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I am reading a translation of Gödel's original paper about in completeness theorem and there are a couple things i don't understand.

Here is the document i am using primarily : http://www.research.ibm.com/people/h/hirzel/papers/canon00-goedel.pdf

In the part "2.6 Undecidability theorem", it is said that (and i have an intuitive feeling) that :

r = subst(q, 19, number(p))

is supposed to mean "x does not prove p(p)" or rather, "p does not prove p(number(p))".

I "feel" the parallel with the so-called class-sign S or $R_q$ in the sketch of the proof in the introduction.

Here is where i get lost. We have :

r = subst(q, 19, number(p))

r = not(proofFor(x, sub(nb(p), 19, nb(nb(p)))))

But since nb(p) is a sequence of "succ" (encoded as '3') with a "0" in the end, there is not '19' to substitute so i end up with r intuitively meaning : "x is not a proof of nb(p)"

r = not(proofFor(x, nb(p))) since sub(nb(p), 19, nb(nb(p))) = nb(p)

and subst(p, 19, number(p)) = forall(17, r) meaning "nb(p) is not provable", where i guess we should have "p(nb(p)) is not provable".

What id i get wrong ?

Thank you.


EDIT1 :

Here is the detail of what i come up with that understanding :

q (17,19) = "not(proofFor(x, sub(y, 19, nb(y))))" with x standing for 17 and 19 standing for 19. I understand q is the number of some formula (so called "relation sign")

p (19) = forall(17, q)

p (19) = forall(x, not(proofFor(x, sub(y, 19, nb(y))))) where 'x' and 'y' denote respectively 17 and 19.

r (17) = sub(q, 19, nb(p))

r (17) = not(proofFor(x, sub(nb(p), 19, nb(nb(p)))))

r (17) = not(proofFor(x, nb(p)))

"x does not prove nb(p)" and :

forall(17, r) : "nb(p) is not provable"

Since : number(n) = succ_n(n, seq(1))

And "succ_n" is defined as follows :

succ n(0, x) = x succ n(n + 1, x) = seq(3) ◦ succ n(n, x)

nb(p) is the number of a sequence of '3' and a '1' in the end, there is no '19' to substitute.


EDIT 2 :

If r was defined as : r (17) = sub(q, 19, p)

r (17) = not(proofFor(x, sub(p, 19, nb(p))))

forall(17, r) = sub(p, 19, p) forall(x, not(proofFor(not(proofFor(x, sub(p, 19, nb(p))))))

I could see how one would interpret it as p(nb(p)) (or p(p) ?) is not provable, but with the given definition i just don't see how

  • 1
    maybe ... anyway my mistake became clear when i read again :

    ¬proofFor κ (x, subst(y, 19, number(y))) ⇒ provable κ (subst(q, 17 19, number(x) number(y)))

    for some reason what i had in mind was : ¬proofFor κ (x, subst(y, 19, number(y))) ⇒ provable κ (subst(q, 17 19, x y))

    Now it's all good ... and now i have to prove the theorem V ... i will probably post another question about that very subject

    – joseph M'Bimbi-Bene Oct 25 '15 at 08:29

1 Answers1

6

Revised edition

I'll use the translation in : Jean van Heijenoort, From Frege to Gödel : A Source Book in Mathematical Logic (1967), page 595-on.


[page 598] A formula of [the system] PM with exactly one free variable, that variable being of the type of the natural numbers (class of classes), will be called a class sign. [...] Let $\alpha$ be any class sign; by $[\alpha; n]$ we denote the formula that results from the class sign $\alpha$ when the free variable is replaced by the sign denoting the natural number $n$ [i.e. the numeral $S(S(\ldots S(0)))$].

[page 600] By $Subst \ a^v_b$ (where $a$ stands for a formula, $v$ for a variable, and $b$ for a sign of the same type as $v$ [a term]) we understand the formula that results from $a$ if in $a$ we replace $v$, wherever it is free, by $b$.

[page 603] The functions $x + y, x.y$, and $x^y$, as well as the relations $x < y$ and $x = y$, are recursive, as we can readily see. Starting from these notions, we now define a number of functions (relations) 1-45, each of which is defined in terms of preceding ones [...]

  1. $x/y \equiv (\exists z)[z < x \land x = y.z]$, $x$ is divisible by $y$.

This function is $/ : \mathbb N \to \mathbb N$, while e.g. $+ : \mathbb N \times \mathbb N \to \mathbb N$.

I.e. they are functions with natural numbers as "inputs" and "output".

[page 604] 13. $Neg(x) \equiv R(5)*E(x)$, $Neg(x)$ is the NEGATION of $x$.

Now we are in the "arithmetized" world of the syntax, but the functions are still numerical ones; thus, we have to read $Neg : \mathbb N \to \mathbb N$ as the function that takes as input the code of a formula $\varphi$ and produces as output the code of the formula $\lnot \varphi$.

I'll stay with the "standard" use of writing $\ulcorner \varphi \urcorner$ for the code of the formula $\varphi$.

  1. $Z(n) \equiv n N [R(1)]$, $Z(n)$ is the NUMERAL denoting the number $n$.

Again, a function that for each number $n$ calculates the code of the numeral $S(S(\ldots S(0)))$, due to the fact that $R(1)=2^1$ is the code of the sequence formed by the single symbol "$0$".

[page 605] $Su \ x^n_y$ results from $x$ when we substitute $y$ for the $n$th term of $x$.

[...] $Sb \ x^v_y$ is the notion $Subst \ a^v_y$ defined above.

$Subst$ is a syntactical operation performed on the formula $\varphi(y)$. The function $Sb$ is an arithmetical one, acting on number. Thus $Sb \ \ulcorner \varphi \urcorner ^v_y$ is the code of the formula $\varphi (v/y)$ obtained performing the $Subst$ operation.

[page 606 ] The fact that can be formulated vaguely by saying: every recursive relation is definable in the system $P$ (if the usual meaning is given to the formulas of this system), is expressed in precise language, without reference to any interpretation of the formulas of $P$, by the following theorem:

Theorem V. For every recursive relation $R(x_1, \ldots, x_n)$ there exists an $n$-place RELATION SIGN $r$ (with the FREE VARIABLES $u_1, u_2,\ldots, u_n)$ such that [...].

Thus, $R$ is a $n$-ary relation between numbers and $r$ is a formula of the formal system $P$.

[page 608] We now define the relation

$$Q(x,y) \equiv \lnot x B_k [Sb \ y^{19}_{Z(y)}].$$

$Q(x,y)$ is a binary numerical relation, defined in terms of previously inrtoduced functions and relations.

By Th.V, it is "expressible" in the system $P$ by a binary relation sign $q$, i.e. by a formula $q(x,y)$.

Let $\ulcorner q \urcorner$ be its code; it is the code number of a formula with two free variables, the two first var in the list of the alphabet, coded with $17$ and $19$ respectively.

Then, $\ulcorner p \urcorner = Gen(17,\ulcorner q \urcorner)$ is the code of the "universal closure" $\forall x q(x,y)$ of the formula $q(x,y)$ with respect to $x$, coded with $17$, due to the fact that the function $Gen(x,y)$ [page 604, n°14] calculates the code number of the GENERALIZATION of the formula coded by $y$ with respect to the VARIABLE coded by $x$.

Thus, the formula $p(y)$ has only one free variable.

Finally, we have $\ulcorner r \urcorner = Sb \ \ulcorner q \urcorner^{19}_{Z(\ulcorner p \urcorner)}$ where $r(x)$ is a new formula with one free variable obtained from $q(x,y)$ substituting the numeral $Z(\ulcorner p \urcorner)$ corresponding to the code for the formula $p(y)$ in place of the second free variable of $q(x,y)$.

This process is called diagonalization :

This is the idea of taking a wff $\varphi(y)$, and substituting (the numeral for) its own code number in place of the free variable. Think of a code number as a way of referring to a wff. Then the operation of ‘diagonalization’ allows us to form a wff that as it were indirectly refers to itself (refers to itself via the Gödel coding). We will use this trick [...] to form a Gödel sentence that encodes ‘I am unprovable in $\mathsf{PA}$’.


Following your translation, we have a formula $Q(x, y)$ that "means" : "$x$ does not prove the formula obtained from the formula with code $y$ by subst of the numeral $number(y)$ in place of the (only) free var".

Thus, $p = forall(17, q)$ means : "for all $x$, $x$ does not prove the formula obtained from the formula with code $y$ by subst of the numeral $number(y)$ in place of its only free variable", i.e. "the formula $\ldots$ is unprovable".


A very good book is :

See page 137 for some illuminating details :

[The expression] ‘$\ulcorner \varphi \urcorner$’ is shorthand for [the] standard numeral for the g.n. [Gödel number] of $\varphi$.

In other words, inside formal expressions ‘$\ulcorner \varphi \urcorner$’ stands in for the numeral for the number $\ulcorner \varphi \urcorner$.

A simple example to illustrate:

  1. ‘$SS0$’ is an expression, the standard numeral for $2$.

  2. On our numbering scheme $\ulcorner SS0 \urcorner$, the g.n. of ‘$SS0$’, is $2^{21}.3^{21}.5^{19}$.

  3. So, by our further convention, we can also use the expression ‘$\ulcorner SS0 \urcorner$’ inside (a definitional extension of) [the language], as an abbreviation for the standard numeral for that g.n., i.e. as an abbreviation for ‘$SSS \ldots S0$’ with $2^{21}.3^{21}.5^{19}$ occurrences of ‘$S$’ !


See also, for useful details : Diagonalization Lemma.

  • I totally understand all you said, but that doesn't help me much. Maybe i was not explicit enough :

    subst(p, 19, number(p)) is supposed to assert its own unprovability, but when i construct, i just understand :

    subst(p, 19, number(p)) means "number(p) is not provable"

    – joseph M'Bimbi-Bene Oct 23 '15 at 10:01
  • I understand that's what we should end up with yet that's precisely what i'm arguing isn't done in the paper.

    "xx does not prove the sentence obtained from the formula (with one free var) with code pp when the code pp is "inserted" in place of the free var"

    I detailed how/why i argue that this is not the case that is being performed.

    I guess i am making a mistake since obviously if Gödel made a mistake his paper wouldn't be celebrated as it is but i need i precise construction or highlight of my mistake

    – joseph M'Bimbi-Bene Oct 23 '15 at 11:36
  • Indeed, using the details i provide, my understanding of the paper :

    "x does not prove the sentence obtained from the formula (with one free var) with code numeral of the code p when the code numeral of the numeral of the code p is "inserted" in place of the free var"

    – joseph M'Bimbi-Bene Oct 23 '15 at 11:41
  • "nb(p)" is the code of the numeral of the code of p.

    From what i understand, there is no substitution to be performed in that.

    – joseph M'Bimbi-Bene Oct 23 '15 at 11:43
  • Thank you for you effort trying to answer me.

    But i still can't wrap my hand around it.

    subst(x, v, y) = subst (nFreePlaces(v, x), x, v, y)

    yet in sub(q, 19, nb(p)) we end up with : not(proofFor(x, subst(number(p), 19, number(number(p))))) we

    – joseph M'Bimbi-Bene Oct 23 '15 at 21:37
  • ok, i see my mistake now. I misread that part :

    ¬proofFor κ (x, subst(y, 19, number(y))) ⇒ provable κ (subst(q, 17 19, number(x) number(y)))

    subst(y, 19, number(y)) is "induced"/"implied" by substituting number(y) as the second free variable of the "relation sign" q.

    – joseph M'Bimbi-Bene Oct 23 '15 at 21:49