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Is it possible to define pairing function (and the inverses) in Presburger arithmetic?

I would guess no but I can't locate a reference nor construct a proof to one way or another.

Rodrigo de Azevedo
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1 Answers1

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The answer is no.

Small comment: I show below that no pairing function is definable in the structure $(\mathbb{N}; +)$. This is strictly stronger than saying that no pairing function can be defined in Presburger arithmetic, which is saying that there is no definition which Presburger arithmetic proves is a pairing function.

Remember that in $(\mathbb{N}; +)$, every definable set is eventually periodic. So building a "sufficiently sparse" set from a pairing function will be sufficient to show that such a function can't be definable in this structure.

So suppose $\langle \cdot,\cdot\rangle:\mathbb{N}^2\rightarrow\mathbb{N}$ were a bijection definable in $(\mathbb{N}; +)$. First, note that the usual ordering $<$ on $\mathbb{N}$ is definable in $(\mathbb{N}; +)$: $$a<b\iff [\exists c(a+c=b)\wedge \forall c(b+c\not=a)].$$ (This assumes $0\in\mathbb{N}$; if not, we can do away with the second conjunct.) Similarly, the minimum function is definable, in the sense that if $\varphi(x, y)$ is any formula of two variables, the function $x\mapsto \min\{y: \varphi(x, y)\}$ is a definable (possibly partial) function in $(\mathbb{N}; +)$.

Now consider the function $$f(x)=\min\{y: \forall a, b<x(\langle a, b\rangle<y)\},$$ and let $$X=ran(f).$$ Clearly $X$ is definable by the considerations above and the fact that the range of a definable function is a definable set; however, it's easy to see that $f$ grows at least as fast as $x^2,$ and so $X$ is too sparse to be definable in $(\mathbb{N}; +)$.

Noah Schweber
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  • Does that mean we cannot decide if two finite sets $A$ and $B$ have same cardinality in Presburger? – Turbo Jun 27 '19 at 02:11
  • So somehow if we introduce multiplication we can compare cardinalities? – Turbo Jun 27 '19 at 02:11
  • @Turbo "Does that mean we cannot decide if two finite sets $A$ and $B$ have same cardinality in Presburger?" How would you even phrase the question in the context of addition alone? "So somehow if we introduce multiplication we can compare cardinalities?" Once we have multiplication, we can talk about numbers representing sets. We can (for example) view a number $n$ as coding the set ${i: p_i\vert n}$, where $p_i$ is the $i$th prime. (cont'd) – Noah Schweber Jun 27 '19 at 02:19
  • So e.g. $12$ would code ${1, 2}$ since $p_1=2$ and $p_2=3$ are the only primes dividing $12$. Now it turns out that we can express "$a$ is in the set coded by $b$" just in terms of $+$ and $\times$, although this isn't obvious at all. And then we can indeed say "the set coded by $a$ has the same cardinality as the set coded by $b$," and many other things besides. This kind of coding is a key component of the proof of Godel's incompleteness theorem. – Noah Schweber Jun 27 '19 at 02:22
  • @Turbo "What if I have many quantifiers?" I don't know what you mean by that. Again, how do you intend to even phrase the question in Presburger arithmetic? – Noah Schweber Jun 27 '19 at 02:23
  • @Turbo No, that's incorrect. Presburger arithmetic refers to the whole thing. (If we expand the language a bit, we get a theory with quantifier elimination so that we only need to look at the quantifier free part, and this might be what you're thinking of.) – Noah Schweber Jun 27 '19 at 02:24
  • I will think more. – Turbo Jun 27 '19 at 02:27
  • 'the set coded by a has the same cardinality as the set coded by b' how would you do this? – Turbo Jul 04 '19 at 18:32
  • @Turbo Like I said, it's not trivial; the comments aren't a good place to discuss it. Basically, though, the idea is that we can use $+$ and $\times$ to talk about finite sequences, and a bijection between two finite sets is given by a finite sequence with certain properties. If you want to see the details you should ask a separate question about it. This whole technique of representing sequences, incidentally, was one of the key innovations in the proof of Godel's incompleteness theorem. – Noah Schweber Jul 04 '19 at 18:45
  • And again, this requires both addition and multiplication; Presburger arithmetic (and again, this refers to the whole thing) cannot do anything like this. – Noah Schweber Jul 04 '19 at 18:46
  • I do not know how to post a proper query. Is there an easy reference to review beforehand? – Turbo Jul 04 '19 at 18:50
  • @Turbo "I do not know how to post a proper query. " I just meant an MSE question. You could cite my comment and ask for details. If you want a reference instead, just read a presentation of Godel's incompleteness theorem, and this will be essentially the first part of showing that "enough" functions are representable in PA (it's enough to do primitive recursive functions, but modern presentations tend to prove the stronger fact that PA-representable = computable). – Noah Schweber Jul 04 '19 at 18:54
  • https://math.stackexchange.com/questions/3283269/comparing-cardinalities-of-sets-in-peano – Turbo Jul 04 '19 at 18:57