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I know this question is a bit trivial, and it is just about the exceptional case in the definitions, but I still want to make things clearer.

  • First, do we require that every language (at least in the context of model theory) must contain an equality symbol? And does this equality symbol must be interpreted as the "real" equality (the equality in the meta-language), or just be treated as an ordinary binary relation? I have some superficial thinking of this issue: To make our model theory more generally, we definitely don't want to impose any additional assumption. But if we don't have equality in the language, properties such as $\kappa$-categorical or strongly minimal, which need to refer to the size of models, would be almost impossible to be satisfied. Since we can duplicate elements in the model arbitrarily, and define their interpretation of relations the same as the original element.
    On the other hand, if we require every language have equality, then what about the set theory itself, which uses $\in$ as the only primitive relation and considers $=$ as a kind of derived relation? Besides, we need to carefully restate theorems such as completeness theorem, since formulas such as "$\forall x\ (x = x)$" was (implicitly) satisfied by every structure, but is not a tautology in syntactic view (and hence can't be derived from empty axiom). Does there any reference discuss this subtle difference?

  • Second, in the definition of quantifier elimination (QE), do we require the eliminated formula has same free variables as the original formula? This change almost doesn't affect the meaning of QE, since if the weaker version of QE holds, we can prove the stronger version for formulas with at least one free variables using the criterion below.

Suppose $\phi(\overline{v})$ is a $\mathcal{L}$-formula. The followings are equivalent:

  • There is a quantifier-free $\mathcal{L}$-formula $\psi(\overline{v})$ such that $T \models \forall \overline{v}\ ( \phi(\overline{v}) \leftrightarrow \psi(\overline{v}))$.
  • If $\mathcal{M}$ and $\mathcal{N}$ are models of $T$, $\mathcal{A}$ is an $L$-structure, $\mathcal{A} \subseteq \mathcal{M}, \mathcal{N}$, then $\mathcal{M} \models \phi(\overline{a})$ if and only if $\mathcal{N} \models \phi(\overline{a})$ for all $\overline{a} \in \mathcal{A}$.

What troubles me is the rest (and the most boring case perhaps) -- the case in which the formula is actually a sentence. If the language has a constant symbol, we can solve this by substituting free variables in the eliminated formula by a constant to make it closed; If the theory is complete, then this sentence has the same satisfiability in every model, so we have $T \models \phi \leftrightarrow \bot$ or $T \models \phi \leftrightarrow \top$ (suppose we have $\bot$ and $\top$ in the formal system). So, the question reduces to does there exist a theory $T$, with no constant symbol and not complete, but has QE?

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    $\begingroup$ These are two totally separate questions so you should ask them in separate posts. $\endgroup$ Apr 13, 2021 at 15:08
  • $\begingroup$ @EricWofsey Because I think the first question may reflect some of my concerns in the second question. $\endgroup$ Apr 13, 2021 at 15:58

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Do we require that every language (at least in the context of model theory) must contain an equality symbol? And does this equality symbol must be interpreted as the "real" equality (the equality in the meta-language)?

Yes and yes. By default, the syntax of first-order logic contains a primitive symbol $=$, and the semantics of first-order logic treats this symbol as referring to true equality in models. The logic without primitive equality is called "first-order logic without equality". Caveat: In my experience, this is the standard modern use of terminology. But one has to be a bit careful, especially when reading older sources, since it has not always been a universal convention that first-order logic includes primitive equality.

To make our model theory more general, we definitely don't want to impose any additional assumption.

But first-order logic without equality sits inside first-order logic with equality, so you don't lose any generality. If you want to do first-order logic without equality, you can just focus on sentences and formulas that don't use the equality symbol. And if you want a symbol for equality which isn't necessarily interpreted as true equality, you can introduce a new binary relation symbol $\hat{=}$.

On the other hand, if we require every language have equality, then what about the set theory itself, which uses $\in$ as the only primitive relation and considers $=$ as a kind of derived relation?

That's not what (standard) set theory does. ZFC is a theory in first-order logic with equality. The extensionality axiom $\forall x\forall y(\forall z(z\in x\leftrightarrow z\in y)\leftrightarrow (x=y))$ includes the equality symbol, and when we talk about models of ZFC, this equality symbol is interpreted as true equality.

Besides, we need to carefully restate theorems such as completeness theorem, since formulas such as "$\forall x\ (x = x)$" was (implicitly) satisfied by every structure, but is not a tautology in syntactic view (and hence can't be derived from empty axiom).

No, $\forall x\ (x = x)$ is a syntactic tautology and can be derived from no hypotheses in any standard proof system for first-order logic. Of course, this sentence can't be derived in a proof system for first-order logic without equality, since such a proof system doesn't have any rules that mention the symbol $=$! But the point is that a proof system for first-order logic without equality is not complete for first-order logic (with equality).

In the definition of quantifier elimination (QE), do we require the eliminated formula has same free variables as the original formula?

Yes.

Does there exist a theory $T$, with no constant symbol and not complete, but has QE?

Well, some formulations of first-order logic also allow $0$-ary relation symbols ("propositional symbols"). A propositional symbol $P$ is a sentence on its own. So for example in the language $L = \{P\}$ where $P$ is a propositional symbol, the theory $T = \{(\exists x\top)\land \forall x\forall y(x = y)\}$ has QE but is not complete, since it does not decide the truth value of $P$.

But on the other hand, we have:

Proposition If $L$ is a language with no $0$-ary symbols (no constants or propositional symbols), then every $L$-theory $T$ with QE is complete.

Proof: Let $\varphi$ be an $L$-sentence. Then there is a quantifier-free $L$-sentence $\psi$ such that $T\models \varphi\leftrightarrow \psi$. But the only quantifier-free $L$-sentences are $\top$ and $\bot$. Thus $T\models \varphi$ or $T\models \lnot \varphi$, so $T$ is complete. $\square$

Even for languages with constant symbols, proving QE is a common strategy for proving that a theory is complete. If a theory $T$ has QE, then it is complete if and only if it decides the truth of all quantifier-free sentences, which comes down to saying that $T$ determines the isomorphism type of the "structure generated by the constants". More precisely, given any model $M\models T$, $M$ has a smallest substructure $M_\emptyset$, the substructure of $M$ generated by the constants. If $T$ has QE, then $T$ is complete if and only if for all $M\models T$ and $N\models T$, the structures $M_\emptyset$ and $N_\emptyset$ are isomorphic. In the case when $T$ has no constants, $M_\emptyset$ is always the empty structure, and when $T$ also has no propositional symbols, any two empty structures are isomorphic.

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  • $\begingroup$ "But the point is that a proof system for first-order logic without equality is not complete for first-order logic (with equality)." Here 'not complete' means the ability of the second deductive system is weaker than the first? $\endgroup$ Apr 13, 2021 at 15:43
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    $\begingroup$ @MinghuiOuyang A proof system is complete for a logic $L$ if every semantically valid sentence in $L$ is provable in the system. Fix a proof system $S$ for first-order logic without equality. Since $\forall x(x = x)$ is semantically valid but not provable in $S$, $S$ is not complete for first-order logic. This means that $S$ is weaker than any proof system $S'$ that is complete for first-order logic, since $S'$ proves more theorems than $S$. $\endgroup$ Apr 13, 2021 at 15:47
  • $\begingroup$ Oh, I get it... $\endgroup$ Apr 13, 2021 at 15:53

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