Nonfirstorderizability

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:

$${\displaystyle \exists X(\exists x,y(Xx\land Xy\land Axy)\land \exists x\neg Xx\land \forall x\,\forall y(Xx\land Axy\rightarrow Xy))}$$

That this formula has no first-order equivalent can be seen by turning it into a formula in the language of arithmetic . Substitute the formula (y = x + 1 v x = y + 1) for Axy. The result,

$${\displaystyle \exists X(\exists x,y(Xx\land Xy\land (y=x+1\lor x=y+1))\land \exists x\neg Xx\land \forall x\,\forall y(Xx\land (y=x+1\lor x=y+1)\rightarrow Xy))}$$

states that there is a set X with these properties:

• There are at least two numbers in X
• There is a number that does not belong to X, i.e. X does not contain all numbers.
• If a number x belongs to X and y is x + 1 or x - 1, y also belongs to X.

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 0, 1, 2, ... as elements. The model is called non-standard otherwise. Therefore, the formula given above is true only in non-standard models, because, in the standard model, the set X must contain all available numbers 0, 1, 2, .... In addition, there is a set X satisfying the formula in every non-standard model.

Let us assume that there is a first-order rendering of the above formula called E. If ${\displaystyle \neg E}$ 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 E exists in first-order logic.

There is no formula A 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.[2] Suppose there is a formula A which is true in all and only models with finite domains. We can express, for any positive integer n, the sentence "there are at least n elements in the domain". For a given n, call the formula expressing that there are at least n elements B_n. For example, the formula B₃ is:

$${\displaystyle \exists x\exists y\exists z(x\neq y\wedge x\neq z\wedge y\neq z)}$$

which expresses that there are at least three distinct elements in the domain. Consider the infinite set of formulae

$${\displaystyle A,B_{2},B_{3},B_{4},\ldots }$$

Every finite subset of these formulae has a model: given a subset, find the greatest n for which the formula B_n is in the subset. Then a model with a domain containing n elements will satisfy A (because the domain is finite) and all the B formulae in the subset. Applying the compactness theorem, the entire infinite set must also have a model. Because of what we assumed about A, the model must be finite. However, this model cannot be finite, because if the model has only m elements, it does not satisfy the formula B_m+1. This contradiction shows that there can be no formula A with the property we assumed.

• The concept of identity cannot be defined in first-order languages, merely indiscernibility.[3]
• 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.