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### $tp^\mathbb{Q}(a/\mathbb N) = tp^\mathbb{Q}(b/\mathbb N)$ iff there is an automorphism $\sigma$ of $\mathbb Q$ fixing $\mathbb N$ and $\sigma(a)=b$

Let $a,b\in\Bbb Q$ such that $\text{tp}^{\mathbb Q}(a/\mathbb N) = \text{tp}^{\mathbb Q}(b/\mathbb N)$. Your proof that $$\forall n\in\Bbb N\quad(n<a\iff n<b)$$ is correct, and similarly \...
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### Prove that divisibility between two natural numbers is not definable in arithmetic

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 ...

### Understanding ZFC

ZFC does generate a “theory”. Formally speaking, a theory is simply a set of sentences in the given formal language. Any set of axioms generates a theory, namely the collection of sentences provable ...

### Axiomatizing ZFC+Con(ZFC) without adding a constant symbol

Use compactness. You get the existence of a model $w$ if every finite subset of ZFC is finitely satisfiable. That can be asserted without adding a constant symbol. For concreteness, for all finite ...

### Is there a first-order sentence defining a cardinal?

Yes, here's one way to do it. It's kind of unsatisfying because it isn't succinct like the first-order characterization of an ordinal. As has been established, being an ordinal is a first-order ...

### $\varphi$ is quantifier free sentence, $T$ a theory, $\vec{c} \not\in T$. $T \vdash \varphi(\vec{c})$ implies $T \vdash \forall x \varphi(\vec{x})$?

The result is known as Generalization on constant. The "trick" is simple: if you have a proof of $\varphi(c)$ and $c$ is not used in the set $T$ of assumptions (axioms), you can replace ...
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### How to prove these three properties about sets?

You are correct about (b), it is likely a typo and they meant A × B ≠ ∅, instead. ( try proving (b) under the assumption that A × B ≠ ∅) As for (a) your proof if fine, it relies on the fact that (p ⟺...