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I have the following problem:

The structure $A$ has a bearer $\Bbb{N}$ (the set of natural numbers) and it is for a language with only one non-logical symbol $p$, which is a predicate interpreted like : $\langle n,k,m \rangle \in p^A \leftrightarrow n^2 = km + 1$ . Prove that $\{0\}$, $\{ 1 \}$ and $\{ \langle n,n \rangle : n \in \Bbb{N} \}$ are definable.

I know that in order to prove that $\{0\}$ is definable in the structure $A$, I have to show that I can build with $p$ a formula $\phi$ such that $\phi(n)$ is true for $\{0\}, \{1\}, \{\langle n,n \rangle : n \in \Bbb{N}\}$. I wan't to know if this can be considered a valid solution.

For the first I can say: $\phi(m):= \exists n \forall k (n^2 = km + 1)$

For the second: $\phi(n):= \exists k \exists m (n^2= km + 1)$

How about the third one? Do I substitute two variable with n like $\phi(k,m)$ or something like that. I also can't think of an appropriate formula.

Any help is welcomed. Thanks :)

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    `\langle n,n\rangle` looks better than ``: $\langle n,n\rangle$ versus $$. – 2017-01-05
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    I notice that you are using mathematical symbols like squaring and addition in your formula $\varphi$, but if $p$ is the only non-logical symbol of your language, you'd need to use $p$ your formula $\varphi$. Thus, for the first one, you'd get something like $\exists n \forall k p(n,k,m)$. No? – 2017-01-05
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    Writing $\phi_1$ and $\phi_2$ for your two $\phi$s, $\phi_2$ isn't strong enough: $\phi_2(3)$ holds with $k = 2$ and $m = 4$ as the witnesses. You can use $\phi_1$ to specify that $k$ and $m$ must be $0$ and then you'll get a $\phi_2$ that works. – 2017-01-05
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    You should clarify whether "$=$" is allowed - some older texts do not consider it a nonlogical symbol. (Whether "$=$" is allowed determines whether (3) has a really short answer, or a bit of a tricky one.) – 2017-01-05

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First of all, you've written your formulas not in terms of $p$. So you should have

  • $\phi_1(x)=\exists n\forall k(p(x, n, k))$

and

  • $\phi_2(x)=\exists k\exists m(p(m, x, k))$

instead of what you've written (note that I'm introducing a new variable, $x$, here - there's no real need to do this, but I think it makes things easier to read).

But your ideas are right. EDIT: As Rob Arthan pointed out, there's actually a problem with $\phi_2$; you need to use $\phi_1$ to fix it . . .

For $S=\{\langle n, n\rangle: n\in\mathbb{N}\}$, think about it this way: how do you know that $x\in S$? Well, $x\in S$ if $x=\langle n, n\rangle$ for some $n$. Do you see how to phrase this appropriately?

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    Can you do the last part, I still don't get it. – 2017-01-05
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    @user3719857 I'm not going to do it out entirely, but here's a further hint. Since $S$ is a set of *ordered pairs*, not just of elements, we're looking for a formula in **two** variables $\varphi(x, y)$ such that $\varphi(a, b)$ holds iff $\langle a, b\rangle\in S$. So $\varphi(x, y)$ should say *what* about $x$ and $y$? – 2017-01-05
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For the third part try $\phi_3(x, y) \equiv \forall k \forall m (p(x, k, m) \Leftrightarrow p(y, k, m))$. (You will need to use $n^2 -1 = (n + 1) (n - 1)$ to show that if $x$ and $y$ are non-zero and distinct, then there are values of $k$ and $m$ that will distinguish them.)

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    To the OP: note that this is only necessary if "$=$" is not considered a logical symbol. (Some older texts do not consider "$=$" to be a logical symbol, but most modern texts do.) – 2017-01-05
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    @NoahSchweber: good point: perhaps you should make that comment on the question. I assumed that the OP was being asked to devise a way of using $p$ to express equality. – 2017-01-05
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    'Tis done.${}{}{}{}$ – 2017-01-05