Smith (2013) $\S$12.1.7 pp.271-2*:

$Lxy:x$ sends an arrow to $y$

1. $\forall x \exists y(Lxy)$
   • This claims that all things send an arrow to another thing.
     A visual representation in Smith (2013) Figure 12.1 $\forall x \exists y(Lxy)$
2. $\exists y \forall x (Lxy)$
   • This claims that there is at least one thing to which every other thing sends an arrow.
     A visual representation in Smith (2013) Figure 12.2 $\exists y \forall x (Lxy)$

They are not equivalent: proposition (1) could be true without (2) being necessarily true. If each thing sends an arrow somewhere, it does not follow that there is some particular thing to which everything must send an arrow.

However if (2) is true then necessarily (1) is true. If everything sends an arrow to one particular thing then a fortiori everything sends an arrow somewhere.

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*Smith, N. (2013). Logic : the laws of truth . Princeton University Press.