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Prove that there is no formula $\varphi(v,u)$ so that $(\mathbb N; 0, 1, +) \models \varphi[x,y]$ iff $x$ divides $y$.

Here is what I have so far: assume on the contrary that we do have such $\varphi$. Then my class has proven during class that $(\mathbb N; 0,1,+,<,\text{div}_1,\text{div}_2,\dots)$ admits a quantifier elimination, where $\text{div}_k(a)$ iff $k$ divides $a$. Thus, by our assumption, we have that $(\mathbb N; 0, 1, +, <)$ admits a quantifier elimination. If we can show a formula in $(\mathbb N; 0, 1, +, <)$ that does not admit a quantifier elimination, we will be done.

My friend was also suggesting that we could instead try to extend our model to $\mathbb Z$ and consider what's called a Presburger arithmetic; however, I am not sure how that would go about.

I would be grateful for your guidance.

Squirrel-Power
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    Suppose that there was a formula $\varphi$ that worked. What would quantifier elimination tell you about $\varphi$? – TomKern Nov 26 '23 at 23:25
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    Thanks @TomKern, a quantifier free formula that would hypothetically be equivalent to $φ[x,y]$ would be of the form $\bigvee \bigwedge \theta_{ij}$ where $\theta_{ij}$ is one of the following's equivalent form or its negation: $y=0,ax=bx,ax=cy,cy=dy,ax+n==bx+m,cy+n=dy+m,ax+n=cy+m,ax+n<cy+m,ax+n<bx+m,cy+n<ax+m$... Basically, any $\theta_{ij}$ either expresses equality or something of the form $0 < px + qy + r$ for some integers $p,q,r$. However, I am unsure how I can deduce a contradiction. – Squirrel-Power Nov 27 '23 at 01:26
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    what tools have you learned about in class? have you seen other descriptions of semilinear sets? – TomKern Nov 27 '23 at 01:55
  • I just found https://math.stackexchange.com/questions/4759634/cant-define-div/4759644 while browsing around. I'm gonna vote to merge as a duplicate. – Greg Nisbet Nov 28 '23 at 00:44
  • It does not follow from definability of divisibility in $(\mathbb{N};0,1,+)$ and quantifier elimination for $(\mathbb{N};0,1,+,<,(\mathrm{div}_n)_n\in \mathbb{N})$ that $(\mathbb{N};0,1,+,<)$ admits quantifier elimination. To conclude this, you would need to know that the definition of divisibility in $(\mathbb{N};0,1,+)$ was itself quantifier-free. (By the way, the phrase is "admits quantifier elimination", not "admits a quantifier elimination".) – Alex Kruckman Nov 28 '23 at 18:22

1 Answers1

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We can do this by showing that the resulting theory is Peano Arithmetic and thus incomplete, while Presburger Arithmetic is complete.

We'll do this by showing that we can define multiplication if we are given access to a divisibility predicate. We can do this because we can express the concept of the square of one number dividing another. We can square numbers because $x(x+1) = \text{LCM}(x, x+1)$ and that gives us enough to get the ball rolling.

Let $L(x, y, z)$ be the relation LCM. It is defined as follows:

$L(x, y, z)$ is $(x\mid z) \land (y \mid z) \land [\forall k]((x \mid k) \land (y \mid k) \to (z \mid k))$

Let $G(x, y, z)$ be the relation GCD. It is defined as follows:

$G(x, y, z)$ is $z \mid x \land z \mid y \land [\forall k](k \mid x \land k \mid y \to z \mid k)$

Let $x \mid^2 y$ be the relation "divides twice". It is defined as follows:

$x \mid^2 y$ is defined as $[\exists k](L(x, x+ 1, x+k) \land (k \mid y))$

Let $R(x, y, z)$ be the product relation.

The product of $x$ and $y$ is the least multiple of the least common multiple such that $\text{GCD}(x, y)$ divides it twice.

It is defined as the conjunction of the following.

  1. $(x \mid z)$
  2. $(y \mid z)$
  3. $[\exists k](G(x, y, k) \land k \mid^2 z)$
  4. $[\forall w](x \mid w \land y \mid w \land [\exists k](G(x, y, k) \land k \mid^2 w) \to w \mid z) $

This shows that multiplication is definable.

Therefore, Presburger Arithmetic + divisibility is equivalent to Peano Arithmetic and thus incomplete. However, Presburger Arithmetic is complete, which is a contradiction.

Greg Nisbet
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  • Thank you, I am a little lost on how $xy$ can be defined as the conjunction of the four formulas. Can you elaborate, with an example? Also, why can $x \mid^2 y$ be defined with the formula you just described? I do not get that one also – Squirrel-Power Dec 13 '23 at 00:06