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Consider $(N, S)$ with S injective and not surjective and suppose the induction principle holds, where the zero in inductive principle is an element which is not in $S(N)$.

Can I prove that $0$ is unique? In other words can we delete explicit reference of zero in Peano's axioms?

Update My attempt of proof.

If $0' \ne 0$ and $0'$ not in $S(N)$ define $A=N-\{0'\}$ so we have:

1) $0 \in A$;

2) if $n \in A$ then $S(n) \ne 0'$, beacuse $0' \notin S(N)$, so $S(n) \in A$.

Hence by induction we have $A = N$ which is a contradiction.

Is my proof correct?

I think this question is the same as the intro of the following:

Prove that the system $(P, S, 0)$ satisfy Peano Axioms.

asv
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    See Peano axioms : 1) $0$ is a natural number. 8) For every natural number $n, S(n) = 0$ is false. That is, there is no natural number whose successor is $0$. – Mauro ALLEGRANZA Dec 04 '19 at 14:17
  • $\mathbb{N}\times {0,1}$ with $S(n,m)=(n+1,m)$ satisfies all axioms that don't involve $0$. Now, you said, using induction. What will be the formulation of induction without $0$? – conditionalMethod Dec 04 '19 at 14:17
  • Every non-zero number can be achieved by repeated application of the successor-map. If $\ 0\ $ were not unique, we would arrive this way at $\ 0\ $ again which would mean that there is a number with successor $\ 0\ $. This is not the case, hence $\ 0\ $ is unique. – Peter Dec 04 '19 at 14:20
  • So you are right; if there is an $a \in \mathbb N$ which is a "new zero", i.e. $S(n) \ne a$ for every $n$ and $a \ne 0$, we have that $a$ "lays nowhere" in the succession, because it is not the first one and it is not a successor: thus, we cannot "reach it" by induction. – Mauro ALLEGRANZA Dec 04 '19 at 14:22
  • I made an update. What do you think about that? – asv Dec 04 '19 at 15:11

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I tried to formalize what (I think) you are trying to do.

As I understand it, you are trying to remove the explicit reference to $0$ in the Peano Axioms, and instead say that there is just something that is outside the range of $s$. So, I figure you want to replace the usual axiom:

$$\forall x \ s(x) \neq 0$$

with something like:

$$\exists z \forall x \ s(x) \neq z$$

And, to remove the explicit reference to $0$ from the typical inductive schema:

$$\forall w_1 ... w_n ((\varphi(0) \land \forall x (\varphi(x) \to \varphi(s(x)))) \to \forall x \ \varphi(x))$$

you would get something like:

$$\forall w_1 ... w_n ((\exists z (\forall x \ s(x) \neq z \land \varphi(z)) \land \forall x (\varphi(x) \to \varphi(s(x)))) \to \forall x \ \varphi(x))$$

But the question now is: is there only one such $z$ that is not covered by the $s$ function, i.e. is it true that:

$$\exists z \forall y (\forall x \ s(x) \neq y \to y = z)$$

Well, below is a formal proof, using an instance of the new inductive scheme with:

$$\varphi(y) := \forall x \ s(x) \neq y \to y = w$$

enter image description here

So yes, the $z$ is indeed unique.

Bram28
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  • Thank you, is it an automatic proof or assistant proof? What is the program? Did you find my proof correct? – asv Dec 04 '19 at 18:37
  • @asv You're welcome! It's called Fitch, which comes with the commercial texbook/software package "Language, Proof, and Logic". There is an online version of it though: https://proofmood.mindconnect.cc/en/Fitch/fitch.php You have to write the whole proof yourself, but the program will check the validity of the steps. – Bram28 Dec 04 '19 at 18:39
  • Thank you very much. What do you think about my proof? – asv Dec 05 '19 at 07:32
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    @asv Your proof is good! – Bram28 Dec 05 '19 at 18:42