# Which combination of Peano axioms shows that $0\neq 1?$ [closed]

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.

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?

• Don't we also have to assume a set theory as well? In ZFC for instance we know that $0\neq 1$ because $x\neq \{x\}$ by the axiom of regularity. Sep 21, 2022 at 11:48
• @JMoravitz I am not familiar with the set theory formulation of Peano's axioms. Just wikpedia's "standard" approach. I know set theory approach is probably more formal or more correct or whatever, but I am not familiar with it. Sep 21, 2022 at 11:56
• I don’t get all the downvotes. I tried deleting this as soon as the answers came, but you’re not allowed to… Maybe I will ask this as a question on meta Sep 22, 2022 at 17:13
• I’m voting to close this question because I don't want to continue to receive downvotes. Sep 22, 2022 at 22:23
• After thinking about your meta post for a few days, I decided to upvote this. I believe it is a natural Question to ask about where $0\neq 1$ is implied by the Peano axioms, in spite of it being a central consideration. You showed some effort in stating the problem by bringing up a trivial model $0=1=2=...$, so with my upvote you now have net positive effect on your reputation. Not that you should be worried. The two succinct Answers are also worth keeping! Sep 24, 2022 at 17:34

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 $$1 \neq 0$$.
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.