Talk:Gödel's completeness theorem

message: The paragraph contrasting with the incompleteness theorem strikes me as misleading.  It isn't clear to me that the meaning of "complete" changes for the incompleteness theorems.  To put things as simply as possible completeness is just "if truth then proof".  For propositional logic and first-order predicate logic the "truths" in the system are basically just tautologies, and every tautology can be proven deductively.  For the more expressive systems that are incomplete, however, you get more truths than you can prove.  That's the whole point of the Godel sentence; it "says of itself" that it is not derivable, hence is true and not derivable.  The relevant difference to emphasize is that the incomplete systems are more expressive, not that there's a different meaning of "complete" involved in the two theories.  Also, in the same paragraph "logical consequence" is used in a confusing/unclear way.  This paragraph needs to be completely scrapped and rewritten, imo.  Maybe I'll come back and do it when I have time.
173.30.27.245 (talk) 16:43, 11 May 2009 (UTC)
replies: text: The issue is that every first order theory, including e.g. Peano arithmetic, is complete in the sense that any statement that is true in every model of the theory is provable in the theory. This sense of completeness is a property of first-order logic itself. In another sense of the word "complete", there are sentences that are true in the standard model of Peano arithmetic that are not provable. So the sense in which that is "incompleteness" is indeed different than the sense in which first-order logic is complete. — Carl (CBM · talk) 18:32, 11 May 2009 (UTC)
replies: text: So in what model the sentence P="P is not provable in Peano arithmetic" is false? I see why you can add P as an axiom to the system and still get a consistent model, but if you add its negation instead, we have the Peano axioms and another axiom (not P) which states that its negation(P) is provable from the Peano axioms... oops. Bbbbbbbbba (talk) 12:20, 10 November 2010 (UTC)
replies: text: Yes, that is possible. It's a consequence of the incompleteness theorems. — Carl (CBM · talk) 13:38, 10 November 2010 (UTC)
replies: text: Sorry, but I can't see how your answer makes sense. You said "that is possible" but I didn't (even implicitly) state the impossibility of anything. If I did, then it's the impossibility of a model in which P is false (sorry for mistakenly typing in "true" in the original question; I have corrected it). And how is the existence of such a model justified by the incompleteness theorems?  —Preceding unsigned comment added by Bbbbbbbbba (talk • contribs) 13:59, 10 November 2010 (UTC)
replies: text: You should read the article Non-standard model of arithmetic. The point is that, because the Goedel sentence ("I am not provable") is not provable, there are indeed models in which it is false, by the completeness theorem. — Carl (CBM · talk) 14:17, 10 November 2010 (UTC)
replies: text: But that's where I am confused. Suppose we have a model where the Peano axioms (denoted by Q) are true, and the Godel sentence P is false, then P is provable from the Peano axioms (because P is false). So altogether we have: Q is true, P is false, yet P can be proved from Q. I find that situation weird. Bbbbbbbbba (talk) 14:33, 10 November 2010 (UTC)
replies: text: P refers to a formalized provability predicate, which in turn quantifies over natural numbers. In a non-standard model, that quantifier ranges over non-standard numbers. So a non-standard model can think something is "provable" without that thing actually being provable in the real world. It's necessary to keep a firm separation between actual provability, in the metatheory, and formalized provability, in the object theory. — Carl (CBM · talk) 14:38, 10 November 2010 (UTC)
thread_title:  "different meaning of 'completeness'" 

message: Could someone please clarify what "completely effective manner" means in "The completeness theorem and the compactness theorem are two cornerstones of first-order logic. While neither of these theorems can be proven in a completely effective manner" Cheers, 128.232.228.79 (talk) 18:41, 15 January 2010 (UTC)
replies: text: It means effective in the sense of computable. The details are in the last paragraph of that section. Algebraist 18:54, 15 January 2010 (UTC)
replies: text:  I started to tag that statement with 'citation needed' but then read this exchange and the latter part of the section; nonetheless do we need to say something in the article about what sense of 'effective' is meant (i.e. not that "the theorems cannot be proven in a manner that will convince college freshmen") or do we assume sophistication on the part of the reader to know what's meant? BrideOfKripkenstein (talk) 15:30, 5 October 2010 (UTC)
replies: text: I don't think we assume sophistication: there are (at least) two senses of effective in which the statement is true, and both are described in that same section. The sentence in question is a summary sentence for the entire section; I don't think it's excessive to ask the reader to actually read the section before asking what is intended there.  But the sense meant is not really "computable", it has more to do with constructive proof. The meaning of effective in the last sentence is about computability. — Carl (CBM · talk) 16:39, 5 October 2010 (UTC)
thread_title:  clarification 

message: Hello! first i don't speak english perfectly, please forgive me for the mistakes...
in the chapter:  statement the theorem 
in the sub chapter: Model existence theorem
It's written:
Alternatively, given Henkin's theorem, the completeness theorem can be proved as follows: If \Phi\models A is valid, then \Phi\cup\lnot A does not have models. By the contrapositive of Henkin's, then \lnot A is an inconsistent formula. But, by the definition of consistency, if \Phi\cup\lnot A is inconsistent then it's possible to build a proof of \Phi\vdash A.
I believe there is a little mistake: should it be written:
By the contrapositive of Henkin's, then \Phi\cup\lnot A is an inconsistent formula.
thanks --Nicobzz (talk) 10:27, 26 September 2018 (UTC)
replies: text: I guess you are right. Moreover, (the current version of) the article Henkin's theorem requires Henkin witnesses for all existentially quantified formulas which are not necessarily available in the setting of the completeness theorem. - Jochen Burghardt (talk) 12:55, 26 September 2018 (UTC)
thread_title:  Is there a little mistake in the abstract of the proof?

message: Should the term "Strong completeness theorem" redirect here? I'm translating some other pages and wonder whether "Strong completeness theorem" and "Gödel's completeness theorem" are the same concept or not. Thanks for helping! Puresnower (talk) 05:04, 14 June 2019 (UTC)
replies: text: Do you have a reference? - Jochen Burghardt (talk) 09:09, 14 June 2019 (UTC)
thread_title:  About strong completeness theorem