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Usually, in model theory, one presupposes that structures (models) are non-empty. I don't like this (related: What's the deal with empty models in first-order logic?). So let us explicitly permit empty structures.

My requests are:

  1. Can you give a complete calculus of first-order logic that works for empty structures too? By "working for empty structures too", I mean that if the demanded calculus proves a sentence, then this sentence should hold in all structures, also in the empty structures. 2. Can you give a completeness proof for this calculus?

You may wonder: why don't you look in the standard logic texts, where a complete calculus + proof of the completeness should be given? But my problem is: In every logic text which allows empty models, they do not give a complete calculus and proof the completeness. Instead, they prove the compactness theorem and other standard results with purely model theoretic methods, rather than proof theoretic methods. Examples of such texts:

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eueuezezeze
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    In first order logic, nonempty universe assumptions usually sneak in from the $\forall$ elimination inference as $$(\forall x ~:~ \text{False}) \vdash \text{False}$$ which works in every case except the empty universe. – DanielV May 31 '16 at 04:12
  • It strongly depends on which first order logic you use, but changing the inference $$t \not \in \Gamma \implies (\Gamma,~\forall x ~:~ M \vdash M[x \to t])$$ into $$t \not \in \Gamma \implies (\Gamma,~\forall x ~:~ M \vdash M[x \to t] \lor \forall x ~:~ \text{False})$$ makes the inferences work in empty structures. It is effectively making the empty universe a special case that has to be checked specially every time. And it's ugly. – DanielV May 31 '16 at 04:39
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    @DanielV: It's true that empty models makes Hilbert-style deduction ugly. But with Fitch-style natural deduction it's actually natural. For example, if we add to Fitch-style propositional logic the rules I gave at http://math.stackexchange.com/a/1684204/21820, we get a calculus that is sound and complete iff we allow empty models. Completeness can either be proven directly in the same manner as the usual one (first extend to a complete theory, and then take unique existential sentences over provable equivalence as the objects in a model). – user21820 May 31 '16 at 09:33
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    I also don't like the arbitrary restriction that a model be non-empty. It is nothing more than an artifact of the choice of deductive system. However, this issue usually never arises because most axiomatizations of structures have constants or existential axioms, which already make all models necessarily non-empty. – user21820 May 31 '16 at 09:47
  • @user21820 - what do you mean with "take unique existential sentences over provable equivalence as the objects in a model" ? no existential sentence is true in the empty domain. – Mauro ALLEGRANZA Jun 02 '16 at 17:12
  • @MauroALLEGRANZA: The standard proof of completeness goes by showing that consistency implies existence of a model. I made an error in saying "unique", but we first extend to a complete system by adding independent sentences one at a time, and then we take all provable existential sentences of the form $\exists x\ ( φ(x) )$, add one constant $c_φ$ for each of them and the axioms $φ(c_φ)$, and repeat all the way. At each step the system is still consistent, and the collection of all the provable existential sentences in the final limit system is a countable model. – user21820 Jun 03 '16 at 00:06
  • @MauroALLEGRANZA: Now if the system doesn't have any existential axiom, then this method produces no provable existential sentence at every step, so the resulting model is precisely the empty model. There is really no difference in the proof than in the case where we disallow empty models, and the completeness theorem follows in the same way. – user21820 Jun 03 '16 at 00:10
  • @MauroALLEGRANZA: Sorry of course you need to take the collection of all the equivalence classes of constants over provable equality. – user21820 Jun 03 '16 at 00:14

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There is a full treatment of empty-domain models in MENDELSON: Introduction to Mathematical Logic. You will see there why they are a special case of First Order Logic.

Fernando Cano
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