In proving Gödel's incompleteness theorem, why does he needed the Diagonal Lemma or the Fixed Point Theorem for building a formula $\phi$ that spoke about itself? Can't this formula be built this way:

Diagonal Function. Let $D(x, y)$ be the diagonal function, such that $D$ returns the result obtained by replacing the formula with Gödel number $y$ for all free occurrences of $a$ in the formula with Gödel number $x$.

Example. Let $\psi(a)$ be a formula that affirms that some formula with the Gödel number of $a$ is closed (has no free variables) and $k$ be it's Gödel number. Using Diagonal Function to construct a formula, by applying $D(k, k) = j$, the Gödel number $j$ will already be the Gödel number of a formula affirming that the formula itself has no free variables. I mean, $\ulcorner \psi(\ulcorner \psi(a) \urcorner) \urcorner = j$, or $\ulcorner \psi(\overline{k}) \urcorner = j$.

What i am missing here?

  • $\begingroup$ Are you referring to a particular exposition of Gödel's work? His own original article does not name any "Diagonal Lemma" or "Fixed Point Theorem" -- its lemmas and theorems are simply numbered. $\endgroup$ Jun 2 '12 at 15:01
  • $\begingroup$ You are right, i'm referring to Mendelson's (Introduction to Mathematical Logic) exposition on the subject. $\endgroup$
    – felipegf
    Jun 2 '12 at 15:05
  • $\begingroup$ So you're using $D(u,u)$ for what Mendelson (fourth edition, section 3.5) just calls $D(u)$? $\endgroup$ Jun 2 '12 at 15:10
  • $\begingroup$ I didn't wanted to define $sub$, but i wanted everyone to see that in fact $D(x,y)$ just replaces all free occurrences of a free variable $a$ (in the $sub$ predicate the Gödel number of the free variable is passed as an argument) in the Gödel number of a formula $x$ for the formula (whose Gödel number is) $y$. In fact, $D(u, u)$ is equivalent to $D(u)$. $\endgroup$
    – felipegf
    Jun 2 '12 at 15:18

If I understand your notation correctly, the formula (whose Gödel number is) $j$ does not assert that $j$ itself has no free variables, merely that $\psi(a)$ (which is different from $j$ itself) has no free variables. Which, incidentally, is false because it has $a$ as a free variable.

  • $\begingroup$ You are right, thank you very much. Do you have any reference on why the diagonal lemma is named after Cantor's diagonal argument? $\endgroup$
    – felipegf
    Jun 4 '12 at 4:05

There are two questions:

  1. Why does Gödel's proof needs the Diagonalization Lemma"
  2. Could the diagonal function be defined as suggested?

My answer to 1 is twofold: Some proofs of Gödel's First Incompleteness Theorem do not use the Diagonalization Lemma (like the one suggested by Turing in his celebrated and seminal article) and a proof by Kripke reported in the following article:

Hilary Putnam: Nonstandards models and Kripke's proof of the Gödel Theorem, Notre Dame Journal of Formal Logic, Volume 41, Number 1, pages 53-58, 2000

Thew following might help understand Gödel's proof and thus better answer the first question:

The "diagonal lemma" (also called "diagonalization lemma", "self-referential lemma” and “fixed-point lemma”) is a generalization (see below (Carnap 1934)) of Gödel's argument. Gödel attributed that generalization to Carnap in the references (Gödel 1934) and (Gödel 1986) given below. Gödel proved the special case of that lemma where the unary relation is "not Bew(x)".

R. Carnap: Logische Syntax der Sprache, Vienna: Julius Springer, 1934

K. Gödel: On Undecidable Propositions of Formal Mathematical Systems, lecture notes taken by S. Kleene and J. Rosser, 1934

K. Gödel: Review of Carnap 1934, in: Gödel: Collected Works I. Publications 1929–1936, S. Feferman et al. editors, Oxford University Press, 1986 p. 389

  • 1
    $\begingroup$ I am not sure that I see how this answers the question---could you perhaps clarify that a bit? By my reading, the original asker wants to understand why the diagonal lemma is needed. As far as I can tell, you have provided references which explain what the diagonal lemma is. The references may be relevant, but they seem orthogonal to the question itself. $\endgroup$
    – Xander Henderson
    Mar 10 '19 at 14:15

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