Leaving out one of the Peano Axioms

What happens if you leave N4 (from Ross' book) out of the Peano axioms which states that if $n$ and $m$ in $\mathbb{N}$ have the same successor, then $n = m$?

• Some discussion of this, together with pictures illustrating Peano-like systems after omitting one axiom, is in H. Enderton: Elements of Set Theory, page 70. – Martin Sleziak Jan 27 '14 at 8:35
• Can you state what the remaining (extralogical) axioms are, expecially are there still axioms left that contain an = ? – Willemien Jan 27 '14 at 12:08

This is the axiom that forces $\mathbb{N}$ to be infinite; if you drop it then you lose the requirement that your 'set of natural numbers' be infinite.

$\mathbb{N}$ would satisfy the remaining axioms, but so would the set $\{ 0, 1 \}$, where we define $\mathtt{succ}(0)=1$ and $\mathtt{succ}(1)=1$. In fact any set of the form $\{ 0, 1, \cdots, n \}$ (for $n \ge 1$) would satisfy the remaining Peano axioms if we define $\mathtt{succ}(k)=k+1$ for $k < n$ and $\mathtt{succ}(n)=n$.

• @SidRohan: Why do you say that? You just dropped precisely the axiom that tells you that has to be the case. – Clive Newstead Jan 27 '14 at 4:31
• Right, so we're defining what '+1' means. I wrote '$\mathtt{succ}$' to disambiguate: where I write $+1$ I mean the standard successor operation on the natural numbers, and where I write $\mathtt{succ}$ I mean the new successor operation on our new set which satisfies the remaining axioms. If I were to use this '+1' notation then what I'm saying is that we actually define $n+1$ to be equal to $n$... and the set we obtain would satisfy all the Peano axioms except the one you're omitting. – Clive Newstead Jan 27 '14 at 4:35
• @Sid: No, it is "If $n$ is an element of the universe then the successor of $n$ is an element of the universe". There's nothing there that requires that "successor" means "the number plus one, computed with the addition we already know". If we already know how addition works, then what would we need axioms for? – Henning Makholm Jan 27 '14 at 4:37
• @HenningMakholm: Thanks, that's a much clearer way of saying what I was trying to say. – Clive Newstead Jan 27 '14 at 4:38
• @TobiasBrandt: That would violate the axiom that 0 is not the successor of any number. – yatima2975 Jan 27 '14 at 10:06

If you omit the axiom $\, Sm = Sn\,\Rightarrow\, m=n\,$ then $S$ need no longer be $1$-$1,\,$ so the axioms now admit finite "dipper" semigroup models, where the orbit of $S$ starting from $0$ need not be an infinite half-line, but may be shaped like the big dipper (i.e. shaped like the letter $\rho),\,$ with an initial preperiod followed by a periodic part. If $\, S^{j+k}0 = S^j0\,$ with $\,j,k\,$ minimal then $\,0,S0,\ldots,S^{j-1}0\,$ is the preperiod, and a periodic cycle of length $\,k\,$ starts at $\,S^j0 = S^{j+k}0 = S^{j+2k}0 =\, \ldots \,$

If you omit the axiom $Sm = Sn \to m =n$ then you loose a equalness there is just N3 left $\forall x \lnot ( 1 = Sx)$ butthat doesn't really say anything about equalness , how are you going to proof that 1 = 1?

Leaving out only N4 of your Peano axioms, you would admit the possibility of a finite structure with $N=\{1,2,3, ... n\},$ $n>1$ and $S(k)=k+1$ for all $k< n$, and $S(n)=m$ for some $m,$ $1<m\leq n,$ i.e. a finite loop not including $1$ through $m-1.$

Leaving out only N3 ($S(x)\neq 1$) would admit the possibility of a finite structure with $N=\{1,2,3, ... n\},$ and $S(k)=k+1$ for all $k< n$, and $S(n)=1,$ i.e. a finite loop including all of $N.$ It would also admit the possibility of a chain going off to infinity in both directions (like the integers).

Leaving out only N5 (induction) would admit the possibility of any number of disjoint finite side loops or chains going off to infinity in both directions (like the integers), each of which would be not be connected to the set we want. See for example where the dominoes represent elements of $N$. N5 filters out all such "junk terms".

See "What is a number again?" at my math blog.

BTW, your reference makes use of the + operator in N2 and N5. It hasn't been defined. Instead of $n+1,$ the author should have written "the unique successor of n."