Which combination of Peano axioms shows that $0\neq 1?$

Question

Please vote to close this question. It's really dumb as when I was reading the Peano axioms, axiom 8 didn't register. Don't waste your time reading this question.... I also cannot delete it (I have tried but it won't let me).

$1$ is defined as $S(0).$

An essentially equivalent question to the one I am asking is, "Why isn't $0\neq 1$ a Peano axiom?"

If we accept Wikipedia's $9$ Peano axioms, as well as the two axioms of addition, how do we show then that $0\neq 1?$ Or do we need to accept the multiplication axioms as well in order to show that $0\neq 1?$ Or is there another axiom I am not seeing that you need, in order to show that $0\neq 1?$

Basically, as far as I can tell, none of the axioms tell us that we cannot have $0=1=2=3=\ldots.$

I think that what I am referring to is (essentially?) the group (semigroup? ring?) $\langle \{0\}, + \rangle\ $ along with $=$ defined as an equivalence relation, and an axiom that enables substitution.

$$$$ $0 \neq 1$ not provable in axiomatic arithmetic?

I looked here, but I'm not sure the question is the same as my one and I don't understand the answers. I also don't think the answers relate to wikipedia's version of the Peano axioms, but maybe I am wrong?

Answer 1

Axiom 8 in the Wikipedia list of axioms says that $0$ is not the successor of any natural number. Since $1 = S(0)$, we have $1 \neq 0$.

Answer 2

You only need two of the axioms:

1. Axiom 1: $0$ is a natural number.
2. Axiom 8: For every natural number $n$, $S(n)=0$ is false.

In particular, the statement of axiom $8$ this is also true for $n=0$ (because, from Axiom 1, we know $n$ is a natural number). Therefore, $S(0)=0$ is false.