Since multiple answers here mention equality being undefinable in first-order logic, I want to say that that’s only true if there are an infinite number of constants/functions/predicates in the language. Since set theory can be formalized using only a single predicate, it is possible to define equality this way:
x=y≡∀z((x∈z↔y∈z)∧(z∈x↔z∈y))
(Where x and y can be the same variable, but z must be a different variable from them both)
By simply replacing every instance of the formula “x=y” with this definition, set theory can be formalized in first-order logic without equality.
Right, but models of set theory without equality can contain many indistinguishable “copies” of the same set, and in this sense the “extensionally equivalent” predicate does not define equality (in the sense of the identity relation on the universe of the model).
That’s also true in first-order logic with equality, since nothing except convention stops us from considering models where multiple objects are equal according to the equality predicate. The choice to exclude models which include duplicate objects is just a side-condition used to filter out inconvenient models when studying semantics. We can include such a side-condition when considering the semantics of set theory without equality, too, so it doesn’t seem fair to me to single it out as being uniquely incapable of defining equality.
(In fact, I’d argue that this also applies to second-order logic. Second-order logic can be given Henkin semantics, which have all the same idiosyncracies as first-order semantics. Using these semantics, we can get models of second-order logic with duplicate objects, just like first-order logic. I’d argue that standard second-order semantics are more or less just using a side-condition to filter out models which have missing subsets. But if we wanted to we could include similar side-conditions when considering models of first-order set theory, too, so it doesn’t seem fair to me to to say second-order logic can define equality while first-order logic can’t.)
Since multiple answers here mention equality being undefinable in first-order logic, I want to say that that’s only true if there are an infinite number of constants/functions/predicates in the language. Since set theory can be formalized using only a single predicate, it is possible to define equality this way:
x=y≡∀z((x∈z↔y∈z)∧(z∈x↔z∈y))
(Where x and y can be the same variable, but z must be a different variable from them both)
By simply replacing every instance of the formula “x=y” with this definition, set theory can be formalized in first-order logic without equality.
Right, but models of set theory without equality can contain many indistinguishable “copies” of the same set, and in this sense the “extensionally equivalent” predicate does not define equality (in the sense of the identity relation on the universe of the model).
That’s also true in first-order logic with equality, since nothing except convention stops us from considering models where multiple objects are equal according to the equality predicate. The choice to exclude models which include duplicate objects is just a side-condition used to filter out inconvenient models when studying semantics. We can include such a side-condition when considering the semantics of set theory without equality, too, so it doesn’t seem fair to me to single it out as being uniquely incapable of defining equality.
(In fact, I’d argue that this also applies to second-order logic. Second-order logic can be given Henkin semantics, which have all the same idiosyncracies as first-order semantics. Using these semantics, we can get models of second-order logic with duplicate objects, just like first-order logic. I’d argue that standard second-order semantics are more or less just using a side-condition to filter out models which have missing subsets. But if we wanted to we could include similar side-conditions when considering models of first-order set theory, too, so it doesn’t seem fair to me to to say second-order logic can define equality while first-order logic can’t.)