I am wondering about the proof of the "Fixed-Point Lemma"

$\text{Mod } \Sigma$ is the class of all models of $ \Sigma$. $\text{Th Mod } \Sigma$ is the set of all sentences which are true in all models of $\Sigma$. This however is just the set of all sentences logically implied by $\Sigma$. We call this set the set of consequences of $\Sigma$ or $Cn \Sigma$. Thus we have that $Cn \Sigma = \{\sigma | \Sigma | \sigma \} = Th\space Mod \space \Sigma$.

Now we let $A$ be a set of axioms for our system over the language of $R$ which is a set of relations, constants, and formulas which create a number theory. Now if we have some set of consequences $Cn\Sigma$ due to our set of axioms $\Sigma$ we create a theory $T$ which models the set of consequences $Cn \Sigma$. Now referring to Wikipedia, as it states:

"Let $T$ be a first-order theory in the language of arithmetic and capable of representing all computable functions. Let $ψ$ be a formula in the language with one free variable. The diagonal lemma states that there is a sentence $φ$ such that $φ \Leftrightarrow ψ(\#(φ))$ is provable in T. Intuitively, $φ$ is a self-referential sentence saying that $φ$ has the property $ψ$. The sentence $φ$ can also be viewed as a fixed point of the operation assigning to each formula $θ$ the sentence $ψ(\#(θ))$. The sentence $φ$ constructed in the proof is not literally the same as $ψ(\#(φ))$, but is provably equivalent to it in the theory $T$.

Let $f\colon \mathbb{N}\rightarrow \mathbb{N}$ be the function defined by: $$f(\#(θ)) = \#(θ(\#(θ)))$$

for each $T$-formula $θ$ in one free variable, and $f(n) = 0$ otherwise. The function $f$ is computable, so there is a formula $δ$ representing $f$ in $T$.

--> I don't quite understand the above sentence. I know that since $\theta$ is a $T$-formula it is a consequence of the theory $T$, so we are only looking at functions $f(n)$ which operate on consequences of $T$ (which $\theta$ is one of). Now it is claimed that since $f$ is computable there is some $\delta$ representing $f$ in $T$. I am not sure what that is supposed to mean/what $\delta$ is really doing. Now I know that we are representing the sentences $\theta$ which are the consequences of the theory $T$. Also for $f(\#(θ)) = \#(θ(\#(θ)))$ is this just saying that we are defining $f$ to be some sort of assignment of a Gödel number to a Gödel number (the $\#(θ(\#(θ)))$)?

Wikipedia goes on to say:

Thus for each formula $θ$, $T$ proves: $$(\forall y) [ δ(\#(θ),y) \Leftrightarrow y = f(\#(θ))]$$

If I can understand the above it is possible that I can crack the rest of the proof - but will try and get it posted here so that others can see the walkthrough of the proof.

Thanks much in advance,


P.S. I was also hoping to gain some understanding beyond just understanding the proof of the lemma. I see a little but here:

Gödel's Incompleteness Theorem - Diagonal Lemma

and am guessing that somehow we are generating some sentence (similar to another real number in Cantors diagonalization argument) in which ?? I don't know how to put the rest in to words as I don't understand exactly what diagonalizing the sentances does (I do a little bit by intuition but can not formalize what is going on). Any thoughts on this would be very welcome. I also am wondering how this fits in with the $\delta$ above.

  • $\begingroup$ You are given as an assumption that computable functions are representable in $T$. You need to review what this means: For any computable $f$ there is a formula $\delta(x,y)$ such that $T$ proves $\delta$ is the graph of a partial function (that is, $T$ proves "for all $x$ there is at most one $y$ such that $\delta(x,y)$") and, for any $n,m\in\mathbb N$, $T$ proves "$\delta(n,m)$" iff $f(n)=m$. $\endgroup$ – Andrés E. Caicedo Jul 14 '13 at 20:27
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    $\begingroup$ The point is that, in order to talk about $f$, $T$ needs to be able to refer to it, since $f$ is not part of the language of $T$. This is what $\delta$ does: It provides us with a surrogate that we can use in place of $f$ when arguing (formally) about it. $\endgroup$ – Andrés E. Caicedo Jul 14 '13 at 20:29
  • $\begingroup$ Thanks Andres! Now, as you say that "$T$ proves "for all $x$ there is at most one y such that $\delta(x,y)$, I am guessing that this was somehow derived from the axioms of T? I don't see the formulation - but also in enderton, (while I can't find an online listing of the 11 axioms he uses in his definition of $A_{E}$- on Pg. 203, which is what I call $T$) he however says that $T ⊬ \forall y(y \neq \textbf{0} \rightarrow \exists x \space y = \textbf{S}x)$. Where $\textbf{S}$ is the sucessor function. So how does T prove $\delta$? $\endgroup$ – Relative0 Jul 15 '13 at 5:54
  • $\begingroup$ The specific details should be given in the source you are reading: There are three versions of representability, I'm not sure if the version being discussed requires this or not. Anyway, yes, if this is one of the requirements in your definition, then it should follow from the axioms of $T$. This is saying that $T$ knows $\delta$ codes a partial function, which is weaker from saying that $T$ proves that $\delta$ is a total function, so the statement $\forall x\exists y\delta(x,y)$ may fail to be provable in $T$. $\endgroup$ – Andrés E. Caicedo Jul 15 '13 at 6:15

The $\delta$ in the proof statement is a formula. In the language of $\text{PA}$, there is no symbol for the function $f$; but $\text{PA}$ is strong enough so that there is some formula $\delta(x, y)$ such that the following two things hold: $$f(n) = m \quad \text{ iff } \quad \text{PA} \vdash \delta(\underline{n}, \underline{m})$$ $$f(n) \neq m \quad \text{ iff } \quad \text{PA} \vdash \neg\delta(\underline{n}, \underline{m})$$

where $\underline{k}$ is the numeral for the number $k$. (A weaker statement would be to only assert the first biconditional; this weaker property of "weakly" representing a function holds of all $\Sigma_1^0$-functions).

Now, given that $f(\#(\theta)) = \#(\theta(\#(\theta)))$, where $\theta(x)$ is a formula in the language of $\text{PA}$, define a new formula as follows:

$$\alpha(x) = \exists y (\delta(x, y) \wedge \psi(y))$$

Finally, define $\varphi = \alpha(\#(\alpha))$. Then by our definitions:

$$\begin{align} \text{PA} \vdash \varphi & \leftrightarrow \alpha(\#(\alpha)) \\ & \leftrightarrow \exists y (\delta(\#(\alpha), y) \wedge \psi(y)) \\ & \leftrightarrow \exists y (y = \#(\alpha(\#(\alpha))) \wedge \psi(y)) \\ & \leftrightarrow \psi(\#(\alpha(\#(\alpha)))) \\ & \leftrightarrow \psi(\#(\varphi)) \end{align}$$

The step from $\exists y (\delta(\#(\alpha), y) \wedge \psi(y))$ to $\exists y (y = \#(\alpha(\#(\alpha))) \wedge \psi(y))$ is (if I'm not mistaken) the other step you mention. That follows from the fact that $\text{PA}$ is representing the function $f$ with $\delta$. Hence, for all $k$, if $f(n) = k$, then $\text{PA} \vdash \forall y (\delta(\underline{n},y) \leftrightarrow y = \underline{k})$.

Simiarly, if $T$ extends $\text{PA}$, it must also strongly represent all recursive functions, and so this applies to $T$ as well.

  • $\begingroup$ Thanks Alex! It is making much more sense! I wonder though - what are we trying to do with $\psi (y)$? Now if I understand correctly $\delta(\underline{x},y)$ as a representation of $f$ and saying that there is $\delta(x,y)$ iff $f(x) = y$ (per Andres above). Now am I correct in saying that x,y are the Godel numbers? I see that $\delta(\underline{x},y)$ is underlined. So we are constructing a diagonal for all $f(x) = y$? Thus somehow since $y = (\#(\alpha(\#(\alpha))))$ and that is self referential there will always be another n? I am not quite seeing why this would be - but also: $\endgroup$ – Relative0 Jul 15 '13 at 5:20
  • $\begingroup$ while I see why $\psi(y) = \#(\alpha(\#(\alpha))$ and that due to the $\wedge$ we have to have $\psi(y)$ furthermore since $y = \#(\alpha(\#(\alpha))$ we have $\leftrightarrow \psi(\#(\alpha(\#(\alpha))))$ but why $\psi(y)$ in the first place? $\endgroup$ – Relative0 Jul 15 '13 at 5:25
  • $\begingroup$ That's not quite the idea. I simply underlined the numbers to indicate that they were numerals, i.e. terms in the language of PA, not numbers (which are the objects we're theorizing about). Also, note we're not saying "$f(x) = y$ iff $\delta(\underline{x},\underline{y})$ is true"; we're saying "... iff PA proves $\delta(\underline{x},\underline{y})$". The difference between truth and what PA can prove is important. The former claim wouldn't really be informative, but the latter is key to the proof. Regarding your last question: $\endgroup$ – Alex Kocurek Jul 15 '13 at 8:02
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    $\begingroup$ ...we mention $\psi(y)$ because the theorem states that for any formula $\psi(y)$, there exists a sentence $\varphi$ with the desired properties. So somehow, the $\varphi$ we construct has to mention $\psi(y)$. $\endgroup$ – Alex Kocurek Jul 15 '13 at 8:03

For another freely available exposition of the diagonalization lemma, try Sec. 39 of my notes Gödel Without Tears. That should hopefully be rather clearer.

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    $\begingroup$ Wow! Astronomical! Thanks Peter - Alas I shall shed tears of joy instead of tears of sadness. $\endgroup$ – Relative0 Jul 15 '13 at 5:27

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