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By Gödel's incompleteness theorems, we can get a true but unprovable sentance $\psi$. However, we know it is true since its falsity implies contradiction. Then, why couldn't we accept this "proof" since it shows that $\psi$ must be true? Does it mean that the law of excluded middle could not be written as an axiom of finite length?

Lwins
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    Godel's incompleteness theorem doesn't give you a "true but unprovable sentence". What it does give you is an undecideable sentence $\psi$ -- one that can be neither proven nor disproven from the axioms of Peano arithmetic (PA). Asking about its truth only makes sense when interpreted in a model of PA. By Godel's completeness theorem, that means there exists a model where $\psi$ is true and there exists a model where $\psi$ is false. –  Nov 12 '17 at 20:39
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    @Hurkyl When people say the Godel sentence is true, they mean it is true in the intended interpretation, the natural numbers. – bof Nov 12 '17 at 20:45
  • Can you explain what exactly you have in mind when you say "its falsity implies contradiction"? You have some serious misunderstandings, but I'm not sure exactly what your misunderstandings are from what you've written. – Eric Wofsey Nov 12 '17 at 21:04
  • @bof: IMO, it's mainly people who understand the subject mean that. People who don't understand the subject often don't even realize the issue I mention exists. –  Nov 13 '17 at 03:49
  • @Hurkyl No it really does give you a demonstrably true, but not provably true sentence, and it doesn't depend on model theory. It is a very clever constructive argument (called the diagonalization lemma) that lets you construct a statement of the form $$G \iff Q(g)$$ where $g$ is the numerical mapping of expression $G$. The $G$ sentence is in fact neither provable nor unprovable as you say, but it is the $Q(g)$ statement that is demonstrably true but not provably true. – DanielV Nov 13 '17 at 05:58
  • @EricWofsey It is correct to say "the falsity leads to a contradiction" (although we can't be sure he means it correctly). If $P_g(n)$ is the claim that the number is $n$ is the proof of the expression $g$, then a statement of the form $$G \iff \forall n~\lnot P_g(n)$$ is constructed. Assuming $\exists n~P_g(n)$ corresponds to the existence of a proof of $G$ which implies $\lnot \exists n~P_g(n)$ by the left to right of the constructed statement. I'd guess that is the contradiction he is referring to. – DanielV Nov 13 '17 at 06:13
  • Um, this is clearly not a duplicate question. The other question is asking "what does true but unprovable mean" and this question is asking "why doesn't a proof in the meta logic also apply as a proof in the inner logic, is there some kind of infinite encoding that results?" Please unmark this as duplicate. – DanielV Nov 13 '17 at 20:33
  • The reason is that Godel's incompleteness theorem assumes that the theory is consistient. So you can show Con(T) -> Godel statement of T, but nothing more. – Christopher King Nov 13 '17 at 20:59

2 Answers2

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We start with an appropriate theory $T$ and make the Gödel sentence $G_T$. When we argue that $G_T$ is true, we do not make that argument within $T$. We generally need to assume something beyond $T$ - such as "$T$ is consistent" - in order to show that $G_T$ is true. But, if $T$ is an appropriate theory, the incompleteness theorems show that $T$ does not prove "$T$ is consistent" and so this argument can't work within $T$.

So the real issue with the phrase in the question is not with the word "true", the issue is with the word "unprovable". In the motto "true but unprovable", the term "true" refers to the standard model, while the phrase "unprovable" only means "unprovable in $T$", not "unprovable in any system whatsoever". Every statement is provable in some system, such as a system that includes the statement as an axiom.

Carl Mummert
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In Godel's incompleteness theorem, completely decidable (strongly representable) predicate $P$ is created with the properties:

$$\forall n (\vdash P(n))$$ $$\not \vdash \forall n ~P(n)$$

This $P$ is not the Godel sentence $G$, neither is $\forall n~P(n)$. The Godel sentence is the statement made by a diagonalization argument based on $P$.

So when the claim is made that $\forall n~P(n)$ is true, it means true in the demonstrable sense. That means there is a proof for $P(0)$, and for $P(1)$, etc. Any number you pick, there is a proof. But when it is said $\forall n ~ P(n)$ is unprovable, they mean that despite there being a proof for each instance of $P$, no matter how cleverly you picked the rules of inference concerning $\forall$ to be, once the entire logic is known, such a $\forall n~ P(n)$ can be constructed that has no proof.

Then, why couldn't we accept this "proof" since it shows that $\psi$ must be true?

There are 2 logics that go into Godels incompleteness theorem. There is the logic that the theorem is referring to, let's call it $L_1$, and there is the logic or generally reasoning used to construct the proof of the claim about $L_1$, call it $L_2$. It is common for $L_2$ to be just model theoretic reasoning, but it can also be completely reliable constructive algorithmic reasoning, since the proof itself is of an existential claim (that such a $P$ can be constructed).

When it is said that we know $\forall n \vdash P(n)$, that is a construction done in $L_2$, not $L_1$. So when you say "couldn't we accept this proof", the answer is yes we can, and yes we do, but we do so in the logic of $L_2$. But the entire point is that the statement isn't provable in $L_1$. So you might ask "but then doesn't Godel's incompleteness theorem not apply to $L_2$?" and the answer would be "it doesn't apply to $L_2$ for that choice of $P$, but it does apply for another choice of $P$, which could then be proven in a more presumptuous logic $L_3$".

DanielV
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  • Disclaimer: I am not an expert with Godel's incompleteness theorem; but I found https://plato.stanford.edu/entries/goedel-incompleteness/ to be a very reliable and understandable source. – DanielV Nov 13 '17 at 06:23
  • You are hiding a critical subtlety in your quantification; sometimes in your post you often quantifying externally over the standard natural numbers. To wit, there exists a model and a number $n$ in that model for which $P(n)$ is false –  Nov 13 '17 at 07:13
  • @Hurkyl Are you trying to make a reference to nonstandard models of natural numbers? – DanielV Nov 13 '17 at 07:44