Constructiveness of Proof of Gödel's Completeness Theorem

Question

As a mathematician interested in novel applications I am trying to gain a deeper understanding of (the non-constructiveness of) Gödel's Completeness Theorem and have recently studying two texts: Mathematical Logic for Mathematicians (by Y.Manin) and the book on Reverse Mathematics by Simpson [2009]. Having read Mendelsohn and Crossley (and other works) I am actually surprised by the non-constructivism in this proof, due to the fact that many classic logic texts do not emphasis it.

However Manin emphasises that the proofs (and he presents 4 proofs for completeness/compactness) are all via the Axiom of Choice.

Meanwhile in the Reverse Mathematics work we learn that the non-constructiveness is identified as the "Weak Konig's Lemma" (WKL) assumption. I see from reviewing the online summaries of Gödel's original proof that Gödel used Konig's Lemma itself. Konig's Lemma lies in a higher "non-constructivity class" ACA in Reverse Mathematics, than does WKL.

Now the problem I have with this non-constructiveness is that both Axiom of Choice and WKL are not necessary truths: we can (I think) have a mathematical universe in which they are negated. However I see that WKL is provable in ZF. Hence Gödel's Completeness theorem is not a necessary truth either - but to be a logic theorem it needs to be a necessary truth, not conditional on another assumption surely?

In this paper Gödel ends his dissertation with :

"essential use is made of the principle of the excluded middle for inﬁnite collections... It might perhaps appear that this would invalidate the entire completeness proof."

In summary has this invalidation has not been suggested in recent years by the recognition of the following facts not known to Gödel or in 1930 ?:

Konig's Lemma is a consequence of the Axiom of Choice
The proof can be reduced to Weak Konig's Lemma, but no further
Axiom of Choice is not a necessary logical truth (merely a mathematical convenience)
The theorems of Turing Computability theory (which play a role in understanding the Lindenbaum Lemma non-constructivity from another perspective)

or does it all depend on whether WKL (as opposed to AC) actually is classically refutable?

EDIT: I have a linked question which contains discussion on a related topic.

I shall add a summary of some of the comments from below, to make the question more self-contained. Also I hope that it will make answering it easier if I add some more background.

In this question I am viewing Logic and Mathematics as separate, but overlapping topics: there are aspects of Logic not relevant to Mathematics and vice versa. This question is primarily about its title: the Constructiveness (or otherwise) of Gödel's Completeness Theorem (GC) and the consequences of that. Mathematical topics and mathematical philosophy are relevant only if they are directly connected, as I see this as a question about Logic.

Historically there is a particular timeline, which contains some unexpected twists:

Gödel (1929) Dissertation (Unpublished) --- Contains qualms about non-constructiveness in GC

Gödel (1930) GC paper Published --- All qualms removed, giving initial appearance of constructiveness. Reinvents Konig's Lemma.

Kleene (1952) Intro. to MetaMathematics --- Proves GC non-constructively using Gödel-Henkin proof. This uses "Lindenbaum Lemma". Claims that AC is not required in his proof.

Later texts (e.g. Boolos-Jeffrey) --- No reference to non-constructiveness in GC

Manin "Logic for Mathematicians" --- "we shall have to use Zorn's Lemma"

Reverse Mathematics --- Weak Konig Lemma for GC

So which of all these is correct? If an Answer were to claim that GC was constructive, then the Answerer will have to justify that answer. If an Answer were to claim that GC was non-constructive then we have at least two distinct possibilities: WKL or Zorn. Furthermore the latter case re-opens Gödel's original non-Constructiveness qualms. Do these qualms have any validity - for example is there a sense in which it is not "universally valid"?

Note also that GC is exceptional amongst Gödel's work in that he generally worked within constructive frameworks as with his Incompleteness Theorem, etc.

I have now studied another proof of GC by Hilbert-Bernays(1939). This theorem only applies in the Arithmetic domain (not Sets) and is again non-constructive. The "price" that has been paid for the non-constructivity is that the system will be $\omega-$inconsistent (rather than incomplete- since it is a completeness theorem). Summarising: a "true" statement (for all numbers) can be falsified by a proof of its negation.

Now $\omega$-inconsistency does not exist for general domains, so is the non-constructive proof still considered entirely valid? Put simply: is there a new type of Gödel Incompleteness?

"Logical truth" here is shorthand for the fact that Cohen independence makes AC not a necessary consequence of ZF. For a statement which is a consequence of ZF (like WKL) the question is whether ZF is logically necessary (at least WKL). In each case can one construct a counterexample? IE False = a counter-example exists. — Roy Simpson, Feb 02 '14 at 23:25
So "truth" means a consequence of $\sf ZF$? Are you trying to find out consistent counterexamples for the completeness theorem in $\sf ZF$ or...? — Asaf Karagila, Feb 02 '14 at 23:26
The answer of Carl and Asaf give all the details about mathematical content of the Theorem : it "speaks of" all models, included the infinite ones; so we need a tool to handle with them ($ZFC$). About your consideration regarding logical or necessary truth, in spite of recent effort to "revamp" logicism - see (Logicism and Neo-Logicism)[http://plato.stanford.edu/entries/logicism/] - it is hard to believe that a "reasonable" interpretation of logical truth can support all the complex "mathematical building". 1/2 — Mauro ALLEGRANZA, Feb 03 '14 at 07:46
The original conception of the founding-fathers of Logicism (Frege and Russell) was that logic is the "most general" science; its truths are "general" because they are "topic-neutral", i.e. they are applicable to everything. This conception ruined against the difficulty of considering some axiom (like AC or Reducibility Axiom) "logical truth". Today, at least in math, the more likely candidate to a "theory of everything" is $ZFC$, and it still need "logically suspect" axioms, i.e.axioms that is not easy at all to justify a priori. 2/2 — Mauro ALLEGRANZA, Feb 03 '14 at 07:51
Mauro, the problem with ZFC from the perspective of this question is that we have (Cohen) Independence- non-well ordered sets exist. A mathematical theorem might well be justified in its use of ZFC all the same. However every other logic theorem at this level (soundness, finite propositional logic completeness, etc) do not need ZFC let alone AC. (Although admittedly sets are required as models). The issue for me is the "extralogical" nature of AC - and thereby the possibility of some kind of counter-example- rather than its "non-constructive" nature in a basic logic theorem. — Roy Simpson, Feb 03 '14 at 10:06
The mere possibility of a counterexample means that the Completeness theorem is not universally valid - and I would like to understand that better. Also I am not claiming Logicism - I am not claiming that mathematics is logic: the nearest I am claiming is that Logic is not Mathematics. — Roy Simpson, Feb 03 '14 at 10:09
@Roy Simpson: of course the completeness theorem is not universally valid. As I pointed out, essentially no theorem of interest - in logic or in mathematics - is universally valid. The axiom of choice is a (tempting) red herring in that respect. — Carl Mummert, Feb 03 '14 at 10:45
@Carl Mummert: what is not universally valid about finite propositional logic completeness; or the FOL soundness theorem? Are there counter-examples to these? — Roy Simpson, Feb 03 '14 at 11:02
Have you examined the proof of these theorems - can you name a formal system in which they are provable? Non-logical axioms are required. For example, completeness even for finite theories in propositional logic require non-logical axioms to manipulate truth assignments and sets of formulas, and non-logical induction axioms. And of course this is just a special case of the actual completeness theorem for propositional logic. Similarly the soundness theorem for a deductive system will require non-logical induction axioms. By definition, every non-logical axiom has counterexamples. @Roy Simpson — Carl Mummert, Feb 03 '14 at 11:11
Relevant: http://mathoverflow.net/questions/52215/soundness-theorem-in-reverse-mathematics — Carl Mummert, Feb 03 '14 at 11:15
@Roy and Carl - my reference to Logicism is aimed at the discussion about the (logical) nature of axioms. As Carl says, you can do quite nothing without induction, This was the objection of Poincaré to Russell's logicism: induction (according to P) was not reducible to logical principles, but it is unavoidable in mathematics, and so (I say) also in metamathematics that is mathematics. So, in the end, Godel's Completeness Theorem is mathematics. — Mauro ALLEGRANZA, Feb 03 '14 at 11:47
@CarlMummert: I admit that I am influenced by the Reverse Mathematics Philosophy - that RCA is basic - that recursion is basic. The Soundness theorem is an RCA theorem - as stated in the first para of the linked answer. Non-RCA theorems ought to have counter-examples? Yes-No? — Roy Simpson, Feb 03 '14 at 11:50
@MauroALLEGRANZA: Induction appears in the basic RCA model of Reverse Mathematics - so yes it is necessary. But there are different "strengths" of Induction: Peano Induction is related to the ACA level and is thus is more non-constructive than Godel's Theorem. But we know that there are issues with Peano Induction and its association with non-standard models. So are there "non-standard models" wrt Godel Completeness too? See Asaf's construction of an Amorphous Set model (below) for the kind of thing I am asking. — Roy Simpson, Feb 03 '14 at 13:01
@Willemien: Where is the incompleteness part? Recall that completeness and incompleteness are essentially unrelated theorems. — Asaf Karagila, Feb 03 '14 at 16:26
@AsafKaragila not sure anymore, you may remove the incompleteness tag, thought at some time I thought they were related — Willemien, Feb 03 '14 at 21:23
Roy, I honestly don't see what the issue here. Some people treat the completeness theorem as a general statement for any language, and this needs (over $\sf ZF$) the ultrafilter lemma to hold; whereas some places (e.g. reverse mathematics) treat it as a theorem about countable languages, where only $\sf WKL$ is required. — Asaf Karagila, Feb 04 '14 at 16:49
(Also please don't throw the incompleteness into the mix. You have a concrete question about the completeness theorem. Figure it out first, then perhaps try to move on to the incompleteness theorem. One question per thread, or at least one topic per thread, is important.) — Asaf Karagila, Feb 04 '14 at 16:53
@AsafKaragila"Incompleteness" - I agree that getting the answers I want, and not a lot of irrelevant discussion is not easy - and I am taking a risk introducing that word in this Question. I will remove it if it looks like it is causing irrelevant discussions... — Roy Simpson, Feb 04 '14 at 17:00
Roy, to quote Hakugare, "It is bad when one thing becomes two." — Asaf Karagila, Feb 04 '14 at 17:03
@RoySimpson & Asaf - I think that there is a connection ... In Godel's paper, the first paragraphs are devoted to describe the "completeness issue" : to prove that (I'm using only my memory) axioms and rules of the "restricted functional calculus" extracted from Principia are adequate to derive all (my word) "distinguished" formulas: the valid ones (and Godel point explicitly at the result already proved for the propositional fragment by Bernays (but also Post). So, I think, his subsequent Incompleteness result can be "read" in the same scenario : are axioms and rules of... 1/2 — Mauro ALLEGRANZA, Feb 05 '14 at 08:46
..formalized arithmetic adequate to derive all arithmetical "distinguished" formulas, where the (not little) difference is that now they are the true (in the intended model) ones. So, the misunderstandig made by students about Compl and Incompl is not (I think) due only to the "unlucky" circumstance that the two theor were discovered by the same guy... I would like to post a question about the source of the validity concept; I know that this site is not the proper place for "historical" research, but Roy's question points to an issue I've already try (unsuccessfully) to address. 2/2 — Mauro ALLEGRANZA, Feb 05 '14 at 08:52

score 19 · Answer 1 · answered Feb 02 '14 at 23:48

19

Here is a partial answer.

Godel's theorem connects logic and set theory. Syntax is the part of the logic, it's where the formulas and proofs live; set theory is the part of the semantics, where the interpretations and models live. Of course one can have them relocated to other contexts, but classically I think that it's a theorem about logic and set theory.

It is provable without the axiom of choice that if we have an infinite tree whose nodes are the natural numbers, and every level of the tree is finite, then there is a branch in that tree. If we omit the requirement that the nodes of the tree are the natural numbers, it won't be true anymore and counterexamples do exist (by which I mean, consistent with $\sf ZF$, of course).

Similarly, given a countable language we write the proof of the completeness theorem for that language, and all our uses of the axiom of choice will be mitigated by the fact that we have a uniform enumeration of all the objects of interest. Therefore some fragment of the completeness theorem is in fact provable without any appeal to the axiom of choice.

On the other hand, the completeness theorem is not about countable languages. It is about any language of first-order logic. So we can have a lot of function symbols, or constants, or so on. So what about the above proof? Well, it can be shown that if the language is well-orderable, then indeed we can repeat the same arguments and everything is well-orderable uniformly and everything is fine.

But alas, without the axiom of choice, not every set can be well-ordered. And that where the non-constructive nature of the proof sneaks in. Let me give a particular counterexample.

Example:

We say that a set $A$ is amorphous if it is infinite, and cannot be written as a disjoint union of two infinite sets. One can show that if $A$ is an amorphous set, then there is no way to linearly order $A$. That is, there is no $R\subseteq A\times A$ such that $(A,R)$ is a linearly ordered set.

Suppose that $A$ is an amorphous set (and it is consistent that such set exists), and let $\mathcal L$ be the language in which $<$ is a binary relation symbol, and for every $a\in A$ there is a constant symbol $c_a$.

We write the axioms, $\varphi$ is the conjuction of the statements $<$ is a linear order, and for every distinct $a,b\in A$ we write $\varphi_{a,b}:=c_a\neq c_b$. This theory is consistent, because if it were to prove a contradiction there would be a finite number of $a\in A$ whose constants appear in the proof, but then we can construct a finite linear order which will be an interpretation of these axioms. This is impossible of course.

However, the theory as a whole, despite being consistent, does not have a model. Because in any model of the theory, reducing just to the constants will have defined a linear order of $A$, and $A$ itself is an amorphous set, so it cannot be linearly ordered.

So where is the axiom of choice coming into the play? When we construct models we need to make some arbitrary choices along the way. When the language is well-ordered then we can delegate these choices to the well-ordered and uniformly make them all; but when the language is an arbitrary set which may not be well-orderable to begin with, then the axiom of choice is needed.

In fact, one can prove that the completeness theorem (and its equivalent, the compactness theorem) is equivalent to the ultrafilter lemma, stating that every filter can be extended to an ultrafilter. These all are also equivalent to many other weak choice principles such as Tychonoff theorem for Hausdorff spaces, or the Boolean Prime Ideal theorem.

So what is missing from this answer? Well, I don't know about contexts where the WKL is not provable, so I cannot help in that aspect. But hopefully this answer shed some light on where the theorem is using fragments of choice and non-constructiveness.

answered Feb 02 '14 at 23:48

Asaf Karagila

393,674

Thanks for this answer. I had not encountered Amorphous Sets before - which WP claims are Dedekind-finite. There is a related question, though on "Countable Languages" which I might ask. My only question on this Answer is whether it is consistent to have the negation of AC in one place and use AC somewhere else in such an overall modeling building construction? I appreciate that the corresponding WKL question remains open - that is the one which would give a full counter-example to Godel Completeness if it can exist. – Roy Simpson Feb 03 '14 at 12:52
Roy, one doesn't usually encounter amorphous sets. As for your question, it's unclear to me. It is consistent that some fragments of the axiom of choice fails, but other fragments hold. If that's what you're asking. (Strongly amorphous sets, however, are generally strongly counterexamples for many choice principles...) – Asaf Karagila Feb 03 '14 at 16:18
As far as AC fragments are concerned do amorphous sets live in the space "WKL + not AC" ... or "not WKL + not AC". If it is the former then do we have sets (or other constructs) in "not WKL"? I have asked my related question, which gives another angle on this entire issue. – Roy Simpson Feb 03 '14 at 16:24
Well, since WKL is provable from ZF, and the whole context here is ZF, then amorphous sets live certain models of WKL+not AC. If you want something which cannot prove WKL you are going to need to weaken your system by far more. – Asaf Karagila Feb 03 '14 at 16:26
ZF-I ZF minus the Infinity axiom might do the trick. – Roy Simpson Feb 04 '14 at 14:45
Roy, perhaps, but in this system you can't really talk about infinite sets anymore. Moreover if we replace the Infinity axiom by its negation, we get something bi-interpretable with $\sf PA$. – Asaf Karagila Feb 04 '14 at 15:43
I dont mean "add negation I" - just leave out I. Then ZF-I will have infinite models, even although we cannot prove the existence of one. One such infinite "set" (e.g. graph) could then fail WKL. Remember the purpose here is to see whether we can find a Set-like counter-example to GC. – Roy Simpson Feb 04 '14 at 15:50
@Roy: That's not quite true. In models with infinite sets, the theory is actually $\sf ZF$ so $\sf WKL$ holds; in models without infinite sets we have that $\sf WKL$ holds vacuously (there are no infinite trees). Therefore $\sf WKL$ is provable. – Asaf Karagila Feb 04 '14 at 15:57
"in models with infinite sets...theory ZF" - Doesnt the inverse relation between theories and models apply in set theory? That is more axioms less models, less axioms more models including all models seen earlier? – Roy Simpson Feb 04 '14 at 16:42
@Roy, there are more models, but the new models satisfy $\sf WKL$. Simply because $\sf WKL$ says "Every infinite tree bla bla bla has a branch" where infinite means internally infinite. The new models have no internally-infinite sets. So the statement is vacuously true. – Asaf Karagila Feb 04 '14 at 16:43
Although it is a side-issue I dont see why ZF-I (as I defined it 4 comments ago) has no "internally infinite" sets. Perhaps I am misunderstanding the way set theory models work. Certainly all internal sets must obey the axioms, but there will be other sets which have additional properties - such as well ordered sets, infinite sets, etc? – Roy Simpson Feb 04 '14 at 17:33
Roy, given a model of $\sf ZF-I$ there are two options. Either it satisfies the axiom of Infinity, in which case it is a model of $\sf ZF$. Or it satisfies the negation of the axiom of Infinity, in which case it has no "internally infinite" set. Since $\sf ZF$ proves $\sf WKL$, and models without internally infinite sets satisfy it vacuously, it follows that every model of $\sf ZF-I$ satisfies $\sf WKL$. – Asaf Karagila Feb 04 '14 at 17:35
If I understand this then we would need to remove another axiom from ZF giving ZFm say. Then maybe ZFm does not prove WKL So what ZF axiom is really responsible for WKL? I could add this as a new Question if you wish.... Is the Power Set axiom implicated in any way? – Roy Simpson Feb 04 '14 at 17:44
That might be a good question, but you might want to pace yourself a little bit. Give the current questions time to sit, people to comment. Generally it seems that the answer lies in sitting through the proof of $\sf WKL$ and checking which axioms we're using. My guess, however, would be that the power set axiom is responsible for the whole thing. It's quite an axiom, you know! – Asaf Karagila Feb 04 '14 at 17:54
OK I will wait patiently for things to happen. In the meantime I need to read a paper on introduction to model theory for ZF... – Roy Simpson Feb 04 '14 at 17:57
Any modern set theory book should suffice (Jech, Kunen, probably a couple others). But it's not an easy subject to get into. – Asaf Karagila Feb 04 '14 at 18:05
Note that WKL ("Every infinite subtree of $2^{<\omega}$ has an infinite branch") is provable in weak subsystems of second-order arithmetic such as $\mathsf{ACA}_0$. The key idea in the proof is that you replace the tree $T$ with the tree $T^{ext} = { \sigma \in T : \sigma \text{ has infinitely many extensions in } T}$. Certainly the root of $T$ is in $T^{ext}$; you can prove by induction that $T^{ext}$ is infinite, and since $T^{ext}$ has no dead ends it's trivial to construct a path. One real source of nonconstructivity (among others) is that membership in $T^{ext}$ is often not decidable. – Carl Mummert Feb 04 '14 at 20:18
@Carl: Maybe you want to ping Roy with that comment too? – Asaf Karagila Feb 04 '14 at 20:20
@Asaf: Yes, thanks, but I was short on space. The upshot is that power set is not used in any serious way – Carl Mummert Feb 04 '14 at 20:21
@Roy Simpson: the main sources of nonconstructivity in WKL are in using classical logic to form the original tree, and then in using nonconstructive set existence axioms to form the subtree of extendible nodes; see my comment above. – Carl Mummert Feb 04 '14 at 20:22
@Carl: It seems to me that one should perhaps point out that in the context of systems of arithmetic, this is equivalent as proving $\sf WKL$ over models of $\sf ZF-Inf+\lnot Inf$ for class-trees (i.e. definable trees, which are infinite, but not "internal sets"). And I was talking about the context of $\sf WKL$ as a theorem in $\sf ZF$ where one is talking about sets, rather than classes. – Asaf Karagila Feb 04 '14 at 20:23
@Asaf: Were you talking about WKL as "Every infinite subtree of $2^{<\omega}$ has an infinite branch"? – Carl Mummert Feb 04 '14 at 20:24
@Carl: Well, yes, but when you replace $\sf Inf$ by its negation, this suddenly becomes a schema rather than a single theorem. And in the context of $\sf ZF$, this is a single theorem. And it can be phrased as "Every countable set $T$ and an order $<$ such that $(T,<)$ is a tree of height $\omega$ without terminal nodes, has a branch". Being a countable set can be phrased as "Infinite, and every infinite subset is equipotent to the whole set", and so $\sf WKL$ can be phrased as a single theorem without referring to $\omega$, and that statement is vacuously provable from the above theory. – Asaf Karagila Feb 04 '14 at 20:32

score 12 · Answer 2 · edited Feb 04 '14 at 19:04

12

If the axiom of choice is not a "necessary logical truth", but only a "mathematical convenience", the same can be said for every axiom of ZFC. None of them is a logical truth, after all. Even the axiom of induction for natural numbers (in Peano arithmetic) is not a logical truth.

In short, almost no interesting theorem of mathematics or logic is a logical truth, unless it is stated in the form $H \to C$, where $H$ is the conjunction of all the axioms needed to prove some theorem $C$. Usually, we just state "$C$" as a theorem, rather than stating "If these axioms hold, so does $C$".

Avigad explains in his paper (linked in the question) what Gödel means by "invalidate". Avigad writes,

I wish to highlight Gödel’s remarkable sensitivity to the question as to what metamathematical methods are necessary to obtain the requisite results, and the impact these methods have on the epistemological consequences

The key word there is "epistemological consequences". The use of the axiom of choice to prove the completeness theorem in ZFC, or the use of weak König's lemma to prove it in second-order arithmetic, in no way diminishes the mathematical correctness of the theorem. But the study of what axioms are needed to prove a particular theorem does influence how we interpret the theorem.

Finally, let me emphasize that the axiom of choice is not very tightly related to constructivism in mathematics. There are fully constructive systems that include the axiom of choice, and fully nonconstructive systems (such as ZF set theory) that do not. So it is often a red herring to focus on whether a proof uses the axiom of choice, because many classical proofs are already nonconstructive long before they try to use that axiom.

edited Feb 04 '14 at 19:04

answered Feb 03 '14 at 00:26

Carl Mummert

81,604

What would be an example of a fully constructive system that includes AC? – Michael Greinecker Feb 03 '14 at 00:32
2

@Michael Greinecker: Here are two. Some forms of Martin-Löf intuitionistic type theory actually prove the relevant form of the axiom of choice as a theorem - and these systems have inscrutable "constructive" bona fides. Some theories of second-order or higher-order intuitionistic arithmetic include the relevant form of the axiom of choice as an axiom scheme. (In contrast, Diaconescu's theorem shows that the axiom of choice implies classical logic in many constructive set theories.) – Carl Mummert Feb 03 '14 at 00:43
Thank you, that are pointers I should be able to work with. – Michael Greinecker Feb 03 '14 at 00:47
1

@Michael Greinecker: The following article by Per Martin-Löf has a nice discussion of how this phenomenon occurs: "100 years of Zermelo’s axiom of choice: what was the problem with it?" at http://www.math.kth.se/~kurlberg/colloquium/2005/MartinLooef.pdf – Carl Mummert Feb 03 '14 at 00:47
@CarlMummert: I find Asaf's answer closer to what I require. I shall review your answer by paragraph to clarify any misconceptions. Para1: I agree. Para2: For general mathematics yes, but this question is not about mathematics - it is about the special (and to me surprising) nature of the proof of Godel Completeness/Compactness -in contrast to (a) other theorems in (two valued) fundamental logic; and (b) the fact that older logic textbooks are not emphasising this. The proofs of soundness are a matter of just checking that the rules work, likewise many proofs of completeness in limited logics. – Roy Simpson Feb 03 '14 at 11:13
Para3: I only read the Avigad paper in preparing the question. Re-reading it I see that there is a reference to further comments in Godel's dissertation. I have not seen Godel's collected works and I believe that these dissertation qualms were removed from the 1930 published paper - giving the impression of Constructiveness? Did Godel change his position between Dissertation and Publication, as I have seen suggested (by Beeson)? – Roy Simpson Feb 03 '14 at 11:23
Para5: "... diminishes the mathematical correctness.." I dont quite understand this. Do you mean that "AC => GC" is mathematically correct (I agree)? If so then my question is: do we have examples of "not AC + not GC"? Is the existence of such examples somehow absurd, or do they exist. At the centre of my question is the Cohen AC independence result - we dont need to assume AC - but it has been assumed in GC. So some structure might exist as a counter-example to GC (without the AC precondition). Para6: Martin-Lof's use of AC is peculiar to that theory - in fact AC "holds" in a recursive model. – Roy Simpson Feb 03 '14 at 11:30
Of course there are models of "ZF + not AC + not GC". But if the concern is nonconstructive or nonlogical axioms, there is no reason to focus on the axiom of choice in particular. In other words, there is an premise underlying the question that the axiom of choice is particularly an issue with the completeness theorem; that premise is not justifiable from the contemporary understanding. Some older books had a myopic focus on the axiom of choice, but that is misleading. For example, if we want to look at constructiveness, the first step is to limit ourselves to countable theories. @Roy Simpson – Carl Mummert Feb 03 '14 at 11:43
@CarlMummert: your last sentence brings up another issue I mention at the end of my Question, and for which I am considering a second Question. This second is broader than GC so I want to leave it for now. AC and WKL are appearing in my question because they are explicit preconditions for GC. Perhaps if I was to read tomorrow that GC in fact depends on the existence of Mahlo Cardinals, I might ask a related question. – Roy Simpson Feb 03 '14 at 12:02
Sorry to be dense, but is "of the use of weak König's lemma" supposed to read "or the use of weak König's lemma"? – Feb 04 '14 at 16:56
@Byron Schmuland: yes; that is a typo – Carl Mummert Feb 04 '14 at 17:00

Constructiveness of Proof of Gödel's Completeness Theorem

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