# Peano axioms: 3 or 5 axioms?

The Peano axioms are usually stated as follows:

1. Zero is a number.

2. If a is a number, the successor of a is a number.

3. zero is not the successor of a number.

4. Two numbers of which the successors are equal are themselves equal.

5. (induction axiom.) If a set S of numbers contains zero and also the successor of every number in S, then every number is in S.

One can formalize Peano arithmetic as a theory of second-order sentences over the signature $$\{0, s\}$$ just by formalizing the axioms 3 to 5. So are 1 and 2 really axioms or are they just type declarations?

• Well, (1) says that the empty set is not a model of PA (in fact, (3-5) still hold for a model without objects). – user228113 Sep 14 '16 at 20:04
• if one has a constant symbol in a signature, we exclude empty models – user368958 Sep 14 '16 at 20:05
• The empty set is never a model of any theory. That is a part of the definition of a model. I agree with OP, 1 and 2 are not axioms. $0$ is a constant in the language and $S$ is a function in the language. 1and 2 use the word "number" this is not part of PA. – Rene Schipperus Sep 14 '16 at 20:07
• Sorry, I'm not an expert of logic, so I guess I'd better not bother others with my misconceptions. I just thought it was a sensible observation (which turned out to be wrong). – user228113 Sep 14 '16 at 20:08
• Well I guess OP uses number in 3,4,5 but they can be rephrased to avoid this and still say something. – Rene Schipperus Sep 14 '16 at 20:09

Peano's original paper had nine axioms, actually. He included axioms such as "for every number $x,$ $x = x$". At the time (1889) there was no well established system of logic, so Peano was starting, essentially, with nothing.
Today, the properties of equality and the fact that function symbols are interpreted by total functions are taken as part of the underlying "logic", apart from the "theory" that is being studied. So axioms 1 and 2 are not needed as axioms. It is necessary, though, to state the signature of the theory. So, if axioms 1 and 2 are left out, it would still be necessary to say that that theory has a constant symbol $0$ and a unary function symbol $s$.