Disclaimer : the question is very vague.

Today as I was thinking about the construction of a model for a set of formulas in Gödel's completeness theorem and I sort of felt that the construction is 'canonical' in some sense. I was thinking that is there any result connecting Gödel's model (call it G) to any other model of a set of formulas. Maybe something like Gödel's model can be embedded into/mapped to any other model such that the image satisfies the set formulas on their own.

My intuition is if we consider only canonical models, i.e. models where there is a corresponding closed term for each element in the model, then for any canonical model C there is a map f : G -> C. I think this is true for the Herbrand model for zeroth order logic.

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    $\begingroup$ "the question is very vague" - I agree. $\endgroup$
    – Peter
    Jan 21, 2020 at 18:09
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    $\begingroup$ I strongly disagree with the vote to close - I think this is a question which is by its nature difficult to articulate, and is clear enough to admit a good answer. I've upvoted and voted to reopen. $\endgroup$ Jan 21, 2020 at 19:55
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    $\begingroup$ To the OP, note that we need to fix an ordering on the formulas in our language to start with. So that's a non-canonical feature right there. $\endgroup$ Jan 21, 2020 at 19:56
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    $\begingroup$ In spite of the arbitrariness of the order you start with, there's surely something canonical about the models Henkin constructs, and so I'm curious to hear what a model theorist might have to say about in what sense such models are special. Seems like a good question to me, even if it turns out that there's nothing special about such models after all. $\endgroup$ Jan 22, 2020 at 1:37

1 Answer 1


Henkin models (Godel's original proof was quite different) do indeed satisfy a kind of "minimality" condition, but there are subtleties.

A good further source is the (sadly hard to find) book Henkin-Keisler models.

For any theory $T$, the set of closed terms in the language of $T$ modulo $T$-provable equality yields a structure in the language of $T$ in a natural way which I'll call "$Term(T)$" (ignoring for the moment the issue that there may be no closed terms at all; if this bothers you, either work in the version of FOL which allows empty structures or assume our language contains at least one constant symbol). In general, $Term(T)$ need not satisfy $T$; however, if $T$ is complete and has the strong witness property, we will have $Term(T)\models T$. In this situation we do indeed have a very nice "minimality" property:

Suppose $T$ is complete and has the witness property, and $M\models T$. Then there is a unique elementary embedding of $Term(T)$ into $M$, and the image of this embedding is exactly the set of definable elements of $M$.

Note that this is stronger than merely saying that $Term(T)$ is the prime model of $T$.

This says that "Henkin models" do enjoy a kind of minimality. However, there are subtleties here:

  • The above ignores the part of the proof where we pass from our initial consistent theory $T_0$ to a theory $T\supseteq T_0$ (in a possibly larger language) which is complete, consistent, and has the strong witness property. There are many ways to do this, essentially depending on how we order the formulas in our language, and these different approaches generally yield different theories (and hence different models at the end of the day).

  • Moreover, even fixing a single appropriate $T\supseteq T_0$, we still have the issue that the model of $T_0$ we produce at the end of the day isn't $Term(T)$ but rather the reduct of $Term(T)$ to the original language of $T_0$, and this restriction of language will in general kill the minimality observation above.

Let me end by mentioning a follow-up to the first bulletpoint above which I haven't seen treated before. Roughly speaking, given our starting theory $T_0$ we get an equivalence relation on orderings of formulas given by "yield the same model" (there are actually a couple inequivalent approaches here, but ignore that for now). Intuitively, there is a connection between how coarse this relation is and how close to being complete and having the witness property $T$ is (in particular, if we start with a theory which already is complete and has the witness property, then the way we order formulas shouldn't impact anything since there's nothing we need to change). We have various tools for treating that sort of thing (e.g. descriptive set theory), so in principle there could be some interesting dividing lines to be drawn here. However, I haven't seen anything along these lines, and so I don't know if there's actually anything here.

  • $\begingroup$ What is the witness property? $\endgroup$
    – Arka Ghosh
    Jan 28, 2020 at 18:09
  • $\begingroup$ @ArkaGhosh One way to phrase it is the following: for every formula of the form $\exists x_1,...,x_n\theta(x_1,...,x_n)$, there are constant symbols $c_1,...,c_n$ such that the theory prove $\theta(c_1,...,c_n)$. We could replace "constant symbol" with "closed term," or alternatively talk about explicit Skolem functions for all appropriate formulas, and get the same ultimate picture. $\endgroup$ Jan 28, 2020 at 18:11
  • $\begingroup$ Aha! That is the exact condition I thought when I was trying to prove that there is a mapping from the Henkin model. But I thought that it was too strict. $\endgroup$
    – Arka Ghosh
    Jan 28, 2020 at 18:14
  • $\begingroup$ @ArkaGhosh Nope, it's true (as long as the theory we use is "good enough" of course)! $\endgroup$ Jan 28, 2020 at 18:15
  • $\begingroup$ By too strict I meant that it kind of guarantees the mapping. $\endgroup$
    – Arka Ghosh
    Jan 28, 2020 at 18:17

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