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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 $\forall x \in X, \exists y \in Y$.

Alexandre Khoury
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    Why would you expect there to be a different between bounded and unbounded quantifiers in this case? – fgp Jan 05 '16 at 14:02
  • For example in set theory with $P(x,y)$ defined as $x\in y$, the statement $(\exists x)(\forall y) P(x,y)$ is false and $(\forall y)(\exists x) P(x,y)$ is true (by taking $x={y}$, for example). – Hagen von Eitzen Jan 05 '16 at 18:00
  • @HagenvonEitzen: It would need to be $y\in x$ for that to work -- $(\forall y)(\exists x)x\in y$ is definitely not true. And even so, it is not trivial that there is no set of everything. – hmakholm left over Monica Jan 05 '16 at 19:02

3 Answers3

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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.

hmakholm left over Monica
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  • Nice example. Having the universe consisting of the only two objects would be simpler, but your example is somehow more natural. See the proof in my answer. – Dan Christensen Jan 05 '16 at 17:50
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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.

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 $c\neq c$.

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:

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

Dan Christensen
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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. $\forall x \exists y(Lxy)$
  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.

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

Daniel Maher
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