# Can we replace these two Peano axioms with this single axiom?

Peano Axioms from Mathworld:

1. Zero is a number.
1. If $$a$$ is a number, the successor of $$a$$ is a number.
1. zero is not the successor of a number.
1. Two numbers of which the successors are equal are themselves equal.
1. (induction axiom.) If a set $$S$$ of numbers contains zero and also the successor of every number in $$S$$, then every number is in $$S$$.

Q: After defining $$1$$ to be the successor of $$0$$, $$2$$ to be the successor of $$1$$, etc, can we replace Axioms 3 and 4 with

Axiom 6. $$0$$, $$1$$, $$2$$, etc. are (pairwise) distinct.

• Jul 5, 2023 at 5:50
• "etc." is a meta-formulation that we cannot use in a formal mathematical language. This is why formal logic is so cumbersome and proofs are usually not formulated in such a language , although they could be formalized. Jul 5, 2023 at 6:18
• @Peter: But then how does one define $1$, $2$, $3$, "etc."? In this context, Tao does use "etc." and so did Peano
– user986614
Jul 5, 2023 at 6:21
• [1] Authors can & will use "ETC" in Discussions , not in Axioms. Hence Axiom 6 is not valid. [2] More-over , that Axiom 6 will allow Natural numbers to be like 1,2,3,4,1,2,3,4,1,2,3,4,3,4,3,4,.... , where we have Pair-wise Distinct Elements.
– Prem
Jul 5, 2023 at 6:39
• My Point [1] , you say "ETC" is there in Axioms , It might be good to include that Part in your Post. My Point [2] still stands , though !
– Prem
Jul 5, 2023 at 6:53

Working in second-order logic, consider a language with a constant $$0$$ and a unary function symbol $$S$$. Let $$\Bbb{N}$$ be the set of natural numbers in the meta-theory. Looking at the following axiom systems.

System 1

1. $$\neg\exists x \,Sx = 0$$
2. $$\forall x\forall y \,(Sx=Sy\rightarrow x=y)$$
3. $$\forall U(0\in U\land (\forall x\, (x\in U\rightarrow Sx\in U)))\rightarrow (\forall x\,x\in U)$$, where $$U$$ quantifies over subsets the universe.

System 2

• $$(An,m)$$ an axiom schema. For all $$n,m\in\Bbb{N}$$ such that $$n\ne m$$, $$S^n 0\ne S^m 0$$, where $$S^n$$ is a term applying $$S$$ $$n$$ times (and $$S^0 0=0)$$.
• $$(B)$$ an axiom same as axiom 3 above.

Assume System 1. To establish the axioms of System 2, we just need to prove $$(An,m)$$ for all $$n,m\in\Bbb{N}$$ such that $$n\ne m$$. Let $$n$$ be the least natural number such there exists an $$m>n$$ such that $$S^n 0 = S^m 0$$.

• If $$n=0$$ then $$0 = S^m 0$$, so $$0$$ is a succesor - in contradiction to axiom 1.
• Otherwise, $$S(S^{n-1} 0) = S^n 0 = S^m 0 = S(S^{m-1} 0)$$, so by axiom 2, $$S^{n-1} 0 = S^{m-1} 0$$ in contradiction to $$n$$ being minimal.

Notice I didn't even have to use the induction axiom 3 to prove this, just that $$\Bbb{N}$$ is well-ordered in the meta-theory.

Now assume System 2. To establish the axioms of System 1, we need to prove axioms 1 and 2. Let $$U=\{S^n 0:n\in\Bbb{N}\}$$ in the meta-theory. Since this is second-order logic, $$U$$ satisfies the assumption of axiom B/axiom 3, so $$U$$ is the whole universe of the model.

If $$0=Sx$$ for some $$x$$ in the universe then there is some $$n\in\Bbb{N}$$ such that $$x=S^n 0$$ and then $$0=S^{n+1} 0$$ which contradicts the axiom $$(A0,(n+1))$$.

If $$x\ne y$$ for some $$x,y$$ in the universe, there are $$n,m\in\Bbb{N}$$ such that $$x=S^n 0, y=S^m 0$$ and necessarily $$n\ne m$$. But then $$s^{n+1} 0\ne S^{m+1} 0$$ by axiom $$(A(n+1),(m+1))$$, so $$Sx\ne Sy$$.

The above proof shows that System 1 and System 2 are equivalent in the sense of $$\models$$. I'm not sure whether they're equivalent in the sense of $$\vdash$$, as I'm unfamiliar with deductive systems for second-order logic and they're not complete anyway.

Addendum: System 1 ($$S1$$) is finite while System 2 ($$S2$$) is an infinite schema. Any finite subset $$S2'\subseteq S2$$ has a finite model of a circular buffer. Since any model of $$S1$$ is isomorphic to $$\Bbb{N}$$, it follows $$S2'\not\models S1$$. Assuming that a deductive system of second-order logic allows a proof to only use finitely many axioms - it follows that $$S2\not\vdash S1$$. On the other hand, the proof that $$S1\models S2$$ does not use the second-order induction axiom, so it's valid in first-order logic. Since first-order logic has a complete deductive system, $$S1\vdash S2$$ in first-order logic. Assuming a deductive system of second-order logic can do everything a deductive system of first-order logic can do then $$S1\vdash S2$$ is second-order logic.

Here's a summary if you don't know formal logic

• Axiom $$6$$ is actually a set of infinitely many axioms.
• Axioms $$1+2+3+4+5$$ together have only one model up to isomorphism which is $$\Bbb{N}$$.
• Axiom set $$1+2+5+6$$ together has only one model up to isomorphism which is $$\Bbb{N}$$.
• Each and every axiom in $$6$$ is provable (with a separate finite proof for each) from axioms $$1+2+3+4$$ together.
• No finite subset of axioms from $$1+2+5+6$$ can prove either $$3$$ or $$4$$ in a finite proof.

So the system consisting of $$1+2+3+4+5$$ is stronger than $$1+2+5+6$$ in terms of what you can prove using finite proofs. However, they both categorically model the natural numbers.