In my logic class, we explored whether various quantified statements implied others. We agreed that the statement "Everyone loves everyone" implies that "Everyone loves someone":
$Px$ = $x$ is a person
$Lxy$ = $x$ loves $y$
$$1. (∀x)(Px \implies (∀y)(Py \implies Lxy))$$
Implies that:
$$2.(∀x)(Px \implies (∃y)(Py ∧ Lxy))$$
We didn't go into it in class however it seems to me that this form only holds try when the inner and outer predicates are the same.
If I change the inner predicate to something else:
$Kx$ = $x$ is a kitten
$$3.(∀x)(Px \implies (∀y)(Ky \implies Lxy))$$
No longer implies that:
$$4.(∀x)(Px \implies (∃y)(Ky ∧ Lxy))$$
The contradictory case is that there are people, but in fact no kittens exist. In this case it is true that for every person, for everything else, its kitten-hood implies the person's love for it (antecedent $Kx$ is always false), but it is not the case for every person (anyone actually) that there exists a kitten that they love.
It worked, however, when the predicates were the same, because if there were people to do the loving then there were people available to be loved even if it was just a solitary narcissist.
Is my understanding correct? It seems counter-intuitive to reject 3 -> 4 when the form is so similar to 1 -> 2. Is there more to it than this?
OP, your reasoning is correct. Imagine four worlds, containing:
People and kittens
People, but no kittens
Kittens, but no people
Neither kittens nor people
Suppose we assume that all people always love all people and all kittens, on every world. Your first three statements hold true on all four worlds, but the fourth fails on the second world. Since there are no kittens on the 2nd world, but there are people, there will be some person without a kitten to love. Your second statement doesn't care about kittens, only people. On the first two worlds, people exist, and everyone loves at least themselves, so there are people to love. On the last two worlds, there are no people, so the second statement holds vacuously.