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I am taking a course where we are covering a bit of logic, and I am trying to understand a some nuances of Gödel's theorems of completeness and incompleteness.

Q1) Is it correct to say that Gödel's completeness theorem refers to the completeness of the deductive system where as Gödel's incompleteness theorem refers to, an unrelated, concept of completeness of a set of axioms?

I am curious because I have been told that Gödel's incompleteness theorem states that a deductive system is either unsound or incomplete. Which I believe to be wrong.

Q2) In first order logic it is my understanding that you cannot reference a sentence inside of itself. Such as "This sentence is not true" or "This sentence has no proof".

However, I am given the sentence "$\alpha(j, A) = \forall i$ $\ i$ is not the Gödel number of a proof of the sentence whose Gödel number is j, where the proof uses only premises in $A$". ($A$ is recursively enumerable set of axioms).

Then we let $\sigma = \alpha(\#\sigma, A)$ where $\#\sigma$ is the Gödel number of $\sigma$. Can we really say that $\sigma$ is not self referential in this situation? How is this different from the statement "This sentence has no proof"? This is from Introduction to Artificial Intelligence by Russel and Norvig.

Q3) Finally, is the sentence $\sigma$ independent of the set of axioms $A$, i.e. is both $A\wedge \sigma$ and $A \wedge \neg \sigma$ both satisfiable?

EDIT: The following argument is nonsense. Just leaving here to recall my train of (incorrect) thought.

It seems like it must be the case that both are satisfiable. Otherwise, if $\sigma$ is always a true statement, then that implies $A \models \sigma$, so $A \vdash \sigma$ by completeness, so now we have a proof, so $\sigma$ is false.

But if $\sigma$ is always false, then $A \wedge \neg \sigma$ is unsatisfiable, so $A \models \sigma$. Then $\sigma$ is true.

It has been a while since I have thought about this, can anyone clarify a bit?

zrbecker
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  • Incompleteness refers to the fact that the theory of a (strong enough) set of axioms is incomplete: there will be unprovable statements. Completeness says that any true statement is provable, and vice versa. (Or false statement; just stick a negation in front of it.) So together, they say that for axioms systems that are strong enough, there will be statements whose truth value cannot be known! –  Nov 16 '13 at 02:18
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    For a fixed recursive set of axioms, say the axioms of first-order PA, your sentence $\sigma$ is, via representability, an ordinary though extremely messy sentence in the language of PA. The non-provability non-refutability (under the assumption of consistency) is a theorem whose proof is motivated by the partial self-referentiality, but in now way depends on it. As to the last question, sure, both PA plus $\sigma$ and PA plus $\lnot\sigma$ have a model. – André Nicolas Nov 16 '13 at 02:22
  • @AndréNicolas How do we distinguish between partial self-reference, and a full self-reference, i.e., why is $\sigma$ not the same as "This sentence has no proof", or is it the same? – zrbecker Nov 16 '13 at 02:27
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    I should not have said partial self reference, as a sentence of PA it has no self reference at all. The informal interpretation has self-referential content. Your English sentence "This sentence $\dots$" is not a sentence in a formal language, and certainly not a sentence of PA. – André Nicolas Nov 16 '13 at 02:31
  • I'm afraid I don't see why PA plus $\neg \sigma$ could have a model. Why can't the faulty argument Danikar gave work? Why can't the implication "$\neg\sigma$ implies $\sigma$ is provable" be written in the language of PA? – Holden Lee Jan 31 '14 at 10:38

1 Answers1

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Is it correct to say that Gödel's completeness theorem refers to the completeness of the deductive system where as Gödel's incompleteness theorem refers to, an unrelated, concept of completeness of a set of axioms?

Yes. Though the two uses of the terms "complete/incomplete" aren't entirely unrelated (the double use isn't a case of terminological perversity -- even though logicians can be guilty of that). For note that a semantically complete deductive system is one such that, if you add a new logical axiom that can't already be derived (or a new rule of inference that can't be established as a derived rule of the system) then the logic becomes useless by virtue of warranting arguments that aren't semantically valid. Likewise negation-complete theory is one such that, if you add as a new axiom some proposition that can't already be derived in the theory, then the theory becomes useless by virtue of becoming inconsistent.

In first order logic it is my understanding that you cannot reference a sentence inside of itself. Such as "This sentence is not true" or "This sentence has no proof".

No, there's nothing that stops you having an interpreted first order language which is such that, among the constants [or among other terms], there are some which are interpreted to denote the sentences of that very language. A language in which self-reference is possible doesn't thereby stop being first-order.

Can we really say that $\sigma$ is not self referential in this situation?

You have defined $\sigma$ as a sentence of mathematical English, and mathematical English can refer to proofs and indeed be self-referential; but this isn't the Gödel sentence you are looking for, which will belong to the formal language of the formal theory $A$, whichever that is. Perhaps you meant that $\sigma$ is supposed to be a sentence of $A$ -- Peano Arithmetic, for example -- which on interpretation we see is true iff and only if no $A$-proof is a proof of it. But such a wff isn't strictly speaking self-referential: it will be a long wff of formal arithmetic, which is about numbers, not about wffs.

The usual textbooks (mentioning no names, ahem) should explain this sort of thing clearly.

Peter Smith
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