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Consider the following axioms which I'll call the "Ultrafinite Peano Axioms":

  1. $0 \in \mathbb N$.

  2. $\forall x,y \in \mathbb N,$ $S(x)=S(y)$ implies $x=y$.

  3. $\forall x \in \mathbb N, S(x) \neq 0$.

  4. There exists a $Z \in \mathbb N$ (a zillion) such that $S(Z) \notin \mathbb N$. $\forall x \in \mathbb N - \{Z\}, S(x) \in \mathbb N$.

  5. Suppose $K$ is a set such that $0 \in K$ and ($\forall x \in \mathbb N-\{Z\}$, $x \in K$ implies $S(x) \in K$). Then $\mathbb N \subseteq K$.

Axioms 1 to 3 are no different than the their corresponding three Peano Axioms. Axiom 4 says that every number except $Z$, a zillion, has a successor and $Z$ has no successor. Axiom 5 is the induction axiom with the restriction that $x \neq Z$. These axioms model the ultrafinite view of math, that there are only a finite number of numbers, so not every number has a successor.

Gödel's Theorem says that the Peano axioms cannot be proven consistent. My question is can we prove that these axioms are consistent?

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  • $\begingroup$ The consistency of this system is easily provable in ZF since orered finite sets function as your axioms state. The second incompleteness theorem applied to Peano Arithmetics says that the sentence in the language with $0,+,.,S$ which means "PA is consistent" can't be proven by PA. Since your theory is not a first order theory over some language, it makes little sense to talk about sentences in the language with $0,S$ meaning it is consistent, and the little sense I get makes me think this sentence does not exist. $\endgroup$
    – nombre
    Oct 9, 2015 at 17:11
  • $\begingroup$ Then add the normal recursive definitions of addition and multiplication to these axioms. $\endgroup$
    – Craig Feinstein
    Oct 9, 2015 at 17:37
  • $\begingroup$ Let's say we do that. Because proofs can be of lenght superior to any fixed bound $B$, we still can't talk about the consistency of this system of axioms. I am not familiar with second order logic so I can't really tell you more. Your system of axiom is of second order, you can make it first order by fixing $B$ as the number of elements in any model of the theory, isolating it (adding a constant symbol for it), dropping axiom 5, setting $S(B) = 0$ and adding "except when $x= B$" wherever it is necessary. $\endgroup$
    – nombre
    Oct 9, 2015 at 17:53
  • $\begingroup$ What happens when it is made first order? Is it consistent? $\endgroup$
    – Craig Feinstein
    Oct 9, 2015 at 18:24
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    $\begingroup$ Godel's theorem does not say PA can't be proved consistent - it says that if PA is consistent, then PA can't be proved consistent in PA itself. $\endgroup$
    – Noah Schweber
    Oct 10, 2015 at 23:41

1 Answer 1

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Your theory is very easy to prove consistent, even in very weak systems, since nothing prevents us from taking $Z=0$ and $\mathbb{N}=\{0\}$, with $S(0)\neq 0$. This system will satisfy all your axioms. As Nombre notes, we could take $Z$ to be any finite number and build a model of your axioms this way, but I find the case $Z=0$ to be especially relevant.

Although you may have had in mind that $Z$ would be very large, you didn't include any axioms that ensure this.

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