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Gödel's completeness theorem: Given a set of axioms, if we cannot derive a contradiction, then the system of axioms must be consistent.

Gödel's incompleteness theorem:'Given any consistent, computable set of axioms, there's a true statement about the integers that can never be proved from those axioms'.

Since the incompleteness theorem assumes that the given set of axioms is consistent, then, by using the completeness theorem shouldn't all statements derived from this set of axioms be true thereby rendering the claim that 'there exists a true statement that cannot be proved' false?

marked as duplicate by Andrés E. Caicedo, MathOverview, amWhy, azimut, Lord_Farin Jul 4 '13 at 18:59

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  • 5
    There are true statements that you cannot derive. The class of true statements includes non-derivable statements. – Daniel Fischer Jul 4 '13 at 9:47
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    That isn't Godel's completeness theorem. That's just the definition of "consistent". – Chris Eagle Jul 4 '13 at 11:09
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    Also, both steps of your argument appear to be nonsense: the thing you call the completeness theorem doesn't say anything about "all statements derived from this set of axioms [being] true" and even if it did, that wouldn't refute the possibility of a true unprovable statement. – Chris Eagle Jul 4 '13 at 11:10
up vote 10 down vote accepted

No. You seem to confuse between "true" and "derivable" (or "provable"). The incompleteness theorem tells us that givens a theory with certain constraints, if it is consistent then there are things it cannot prove.

Furthermore, the completeness theorem tells us that a statement is provable from a theory if and only if it is true in all the models of the theory. So by combining these two we simply have the following statement:

Given a theory which satisfies the requirements of the incompleteness theorem, there will be a statement about the integers which is true in some models, but false in other models.

One finer point here is that when we say that a statement about the integers is "true", we mean to say that it is true in their standard model, $\Bbb N$. But the integers have other models, which are called non-standard models. So if a statement is true in the standard model of the integers it tell us nothing about its truth value in other models. In fact, if the statement is not provable it will be false in some of these models.

  • You seem to be answering as if the OP had said "Completeness says that all true statements are provable. Incompleteness says that some true statements are unprovable. Isn't that a contradiction?" Alas, what they said makes far less sense than that. – Chris Eagle Jul 4 '13 at 11:14
  • @Chris: I tried to revise. – Asaf Karagila Jul 4 '13 at 13:27

To add to Asaf's useful answer.

  1. "Gödel's completeness theorem: Given a set of axioms, if we cannot derive a contradiction, then the system of axioms must be consistent." No, that would trivialize the result, since consistency is defined as failing to entail a contradiction! The completeness theorem says that, if a bunch of sentences is consistent in the sense of not entailing a contradiction [in a standard system of first-order logic] then it has a model.

  2. "Gödel's incompleteness theorem:'Given any consistent, computable set of axioms, there's a true statement about the integers that can never be proved from those axioms'." Better, 'Given any consistent, computable set of axioms, there's a statement $\varphi$ in the language of arithmetic such that the axioms prove neither $\varphi$ nor $\neg\varphi$' Since, given an interpretation, one of $\varphi$ and $\neg\varphi$ will be true, your statement follows. But the fundamental result doesn't involve the notion of truth but is purely syntactic.

  3. With those clarifications, we can see better why there is no tension between them. The completeness theorem tells us that, given a set of first-order axioms in a language that includes the language of arithmetic, there is an interpretation which makes all those axioms [and their consequences] true. But nothing at all follows from that about whether the axioms will suffice to prove one of $\varphi$ and $\neg\varphi$ for any arithmetical sentence $\varphi$. It only shows that there is a model for all the sentences which it can prove.

  • Usually your additions come in the form of "footnotes to ..." :-) – Asaf Karagila Jul 4 '13 at 17:56

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