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Model theory is the branch of mathematical logic which deals with the connection between a formal language and its interpretations, or models.
10
votes
Accepted
What is an axiomatization of the equality-free theory of antisymmetric relations?
Your proposed sentence is not strong enough. Consider, for example, the "distance-$<17$" relation on $\mathbb{R}$ with the usual metric.
The issue is that we need "transitivity within reflexivity regi …
4
votes
Accepted
History of unstable formulas
I disagree with the claim that the order property isn't intuitive.
When I took my first model theory class, types were introduced to me as a generalization of Dedekind cuts. So the very first example …
3
votes
1
answer
248
views
Is there a maximal fragment of FOL with "no negation at all?"
Say that a logic $\mathcal{L}$ is nowhere-negative iff for every $\mathcal{L}$-theory $T$ there is a structure $\mathfrak{A}$ such that $$\mathit{Th}_\mathcal{L}(\mathfrak{A})=\mathit{Ded}_\mathcal{L} …
4
votes
If $\mathcal{A} \equiv \mathcal{B}$ and $\mathcal{A} \not \cong \mathcal{B}$, is it possible...
Here's another example, with a computability-theoretic motivation and a set-theoretic proof: for a countable linear order $\alpha$, let $H(\alpha)$ be the linear order $\alpha+\alpha\cdot\eta$, where …
5
votes
Accepted
A (seem to be) elementary logic question
http://en.wikipedia.org/wiki/Beth_definability
5
votes
Undecidable completion of undecidable theory, and pairs of RCF
Even given Stefan's reformulation, the answer is still "no."
Let $L$ be a language consisting only of countably many constant symbols $c_0$, $c_1$, $c_2$, . . . and a single unary relation symbol $U$ …
5
votes
Accepted
Categorical set theories that are not extensions of second-order ZFC
Answering your more specific question: yes, it$^*$ can be so characterized, in the following way:
First, start with the axioms of second-order Zermelo set theory with choice (without Replacement). N …
6
votes
0
answers
230
views
Is there a nice(r) counterexample to this strengthening of Tarski's theorem?
Given a regular logic $\mathcal{L}$ and a structure $\mathfrak{A}$ (in a finite relational language $\Sigma$ for simplicity), let $\overline{Th_\mathcal{L}(\mathfrak{A})}$ be the structure defined as …
3
votes
Accepted
An extension of Woodin's star axiom
This principle is inconsistent, even if we just look at $(H_{\omega_2};\in)$. This is because - for example - the continuum hypothesis is expressible as a sentence in this structure ("There is an $\om …
11
votes
0
answers
512
views
Using Lindstrom's theorem to prove Craig interpolation
[EDIT: The theorem I call "Beth definability" below is apparently not generally called that (wikipedia notwithstanding; see https://math.stackexchange.com/questions/288450/two-forms-of-beths-theorem). …
5
votes
1
answer
186
views
Persistent finite axiomatizability, relational edition
Say that a finitely axiomatizable first-order theory $T$ (in a finite language $\Sigma$) is persistently finitely axiomatizable iff the set $T_{\mathsf{w/o=}}$ of equality-free $T$-theorems is finitel …
4
votes
Why every $\Sigma_1^1(\mathrm{mon})$ sentence true of $\omega$ is also true of $\omega+\zeta$?
This is just a partial answer; I want to point out how a very "coarse" analysis solves one of your questions and gives a weak positive result along the lines of the other. I'll also point out why such …
7
votes
0
answers
283
views
Is there a natural logic in which equality has this weird behavior?
Below, by "logic" I mean "regular logic without equality;" see Badia/Caicedo/Noguera, Maximality of logic without identity.
Given a logic $\mathcal{L}$, a structure $\mathfrak{M}$, and an equivalence …
4
votes
0
answers
144
views
On self-reference in a weak structure
Below, all structures are countable and in finite languages. This question is motivated by the following: "To what extent is self-reference possible when nothing like the diagonal lemma holds?"
Somew …
15
votes
Are integers conservatively embedded in the field of complex numbers?
I don't know a reference, but here's a (slightly overkill) proof:
It's known that $(\mathbb{Z};+,\times)$ interprets an algebraically closed field $K$ of infinite transcendence degree and characterist …