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The Prenex normal form is a canonical form for logical formulas in 1st & 2nd order logic. I've read that higher order logic has no such thing that can be computed. However, I'm wondering what the best is that can be done. The proofs that there exists no such forms in general is via (un)decidability arguments, etc. But I would like to examine a concrete example where such a normal form cannot be computed. I want to see exactly why for that particular example it's impossible. Can you provide a simple example?

https://en.wikipedia.org/wiki/Prenex_normal_form

Here is the only evidence I can find to support the claim that 2nd-order logic has a Prenex normal form:

In first order logic it was observed very early on that formulas can be brought into a logically equivalent Prenex Normal Form in which all quantifiers occur in the beginning of the formula. This is possible also in second-order logic and the proof is essentially the same.

https://plato.stanford.edu/entries/logic-higher-order/#PropSecoOrdeForm

Daniel Donnelly
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1 Answers1

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The intention of this note is not to provide an answer, but to set some issues straight regarding the question.

The notion of Prenex Normal Form (PNF) of first-order logic can be meaningfully extended to second-order and third-order logic.

Notice that analytical hierarchy can be set out for third-order properties of natural numbers, the levels of which are denoted by $\Sigma^{2}_{n}$ and $\Pi^{2}_{n}$.

Also notice that third-order logic over a base model is actually first-order logic over the cumulative hierarchy of models. Starting with $N_{1}=\langle\mathbb{N},+,\cdot\rangle$, we can observe that:

$3$rd-order logic over $N_{1}=\langle\mathbb{N},+,\cdot\rangle\iff$ $2$nd-order logic over $N_{2}=\langle \mathcal{P}(\mathbb{N}), E, \mathbb{N}, +,\cdot\rangle$,

where $\mathcal{P}$ is the powerset and $E$ is the binary membership relation such that $E\subseteq\mathbb{N}\times\mathcal{P}(\mathbb{N})$ and $E(n, r)\leftrightarrow n\in r$;

$2$nd-order logic over $N_{2}=\langle \mathcal{P}(\mathbb{N}), E,\mathbb{N}, +,\cdot\rangle\iff$ $1$st-order logic over $N_{3}=\langle \mathcal{P}(\mathcal{P}(\mathbb{N})),E',\mathcal{P}(\mathbb{N}), E, +,\cdot\rangle$,

where $E′$ is the binary membership relation such that $E′\subseteq \mathcal{P}(\mathbb{N})\times \mathcal{P}(\mathcal{P}(\mathbb{N}))$ and $rE′X\leftrightarrow r\in X$.

But, whereas monadic first- and second-order logic is decidable, monadic third-order logic is undecidable. We can get a grip on this difference by considering a domain consisting of $x, y$. Monadic second-order logic allows predicates to have values $\{x\}, \{y\},\{x,y\}$. However, by third-order, ordered pairs by Kuratowski's construction become definable, such as $\{\{x\},\{x,y\}\}$, which would act like dyadic predicates (see MJD's answer for a nice explanation of the construction of ordered pairs).

Tankut Beygu
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