Geach–Kaplan Sentence

In formal logic, nonfirstorderizability is the inability of a natural-language statement to be adequately captured by a formula of first-order logic. Specifically, a statement is nonfirstorderizable if there is no formula of first-order logic which is true in a model if and only if the statement holds in that model. Nonfirstorderizable statements are sometimes presented as evidence that first-order logic is not adequate to capture the nuances of meaning in natural language.

The term was coined by George Boolos in his paper "To Be is to Be a Value of a Variable (or to Be Some Values of Some Variables)". Reprinted in
Quine argued that such sentences call for second-order symbolization, which can be interpreted as plural quantification over the same domain as first-order quantifiers use, without postulation of distinct "second-order objects" (properties, sets, etc.).

Examples

Geach-Kaplan sentence

A standard example is the '' Geach– Kaplan sentence'': "Some critics admire only one another."
If ''Axy'' is understood to mean "''x'' admires ''y''," and the universe of discourse is the set of all critics, then a reasonable translation of the sentence into second order logic is:
$\exists X \big( (\exists x \neg Xx) \land \exists x,y (Xx \land Xy \land Axy) \land \forall x\, \forall y (Xx \land Axy \rightarrow Xy)\big)$
In words, this states that there exists a collection of critics with the following properties: The collection forms a proper subclass of all the critics; it is inhabited (and thus non-empty) by a member that admires a critic that is also a member; and it is such that if any of its members admires anyone, then the latter is necessarily also a member.

That this formula has no first-order equivalent can be seen by turning it into a formula in the language of arithmetic. To this end, substitute the formula $( y = x + 1 \lor x = y + 1 )$ for ''Axy''. This expresses that the two terms are successors of one another, in some way. The resulting proposition,
$\exists X \big( (\exists x \neg Xx) \land \exists x,y (Xx \land Xy \land (y = x + 1 \lor x = y + 1)) \land \forall x\, \forall y (Xx \land (y = x + 1 \lor x = y + 1) \rightarrow Xy)\big)$
states that there is a set with the following three properties:
* There is a number that does not belong to , i.e. does ''not contain all'' numbers.
* The set is inhabited, and here this indeed immediately means there are at least two numbers in it.
* If a number belongs to and if is either or , then also belongs to .
Recall a model of a formal theory of arithmetic, such as first-order Peano arithmetic, is called ''standard'' if it ''only'' contains the familiar natural numbers as elements (i.e., ). The model is called non-standard otherwise. The formula above is true only in non-standard models: In the standard model would be a proper subset of all numbers that also would have to contain all available numbers (), and so it fails. And then on the other hand, in every non-standard model there is a subset satisfying the formula.

Let us now assume that there is a first-order rendering of the above formula called . If $\neg E$ were added to the Peano axioms, it would mean that there were no non-standard models of the augmented axioms. However, the usual argument for the existence of non-standard models would still go through, proving that there are non-standard models after all. This is a contradiction, so we can conclude that no such formula exists in first-order logic.

Finiteness of the domain

There is no formula in first-order logic with equality which is true of all and only models with finite domains. In other words, there is no first-order formula which can express "there is only a finite number of things".

This is implied by the compactness theorem as follows. Suppose there is a formula which is true in all and only models with finite domains. We can express, for any positive integer , the sentence "there are at least elements in the domain". For a given , call the formula expressing that there are at least elements . For example, the formula is:
$\exists x \exists y \exists z (x \neq y \wedge x \neq z \wedge y \neq z)$
which expresses that there are at least three distinct elements in the domain. Consider the infinite set of formulae
$A, B_2, B_3, B_4, \ldots$
Every finite subset of these formulae has a model: given a subset, find the greatest for which the formula is in the subset. Then a model with a domain containing elements will satisfy (because the domain is finite) and all the formulae in the subset. Applying the compactness theorem, the entire infinite set must also have a model. Because of what we assumed about , the model must be finite. However, this model cannot be finite, because if the model has only elements, it does not satisfy the formula . This contradiction shows that there can be no formula with the property we assumed.

Other examples

* The concept of identity cannot be defined in first-order languages, merely indiscernibility.
* The Archimedean property that may be used to identify the real numbers among the real closed fields.
* The compactness theorem implies that graph connectivity cannot be expressed in first-order logic.

See also

* Definable set
* Branching quantifier
* Generalized quantifier
* Plural quantification
* Reification (linguistics)

References

{{reflist

External links

Printer-friendly CSS, and nonfirstorderisability by Terence Tao

Logic