What's the difference between $∀x\,∃y\,L(x, y)$ and $∃y\,∀x\,L(x, y)$?

Question

Everybody loves somebody. $∀x\,∃y\,L(x, y)$
There is somebody whom everybody loves. $∃y\,∀x\,L(x, y)$

What's the difference between these two sentences? If they are same, can I switch $\exists y$ and $\forall x$?

Suppose there are three people in the world: Ann, Bob, and Chuck. Suppose that Ann loves Bob, Bob loves Ann, Chuck loves Ann, but nobody loves Chuck. Is statement 1 true? Is statement 2 true? — bof, Oct 23 '16 at 09:12
It's even possible with only two people: $A$ and $B$, each loving the other only. — Patrick Stevens, Oct 23 '16 at 10:42
There's a difference between one for all and all for one: https://en.wikipedia.org/wiki/Unus_pro_omnibus,_omnes_pro_uno — Michael Hoppe, Oct 23 '16 at 10:53
@AsafKaragila What a sad world. Couldn't you assume a better world? — Simply Beautiful Art, Oct 24 '16 at 14:22
Suppose the variables range over real numbers instead of people, and $L(x,y)$ means "$x\lt y.$ Is the first sentence true or false? Is the second one true or false? — bof, Oct 25 '16 at 07:52
@SimpleArt How about a world where everybody loves everybody else but not himself/herself? — Andreas Blass, Oct 25 '16 at 18:25
@AndreasBlass Lmao, I love everyone here and myself. That's the world I wanna live in. — Simply Beautiful Art, Oct 25 '16 at 18:40
@Ned Stack : for every person, there is a jacket that has the suitable size for that person. However, it is not true that there is a jacket that suits every person. — Watson, Nov 20 '16 at 17:05

score 36 · Answer 1 · 2016-10-23T09:28:28.463

36

In 1, everybody loves someone, be it $y$ or $z$. In 2, everybody loves $y$.

2 is stronger than 1: 2 implies 1 but not conversely.

edited Oct 23 '16 at 09:28

answered Oct 23 '16 at 09:18

1

Very clear example! :D – Jeel Shah Oct 23 '16 at 23:03
4

In either case, I predict a drama between the four people involved. Why can it never be as simple as $x \leftrightarrows y$ and $z \leftrightarrows t$? – Jeppe Stig Nielsen Oct 24 '16 at 13:19
@JeppeStigNielsen: $\forall x,\exists y:L(x,y)\land L(y,x)\land\forall z\ne x,y:\lnot L(x,z)\land\lnot L(y,z)$ maybe. – Oct 25 '16 at 21:19

score 13 · Answer 2 · answered Oct 24 '16 at 06:05

13

I'm late to the party but: replace "loves somebody" with "has someone as a mother".

Everbody has a mother.

vs.

There is somebody who is everybody's mother.

?????

answered Oct 24 '16 at 06:05

fleablood

130,341

score 10 · Answer 3 · answered Oct 23 '16 at 09:29

10

That's a great example of why quantifiers don't commute! For the sake of simplicity, assume everybody in the world is married, and everybody loves his spouse. Then the first formula is satisfied. However, there is no reason to think that the second formula is satisfied; in fact, it could be that people only love their spouses, so that there is nobody in the world who is loved by everybody.

answered Oct 23 '16 at 09:29

Mikhail Katz

47,573

1

At first, I wanted to joke that your counterexample doesn't hold in a world with extreme polygamy, but then I realized that that would require people to be able to marry themselves. – Kevin Long Oct 23 '16 at 19:41

David E · Answer 4 · 2016-10-24T17:02:22.333

There are already some fantastic answers here. But I wanted to add a discussion about why quantifiers don't commute, and what we even mean by quantifiers.

So what doe we mean by: $\forall a \ \exists \, b \ P(a, b)$? Well, it means that if I choose to fix any $a$, then given this information, I can find a $b$ which has $P(a,b)$ being true. That is, $b$ is dependent on $a$. To make this clear, we sometimes write:

$$ \forall a \ \exists \, b(a) \ \ P(a, b) \qquad \text{or alternatively} \qquad \forall a \ \exists \, b_a \ \ \ \ P(a, b) $$

Conversely, $\exists b \ \forall a \ P(a, b)$ means, that with no knowledge of $a$, I can find a $b$ which satisfies $P(a, b)$ - or in other words, I can find a $b$ working for all $a$.

In essence: In the first case, you can think of $b$ as a function of $a$ - it could be a different $b$ for each $a$. In the second case, we must have that $b$ can be constant, that is, independent of the choice of $a$.

score 4 · Answer 5 · answered Oct 23 '16 at 09:33

4

Everyone likes some sweet.

There is one sweet liked by everyone.

First one allows the interpretation which sweet is liked by whom is individual choice.

Second sentence says there is a universally liked sweet.

answered Oct 23 '16 at 09:33

P Vanchinathan

20,267

score 2 · Answer 6 · answered Oct 23 '16 at 09:18

If $L$ satisfies 2., then it necessarily satisfies 1. Therefore you can switch $\exists y$ and $\forall x$ to go from 2. to 1., but not the other way around. Counterexample: let $L$ be a relation over set $S=\{a,b,c\}$, and suppose $L(a,b)$, $L(b,c)$, $L(c,a)$. You can easily verify that 1. holds here, but 2. does not.

What's the difference between $∀x\,∃y\,L(x, y)$ and $∃y\,∀x\,L(x, y)$?

6 Answers6