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I-Does it true that having or not having elimination of quantifier property, say $EQ $, for structures depends on the language being used for describing the structure? If so, so one can "always" expand the language $L$ to $L'$, then the completion of a theory $T$, $T'$ will have $EQ$, isn't it? (I came across to this when I was reading $EQ$ for Discrete linear Orders and as far as I remember this method for proving $EQ$ has been used for some other theories too.)

II-May you please tell me an example to clarify this sentence" $EQ$ is equivalent to substructure completeness"?

KReiser
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Maryam Ajorlou
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Re: I, yes, quantifier elimination is extremely language dependent. Given any structure $\mathcal{M}$, there is a natural expansion $\mathcal{M}_{def}$ gotten by adding an $n$-ary relation symbol $R_\varphi$ for every $n$-ary formula $\varphi$ in the language of $\mathcal{M}$ which we interpret accordingly: $$R_\varphi^{\mathcal{M}_{def}}=\{(a_1,...,a_n)\in \mathcal{M}^n: \mathcal{M}\models\varphi(a_1,...,a_n)\}.$$ In one sense $\mathcal{M}_{def}$ doesn't add any essential strength to $\mathcal{M}$ - e.g. they have the same definable relations - but $Th(\mathcal{M}_{def})$is guaranteed to have QE whether or not $Th(\mathcal{M})$ has QE.

Re: II, the equivalence between substructure completeness and QE is treated here. Simply put, a first-order theory has QE iff it is substructure complete.

Noah Schweber
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  • Thank you so much. For $I$- well, doesn't this diminish the value of Elimination of Quantifier property ?? (Always, there is an expanded language somewhere which leads to having Elimination of Quantifier property for every given theory). (I ask because I've seen model theorists discuss this property emphatically) – Maryam Ajorlou Dec 18 '19 at 21:51
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    @MaryamAjorlou It depends on what you mean by "value" - what are you trying to prove about the theory in question? For example, consider the proof that the theory ACF of algebraically closed fields is decidable. This has two steps, namely proving that the quantifier-free fragment of ACF is decidable and proving that ACF has quantifier elimination. That first step is sensitive to the specific language used: if we'd started by shifting attention to the "trivially QE"-version ACF$^{def}$, it wouldn't work. – Noah Schweber Dec 18 '19 at 22:12
  • So in proving decidability, QE (or model completeness or similar) is useful because it tells us that we only need to understand a small fragment of the language involved - which is only useful if that fragment is actually easy to understand! Replace the language with a richer one and the quantifier-free fragment suddenly becomes much larger. – Noah Schweber Dec 18 '19 at 22:13
  • This plays into the bigger point that QE is generally something you use to study a theory you already care about: e.g. if you have a theory $T$ you care about and you want to understand what the definable relations in models of $T$ look like, knowing that $T$ has QE will help a lot if it is easy to understand the quantifier-free fragment of $T$ (as above, if we shift from $T$ to $T^{def}$ this "reduction" winds up not being useful). See also the discussion here. – Noah Schweber Dec 18 '19 at 22:15
  • For the value, I mean the Elimination of Quantifier is a desirable property. There are some theorems or practical tests to identify whether a theory has or has not QE. For example, (roughly speaking ) having algebraically prime models for T + having simply closed substructures implies QE. – Maryam Ajorlou Dec 19 '19 at 13:43
  • Or there would be some other nice properties which can be obtained by QE, for example, T has QE implies T is model complete, or as you referred to the link, T has QE is equivalent to T is a substructure complete theory. But under the Morleyziation method T admits QE, and then those theorems, which need QE assumption, hold automatically. – Maryam Ajorlou Dec 19 '19 at 13:44
  • Regarding to your example of decidability for ACF, it became clear to me somewhat. Though two kinds of QE shaped in my mind, weak QE and strong QE which obtains without the Morleyiziation method. – Maryam Ajorlou Dec 19 '19 at 14:15
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    @MaryamAjorlou There are two key points to understand about Morleyization: (1) In a situation where we want to prove a theorem about an arbitrary first-order theory $T$, often what we want to prove is not language-dependent. In such a situation, we might as well assume that our theory has QE, since we can always Morleyize it. (2) If, instead, we want to prove something about a particular first-order theory $T$, it can be very helpful to know that $T$ has QE. In such a situation, Morleyizing is not so valuable: e.g. one of the main ways to apply QE is to understand ... (continued) – Alex Kruckman Dec 19 '19 at 22:35
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    ... all the formulas as Boolean combinations of atomic formulas. And this is only useful if we already understand all the atomic formulas. If we Morleyize, we introduce many new atomic formulas, which are just as complicated as arbitrary formulas. So after Morleyizing, we have QE, but it's not so useful! – Alex Kruckman Dec 19 '19 at 22:37
  • Now, do you think can one make (a) differences between those theories which admit QE innately and those admit QE by this method? for a while, I felt the first group would be richer ( richer, at least aspect of having other good properties ). This may seem crazy view, because as Noah Schweber said the construction doesn't add any essential strength. ( As two random examples, I glanced QE for the real fields(Tarski), QE for the p-adic fields (Macintyre), both prove QE in expanded languages). – Maryam Ajorlou Dec 20 '19 at 13:37
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    @MaryamAjorlou Well, theories that admit QE innately certainly have some special properties - for example, they are substructure-complete, as noted in your question. But any property that only depends on the category of definable sets cannot distinguish between theories with and without QE, since Morleyization does not change the definable sets, as noted by Noah in his answer. – Alex Kruckman Dec 23 '19 at 06:41
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    In practice, if you're interested in a theory "in the wild", it's very lucky indeed to discover that it has QE in its most natural language. Often you have to make a definable expansion of the language (as in the case of real closed fields that you refer to) to obtain QE. But Morleyization is also a definable expansion of the language. So there's a question of where to draw the line: which definable expansions are "natural" and which aren't? Some useful measures of how drastic a definable expansion is are the quantifier complexities of the formulas involved. – Alex Kruckman Dec 23 '19 at 06:43
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    In the example of real closed fields, the ordering relation $x \leq y$ is definable by both an existential formula ($\exists z, (z^2 = y-x)$) and a universal formula ($\forall z, (z = 0 \lor \lnot (z^2 = x - y))$). The theory of $(\mathbb{R};0,1,+,\times)$ does not have QE, but the theory of $(\mathbb{R};0,1,+,\times,\leq)$ does have QE, and the form of the definability of $\leq$ implies that the theory of $(\mathbb{R};0,1,+,\times)$ is model-complete (which is almost as good as QE for many purposes). – Alex Kruckman Dec 23 '19 at 06:46
  • @AlexKruckman, -I thought the method that we add constants to the language for some certain goals in some proof differs from the Morleyziation, but as today I saw it is, indeed, a case of Morleyziation. Besides Morleyziation, does exist any other construction deal with language? – Maryam Ajorlou Dec 28 '19 at 19:18
  • -My question may seem meaningless, but if we say definable sets don't change under this construction because those formulas who are satisfied in the model don't change, and so what is preserved during Morleyziation is the satisfiability of the formulas. What are other components like satisfiability which model theorists may like to define a construction for preserving them for some certain goals? – Maryam Ajorlou Dec 28 '19 at 19:18
  • @MaryamAjorlou What does it mean to "deal with language"? Arguably Skolemization could count, but in general adding Skolem functions changes the behavior of the theory in very strong ways (contra Morleyization). – Noah Schweber Dec 28 '19 at 19:19
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    @MaryamAjorlou "What are other components like satisfiability which model theorists may like to define a construction for preserving them for some certain goals?" Tameness properties form a natural class of candidates. By "tameness" here I mean general facts about the "shape" of the definable sets; for example, minimality and its variants (e.g. o-minimality). So for example a huge result in this context is that the expansion $\mathcal{E}$ of $\mathcal{R}=(\mathbb{R};+,\cdot,0,1)$ by adding exponentiation $x\mapsto e^x$ is o-minimal. (cont'd) – Noah Schweber Dec 28 '19 at 19:21
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    Of course $\mathcal{E}$ has more power than $\mathcal{R}$ since the exponential function is not definable in $\mathcal{R}$, but the fact that we retain o-minimality means that lots of nice arguments applicable to $\mathcal{R}$ are also applicable to $\mathcal{E}$. In some sense, the exponentiation function is "tame over $\mathcal{R}$." – Noah Schweber Dec 28 '19 at 19:22
  • @NoahSchweber, I meant Morleyziation is a construction that " adds "some predicates to the" language", to preserve a certain property. So the construction at the first step deals with the language. I was searching to know more about any other kinds of constructions in Model Theory to compare . – Maryam Ajorlou Dec 28 '19 at 19:24
  • @MaryamAjorlou Well, per my previous comment, would you count Skolemization? It begins by changing the language, but then in general is forced to distort things greatly. – Noah Schweber Dec 28 '19 at 19:26
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    If you allow changing the domain as well, and look beyond pure model theory, an interesting "opposite" example occurs in computable structure theory: namely, Marker extensions (named for David Marker). Here the idea is to take a simple relation and make it more complicated. For example, suppose I have a structure $\mathcal{A}$ whose language includes a unary predicate $U$. Then $U^\mathcal{A}$ is "given immediately" in any copy of $\mathcal{A}$. But now consider the structure $\mathcal{B}$ defined as follows. We remove $U$ from the language, add a new unary predicate $V$, and a binary $E$. – Noah Schweber Dec 28 '19 at 19:29
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    We set $V^\mathcal{B}=\mathcal{A}$ (without $U$), and add for each $a\in\mathcal{A}$ a new set $X_a$ of infinitely many new objects (with $X_a\cap X_b=\emptyset$ for $a,b\in\mathcal{A}$ distinct). The relation $E$ connects the elements of $X_a$ and $a$: we have $aEx$ iff $a\in V$ and $x\in X_a$. Finally, we add a new unary predicate $\hat{U}$ such that $\hat{U}$ holds of exactly one element of $X_a$ for each $a\in U^\mathcal{A}$. Now in a copy of $\mathcal{B}$, the original relation $U$ is "computably enumerable but not (in general) computable" in any copy of $\mathcal{B}$ a precise sense. – Noah Schweber Dec 28 '19 at 19:31
  • Anyways, I think this comment section is going on too long; maybe ask a new MSE question about this? – Noah Schweber Dec 28 '19 at 19:32
  • @NoahSchweber, Thank you very much for saying about the class and that nice example. Both constructions add some predicates to the language. But what do they want to preserve under Skolemization? You are totally right comments have been too long. – Maryam Ajorlou Dec 28 '19 at 19:46
  • Let us continue this discussion in chat. – Maryam Ajorlou Dec 28 '19 at 19:46