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Background: I'm writing up the answers to the end-of-chapter exercises in the forthcoming second edition of my Introduction to Formal Logic. Those exercises and all the extensive worked answers will be freely available online, and could be useful (I hope!) to people following other texts. Which is why it seems fair play to consult the wisdom of the crowd here!

One of the slightly nastier natural deduction exercises asks the reader to prove that $$\forall x\exists y(Fy \to Gx) \therefore \exists y\forall x(Fy \to Gx)$$ is valid by producing a derivation from the premiss to the conclusion. This is initially a rather surprising result because of course it is usually a quantifier shift fallacy to argue from $\forall x\exists y$ to $\exists y\forall x$. But in this case the inference is OK.

A bit of brute force applying familiar tricks will get the natural deduction proof done in twenty-plus lines (using familiar rules, as e.g. in Magnus forallx). But what I'm having trouble with (a failure of imagination?!) is coming up with a neat version of the preliminary chat:

What if anything can we say (in a brief arm-waving way), before giving the formal proof, to motivate the claim that -- on second thoughts -- this inference intuitively ought to be valid (so this case isn't a quantifier-shift fallacy)?

Any good suggestions gratefully received (and will be equally gratefully acknowledged)!

MJD
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Peter Smith
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  • Here's what it looks like when put into a proof tree generator. I hope this helps. – Shaun Apr 02 '20 at 19:43
  • Is that truth tree correct? Something odd seems to happen at the bottom. The hand-done treeI did for interest is nastier ...and doesn't really help in generating intuitions! – Peter Smith Apr 02 '20 at 19:49
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    One possibility for an outline: if $Fy_0$ is false for some $y_0$, then $\forall x (F y_0 \rightarrow G x)$. Otherwise, $Fy$ is true for all $y$, then the premise implies $\forall x (G x)$, and any arbitrary $y_0$ (from the implicit assumption that universes are non-empty) will satisfy $\forall x (Fy \rightarrow Gx)$. – Daniel Schepler Apr 02 '20 at 19:55
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    The argument given by @DanielSchepler is how I convinced myself that the inference is correct. If your readers already know about prenex form, then you could also point out that both the assumption and the conclusion are prenex forms of $(\forall y,Fy)\to(\forall x,Gx)$. And then you could ask whether these two approaches are really different. – Andreas Blass Apr 02 '20 at 21:31
  • At the intuitive level, the trick is that $Fy$ and $Gx$ express "unrelated" properties and $\to$ does not express a "link". Consider: "either $y$ is not a French or $x$ is a Geographer"; we can move around the quantifiers freely. – Mauro ALLEGRANZA Apr 03 '20 at 07:16
  • As a student myself, I agree that the explanation given by @DanielSchepler is quite intuitive, but I have a naive concern: If our context is some universe with no objects at all, the first statement would be true, but the second false, this to me adds quite a bit of confusion – Carlyle Dec 21 '23 at 22:18
  • I also think Daniel's argument is a lot more clear if you combine it with what is said by @MauroALLEGRANZA, i.e. work with $\lor$ instead of $\rightarrow$, then you avoid the rule of "false implies anything" which sometimes confuses students too – Carlyle Dec 21 '23 at 22:22

2 Answers2

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This is not valid unless your logic allows you to prove some $y$ exists. I'm not ecstatic to renew the perennial flame war over whether to formulate logic so that $\exists x (\top)$ is provable, especially with an expert in logic, but I would direct your attention to Kruckman's thoughts on the matter if it's something you've never considered before. It's also straightforward to see that this proposition is not constructively provable, so some nastiness is all but inevitable.

This problem can be solved using the Drinker paradox. Using the Drinker paradox, take $y$ such that $F(y) \to \forall z (F(z))$; this is the $y$ we need. For consider some $x$; take $z$ such that $F(z) \to G(x)$. Then we have $F(y) \to F(z)$, and $F(z) \to G(x)$; thus, we have $F(y) \to G(x)$.

Mark Saving
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As Blass had pointed out, it is the simpler $(\forall y Fy)\to (\forall x Gx)$ which is equivalent to both

$\forall x\exists y (Fy\to Gx)$

and

$\exists x\forall y (Fy\to Gx)$ and $\exists y\exists x (Fy\to Gx)$

And we get to the former through a lengthy sequence of basic manipulations.

I think what obfuscates it is that in the $\forall x \exists y$-association statement, the $x$ in the matrix is actually placed last. The $\exists y \forall x$ variant is the one which is "in order", but then we need to make upfront distinctions about the possible scenarios for $y$. The thing somewhat resolves by performing just one contraposition. Let me define $T=\neg G$ and $L=\neg F$ and state the question again.

Question: Why is

$\forall x\exists y (\neg Ly\to \neg Tx)$

$\exists y\forall x (\neg Ly\to \neg Tx)$

Answer: The are both equivalent to $(\exists x Tx)\to (\exists y Ly)$.

Say "If there is a teacher, then there's also a learner."

Given this, I'll vocalize the more convoluted two (actually equivalent) statements implied by it. Firstly,

$\forall x\exists y (Tx\to Ly)$

expresses that for every person $x$, there's another person $y$ such that if $x$ is a teacher, then $y$ is a learner. If $Tx$ holds, then this is clear, because it's dubious what the "teacher" attribute means if nobody is their student (a learner). But I choose an example so that we can conceive a learner being there even if there's no teacher. As in, if everybody else in the world but me died, I'd not be a teacher but I'd still go to the library to learn.

The original phrasing, now written

$\forall x\exists y (\neg Ly\to \neg Tx)$

is fairly annoying to take apart, because the $Tx$ is at the end. To analyze this we jump to the end and make the distinction: For any given $x$, either $x$ is a teacher or is not. If not, then okay we're done as the conclusion $\neg Tx$ will always hold. As we say there can be a learner without there being a teacher this $\neg Tx$ does not invalidate the case where $Ly$ also holds. Now if on the other hand $Tx$ does hold, then we certainly know that there's some learner in the world ($Ly$, in fact a student), and the implication also holds, now by contradiction with $\neg Ly$.

The other,

$\exists y\forall x (Tx\to Ly)$

expresses that there's person $y$ such that for all persons $x$ does holds that if $x$ is a teacher, then $y$ is a learner. To analyze we ask if there's a teacher at all. If yes, then there's also a learner (again, it's even a student) and so $\exists y Ly$ point blank. Otherwise the implication $Tx\to Ly$ is always trivial also. Lastly, given we have the "big" disjunction regarding $Tx$ here, the analysis of the statement in terms of $\neg Ly\to \neg Tx$ is no different.

Nikolaj-K
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