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The Gödel sentence must be provable or unprovable by itself - you have to resolve all definitions until it only uses the elementary symbols of Peano arithmetic. What is the correct way to resolve definitions of primitive recursive functions so that the Gödel sentence stays Π1?

According to http://www.staff.science.uu.nl/~ooste110/syllabi/peanomoeder.pdf page 10, the formula that corresponds to a primitive recursive function with f(0)=c and f(x+1)=g(f(x),x) is something like this:

∃am( rm(a,m*(0+1)+1)=c ∧ ∀i<x( g(f(i),i)=rm(a,m*(i+1+1)+1) ) ∧ rm(a,m*(x+1)+1)=u )

Informally: There exists a,m such that the remainder of a divided by m*(i+1)+1 is f(i) for all i <= x (the sequence of values of the function up to f(x) is coded into a,m).

The value of f(x) is in free variable u. Resolving the remainder function rm(a,b) needs additional existential quantifiers.

The formula above is a Σ1 formula. Adding a generalization (like in the Gödel sentence) makes it Π2. But the Gödel sentence is said to be Π1.

So is the formula above the wrong way to resolve definitions in the Gödel sentence or are the unbounded existential quantifiers no problem for some other reason? The paper linked above seems to suggest the latter on page 15/16.

Asaf Karagila
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brunni
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    "The Gödel sentence must be provable or unprovable by itself". What precisely do you mean? – Andrés E. Caicedo Apr 20 '14 at 17:15
  • The Gödel sentence cannot contain non-elementary symbols like proofor(), subst() and number() - except when it also contains their definition. But the latter would not be the usual form of the Gödel sentence so all definitions have to be resolved. The function/relation proofor(x,y) knows only about logical inference and the axioms of some theory that can represent Peano arithmetic. prooffor(x,y) does not know about the definitions of prooffor(), subst() or number(). – brunni Apr 21 '14 at 10:28
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    What does it mean for a relation to "know about logical inference" or about axioms? – Andrés E. Caicedo Apr 21 '14 at 13:24
  • That the relations "knows about" is just a figure of speech without exact meaning. I should have put it in quotes. Let me put it another way: If G is the Gödel sentence, then ∀x¬prooffor(x,G). G itself cannot literally use prooffor because prooffor within G would then be a new relation symbol without any axiom. prooffor within G has to be expressed with the elementary symbols of Peano arithmetic. The same holds for subst and number. – brunni Apr 21 '14 at 13:54
  • I think I have tied up the loose ends now and will publish the work that led to this question: http://brunni.de/goedel/the_number_of_the_beast.html – brunni Apr 23 '14 at 10:05

2 Answers2

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The standard Gödel construction makes the Gödel sentence have the form $\forall x\neg Gdl(x)$ where $Gdl$ is a $\Sigma_1$ wff expressing a certain primitive recursive property. Note the negation sign! It is crucial. The negation of a $\Sigma_1$ wff is $\Pi$, and the universal quantification of such a wff is still $\Pi_1$. So the Gödel sentence is indeed $\Pi_1$.

Adding a generalization (like in the Gödel sentence) makes it $\Pi_2$.

Look carefully and you'll see that there is a crucial negation sign! The Gödel sentence says that every number is such that it isn't the number of the proof of such-and-such. [E.g. Check out the details in my Gödel book.]

Peter Smith
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  • Well - that was simple and I did not see it. I wonder what else I overlooked: http://www.brunni.de/goedel/the_number_of_the_beast.html – brunni Apr 20 '14 at 19:03
  • I corrected the article and the code to use "not isvalidproof(x,y) = 1" instead of "isvalidproof(x,y) = 0" – brunni Apr 20 '14 at 20:38
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There are two ways to say that the value of a primitive recursive function $f$ on input $n$ is $k$:

  • One can say that there exists a sequence $a_0, \ldots, a_n$ satisfying these three conditions:

    1. $a_0 = f(0)$,

    2. for all $i < n$, $a_{i+1}$ is computed from $a_i$ using the primitive recursive definition of $f$, and

    3. $a_n = k$.

  • Alternatively, $f(n)= k$ if and only if, for every sequence $a_0, \ldots, a_n$ satisfying (1) and (2) above, $a_n = k$. This is because, for every primitive recursive function $f$ and every $n$, there is exactly one sequence satisfying (1) and (2).

The first bullet will give you a $\Sigma^0_1$ definition, the second bullet will give you a $\Pi^0_1$ definition. More generally, every computable function (not just every primitive recursive function) is definable both by a $\Sigma^0_1$ formula and by a $\Pi^0_1$ formula, using much the same method.

Carl Mummert
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