peano arithmetic - Naturally definable sets of natural numbers (2): Can the circle be broken?

(follow-up to: Naturally definable sets of natural numbers)

Every formula $Psi(x)$ in the first-order language of Peano arithmetic defines a set of natural numbers. Some of these sets are finite, others are infinite. Every finite set $lbrace n_0, n_1, ..., n_k rbrace$ can be defined by an equation $p(x) = q(x)$ with $p(x), q(x)$ finite polynomials in $x$ with natural coefficients. Let in the following $phi(x)$ be such an equation [read "phi" for "finite"]. Infinite sets cannot be described by any $phi(x)$.

Given a formula $Omega(x)$ which defines an infinite set [read "omega" for "infinite"]. Then every formula of the form $Omega(x) vee phi(x)$ or $Omega(x)wedge negphi(x)$ defines an infinite set, too.

The motivation of the following definition is this: A formula defining an infinite set shall be called arbitrary if it is derived from a natural (= non-arbitrary) formula by adding or removing finitely many arbitrary elements.

Definition (wannabe): A formula $Omega(x)$ is arbitrary iff it defines an infinite set and is equivalent

1. to a formula $omega(x) vee phi(x)$ with $phi(x) notrightarrow omega(x)$ or


2. to a formula $omega(x) wedge neg phi(x)$ with $omega(x) notrightarrow negphi(x)$

where $omega(x)$ is not arbitrary. (Of course, $omega(x)$ defines an infinite set.)

On first sight, this definition seems circular:

Let $Omega(x) equiv omega(x) vee phi(x)$ with $phi(x) notrightarrow omega(x)$.

Then $omega(x) equiv Omega(x) wedge negphi'(x)$ with $Omega(x) notrightarrow negphi'(x)$.

Then $Omega(x)$ is arbitrary iff $omega(x)$ is not arbitrary.

Might this seemingly vicious circle not be in fact a (hidden) recursive definition (by something like "(abstract) length of formulas")?

Cannot this circle be broken? What about the intuition, that $(exists y) x = 2 cdot y$ is a non-arbitrary formula, but that $(exists y) x = 2 cdot y vee x = 17$ is an arbitrary one?

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