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.
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 19:51FR(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 19:55Elf(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 21:06In my understanding, n Gl x can never return the number of a sequence
– joseph M'Bimbi-Bene Oct 16 '15 at 21:11let 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 21:47let 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 17:45then 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