Would this be a valid definition of natural numbers?

Question

We represent the "Set of Natural numbers" as a logical predicate that matches all natural numbers.
Let us define the property of a property being natural-like as follows:

$$\begin{array}\\ \lambda(n) =:&\exists x. \text{zero}(x) \land n(x)\\ &\forall x, y. \text{succ}(x, y) \land n(x) \implies n(y)\\ &\forall x, y. \text{zero}(y) \land \text{succ}(x,y) \implies \lnot n(x)\\ &\forall x, y, z. n(x) \land \text{succ}(x,y) \land \text{succ}(x,z) \implies y = z\\ &\forall x, y, z. n(x) \land n(y) \land \text{succ}(x,z) \land \text{succ}(y,z) \implies x = y \end{array}$$

We further define that $\text{zero}(x) \land \text{zero}(y) \implies x = y$. Note that $=$ is assumed to be transitive, reflexive and commutative.

Defining that $$\phi \subseteq \psi =: (\forall x. \phi(x) \implies \psi(x))$$

Can we then consider $\mathbb{N}$ to be any set with the property $\mu$, where... ?

$$\forall n. \mu(n) \implies \lambda(n)$$ $$\forall n, m. \mu(n) \land \lambda(m) \implies n \subseteq m$$

That is, is $\mu$ sufficiently equivalent to the Peano definition of natural numbers?

More specifically, if $\mu(n)$, is there an $x$ with properties clearly not typical of a natural number that still satisfies $n(x)$?

Answer 1

Here is my intuition.

First, note that if $\lambda(n)$, then $n$ does not necessarily characterize natural numbers. For example, define a predicate $\zeta(x)$ to mean $x \in \{ 0 \}$. Then it still holds that $\lambda (\zeta)$, because $\lambda$ does not include a rule stating that successors exist. But say we fix this so that $\lambda(n)$ indeed means what you intend.

Next, I assume a property of properties, $\mu$, such that the following two hold.

• (1) : $\mu \subseteq \lambda$
• (2) : $\forall m, n. \mu(n)\land\lambda(m) \implies n \subseteq m$

Then say there is a set $S$ with the property $s$, where $\mu(s)$. From the above facts about $\mu$, we get

• (1') : $\lambda(s)$
• (2') : $\forall m. \lambda(m) \implies s \subseteq m$

Result (1') means that $s$ characterizes at least the natural numbers, but maybe more. Result (2') states that if $m$ is a property that characterizes naturals, then every member of $s$ (and therefore of $S$) is also a member of $m$ (and therefore a natural number). In other words, every element of $S$ is a natural number. Statement (2') still allows that $S$ contains a natural number twice. But then (1') forbids it because $\lambda$ poses uniqueness of successors and your system poses uniqueness of zero.

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EDIT: I've given a second thought to this question and I think there's more to say. I believe the above definitions are not sufficient to define natural numbers.

The conclusion of (2') (which restricts the set under consideration to have no more elements than those of $m$'s set) holds for any $m$ such that $\lambda(m)$. If there is no $m$ in the universe that fully corresponds to the natural numbers, then (2') has no "purpose".

So, the above definitions could be sufficient if there actually was a set, say $N$, in the universe such that it contains exactly the natural numbers. Call $n$ the property that fully characterizes the set $N$. Obviously, we have $\lambda(n)$. Then, since result (2') should hold for any property $m$ for which $\lambda(m)$, it should also hold for $n$. It is the existence of such a property $n$ that allows us to replace $m$ with $n$ in (2') and therefore restrict $s$ to actually contain no more than the natural numbers.