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We all love proving a given structure models a given theory but I had a conundrum when needing to show that any non-empty set is a valid structure (in the empty signature). Of course the empty set doesn’t work but how could one prove that for any non-empty $X$, $X\models \varnothing$? My immediate thoughts are that any quantifier-free formula is equivalent to a Boolean combination of equalities of distinct variables but I’m not sure what to do from here—-if “here” is even the right place to start from.

Asaf Karagila
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Lave Cave
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1 Answers1

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$X\models \varnothing$ means: For all $\varphi\in\varnothing$, $X\models\varphi$. Since there are no $\varphi\in\varnothing$, this is true (vacuously).

That's all there is to it. This has nothing to do with whether the signature is empty or not, or quantifier-free formulas, etc. In any context, anything that counts as a structure will be a model for the empty theory, simply because the empty theory imposes no requirements on its models.

Alex Kruckman
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  • The issue for me is that we need to have satisfaction of basic logical sentences like $\exists x\phi$ which is why the empty set doesn’t work. Where I’m come from, all sets can be structures, but the empty one is then proven to not model any theories. I think your argument doesn’t work as such a semantics would mean that the empty structure models the empty theory which is false because there’s still a logic that needs to be satisfied. – Lave Cave Nov 03 '24 at 05:47
  • @LaveCave My argument is literally two lines. If you agree that the definition of $M\models T$ is "for all $\varphi\in T$, $M\models\varphi$", and if you understand vacuous truth, then what's the issue? – Alex Kruckman Nov 03 '24 at 12:00
  • The issue is that this argument would work for the empty structure, if it’s allowed by the semantics. Of course, even if it’s considered, it doesn’t model the empty theory as it fails to satisfy the tautology $\exists x x=x$. My question is about a proof that other sets don’t have similar pitfalls. – Lave Cave Nov 03 '24 at 12:06
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    @LaveCave Why do you think the sentence $\exists x, x=x$ has anything to do with the question of whether $X\models \varnothing$? The empty theory does not contain this sentence... Taking a step back, there are two conventions for semantics of first-order logic. One (the more common in the literature) disallows empty structures. Under this semantics, the statement $\emptyset\models \emptyset$ doesn't even make sense, because $\emptyset$ is not a structure. The other convention allows empty structures. Under this semantics, $\emptyset\models \emptyset$ is true, by the argument in my answer. – Alex Kruckman Nov 03 '24 at 12:12
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    The issue with empty structures is an issue of soundness: Standard proof systems allow us to prove $\exists x, x=x$, and this sentence is false in empty structures, so the soundness theorem would fail. The fix is to adjust the proof system to one that is sound and complete for a semantics including empty structures. See here or here for more discussion. But all of this is totally irrelevant to the question of whether $X\models \varnothing$, which is obviously true whenever $X$ is a structure. – Alex Kruckman Nov 03 '24 at 12:15
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    @LaveCave "there's still a logic that needs to be satisfied". No, this is getting things backwards. The relevant definition only refers to the definition of truth in a structure, which extends just fine to the empty structure should we decide to count it. The "logic" is just the set of statements that are true in every structure, which is different depending on whether "every structure" includes the empty structure. The empty structure doesn't "need" to satisfy the logic, it "will" satisfy it by definition if it is structure, and if it's not, as Alex says, the question is malformed. – spaceisdarkgreen Nov 03 '24 at 17:45