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

Question

1. Everybody loves somebody. $∀x\,∃y\,L(x, y)$

2. 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 $\forall x$?

Answer 1

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.

Answer 2

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.

Answer 3

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.

Answer 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.

Answer 5

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 $a$.

Answer 6

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.

?????