Hilbert-Bernays theorem

Question

As far as I understand, the following result is called the Hilbert-Bernays theorem (is that correct?):

If a (first-order) formal theory $T$ is consistent, then it has a model.

Can anybody give me a reference, preferrably with a proof of this fact?

EDIT. Gentlemen (and ladies), before you close this question:

is it possible that this special proposition has its own name, different from "the completeness theorem"?

C.Smorynski in "Handbook of Mathematical Logic" (edited by Jon Barwise) calls a similar statement "the Hilbert-Bernays theorem" (Theorem 6.1.1 in volume 4), that is why I used this name.

Can anybody explain me the difference between the result that Smorynski mentions (the "true Hilbert-Bernays theorem") and what I initially asked about? And if possible give me a reference where the Hilbert-Bernays theorem is proved.

https://mathoverflow.net/questions/271140/model-existence-theorem-in-topos-theory — Ivan Di Liberti, Mar 25 '18 at 07:06
@IvanDiLiberti does this thread contain a reference for what need? — Sergei Akbarov, Mar 25 '18 at 07:28
This is Godel's Completeness Theorem. No, it's not Hilbert-Bernays. Hint" www.google.com — David C. Ullrich, Mar 25 '18 at 15:52
Essentially every introductory logic text has a proof of this fact - it is one of the fundamental theorems of first-order logic. — Carl Mummert, Mar 25 '18 at 22:16
The theorem stated in the question is not Theorem 6.1.1 of Smorynski, which has both mush stronger assumptions and a much stronger conclusion. That theorem is essentially an arithmeticized version of Gödel's completeness theorem into PA. — Carl Mummert, Mar 26 '18 at 11:04
I think that Smorynski has just made up a name for a theorem to honor Hilbert and Bernays. The theorem he states is not known by that name in general, it is just the formalized version of Gödel's completeness theorem. The difference, in any case, is that the general completeness theorem is proven in an (unspecified) strong metatheory such as ZFC, while Smorynski is stating a formalized version provable in PA. The proof is given by Smorynski: simply formalize Henkin's proof in PA. — Carl Mummert, Apr 01 '18 at 00:34
I don't think there would be any issue with you posting another question specifically about the theorem stated by Smorynski. It would be ideal, in any such question, for you to include a full background of the sources you are looking at and the specific question you have about them. I think that the primary issue in this question was that you posted about a very well known theorem that Smorynski takes as standard background knowledge for anyone reading his paper. — Carl Mummert, Apr 01 '18 at 00:37
@Carl, thank you! It seems to me I found a proper solution for how to formulate this: https://mathoverflow.net/questions/282975/a-formalistic-variant-of-the-g%C3%B6del-completeness-theorem — Sergei Akbarov, Apr 01 '18 at 14:42
@CarlMummert I don't have a copy of the Hilbert-Bernays book handy, so I can't check this, but I believe that book is the first place where the arithmetized completeness theorem was explicitly stated and proved. If so, Smorynski's calling it the Hilert-Bernays theorem is justified. (The Hilbert-Bernays proof may have been more complicated than "just formalize Henkin's proof", since Henkin came along some time after Hilbert and Bernays.) — Andreas Blass, Jul 17 '18 at 16:17
Thanks, @Andreas. I am not familiar with the Hilbert-Bernard text - I should check that. As you said, their proof would not use Henkin techniques. — Carl Mummert, Jul 18 '18 at 19:15

Peter Smith · Answer 1 · 2018-03-25T22:01:23.587

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It is not in general true [as the question assumed in its original form] that if a formal theory $T$ is consistent, then it has a model. (For example, take second-order Peano Arithmetic extended with a constant $c$ governed by the axioms $c \neq 0$, $c \neq S0$, $c \neq S00$, etc. This is consistent but has no model -- as nicely explained in this answer by Henning Makholm)
It is a theorem that if a first-order theory is consistent, then it has a model. This is a version of the completeness theorem for first-order logic that, famously, is due ultimately to Kurt Gödel.
Any standard math logic text will prove this: but there's a good version in Leary/Kristiansen, A Friendly Introduction to Mathematical Logic, which is freely available through the friendly generosity of the authors.

edited Mar 25 '18 at 22:01

answered Mar 25 '18 at 08:27

Peter Smith

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Yes, I had in mind the first order theories, of course. – Sergei Akbarov Mar 25 '18 at 08:33
Yes, this seems to be a well-written book... – Sergei Akbarov Mar 25 '18 at 08:46
To emphasize point 1 differently - I think it is often misunderstood. Second-order Peano arithmetic is consistent with that axiom in the syntactic sense, and so it does have a model, just not a model from the specific collection of full models that someone might like to accept for full second-order logic. We could make first-order logic incomplete in the same way if we only allowed finite models. So any theorem of this kind links "provability of a certain kind" with "model of a certain kind" relative to a "language of a certain kind". All three of the quoted phrases are important. – Carl Mummert Mar 25 '18 at 22:16
Related to this, there is a question: what are some broad notions of "formal theory" for which it is possible to define a notion of "model" so that every consistent theory in this notion has a model in that notion? Certainly Henkin's proof generalize to any system that looks much like Predicate Logic, including higher order logic and some kinds of type theories. Do you know of a general treatment of "formal systems" with this kind of result? I have sometimes wondered if they exist. – Carl Mummert Mar 25 '18 at 22:19
Is it obvious that second-order Peano Arithmetic is consistent? – Pierre-Yves Gaillard Mar 26 '18 at 15:20
@PeterSmith - Thanks! I'd be most grateful if you could you give me a reference for the fact that second-order PA is consistent relative to Zermelo set theory. (Please insert @Pierre-YvesGaillard if you'd like me to be notified.) – Pierre-Yves Gaillard Mar 26 '18 at 17:25
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@Pierre-Yves Gaillard: the consistency of second order PA is provable in ZFC by just confirming that $(\omega, P(\omega))$ is a model. – Carl Mummert Apr 01 '18 at 00:30

score 1 · Answer 2 · answered Mar 25 '18 at 08:11

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This is called the "completeness theorem". P. T. Johnstone's "Notes on logic and set theory" proves this in Chapter 3. I liked this book for having full details while still being quite short.

answered Mar 25 '18 at 08:11

Dap

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Dap, thank you, I'll look at this. The first impression is, however, that this book is written a bit carelessly... – Sergei Akbarov Mar 25 '18 at 08:54

score 0 · Answer 3 · edited May 09 '22 at 08:41

We can see S.C.Kleene, Introduction to Metamathematics (1952), §72. Gödel's completeness theorem (page 389-on) :

Theorem 34C. If a predicate letter formula $F$ is irrefutable (i.e. if $¬F$ is unprovable) in the predicate calculus, then $F$ is satisfiable in the domain of the natural numbers. (Gödel’s completeness theorem for the predicate calculus, 1930.)

And see page 395 :

Theorem 36. The addition to the postulate list for the predicate calculus of an unprovable predicate letter formula $G$ for use as an axiom schema would cause the number-theoretic system as based on the predicate calculus and Postulate Group B (§19 [first-order Peano's axioms]) to become $ω$-inconsistent. (In fact, a certain formula would become refutable which expresses a true proposition of the form $(y)D(y)$ where $D(y)$ is an effectively decidable predicate.) (Hilbert-Bernays completeness theorem, 1939.)

The "usual" modern formulation : "every consistent f-o theory has a model", is later; the source is probably A.Tarski and his US school.

See Leon Henkin, The Completeness of the First-Order Functional Calculus, JSL (1949), [doi:10.2307/2267044}(https://doi.org/10.2307/2267044), jstor :

THEOREM. If $\Lambda$ is a set of formulas in which no member has any occurrence of a free individual variable, and if $\Lambda$ is consistent, then $\Lambda$ is simultaneously satisfiable in a domain of individuals having the same cardinal number as the set of primitive symbols of [the system].

See Henkin's paper (1st page) : "Although several proofs have been published showing the completeness of the propositional calculus (cf. Quine), for the first-order functional calculus only the original completeness proof of Gödel (1930) and a variant due to Hilbert and Bernays [Grundlagen der Mathematik, vol II, 1939] have appeared."

For sure, we can find it into Abraham Robinson, Introduction to Model Theory and the Metamathematics of Algebra (1963, 1st ed.1951 as On the Metamathematics of Algebra), page 12 :

THEOREM (EXTENDED COMPLETENESS THEOREM OF THE LOWER PREDICATE CALCULUS). Every consistent set of sentences $K$ in a language $L$ [the lower predicate calculus] possesses a model.

Hilbert-Bernays theorem

3 Answers3