Do we assume an existence of a particular object when using the universal quantifier?

Question

I'm doing a course on logic. In the beginning of the first section on first-order logic, it is said that when using the universal quantifier, we don't assume there exists a particular object for which a formula inside the universal quantifier holds. See the image below:

Therefore, if we have ∀xP(x), we don't assume existence of a particular object that the predicate P holds to. And so if we have an empty set, ∀xP(x) is true.

But later into the course, I'm introduced into axiom schemata of first-order logic, from which I can infer that ∀xP(x)→∃xP(x). But if that is true, whenever I use the universal quantifier, it actually does imply that there exists a particular object for which the formula inside the scope of the universal quantifier holds.

So do we assume an existence of a particular object when using the universal quantifier in first-order logic?

Thank you for your help.

Answer 1

Most logics do indeed assume the existence of at least one object in the domain that the quantifier is quantifying over. This is known as the Assumption of Existential Import.

And yes, we need the Assumption of Existential Import to be able to infer $\exists x \ P(x)$ from $\forall x \ P(x)$

In free logics, however, this assumption is not made.

But I think that the example from the book is slightly orthogonal to this issue. Their example is: "all dots are blue". That would indeed be true if there are no dots ... but it is also not assuming that there is at least one dot. Rather, the Assumption of Existential Import says that there is at least one object .. but that does not need to be a dot.

Indeed, note that the translation of "All dots are blue" is $\forall x (Dot(x) \to Blue(x))$, and while from that we can infer $\exists x (Dot(x) \to Blue(x))$, that is not the same as $\exists x (Dot(x) \land Blue(x))$, i.e. it is not saying that there is at least one blue dot. And you can't infer $\exists x \ Dot(x)$ either from $\forall x (Dot(x) \to Blue(x))$

If you assume that for every kind of thing that you can talk about there is some object that is part of that (e.g. if we would assume that for any 1-place predicate $P(x)$ there is an object $a$ such that $P(a)$), then we would be making a far stronger assumption than the assumption of existential Import. This much stronger Assumption would be called the Assumption of Categorical Existential Import. Indeed, this latter assumption is not an assumption that typical logics make.

However, the Assumption of Categorical Existential Import is sometimes useful when doing categorical logic, e.g. when doing the kind of categorical syllogisms that Aristotle and the medieval monks studied: some categorical syllogisms are valid without the Assumption of Categorical Existential Import, but some are valid only if you make this Assumption ... these are said to be the 'conditionally valid' categorical syllogisms.

In short, most logics assume the existence of at least one object, but most logics do not assume the existence of a particular (kind of) object.

Answer 2

Statements

The following two statements are true:

1. $∀xP(x)$ does not necessarily imply the existence of any particular object and so the set of objects could indeed be empty.

2. $∀xP(x)$$\implies$∃xP(x) subject to the condition that the quantifiers $∀$ and $∃$ are not referring to empty sets.

Therefore, we do not have a contradiction.

Explanation

The key observation here is that you were missing the additional constraint in statement 2 which allows us to avoid such a conflict. Since we presume we are referring to nonempty sets, this means that there is no implication that we are presuming that an object exists as this is an implicit assumption made in the statement. That is to say that we assume that the set contains some objects before we are able to invoke statement 2.

Example

Referring back to the example that you have provided a picture of, we note that all the dots in the set are blue is true when there are no dots in the set, and so it is clear that this does not necessarily imply the existence of a blue dot (statement 1). However, if we impose the constraint that the set cannot be empty, then there must exist at least one blue dot (which is exactly what statement 2 tells us).