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I'm reading http://www.math.mcgill.ca/rags/JAC/124/GodelsProof.pdf

Near the final pages, they prove it.

Here's the proof as I understand it:

All symbols in the language of arithmetic are given a unique Godel number (which is a whole number)

All formulas in arithmetic are also given a unique Godel number by the method: $2^{a}3^{b}5^{c}7^{d}.....$, where 2,3,5,7... are the primes and $a,b,c,d$ are the Godel numbers of the symbols which appear in the formula in the order of their appearance.

Every proof in arithmetic is given a unique Godel number using the same method, except this time $a,b,c,d....$ are the Godel numbers of the formulas which appear in the proof in the order of appearance.

$Dem(x,z)$ is an arithmetic predicate which takes two Godel numbers $x$ and $z$. It is true if the proof corresponding to Godel number $x$ results in the formula (or proves the formula) corresponding to the Godel number $z$.

$sub(y,16,y)$ is an arithmetic function of $y$ (I don't know why two y's are used in the notation). It gives us the Godel number of the formula obtained by replacing all instances of the symbol corresponding to Godel number 16, in the formula corresponding to Godel number $y$, by $y$.

Now we start with the formula:

$$\forall(x), \neg Dem(x,sub(y,16,y))$$

Suppose this formula has Godel number $n$, and $y$ has been assigned Godel number $16$. We substitute $n$ for $y$ to get another formula:

$$\forall(x), \neg Dem(x,sub(n,16,n))$$

By definition of $sub(n,16,n)$, the above formula has Godel number $sub(n,16,n)$, as it has been obtained by replacing $y$ by $n$ in the formula of Godel number $n$

Also, this formula is saying "The formula of Godel number $sub(n,16,n)$ (i.e. the formula itself) is not demonstrable".

Now we prove that this formula is true. We assume that it is false. This means the formula is demonstrable. But that means its negation, which says "The formula is demonstrable", is also demonstrable. Since this is a contradiction, we arrive at the conclusion that the formula is true and undemonstrable. Have I understood this all correctly?

My problem

My problem is with the $Dem$ predicate. Given two Godel numbers $x$ and $z$, how do we calculate the truth value of $Dem(x,z)$?

To do that, we prime factorize $x$, and look at the power of the largest prime factor. If that power is equal to $z$, then that means the formula assigned to $z$ is the last formula (or the final result) in the proof assigned to Godel number $x$. So that means $x$ proves $z$, right?

But this method regards any sequence of formulas, ending in the formula corresponding to Godel number $z$, as a proof of the formula having Godel number $z$. This doesn't take into account any transformation rules of the axioms, whether the proof is logically valid or not. For example, $Dem(z,z)$ would be true by this method and hence every formula would be demonstrable.

Ryder Rude
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  • Not exactly... $x$ must encode a sequence of formulas. When you unpack it, the power of the last factor must be the code of the formula proved and every power must encode a formula whose sequence is a correct derivation of the last formula (i.e. it must encode an instance of an axiom or...) – Mauro ALLEGRANZA Mar 27 '20 at 07:36
  • @MauroALLEGRANZA But how do we check arithmetically that the sequence is logical? Wouldn't we have to do meta-mathematical reasoning? – Ryder Rude Mar 27 '20 at 07:55
  • Unpack $x$ and get e.g. $(x_1,x_2,\ldots, x_n)$. First step: verify that $x_n=z$. Second step: verify that $x_1$ encode an instance of an axiom (it must). Third step: verify that $x_2$ is: now we have two possibilities: either an axiom or a consequence of $x_1$ by Gen. And so on: this is the gist of Godel's discovery that a proof system can be arithmetized and that the operations with it are recursiv i.e. are "effectively computable". – Mauro ALLEGRANZA Mar 27 '20 at 10:34
  • The operations with G-numbers at "arithmetical" level are only computations with natural numbers. Only at the "meta" level they are interpretable as operations with symbols and expressions and sequences of expressions that "means" formulas and derivations. – Mauro ALLEGRANZA Mar 27 '20 at 10:37

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Here's the basic idea (as Mauro indicates)

To determine whether the relation Dem(m, n) holds, proceed as follows. First decode m (undo the Gödel numbering). That’s a mechanical exercise. Now ask: is the result a sequence of PA wffs? That’s algorithmically decidable (since it is decidable whether the Gödel number decodes into a string which is a sequence of wffs). If it does decode into a sequence of wffs, ask next: is this sequence a properly constructed PA proof? That’s decidable too (check whether each wff in the sequence is either an axiom or is an immediate consequence of previous wffs by one of the rules of inference of PA’s logical system). If the sequence is a proof, ask: does its final wff have the g.n. n? That’s again decidable. Dem(m, n) holds if the sequence coded by $m$ is a proof of the wff coded by $n$.

(So even if you choose a coding where a sequence of one wff gets the same code as the code for that wff, Dem(n, n) won't be true unless $n$ codes an instance of an axiom.)

Putting all that together, there is a computational procedure for telling whether Dem(m, n) holds. Moreover, at each and every stage, the computation involved is once more a straightforward, bounded procedure that can be done with ‘for’-loops. So Dem will be a primitive recursive relation, expessible in PA.

For more on this, see §39 of Gödel Without (too many) Tears, downloadable from https://www.logicmatters.net/igt/godel-without-tears/ Or even better of course the Gödel book of which those notes are the short version.

Peter Smith
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  • In a way, we 've proved the truth of the formula $\forall x, \neg Dem(x,sub(n,16,n))$. But we've shown that it can't be proved within the theory. So what theory does our proof come under? – Ryder Rude Mar 28 '20 at 00:02
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    We are not working "under a theory", we are looking at formal PA from "outside", just using ordinary mathematical reasoning. By ordinary mathematical reasoning , we can show that G [your Gödel sentence] is unprovable in formal PA. And interpreting G, we see it is true if and only if it is unprovable in PA . So it is true. That's just elementary maths, once you have the trick of it (and it was v. important to Gödel that it is so very elementary, for reasons to do with Hilbert's programme). – Peter Smith Mar 28 '20 at 09:38
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    OK, you can now ask, "if I want to go up a level and prove that G is unprovable-in-PA in some formal theory, what will that take?" Well, take PA+Con(PA) [where Con(PA) is the formal sentence that says that PA is consistent]. This is already enough to prove not-Prov(G) [where Prov is the arithmetized provability predicate for PA]. Is that good enough? – Peter Smith Mar 28 '20 at 09:39