Why is addition not completely defined here?

Question

Say for the natural numbers, we define addition this way:

$0 + 0 = 0$, and if $n+m = x$, then $S(n) + m = n+S(m) = S(x) $

Say we have the regular Peano axioms, except we delete the axiom of induction and add the following axiom:

$\forall n \in \mathbb{N} \exists k,l \in \mathbb{N} : n = k+l$

i.e., every natural number can be expressed as the sum of 2 natural numbers.

Conceptually, it seems to me that addition is defined on the entire domain of the naturals, as it seems we could argue the following:

We can guarantee that for any 2 numbers $n,m$ which are in a number system satisfying these axioms, that the addition of $n+m$ is defined- as because both $n$ and $m$ are, by the axiom in the second modification above, the result of addition, they are the result of repeated applications of the successor operation on $0$ ; therefore they are in the scope of the recursion contained in the addition definition.

However, here some users stated that addition was not completely defined, or that the above argument I presented was fallible, in a post I made about whether these modifications are equivalent to the Peano axioms (here I merely ask about the logical propriety of this argument inasmuch as the definition of addition is concerned). I do not understand why exactly I haven't hereby specified an addition function, as it seems we could use the information both contained in the definition of addition, and in an axiom about it, to conclude a fact of it, such as that it is completely defined for the naturals. Is there something illogical here about the interplay between defining addition and having axiom about it?

Answer 1

Consider the following structure:

First, $\mathbb{N}$ and $\mathbb{Z}$ are the "usual" sets of natural numbers (starting from $0$), and integers, constructed in the usual way within Zermelo-Fraenkel Set Theory.

Let $N=\mathbb{N}\times\{0\} \cup \mathbb{Z}\times\{1\}$.

Define a "succesor function" $s\colon N\to N$ by $$s(n,0) = (n+1,0), \qquad\qquad s(z,1) = (z+1,1).$$

Define a partial binary operation $\oplus$ on $N$ by letting $(n,0)\oplus(m,0) = (n+m,0)$ and $(x,1)\oplus (y,1) = (x+y,1)$. The operation is undefined if one operand is in $\mathbb{N}\times\{0\}$ and the other is in $\mathbb{Z}\times\{1\}$.

We now interpret Peano's first four axioms as follows. The set of numbers is $N$. The successor function is $s$. The distinguished number $\mathbf{0}$ is $(0,0)$.

Then we have:

1. Peano's first axiom: $\mathbf{0}$ is a number. True.
2. Peano's second axiom: if $x$ is a number, then $s(x)$ is a number. True.
3. Peano's third axiom: if $n$ and $m$ are numbers, and $s(n)=s(m)$, then $n=m$. True.
4. Peano's fourth axiom: if $m$ is a number, then $s(m)\neq\mathbf{0}$. True.

Now let us look at $\oplus$. It satisfies that $\mathbf{0}\oplus\mathbf{0}=\mathbf{0}$. It also satisfies that if $n$ and $m$ are numbers, and $n\oplus m=k$, then $s(n)\oplus m = n\oplus s(m) = s(k)$.

Thus, $(N,s,\oplus)$ satisfy all conditions you have placed on them. Does this structure satisfy your replacement "axiom"? Yes:

Let $n\in N$. If $n\in \mathbb{N}\times\{0\}$, then $n=n\oplus\mathbf{0}$, so we can take $k=n$, $l=\mathbf{0}$. If $n\in\mathbb{Z}\times\{1\}$, then we can write $n=n\oplus(0,1)$. So we can take $k=n$ and $l=(0,1)$.

But this structure does not satisfy Peano's Fifth axiom. Your replacement, after adding a new primitive symbol $+$ with the properties that we must have $0+0=0$ and that if $n+m=k$ then $s(n)+m=n+s(m)=s(k)$, is not equivalent to Peano.

Note also that the axioms and properties do not even guarantee that you can perform $\oplus$ to any pair of "numbers", so you certainly do not know that addition is "completely defined" (i.e., a complete, rather than a partial, operation).

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The usual way that $+$ is defined under Peano's axioms is to say use the Recursion Theorem. It is the Recursion Theorem that guarantees that you have a function defined on all of $\mathbb{N}$. Absent that, you do not have definition of a function. And the Recursion Theorem requires the induction axiom.

You seem to be trying to define both "natural number" and $+$ "at the same time". You can't do that. Or rather, you cannot guarantee that you have a complete definition without engaging in circularity. In its most charitable interpretation, you are stuck in an ultrafinitistic setting in which the only numbers you can use/guarantee are "numbers" are the ones that you can explicitly write out using only $0$ and the successor function. Addition doesn't get you anything that the successor function doesn't get you, since you can write $n$ as $0+n$.

Answer 2

The problem with construction by the successor function is that it allows your set of numbers to have multiple chains $$0 \rightarrow 1 \rightarrow 2\rightarrow...$$ $$... \rightarrow n \rightarrow S(n) \rightarrow S(S(n)) \rightarrow ...$$ $$...\rightarrow m \rightarrow S(m) \rightarrow S(S(m)) \rightarrow ...$$ $$...$$ where each chain shares no common element with the others. The purpose of the induction axiom is that it extends any true property on the first, standard chain to the entire set of numbers. In such a way the axioms would only allow structures that contains the first chain and other chains that do not violate properties that work on the first chain. However this does not, and cannot, eliminate all the chains because any consistent system that contains the Peano axioms must be incomplete, as shown by the incompleteness theorems. That's why we have nonstandard models of the natural numbers.

On the other hand, your axiom about addition, or any other axiom, cannot replace the axiom of induction. Because the axiom of induction extends any true properties on the first chain to all the chains. Using successor function only allows you to progress on a chain, it does not allow you to progress from one chain to another, and such progression to all the chains is impossible to define because the Peano axioms must be incomplete. Thus extension to all the chains i.e. the set of natural numbers, must either be explicitly stated, or done via an infinite list of axioms.

In the example I gave to you, I constructed an infinite list of chains:

$$(0,0) \rightarrow (0,1) \rightarrow (0,2) \rightarrow ...$$ $$(1,1) \rightarrow (1,2) \rightarrow (1,3) \rightarrow ...$$ $$(2,2) \rightarrow (2,3) \rightarrow (2,4) \rightarrow ...$$ $$...$$

which satisfies your addition and total ordering with my given definition of addition. However, it doesn't satisfy that $\forall x \exists y \quad x \neq 0 \rightarrow y = S(x)$