5

I've learned that $\forall x \forall y \phi$ is equivalent to $\forall y \forall x \phi$, and the same is true for $\exists x \exists y$. However $\exists !x \exists !y$ is not equivalent to $\exists !y \exists !x$. I do not see how this makes sense intuitively.

Additionally, I've been evaluating these expressions as "for all variables x and for all variables y", "there exists variables x and y", so this would be "there only exists one x, y".

It seems that my way of thinking about this is incorrect.

How should I be evaluating these?

user21820
  • 60,745
user13964566
  • 61

2 Answers2

11

To start things off, let's whip up a simple-to-understand example showing that the sort of switching you have in mind can't possibly be legal in general (it sounds like you already know this, but other readers might benefit from its explicit inclusion here):

Imagine a large dining room with lots of tables, where all but one of the tables have multiple chairs and a single table has a single chair. Let $\varphi(a,b)$ be the formula "$a$ is a chair at table $b$." It may well be the case that exactly one table has exactly one chair at it. However, barring some very surprising chair-related shenanigans, it should not be the case that exactly one chair belongs to exactly one table: presumably every chair belongs to exactly one table. In any reasonable instance of the above situation, $\exists!x\exists!y\varphi(x,y)$ should be false but $\exists!y\exists!x\varphi(x,y)$ should be true.

OK, now we know that there's got to be something wonky going on; let's find it.

As is often the case, the culprit lies in the set of our human-linguistic-efficiency-oriented foibles. Despite the notation we use for it, "$\exists!$" should really be thought of as a mixed-quantifier expression: when we write out $$\exists !x\varphi(x)$$ in terms of the basic quantifiers $\exists$ and $\forall$, what we get is $$\exists x\forall x'[\varphi(x)\wedge(\varphi(x')\rightarrow x=x')]$$ (or something similar). In particular, this means that a quantifier prefix like $$\exists!x\exists!y$$ should really be thought of as $$\exists x\forall x'\exists y\forall y',$$ and the switching you have in mind would require passing the $\forall x'$ through the $\exists y$ and passing the $\exists x$ through the $\forall y'$. Each of these two maneuvers is of course illegal in general. Incidentally, the discussion here is relevant.

Finally, as a fun exercise let's dive into the parenthetical remark above. We can also interpret "$\exists!$" as having the form "$\forall\forall\exists$" (phrasing this precisely takes a bit of work, but the idea should be intuitively clear):

Regardless of what $\varphi$ is, the formula $$\exists!x\varphi(x)$$ is equivalent to $$\forall x\forall x'\exists x''[\varphi(x'')\wedge (x\not=x'\rightarrow\neg(\varphi(x)\wedge\varphi(x')))].$$ Note that the "matrix" (= inside-the-quantifiers part) of this latter formula is a Boolean combination of instances of $\varphi$ itself and quantifier-free formulas, so this really is "just" a $\forall\forall\exists$-interpretation of $\exists!$. Note also that the existential bit doesn't interact with the universal bits at all so this also gives a $\forall\exists\forall$ and $\exists\forall\forall$ interpretation; however, each of these is strictly worse than the $\exists\forall$ interpretation above.

Meanwhile it's not hard to show that we cannot interpret $\exists!$ in a purely-existential, purely-universal, or $\forall\exists$ way. So the pair of interpretations we've seen above - $\exists\forall$ and $\forall\forall\exists$ - is in a precise sense optimal.

Noah Schweber
  • 260,658
  • Expanding on the parenthetical remark in the middle of the answer: $∃x;∀y;\big(\varphi (y) ↔ y=x\big)$ is my favourite expansion of $\exists !x;\varphi(x)$ because most succinct. – ryang Apr 21 '22 at 03:17
  • @ryang It's less succinct if you don't have "$\leftrightarrow$" as primitive. :P – Noah Schweber Apr 21 '22 at 03:18
  • It admittedly takes longer (than your above suggestion) to parse/verify. -) – ryang Apr 22 '22 at 17:38
1

Noah’s answer is excellent, but just to add an illustrative example, consider the difference between:

  1. There is exactly one person who has exactly one car. (So every other person has either multiple cars, or no car.)
  2. There is exactly one car that has exactly one owner. (So ever other car is co-owned, or abandoned.)

I think it’s reasonably clear that these mean different things. And to really justify they do, one can come up with examples where one fails and the other holds: Imagine a very rich suburb where almost everyone owns 10 private cars, except for one miser who has just 1. Then it’s true that “there’s exactly one person with exactly one car”, but it’s not true that “there’s exactly one car with exactly one owner” (since there are lots of cars that each have exactly one owner).

Exercise: Come up with a similar example where (2) holds but (1) fails.

Peter LeFanu Lumsdaine
  • 11,200