1
$\begingroup$

I didn't know how to formulate a more clear title for this question:

Take arithmetic to be the structure $\mathcal N= (\mathbb N, \sigma, +,•, 0,1)$ with its standard interpretation. When I use the term "theory of arithmetic" below, I mean the set of first order sentences that are true in this standard structure of arithmetic.

When we want to formally derive first order statements about arithmetic, we usually take the Peano axioms.

However, I am wondering how much we can get from the following alternative approach:


The idea is to more "directly" capture the rules of arithmetic into the structure.

We completely drop the Peano axioms, and instead define the axioms as follows:

We extend the structure with an infinite amount of constants $n_0, n_1, \dots$, where we will interpret the term $n_0$ as the number $0$ in the underlying structure, $n_1$ as $1$, etc.

  • We then have an axiom scheme which gives an axiom for every $n_i$ that $\sigma(n_i)=n_{i+1}$

  • we also have axioms for addition and multiplication which simply capture their definitions in terms of $\sigma$.

How much of the theory of arithmetic can we get from this axiomatization?


Note, I'm not trying to improve on the Peano axioms or anything. I'm just trying to improve my understanding of the formal method.

$\endgroup$
9
  • $\begingroup$ What do you mean by "worse than Peano axioms"? $\endgroup$ – Taroccoesbrocco Mar 31 '18 at 21:03
  • 1
    $\begingroup$ In first order logic (with the usual language of Peano arithmetic), you can't capture the idea that "there are no other numbers than the natural numbers", see nonstandard models of Peano arithmetic. $\endgroup$ – Taroccoesbrocco Mar 31 '18 at 21:06
  • $\begingroup$ well for starters I don't think you can prove anything starting with $\forall x$ except your axiom about zero being the only one without a predecessor (and its trivial variants) $\endgroup$ – mercio Mar 31 '18 at 22:28
  • 1
    $\begingroup$ So... you just take Peano and remove induction? $\endgroup$ – Asaf Karagila Apr 1 '18 at 6:54
  • 2
    $\begingroup$ Each natural number has a defining closed term using $0$ and the successor function. So the constants, while very nice, don't actually add much. $\endgroup$ – Asaf Karagila Apr 1 '18 at 8:45
2
$\begingroup$

Unfortunately being countable is not a first order property. This is proven by the Upward Löwenheim–Skolem theorem which says for a given theory that if there is an infinite model, there is an uncountably infinite model. Hence it's impossible to capture the idea that "there are no other numbers other than the natural numbers".

$\endgroup$
1
  • 4
    $\begingroup$ It is even worse than this: there are nonstandard models of Peano arithmetic with countably infinite elements. The fact that it's impossible to capture the idea that "there are no other numbers other than the natural numbers" in first order logic does not depend on cardinality. $\endgroup$ – Taroccoesbrocco Mar 31 '18 at 23:31

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.