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I am trying very hard to fully understand gödel's paper on the incompleteness theorem.

I have a slight technical question about one of the 45 primitive recursive functions introduced to build the predicate (or "function" ? "relation" ?) provable(x).

http://mrob.com/pub/math/goedel.html

my question is : in the 22nd function

FR(x) ≡ (n){0 < n ≤ l(x) → Elf(n Gl x) ∨ (∃p,q)[0 < p,q < n & Op(n Gl x,p Gl x,q Gl x)]} & l(x) > 0

apparently in that formula we test for all n if the n-th item is an elementary formula or one obtained using the operators '¬', '∨' or '∀' on p,q-th formulae with p,q < n.

My question is : n Gl x ≡ ε y [y ≤ x & x/(n Pr x)y & not x/(n Pr x)y+1] where "n Pr x" is the n-th prime factor of x. And for a prime factor (or number) is associated the gödel number of a symbol only.

from what i understand about gödel numbering, "n Gl x" is only supposed to give a symbol form the alphabat the system is built on, not a formula.

What did i not understand ? How can "Elf(n Gl x)" or "Op(n Gl x,p Gl x,q Gl x)" possibly be true since, once again, from my understanding, "n Gl x" is only supposed to give the number of a symbol

1 Answers1

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Yes, $Bew(x)$ is a predicate; it holds if $x$ is the Gödel number of a provable formula $\varphi$.

In the 22nd formula, the predicate $FR(x)$ "tests" if the number $x$ is the code of a sequence of fomulae (and not only a sequence of "basic" symbols); it "return" $TRUE$ when for all $n : 0 < n \le lenght(x)$ the $n$-th item in the sequence is an elementary formula or it is obtained using the operators $¬, ∨$ or $∀$ from previous elements in the sequence, i.e. when there are $p,q < n$ such that ... .

See previous part of the document; the number of the "basic" symbols are used to assign number to strings of them (i.e. formula) and then to sequences of formulae. Thus, there is not only a uniquely associated natural number for every basic sign, but also for every sequence of basic signs.


The formula 6 :

$n Gl x ≡ ε y [y ≤ x \land x/(n Pr x)^y \land \lnot x/(n Pr x)^{y+1}]$

can be used to "decode" a sequence.

Starting from a number $x$ encoding a sequence, it retrieves the greatest $y$ such that the $n$th prime number raised to $y$ divides $x$.

We can try with some simple examples; unfortunaltely, the original language used by Gödel does not use "$=$" as primitive.

We will assume it as primitive, and modify a little bit the basic encoding :

$0$ is "$=$"; $1$ is "$0$"; $3$ is "$f$"; $5$ is "$\lnot$"; $7$ is "$\lor$"; $9$ is "$\forall$"; $11$ is "("; $13$ is ")"; $17$ is "$x_1$", and so on...

In this way, we can encode the simple formula $0=0$ as : $2^13^05^1=2 \times 1 \times 5=10$.

Thus : $1Gl10=1, 2Gl10=0$ and $3Gl10=1$. Compulsing the alphabet, we can rewrite the string, because $1$ encodes "$0$" and $0$ encodes "$=$".

The encoding mechanism works due to the unique factorization into prime fcators; for every $n$ we can always retrieve with a finite number of divisions the exponents of its prime factors.

The encoding mechanism is used to encode not noly strings of symbols, but also sequences of strings : if $n$ is the code of the string $\sigma$, we encode with $2^n$ the sequence formed by the single string $\sigma$.

With the previous example, $2^{10}$ is the code of the sequence $\langle 0=0 \rangle$.

Now we can go on, considering a very simple example of proof in the system used by Gödel :

1) $\lnot(fx_1 = 0)$ --- Ax.1

2) $x_1 \forall \lnot (fx_1 = 0)$ --- from 1) by generalisation.

Let $n_1=2^53^{11}5^37^{17}11^013^117^{13}$ the code of the 1st formula, and let $n_2$ the code of the 2nd formula.

Thus, the number $n=2^{n_1}3^{n_2}$ is the code of the sequence :

$\langle \lnot(fx_1 = 0), x_1 \forall \lnot (fx_1 = 0) \rangle$.

The number $n$ encodes a proof in the system.

  • I think i was not clear enough : from what i understood, with the defined operations and the way the numbering works, there is no way to attribute the gödel nb of a "string" to a prime factor – joseph M'Bimbi-Bene Oct 16 '15 at 19:29
  • I think i was not clear enough :

    from what i understood, with the defined operations and the way the numbering works, there is no way to attribute the gödel nb of a sequence to a n-th prime factor.

    For a given sequence/number, the power of the n-th prime factor gives the number of the symbol at position n.

    the concatenation operation used throughout the 45 functions doesn't attribute either the gödel nb of a sequence as the power of a prime factor

    – joseph M'Bimbi-Bene Oct 16 '15 at 19:51
  • I totally understand the idea behind the encoding (i think), but back to the 22nd function

    FR(x) ≡ (n){0 < n ≤ l(x) → Elf(n Gl x) ∨ (∃p,q)[0 < p,q < n & Op(n Gl x,p Gl x,q Gl x)]} & l(x) > 0

    taking, just for the sake of it your example, $2^5 3^3 5^1$, n Gl x will "return" either 5, 3 or 1, the nb of a symbol, not of a sequence, so the "tests" Elf(n Gl x) or Op(n Gl x,p Gl x,q Gl x) will necessarily fail

    – joseph M'Bimbi-Bene Oct 16 '15 at 19:55
  • I totally get it, and from what i understand

    Elf(y) must also be applied to numbers that encode formulae, or sequence of symbols, but n Gl x returns the number of a single symbol,

    so Elf(n Gl x) fails necessarily, since a single symbol cannot be an elementary formula.

    The same goes for Op(n Gl x,p Gl x,q Gl x)

    – joseph M'Bimbi-Bene Oct 16 '15 at 21:06
  • And from what i understood about the numbering, the 45 functions and the construction of formulae with those functions, there is no way to have like $x = ... p_k^{nbOfASequence} ...$ where x is the number of some formula and pk is the k-th prime factor.

    In my understanding, n Gl x can never return the number of a sequence

    – joseph M'Bimbi-Bene Oct 16 '15 at 21:11
  • let x = E(¬) = $2^5$

    let y = E( (0=1) ) = $2^11 3^1 ... p_{k}^{13}$ (i'm not ecoding the '=' sign)

    But according to 13 (that uses 8), 8 (x * y) is supposed to return a number such that its first length(x) prime factors have the same power that of the prime factors for x

    (n ≤ l(x) → n Gl z = n Gl x] or (∀n ≤ length(x) . item(n, z) = item(n, x)) )

    and the length(x)-th to length(x)+length(y)-th prime factors have the same power that of the prime factors of y.

    Or what did i get wrong ?

    – joseph M'Bimbi-Bene Oct 16 '15 at 21:47
  • I think i understand now. For example :

    let x = $p_{1}^{x1} p_{2}^{x2} ... p_{lgx}^{xlgx}$ be the gödel number of some formula, where $p_{1}, p_{2}, ... p_{lgx}$ are the first, second, ..., to lg(x)-th prime factor of x.

    $x1 x2 ... xlgx$ being the numbers of the symbols of the associated formula (whose nb is x).

    let y = $p_{1}^{y1} p_{2}^{y2} ... p_{lgy}^{ylgx}$ be the gödel number of some formula, where $p_{1}, p_{2}, ... p_{lgy}$ are the first, second, ..., to lg(y)-th prime factor of y.

    $y1 y2 ... ylgy$ being the numbers of the symbols of the associated formula (whose nb is y).

    – joseph M'Bimbi-Bene Oct 17 '15 at 17:45
  • let form(x) be the formula whose nb is x.

    then the nb of the sequence of formula form(z) = (form(x), form(y)) is (i suppose) constituted using the functions seq(x) or R(x) and the concatenation.

    z = seq(x) * seq(y) where * is the concatenation operator.

    Thank you for the edition of your reply.

    I guess my problem was that i didn't know/understand how to encode sequences of formulae.

    – joseph M'Bimbi-Bene Oct 17 '15 at 17:48
  • i thought the sequence of formulaa (x,y) was constituted only using the concatenation operator, i forgot i/we/one haas to use the function seq(x) or R(x) in order to attribute the formula to a single prime factor. – joseph M'Bimbi-Bene Oct 17 '15 at 18:03