# First order Peano axioms and their intepretation

I am still better trying to understand the first order Peano Axioms and their relation to the standard model. Just for reference, here are the axioms I work with:

$1.\space \forall x (\neg x+1=0)\hspace{2.14cm} 4.\space\forall x\forall y(x+1=y+1 \rightarrow x=y)\\ 2.\space\forall x(x+0=x) \hspace{2.41cm} 5.\space\forall x\forall y(x+(y+1)=(x+y)+1)\\ 3. \space \forall x(x\times0=0)\hspace{2.41cm} 6. \space \forall x\forall y (x\times(y+1) = (x\times y)+x)\\$

As well as the axiom schema of infinite, but countable axioms for induction:

$$7. \space\forall x_1...\forall x_n(\varphi[0/y]\land \forall y( \varphi\rightarrow \varphi[y+1/y]))\rightarrow \forall y\varphi)$$

Now, for correctness to hold, whatever is deduced from these axioms, must hold in every interpretation. But, what I find weird is, it seems these axioms have already defined the function symbols and constants to be equivalent to the structure of $\mathfrak N$, ie. $\{+,\times,0,1\}$, instead of having undefined $f_1,f_2,c_1,c_2$, so as I understand it the mappings are already assigned. How is this allowed, since clearly that's just one interpretation of the functions/constants, while the deduction should apply to every interpretation?

Further if that's the case, do the different possible models of PA concern themselves only with the domain $A$ of the model? I read that there are non-standard models of PA for every cardinality, so what would a model with cardinality of one look like?

Any examples/sources for FO Peano deductions are of course also welcome.

• You seem to have a typo in axiom 1. – Andrés E. Caicedo Aug 12 '17 at 2:58
• With regard to the "models of every cardinality," that's not true. There are models of every infinite cardinality, but PA has no finite models. – Malice Vidrine Aug 12 '17 at 3:14

## 2 Answers

But, what I find weird is, it seems these axioms have already defined the function symbols and constants to be equivalent to the structure of $\mathfrak N$, ie. $\{+,\times,0,1\}$, instead of having undefined $f_1,f_2,c_1,c_2$, so as I understand it the mappings are already assigned.

No, they haven't. The symbols $+$, $\times$, $0$, and $1$ used in the axioms are just symbols. There is no assumption that they refer to the usual meaning of addition, multiplication, zero, and one: $+$ could be interpreted as any binary operation, $\times$ as any binary operation, and $0$ and $1$ as any two constants. If you like, you could rewrite the axioms using $f_1,f_2,c_1,c_2$ as you suggest and the axioms would have the exact same meaning: you're just using different symbols.

I read that there are non-standard models of PA for every cardinality, so what would a model with cardinality of one look like?

This statement is not quite correct. There are non-standard models of PA of every infinite cardinality, but there are no finite models of PA. For instance, you can prove from PA that the elements $0$, $1$, $1+1$, $(1+1)+1$, $((1+1)+1)+1$, and so on are all distinct, so there are infinitely many different elements.

• So I take it that the arguably more common deductions (for example: en.wikipedia.org/wiki/…) in Peano Arithmetic use some other logic (would that count as 2nd order logic?). It seems a bit limiting to not be sure whether one is one :). – Dole Aug 12 '17 at 3:38
• The arguments on that Wikipedia page do not assume that $+$, $\times$, $0$, and $1$ have their usual meanings either. They do assume that $+$ has been defined recursively from a successor operation, which cannot be done in first-order logic and only makes sense in a second-order version of PA. However, if you change these recursive definitions to axioms, then the arguments are perfectly valid in first-order logic. – Eric Wofsey Aug 12 '17 at 3:43

I thought The Incompleteness Phenomenon by Goldstern and Judah captured this well, but there must be other books on the subject as well. You have the Peano axioms and the standard model. A standard construction is to add a countable set of new axioms $c \gt 0, c \gt 1, c \gt 2, \ldots$ Any finite set of these axioms is satisfied by the standard model because you can find a greatest constant that $c$ must be greater than and choose it higher. Compactness now tells you that all the axioms have a model, but the standard model does not satisfy it, so there must be a nonstandard model. They state that all countable nonstandard models are of the form $\Bbb N$ followed by a dense linear order of sets isomorphic to $\Bbb Z$