# When we count the same set, must the number always be the same?

Return to Frege's question, What justifies arithmetic? And consider the ur-proposition that counting a finite set always produces the same number, and ask whether this has a logical justification, that is a justification from assumptions which are necessarily true. Note that few of us actually first believed this assertion because of a logical proof. We accepted it say by authority - we were told it was so - or by experience - we experimented and discovered that counting the same set in two different ways always produced the same. Neither of these methods of justification, of course, necessarily ensure truth.

OK you say, but there is a proof from bedrock principles using induction which is entirely from logical principles. Model a counting of a finite set $A$ as an injective function from an initial segment of the natural numbers (beginning with 1) onto $A$, that is a sequence without repetitions. For the base case suppose a counting of a set produces the number 1. Then there is one thing, and the only way to count the set would produce the number 1. So check for the base case.

Now suppose $m$ is the successor of $n$ and the induction hypothesis holds for $n$ (that is, if we count any set to have the number $n$, then any other counting of the set must also produce the number $n$). Consider a set $A$ where one counting produces the number $m$, and another counting produces the number $v$. It does seem an acceptable (and logical) principle that if $m$ is the successor of $n$, and we have counted to $m$, then we need to have counted first to $n$, and then counted one more. That is, if $\boldsymbol f$ counts $A$ as $m$, then there exists $\boldsymbol f',B,b$ such that $b \in A \land B = A$\ $\{b\} \land\boldsymbol f'$ counts $B$ as $n$. Induction proves that $v \neq 1$ has a predecessor $u$. Letting $\boldsymbol g$ count $A$ as $v$, then there exists $\boldsymbol g',C,c$ such that $c \in A \land C = A$ \ $\{c\} \land \boldsymbol g'$ counts $C$ as $u$. And - we can't apply the induction hypotheses immediately because while $B$ is of size $n$, and $C$ is of size $u$, $B$ and $C$ need to be the same set to apply the hypothesis, but they may be different since $b$ and $c$ may be different.

The way the proof must continue is that (assuming that $b$ and $c$ are different, since if they are the same there is no problem) we assert the existence of a sequence $\boldsymbol g''$, where we replace $b$ by $c$, that is where $\boldsymbol g'(i) = b$, assign instead $\boldsymbol g''(i) = c$ and leave everything else the same. Then by simple logic $\boldsymbol g''$ counts $B$ and indeed counts it as as $u$. Then one can apply the induction hypothesis, and the ur-proposition follows.

Only the assumption that there exists a sequence with the one replacement - that is, given a sequence $abcd...x...$, there must exist a sequence $abcd...y...$, for any $x$ and $y$ - is, or does not appear to be, a logical assumption. The new sequence is not a part of the older one, and its existence is not implicit in the older's existence. The assumption simply does not fall out of the meaning of "sequence." It is in fact very useful - I claim that with it and some other, only necessarily true, assumptions, one can develop arithmetic -, and so it can be given the justification that it leads to results which are useful in the sciences, and therefore it is desirable to make it. But this is different from being logical, and the question must be raised if in some cases reality is better represented without this assumption or perhaps making a different one. Modern physics is full of non-intuitive explanations; perhaps it can be better presented by using a non-intuitive mathematics.

So is the uniqueness of counting necessarily true? That one proof depends on a non-logical assumption, is not conclusive, since after all perhaps there is another, this time logical, proof where the replacement assumption is not needed. I have tried a bit and failed from assumptions which satisfy me, so I conjecture that it is essential in the framework of these assumptions. Anyway, my question is, is it? I formalize the question in the following way.

Consider FOL with equality, with one constant 1, with one two-place predicate $\sigma$ (successoring), and with two types (standard and bold). Standard types are natural numbers, and bold types are finite sequences of natural numbers.

Define:

$\boldsymbol f^D(x)$ abbreviates $\exists y \space \boldsymbol f(x) = y$
This defines $x$ being in the domain of $\boldsymbol f$.

$\boldsymbol f^H(x)$ abbreviates $\boldsymbol f^D(x) \land \forall y (\sigma(x,y) \Rightarrow \neg \space \boldsymbol f^D(y))$
This defines $x$ as the top or high number in the initial segment defining the sequence $\boldsymbol f$.

$\boldsymbol f =_k \boldsymbol g$ abbreviates $\forall x \neq k \forall y(\boldsymbol f(x) = y \iff \boldsymbol g(x) = y)$
This says $\boldsymbol f$ and $\boldsymbol g$ are defined and have the same values for the same $x$, except perhaps at $k$.

Consider the following axioms:

1/ $\forall x \forall y \forall z (\sigma x,y \land \sigma x,z \Rightarrow y = z)$

2/ Induction schema, i.e. $\phi (1) \land \forall n \forall m (\sigma n,m \land \phi (n) \Rightarrow \phi (m)) \Rightarrow \forall n \space\phi (n)$

3/ $\forall\boldsymbol f \space \boldsymbol f^D (1)$

4/ $\forall\boldsymbol f \exists n \space \boldsymbol f^H (n)$

5/ $\forall\boldsymbol f \forall n \forall m \space (\boldsymbol f^H(m) \land \sigma n,m \Rightarrow \exists \boldsymbol f' ((\boldsymbol f')^H(n) \land \boldsymbol f' =_m \boldsymbol f))$

6/ $\forall\boldsymbol f \forall n \forall m \space (\boldsymbol f^D(m) \land \sigma n,m \Rightarrow \boldsymbol f^D(n))$

7/ $\forall \boldsymbol f \forall i \forall c (\boldsymbol f^D(i) \Rightarrow \exists \boldsymbol f' (\boldsymbol f'(i) = c \land \boldsymbol f' =_i \boldsymbol f))$

Now 1/ through 7/ are sufficient to prove:

(UNIQUENESS) $\forall k \forall n \forall \boldsymbol f \forall \boldsymbol g (Img(\boldsymbol f) = Img(\boldsymbol g) \land Injection(\boldsymbol f) \land Injection(\boldsymbol g) \land \boldsymbol f^H(k) \land \boldsymbol g^H(n)\Rightarrow k = n)$,

where
$Img(\boldsymbol f) = Img(\boldsymbol g)$ abbreviates $\forall y(\exists x \space \boldsymbol f(x,y) \iff \exists x \space \boldsymbol g(x,y))$ and
$Injection(f)$ abbreviates $\forall x \forall y (\boldsymbol f(x) = \boldsymbol f(y) \Rightarrow x = y)$

(Indeed I would claim 1/ through 7/ are sufficient to develop a satisfactory arithmetic.)

My question is: Is there a model in which 1/ through 6/ (i.e. eliminate 7/) are true but in which UNIQUENESS is false?

NOTE. Let $\phi (k)$ abbreviate

$\forall n \forall \boldsymbol f \forall \boldsymbol g (Img(\boldsymbol f) = Img(\boldsymbol g) \land Injection(\boldsymbol f) \land Injection(\boldsymbol g) \land \boldsymbol f^H(k) \land \boldsymbol g^H(n)\Rightarrow k = n)$

Then using only 1/ through 6/ I can prove $\phi(1)$ and, for any $k$, $\forall x_1,x_2,...,x_k (\sigma 1,x_1 \land \sigma x_1,x_2 \land \space ... \space \land \sigma x_{k-1},x_k \Rightarrow \phi (x_k))$

UNIQUENESS however of course is $\forall k \phi (k)$, and this I cannot prove, hence my question.

• I think you can save a lot of work if you reduce your problem to showing that the set $[n] = \{1,2,\ldots, n\}$ has size $n$ and that this size is well-defined. – Dirk Nov 29 '17 at 10:28
• I'm not sure the problem reduces, but in any case showing, if the set [n] has a size, that this size must be $n$ presents the same difficulty as proving UNIQUENESS. (In the system you cannot show that [n] has a size, because you can't show any function exists.) – abo Nov 29 '17 at 11:54
• @EmilJeřábek Could you give a reference for the unprovability of the pigeonhole principle in $I\Delta_0(R)$ (or even just a precise statement)? I was completely confused by OP's question, but your reformulation seems to make it interesting, so now I would like to know more. – Gro-Tsen Nov 29 '17 at 15:06
• Nice to see you again, @abo – David Roberts Nov 29 '17 at 20:47
• Just to note that in fact Axiom 3 is superfluous, and can be proven using an easy induction with $\phi(n)$ as $\space(\boldsymbol f^H(n)\Rightarrow \boldsymbol f^D(1))$. (Hi David!) – abo Dec 6 '17 at 13:36

## 1 Answer

Such a model can be constructed using known independence results in bounded arithmetic as follows.

Let $I\Delta_0(f)$ be a theory in the usual language of arithmetic ($0,S,+,\cdot,<$) augmented by a new function symbol $f$, axiomatized by Robinson’s arithmetic plus induction for all bounded formulas in the expanded language. By a result of Ajtai [1,2], $I\Delta_0(f)$ does not prove the bijective pigeonhole principle $\mathrm{PHP}_n(f)$: this means that there exists a model $(\mathcal M,f)\models I\Delta_0(f)$, and a (necessarily nonstandard) element $n\in M$ such that $f$ restricted to the interval $[0,n]_\mathcal M$ defines a bijection $[0,n]\to[0,n-1]$.

Now, let us define a model $\mathcal N$ of the theory from the question:

• the number sort of $\mathcal N$ consists of $[0,n]$, with $\sigma$ being the graph of $S$ restricted to $[0,n-1]$

• the sequence sort consists of all sequences of the form $f\restriction_{[0,m]}$ and $\mathrm{id}\restriction_{[0,m]}$ for $m\le n$

It is straightforward to see that $\mathcal N$ satisfies axioms 1 and 3–6, and the functions $\mathrm{id}\restriction_{[0,n-1]}$ and $f\restriction_{[0,n]}$ witness that UNIQUENESS fails.

In order to see that $\mathcal N$ satisfies axiom schema 2 (induction) as well, notice that under the obvious interpretation of $\mathcal N$ in $\mathcal M$, number quantifiers $\exists x$, $\forall x$ translate to bounded quantifiers $\exists x\le n$, $\forall x\le n$, and sequence quantifiers can be simulated by bounded quantifiers as well: for example, we may code $\mathrm{id}\restriction_{[0,m]}$ as $2m$, and $f\restriction_{[0,m]}$ as $2m+1$. Thus, $\exists\mathbf g$, $\forall\mathbf g$ become $\exists z_\mathbb g\le2n+1$, $\forall z_\mathbb g\le2n+1$, and under this encoding, $\mathbb g(x)=y$ is defined by the bounded formula

$$\exists m\le n\,\bigl((z_\mathbb g=2m\land y=x\land x\le m)\lor(z_\mathbb g=2m+1\land y=f(x)\land x\le m)\bigr).$$

Thus, any formula $\phi$ translates to a $\Delta_0(f)$ formula in $\mathcal M$, and consequently satisfies induction.

References:

[1] Miklós Ajtai: The complexity of the pigeonhole principle, in: Proceedings of the 29th Annual Symposium on Foundations of Computer Science, IEEE, 1988, pp. 346–355, doi: 10.1109/SFCS.1988.21951.

[2] Miklós Ajtai: The complexity of the pigeonhole principle, Combinatorica 14 (1994), no. 4, pp. 417–433, doi: 10.1007/BF01302964.