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I don't know much about model theory, but it was suggested in another math forum that a theory with model $\{ \}$ would be consistent. Is this correct?

Asaf Karagila
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Dan Christensen
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    This is correct. – goblin GONE Jun 02 '16 at 13:40
  • @goblin: Well, is it? – Asaf Karagila Jun 02 '16 at 14:07
  • @AsafKaragila, pretty sure its true - in fact, even under the conventions described in your answer (which I hadn't considered), it is vacuously true that every theory with $\emptyset$ as a model is consistent (because no theory has $\emptyset$ as a model.) – goblin GONE Jun 02 '16 at 14:15
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    @goblin: If your logic has the inference rule $\forall x\varphi\to\exists x\varphi$, then no. The empty theory proves the existence of objects in the universe. – Asaf Karagila Jun 02 '16 at 14:17
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    @AsafKaragila, I think we're reading Dan's question very differently. From my point of view, Dan asks: "If $\emptyset \models T$, can we deduce $\neg(T \vdash \bot)$?" Hence if, by convention, we declare that $\neg(\emptyset \models T)$, then the answer is "yes, vacuously." – goblin GONE Jun 02 '16 at 14:23
  • $\exists x (x=x)$ is valid for the "usual" FOL semantics (that requires not-empty domains) and thus : $\emptyset \vDash \exists x (x=x)$, while of course it is not true in the empty domain. – Mauro ALLEGRANZA Jun 02 '16 at 14:52
  • @MauroALLEGRANZA The theory in question is a set theory that makes no unconditional existential claims, i.e. no axioms of the form $\exists x: P(x)$ and does not assume a non-empty universe. – Dan Christensen Jun 02 '16 at 14:53
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    $∃x(x=x)$ is not an axiom of set theory, but a "logical truth" (with the semantics as above). As per @Asaf answer, $\forall x (x=x)$ is an axiom of FOL with equality; $\forall x \varphi \to \exists x \varphi$ is a valid formula; thus, bt modus ponens $\exists x (x=x)$ is true in every non empty domain, set theory included. Of course, by the discussion above, in a set theory with empty domain, this does not hold. – Mauro ALLEGRANZA Jun 02 '16 at 14:57
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    Interesting: a set theoy that "does not assume a non-empty universe"; in a set theory with an empty universe, there is the empty set ? – Mauro ALLEGRANZA Jun 02 '16 at 14:58
  • See the post what's the deal with empty models in FOL for a discussion. – Mauro ALLEGRANZA Jun 02 '16 at 15:46
  • @MauroALLEGRANZA So, would such a set theory be consistent? – Dan Christensen Jun 02 '16 at 15:52
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    This recent question is also relevant: http://math.stackexchange.com/questions/1806091/complete-calculus-of-first-order-logic-working-for-empty-structures-too – Alex Kruckman Jun 02 '16 at 16:27
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    Yes; if it has a model, it is consistent, because $\bot$ is false in the empty domain and thus, the suitable modified FOL calculus sound for the empty domain does not prove $\bot$. – Mauro ALLEGRANZA Jun 02 '16 at 16:56
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    In the empty domain every formula $(∀x)\varphi(x)$ is true: thus, a set theory for the empty domain has the theorem $(∀x)( x \in x)$, but also $(∀x) \lnot ( x \in x)$. – Mauro ALLEGRANZA Jun 02 '16 at 17:02
  • @MauroALLEGRANZA That doesn't seem right somehow. In my non-existential set theory, I am quite sure that I cannot prove either of these theorems. But you could easily prove that $\forall x:[x\notin U] \implies \forall x \in U: P(x)$for any unary predicate $P.$ – Dan Christensen Jun 04 '16 at 20:15
  • @MauroALLEGRANZA Is not postulating the unconditional existence of any objects the same as postulating that no objects exist? Is not postulating a non-empty universe the same postulating an empty universe? I wouldn't think so in either case. What exactly is meant by the empty set being a model for a theory? – Dan Christensen Jun 05 '16 at 05:02

3 Answers3

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This is really up to convention.

In most cases it is convenient to disallow the empty set from being a model of anything. It helps simplifying a lot of statements, for example: "Every finite partial order has a maximal element" is false if we allow the empty set to be a model of the theory of partial orders; another example would be the inference rule $\forall x\varphi\rightarrow\exists x\varphi$ (which in turns implies that the empty set is never a model of a consistent theory, since $\forall x(x=x)\rightarrow\exists x(x=x)$, so in particular there is some $x$ in the universe).

On the other hand, in cases like ordinals and the likes, it is convenient to have the empty set as a partial order, as it simplifies a lot of things when dealing with ordinals. And certainly there are other examples of this nature elsewhere.

Asaf Karagila
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    +1. For further clarity, let me point out that the moment ambiguity creeps in is when you use the word "inconsistent" - with respect to what proof system? If the proof system is one which is sound for the semantics which allows empty structures, the answer to your question is "yes" - if the proof system is one which is complete for the semantics which does not allow empty structures (and this is by far the most common choice in modern mathematical logic), then the answer to your question is no. – Noah Schweber Jun 04 '16 at 04:19
  • I am thoroughly confused about the downvotes. Especially since the second downvote is really odd. Marco and Dan received downvotes around the same time, Daniel received an upvote and Dan received an upvote a few minutes later. – Asaf Karagila Jul 24 '16 at 07:36
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    I did not downvote this answer, although I disagree with it. I think you establish by example that a consistent theory isn't necessarily modeled by the empty set, but that isn't the question being asked. I don't know what other reasons some may have in downvoting, and I think there is nothing wrong with ignoring downvotes from users who can't be bothered to explain their reason. – DanielV Jul 24 '16 at 11:13
  • @Daniel: Of course, I didn't think that it was you who downvoted. I just remarked on the oddity of the situation. Especially since now all the posts on this page have a -2 and +whatever needs to get to the current score. What was weird is the two votes on Dan's post (I could understand a "lone gunman" voting on all the posts on this page, but two votes come across as odd, as this thread has been inactive for a while). – Asaf Karagila Jul 24 '16 at 11:15
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If an axiom/inference set has any model at all, then it is as consistent as model theory. That includes having a model with an empty universe.

This comes from the fact that if $A$ is consistent, and $A \vdash B$, then $B$ is consistent. Here $A$ is your possibly empty model + model theory, and $B$ is your axioms/inferences.

DanielV
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    This doesn't address the real issue, which is whether empty structures are allowed in the semantics of first-order logic; see Asaf's answer. – Noah Schweber Jun 04 '16 at 04:17
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    @NoahSchweber Sorry, not to be antagonistic but Asaf's answer is simply wrong. It is not a matter of convention. If a theory has a model then it is as consistent as model theory, whether your are dealing with a first order logic or with typed lambda theory or with combinators or anything. He cites theories that don't have empty models, which is completely irrelevant to the question. It is a question about logics that do have empty models. It is possible, not even hard (though not attractive) to create Hilbert and Fitch style inferences and axioms that hold in empty models. – DanielV Jun 04 '16 at 05:12
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    How can my question be wrong if you say that it is a matter of convention, and I begin by answer by saying that it is a matter of convention? Does the meaning of "wrong" is a matter of convention as well? I also said that this is about how your logic is formulated, and that in some cases it is convenient to allow models to be empty whereas in other cases it is less convenient. Can you perhaps point out exactly where I'm saying something wrong? – Asaf Karagila Jun 04 '16 at 12:11
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    I said it was not a matter of convention. You say ""Every finite partial order has a maximal element" is false if we allow the empty set to be a model of the theory of partial orders". There is no such thing as "allowing" the empty set to be a model or not. Either your logic $L_1$ has axioms preventing the empty set from being a model, or your logic $L_2$ does not have axioms preventing the empty model. Usually it is the former. In the case of $L_1$, it is vacuously true by hypothesis that "if the empty set models $L_1$ then it is consistent". – DanielV Jun 04 '16 at 16:40
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    In the case of $L_2$, since the logic has a model it is true that the logic is as consistent as model theory. It is not a matter of convention. Any logic which is satisfied by the empty structure is trivially consistent. It just happens that this is almost never the case, there is almost always some axiom that prevents there from being an empty model. Guessing, perhaps you are conflating the fact that FOL by convention uses axioms/inferences which are not satisfied by the empty model? – DanielV Jun 04 '16 at 16:41
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    I say it is trivial, because this is mostly just the transitive property of entailment at work: $(A \vdash B) \land (B \vdash C) \implies (A \vdash C)$. Here $A$ is model theory restricted to the empty model, $B$ is the logic in question, and $C$ is the false (or inconsistent) statement. Were it possible that the logic could entail an inconsistency, then the model theory itself would have to be inconsistent. Citing examples of theories without empty models does not mean that theories with empty models have ambiguous consistency. – DanielV Jun 04 '16 at 16:41
  • Wait, so this is not a matter of convention, but rather a matter of choice of logic. Right? And the choice of logic is the convention. So it is a matter of convention. – Asaf Karagila Jul 24 '16 at 07:25
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    @AsafKaragila The question "If a logic is modeled by the empty model, is it consistent" is not a matter of choice and it is not a matter of convention (although I will say, whether X models Y is a purposefully informal concept). If a logic is modeled by the empty model, then it is consistent. Since it is true of "all logics", then of course by universal instantiation you can "choose" the logic to which to apply this statement. The point of model theory is to outside of the idiosyncracies of any particular logic, and thus be applicable to all logics, from lambda calculus to FOL etc. – DanielV Jul 24 '16 at 11:10
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    Oh, give me a break. For some reason whenever I talk to my fellow logicians, I never have to mention whether or not the empty set is allowed to be a model because this follows from the context. Perhaps the meaning you are assigning to the word "convention" is very different than what I assign to it, and while I'm not a native speaker, I feel that at least the convention around here is that a convention is what it should be. So pretty please, with of sugar on top, explain to me why is it the case that I never had to tell my colleagues whether or not I am working with one logic or the other. – Asaf Karagila Jul 24 '16 at 11:13
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    I'm not trying to be hard on you....I think we mean the same thing by "convention". It is convention that the empty set is not allowed to be called a "model", to make it easier to say things like "Natural Deduction satisfies all models". From your posts, I take it you study set theory, right? I would guess that the reason you never have to specify which logic you use is that most set theories (ZFC especially) are written as axioms in a FOL, never in anything besides a FOL, and the choice of which FOL is inconsequential. I don't mind talking to you in chat. – DanielV Jul 24 '16 at 11:45
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    No chat is necessary. I'm just trying to figure out how my answer is "simply wrong", and how this is "not a matter of convention". Yes, you are correct, if you move away from FOL and its close relatives, it becomes increasingly more about explicit choices that you make. But it's not even about set theory, the majority of mathematicians (i.e., excluding intuitionists/constructivists/proof theorists/"hands on" logicians) don't specify their logical system anyway. And yes, because FOL can be formulated with and without empty structures, this becomes a matter of convention. – Asaf Karagila Jul 24 '16 at 12:43
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If $T$ is inconsistent then it proves any statement of first order logic and so it proves also $\exists x \exists y. \neg (x=y) $ so the empty set is not a model.

Marco Disce
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    The problem is that it is a common convention in first-order logic to disallow empty structures. If we fix a proof system $S$ for first-order logic which is sound and complete for this semantics, we lose soundness on empty domains. So the question of whether theories are deductively closed or not matters. For example, the theory $T = {\forall x, x\neq x}$ is inconsistent according $S$, but the empty structure is a model of $T$. On the other hand, if we replace $T$ with its deductive closure $T'$ (in $S$), $T'$ contains $\exists x, x = x$, so the empty structure is not a model of $T'$. – Alex Kruckman Jun 02 '16 at 16:23
  • @AlexKruckman - For sure, we have to modify the proof system to manage the empty structure; if suitable modified, we cannot prove $\exists x (x=x)$, and thus - according to the "usual" logic - having found a formula that is not provable, we have consistency, based on the def of inconsistency as : $\Gamma \vdash \bot$. My doubt is: in the modified proof system, have we to modified also the propositional rules, like ; $\bot \vdash \varphi$ ? I think not... – Mauro ALLEGRANZA Jun 02 '16 at 16:29
  • @MauroALLEGRANZA The propositional rules should be no problem, at least for sentences. I suppose there might be issues depending on how your system deals with formulas with free variables. This is probably a better discussion to have in the recent question I linked to above. My point in the comment to Marco's answer was just that if we're not clear about our conventions it might be possible for $\emptyset$ to satisfy all the sentences contained in $T$ while not satisfying all the sentences proved by $T$. – Alex Kruckman Jun 02 '16 at 17:00
  • @AlexKruckman - in the post you have linked above, there is the ref to Mendelson (6th ed) treatment of it; I've browsed it and it "looks good"... (sorry for being presumptuous). Of course, the quantifier axioms are modified and there is no more Gen rule. The modified Mendelson predicate calculus is proved complete. – Mauro ALLEGRANZA Jun 02 '16 at 17:06
  • @AlexKruckman would you say that "$T$ inconsistent $\implies$ $\emptyset$ is not a model" is false? Or that it is true but my argument is invalid? – Marco Disce Jun 02 '16 at 17:31
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    @Marco If our proof system is sound for our semantics, then the statement is obviously true: $T$ inconsistent means $T\vdash \bot$, so $T$ has no models, empty or otherwise (i.e. your argument is completely correct, though there's no need to use a sentence as complicated as $\exists x,\exists y, x\neq y$). But it seems likely that the OP is considering a "standard/conventional" proof system for first-order logic, which is sound for non-empty structures, but then is considering the empty structure anyway. At this level, things get weird, so it's important to be clear about conventions. – Alex Kruckman Jun 02 '16 at 19:53
  • @AlexKruckman So "if the proof system is sound for our semantics, then the statement is obviously true" and the argument holds. On the other hand if the proof system is the "standard/conventional" one for first-order logic the statement "$\forall T$, $T$ is inconsistent $\implies$ empty set is not a model of $T$" is still true but just vacuously true because in this case we have another definition of "model" that explicitly excludes the empty set. Right? – Marco Disce Jun 05 '16 at 06:32