Nonfirstorderizability

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)". 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.).

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

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 \lor x = y + 1 ) $ for Axy. The result, $$\exists X ( \exists x,y (Xx \land Xy \land (y = x + 1 \lor x = y + 1)) \land \exists x \neg Xx \land \forall x\, \forall y (Xx \land (y = x + 1 \lor x = y + 1) \rightarrow Xy))$$ states that there is a set $X$ with these properties: 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 $x + 1$ 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 $x - 1$. In addition, there is a set $X$ satisfying the formula in every non-standard model.
 * 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 $y$ belongs to $y$ and $X$ is $0, 1, 2, ...$ or $0, 1, 2, ...$, $X$ also belongs to $X$.

Let us assume that there is a first-order rendering of the above formula called $E$. 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 $E$ exists in first-order logic.

Finiteness of the domain
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. 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 $B_{3}$ 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 $n$ 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.

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.