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

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 $\forall x$?

• Consider a world where people only love themselves. Commented Oct 23, 2016 at 9:12
• 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
Commented Oct 23, 2016 at 9:12
• It's even possible with only two people: $A$ and $B$, each loving the other only. Commented Oct 23, 2016 at 10:42
• There's a difference between one for all and all for one: en.wikipedia.org/wiki/Unus_pro_omnibus,_omnes_pro_uno Commented Oct 23, 2016 at 10:53
• @AsafKaragila What a sad world. Couldn't you assume a better world? Commented Oct 24, 2016 at 14:22

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.

• Very clear example! :D Commented Oct 23, 2016 at 23:03
• In either case, I predict a drama between the four people involved. Why can it never be as simple as $x \leftrightarrows y$ and $z \leftrightarrows t$? Commented Oct 24, 2016 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.
– user65203
Commented Oct 25, 2016 at 21:19

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.

?????

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.

• 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. Commented Oct 23, 2016 at 19:41

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 $a$.

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.

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.