(Epistemic status: quickly-recounted lightly-edited cached state that I sent in response to an email thread on this topic, that I now notice had an associated public post. Sorry for the length; it was easier to just do a big unfiltered brain-dump than to cull, with footnotes added.)
here's a few quick thoughts i have cached about the proof of löb's theorem (and which i think imply that the suggested technique won't work, and will be tricky to repair):
#1. löb's theorem is essentially just the Y combinator, but with an extra level of quotation mixed in.^{[1]}
Y : (A → A) → A
Y f := let g be (λ s. f (s s)) in g g
löb : □ (□ A → A) → □ A
löb f := let g be (λ s. f (s "s")) in g "g"
the standard proof of löb's theorem (e.g. as it appears in the cartoon guide) is complicated by two points:
1a. one thing we need here is that the variable s
is sensical, and in particular we need some S with S ≃ (□ S → A)
. the existence of such an S follows from the diagonal lemma.^{[2]}
1b. the other thing we need here is to show that quotation and normal proof steps (function application or modus ponens or whatever) interact in the "obvious way", and don't gum each other up. for instance, we need that □(A → B) ∧ □(A)
yields □(B)
, showing that we can "apply quoted functions to quoted arguments", in order for syntax like f (s "s")
to be sensical.
if i were trying to make a guide to löb's theorem today, i'd break it down as above; for me, at least, this breakdown is significantly more intuitive than the cartoon guide (which, iirc, doesn't even go into the diagonal lemma, and just blends the Y combinator together with 1b).
this suggests that you're going to be hard-pressed to do any self-reference without routing through the nomal machinery of löb's theorem, in the same way that it's hard to do recursion in the lambda calculus without routing through the Y combinator--you don't need to use Y per se, but you're unlikely to find a simpler way to do generic recursion.
this also suggests that you're hiding all the work done by the normal proof of löb's theorem in the definition of "zip", and that this proof won't turn out to be simpler when "zip" is cashed out in full.
#2. fun fact: the diagonal lemma is also just an instance of the Y combinator with an extra level of quotation.
in fact, there's a proof of the diagonal lemma that goes by analogy to the proof-sketch of löb's theorem (above). very roughly speaking, the way that you avoid infinite regress is that, in the step 1a, you need an S with S ≃ □ S → Type
, and this S can be written manually because Type (or Proposition or w/e you want to use) is very rich (namely, it contains existential quantifiers and equality).^{[3]}
indeed, you can think of the diagonal lemma as simply an application of löb's theorem where the target A is Type (or Proposition or w/e).
in fact, this analogy is quite tight. one cool corollary is that we can pop open tarski's undefinability of truth, and replace the diagonal lemma with löb's theorem. the result is a new theorem that i call "the undefinability of normalization": no theory can prove that its terms normalize, on pain of contradiction.
(proof, in case it's not obvious: invoke löb's theorem on (not ∘ normalize : □ Bool → Bool)
, now you have syntax for a boolean value that's provably-equal to its own negation.)
#3. fun fact: löb's theorem does not hold for various breeds of normalized proofs (such as proofs with cuts eliminated--though exactly what counts as "normalized" in PA is somewhat subtle). in particular, when □ denotes normalized proofs, □(A → B) ∧ □(A)
does not yield □(B)
, because feeding a normalized argument into a normalized function does not yield a normalized output. and while we (who trust the theory) know that unnormalized-box(B) implies normalized-box(B), no consistent theory can prove its own version of that fact, by the undefinability of normalization!
...which yields the hypothesis that the One True Counterfactuals are those facts which have short normalized proofs.
(recall: i think it was patrick who first set forth the trolljecture that counterfactuals correspond to short proofs of material implications? and this trolljecture was shot down by the existence of spurious counterfactuals, which use löb to deduce arbitrary material implications with a ~constant proof-length. but that counterexample doesn't apply to normalized proofs! so the repaired trolljecture yet stands!)
(though i admit this hasn't risen high enough in my priority-queue for me to actually try to prove nice things about this bold new notion of counterfactuals. also they obviously can't be the whole answer, b/c normalized proofs about large worlds are large, etc.)
slighly more formally: both the Y combinator and löb's theorem are instantiations of a generalization of lawvere's fixpoint theorem to semicategories. in the case of the Y combinator, we use some domain (in the sense of domain theory) that serves as semantics for the untyped lambda calculus. in löb's theorem, iirc we use a semicategory where a morphism from A to B is a proof of □ A → B (tho this is an old cached thought that i haven't checked). that the latter is not a category follows from gödel incompleteness; there's no identity morphism at ⊥.
note: i dislike semicategories, and if you have a more elegant way to unify Y and löb, i'm eager to hear it. ↩︎
in other words: the santa clause sentence is just the natural/obvious type of the variable in the Y combinator (as modified to work with gödel-codes). if you want a better intuition for it, i recommend the exercise of trying to figure out what type the variable s
should have in the normal Y combinator. ↩︎
in fact, according to me, this is one of three keys to löb's theorem.
the first two keys are ofc gödel-codes + the Y combinator, with the caveat that making the Y combinator behave requires the diagonal lemma.
and the diagonal lemma is again just gödel-codes + the Y combinator, except that in this very specific case we can make the Y combinator behave without already needing the diagonal lemma, by hand-coding in a workaround using quantifiers and equalities and whatnot.
and the hand-coded workaround is kinda hacky. very roughly speaking, the trouble is that we want to let G(s)
be F (s "s")
, but showing that s "s"
is sensical relies on the diagonal lemma, which we're in the middle of proving and can't invoke.
and the solution is to let G(s)
instead be ∃ x. just x = try (s "s") ∧ F x
. we don't need to know that s "s"
is sensical, because we're naming a proposition, and we can just bolt on a clause asserting that s "s"
is sensical! we just build our sentence to say "s "s"
is sensical, and also ...", and thus sidestep the problem!
in other words: we're able to build a sentence that would be meaningful (it would mean a falsehood) even if s "s"
wasn't sensical, and this lets us form a sort of "guarded" version of the Y combinator (in the special case where we're producing a sentence) that doesn't depend critically on the sensicality of s "s"
. (which ends up making s "s"
be sensical in the actual use-case of G "G"
, and so the guard ends up redundant and harmless.)
(exercise: why can't you use this technique to avoid the invocation of the diagonal lemma from the proof of gödel's incompleteness theorem, i.e. löb's theorem where the target A is ⊥?)
this technique lets us slip around the need to invoke the diagonal lemma before we've finished proving the diagonal lemma. which is a critical step in the bootstrapping process.
...i don't have this point entirely distilled yet, but hopefully that gives at least some intuition for what i mean when i say that löb is just gödel-codes + the Y combinator + a hacky workaround to get you off the ground. ↩︎
an attempt to rescue what seems to me like the intuition in the OP:
(note that the result is underwhelming, but perhaps informative.)
in the lambda calculus we might say "given (f : A → A), we can get an element (a : A) by the equation (a := f a)".
recursive definitions such as this one work perfectly well, modulo the caveat that you need to have developed the Y combinator first so that you can ground out this new recursive syntax (assuming you don't want to add any new basic combinators).
by a directly analogous argument, we might wish define (löb f := f "löb f"). very direct! very quine-like!
and, this proof works perfectly well. ...modulo the caveat that you need to have developed a modified Y combinator first, so that you can ground out this recursive syntax (assuming you don't want to add any new axioms).
so if you prove the existence of a modified Y-combinator (that works w/ quotations), then you can add recursive syntax (modulo quotation) to your theory, and get a very "direct/quine-like" proof of löb's theorem.
the only hiccup is that the modified-Y-combinator is itself löb's theorem, so the above proof amounts to "once you have proven löb's theorem, there is a very direct and intuitive proof of löb's theorem", which... is perhaps less exciting :-p
(bonus fun fact: in type theory, there's an operator that forms quines. and the proof of löb's theorem is extremely simple, given this operator; you just take a (syntactic) term of type □ A → A, and feed it the quotation of the result of this whole operation, using the quining operator. the hiccup is, ofc, that the quining operator is itself just löb's theorem!)
various attempts to distill my objection:
the details of the omitted "zip" operation are going to be no simpler than the standard proof of löb's theorem, and will probably turn out to be just a variation on the standard proof of löb's theorem (unless you can find a way of building self-reference that's shorter than the Y combinator (b/c the standard proof of löb's theorem is already just the Y combinator plus the minimal modifications to interplay with gödel-codes))
even more compressed: the normal proof of löb is contained in the thing labeled "zip". the argument in the OP is operating under the assumption that a proof can make reference to its own gödel-number, and that assumption is essentially just löb's theorem.
a thing that might help is asking what specific proof-steps the step "1. the gödel-code P of this very proof proves C" cashes out into. obviously step 1 can't cash out into verifying (under quotation) every step of the entire proof (by checking for the appropriate prime-factors), because then the proof would require at least one step for every step in the proof plus three extra steps at the end, which would make the proof infinite (and infinite proofs admit contradictions).
so step 1 has to involve some indirection somehow. (cf: we can't get a fixpoint in λ-calculus by f (f (f ...
, we need some indirection, as occurs in (Y f)
aka (λ s. f (s s)) (λ s. f (s s))
, to make the term finite.)
this works fine, as long as you are ok w/ the gödel-code in step 1 being not P, but some different proof of the same conclusion (analogous to the difference between f (Y f)
and Y f
), and so long as you have the analog of the Y combinator (modified to interplay with gödel-codes).
but the analog of the Y combinator is precisely löb's theorem :-p
if i understand your OP correctly, we can factor it into three claims:
(1) is löb's theorem.
(2) is an indirect proof of löb's theorem, from the assumption of löb's theorem (in direct analogy with how, once we've found syntax Y f
for the fixpoint of f, we can wrap it in f to find an alternate syntax, namely f (Y f))
.
(3) is false (i'm fairly confident)--you can get real close, but you can't quite get there (in direct analogy to how you can get λ-term Y f
that means the same thing as f (Y f)
, but you can't get a λ-term fix f
that is syntactically exactly f (fix f)
, on pain of infinite syntax).
afaict, there's two ways to read this proposal (if we port across the analogy to λ-calculus), according to whether you cash "zip" out to a recursive instance of the whole proposal, vs whether you cash "zip" out to some different proof of löb's theorem:
"i hear that people find the Y combinator confusing. instead of using (Y f)
to name the fixpoint of f, why not use f (f (f (f (...
ad infinitum?"
"i hear that people find the Y combinator confusing. instead of using (Y f)
to name the fixpoint of f, why not use f (Y f)
?"
the answer to (1) is that it doesn't make sense unless you allow infinite syntax trees (which aren't permitted in normal math b/c infinite proofs admit contradictions).
the answer to (2) is that, while (Y f)
may be somewhat tricky to grasp, f (Y f)
is surely no simpler :-p
this suggests that you're going to be hard-pressed to do any self-reference without routing through the nomal machinery of löb's theorem, in the same way that it's hard to do recursion in the lambda calculus without routing through the Y combinator
If by "the normal machinery", you mean a clever application of the diagonal lemma, then I agree. But I think we can get away with not having the self-referential sentence, by using the same y-combinator-like diagonal-lemma machinery to make a proof that refers to itself (instead of a proof about sentences that refer to themselves) and checks its own validity. I think I if or someone else produces a valid proof like that, skeptics of its value (of which you might be one; I'm not sure) will look at it and say "That was harder and less efficient than the usual way of proving Löb using the self-referential sentence and no self-validation". I predict I'll agree with that, and still find the new proof to be of additional intellectual value, for the following reason:
The skeptic response to that will be to say that those peoples' intuitions are the wrong way to think about y-combinator manipulation, and to that I'll be like "Maybe, but I'm not super convinced their perspective is wrong, and in any case I don't mind meeting them where they're at, using a proof that they find more intuitive.".
Summary: I'm pretty confident the proof will be valuable, even though I agree it will have to use much of the same machinery as the usual proof, plus some extra machinery for helping the proof to self-validate, as long as the proof doesn't use sentences that are basically only about their own meaning (the way the sentence is basically only about its own relationship to the sentence C, which is weird).
This sentence is an exception, but there aren't a lot of naturally occurring examples.
No strong claim either way, but as a datapoint I do somewhat often use the phrase "I hereby invite you to <event>" or "I hereby <request> something of you" to help move from 'describing the world' to 'issuing an invitation/command/etc'.
which self-referential sentence are you trying to avoid?
it keeps sounding to me like you're saying "i want a λ-calculus combinator that produces the fixpoint of a given function f, but i don't want to use the Y combinator".
do you deny the alleged analogy between the normal proof of löb and the Y combinator? (hypothesis: maybe you see that the diagonal lemma is just the type-level Y combinator, but have not yet noticed that löb's theorem is the corresponding term-level Y combinator?)
if you follow the analogy, can you tell me what λ-term should come out when i put in f
, and how it's better than (λ s. f (s s)) (λ s. f (s s))
?
or (still assuming you follow the analogy): what sort of λ-term representing the fixpoint of f would constitute "referring to itself (instead of being a term about types that refer to themselves)"? in what sense is the term (λ s. f (s s)) (λ s. f (s s))
failing to "refer to itself", and what property are you hoping for instead?
(in case it helps with communication: when i try myself to answer these questions while staring at the OP, my best guess is that you're asking "instead of the Y combinator, can we get a combinator that goes like f ↦ f ???
?", and the two obvious ways to fill in the blanks are f ↦ f (Y f)
and f ↦ f (f (f (...
. i discussed why both of those are troublesome here, but am open to the possibility that i have not successfully understood what sort of fixpoint combinator you desire.)
(ETA: also, ftr, in the proof-sketch of löb's theorem that i gave above, the term "g "g""
occurs as a subterm if you do enough substitution, and it refers to the whole proof of löb's theorem. just like how, in the version of the Y combinator given above, the term g g
occurs as a subterm if you do enough β-reduction, and it refers to the whole fixpoint. which i note b/c it seems to me that you might have misunderstood a separate point about where the OP struggles as implying that the normal proof isn't self-referring.)
((the OP is gonna struggle insofar as it does the analog of asking for a λ-term x
that is literally syntactically identical to f x
, for which the only solution is f (f (f (...
which is problematic. but a λ-term y
that β-reduces pretty quickly to f y
is easy, and it's what the standard constuction produces.))
At this point I'm more interested in hashing out approaches that might actually conform to the motivation in the OP. Perhaps I'll come back to this discussion with you after I've spent a lot more time in a mode of searching for a positive result that fits with my motivation here. Meanwhile, thanks for thinking this over for a bit.
well, in your search for that positive result, i recommend spending some time searching for a critch!simplified alternative to the Y combinator :-p.
not every method of attaining self-reference in the λ-calculus will port over to logic (b/c in the logical setting lots of things need to be quoted), but the quotation sure isn't making the problem any easier. a solution to the OP would yield a novel self-reference combinator in the λ-calculus, and the latter might be easier to find (b/c you don't need to juggle quotes).
if you can lay bare the self-referential property that you're hoping for in the easier setting of λ-calculus, then perhaps others will have an easier time understanding what you want and helping out (and/or you'll have an easier time noticing why your desires are unsatisfiable).
(and if it's still not clear that löb's theorem is tightly connected to the Y combinator, such that any solution to the OP would immediately yield a critch!simplified self-reference combinator in the λ-calculus, then I recommend spending a little time studying the connection between the Y combinator, löb's theorem, and lawvere's fixpoint theorem.)
I know very little about this area, but I suspect that a writeup like this classic explanation of Godel Incompleteness might be a step in the right direction: Godel incompleteness.
That thing is hilarious and good! Thanks for sharing it. As for the relevance, it explains the statement of Gödel's theorem, but not the proof it. So, it could be pretty straightforwardly reworked to explain the statement of Löb's theorem, but not so easily the proof of Löb's theorem. With this post, I'm in the business of trying to find a proof of Löb that's really intuitive/simple, rather than just a statement of it that's intuitive/simple.
It would kind of use assumption 3 inside step 1, but inside the syntax, rather than in the metalanguage. That is, step 1 involves checking that the number encoding "this proof" does in fact encode a proof of C. This can't be done if you never end up proving C.
One thing that might help make clear what's going on is that you can follow the same proof strategy, but replace "this proof" with "the usual proof of Lob's theorem", and get another valid proof of Lob's theorem, that goes like this: Suppose you can prove that []C->C, and let n be the number encoding a proof of C via the usual proof of Lob's theorem. Now we can prove C a different way like so:
Step 1 can't be correctly made precise if it isn't true that n encodes a proof of C.
Yes to both of you on these points:
(I'll write a separate comment on Eliezer's original question.)
Thanks for your attention to this! The happy face is the outer box. So, line 3 of the cartoon proof is assumption 3.
If you want the full []([]C->C) to be inside a thought bubble, then just take every line of the cartoon and put into a thought bubble, and I think that will do what you want.
LMK if this doesn't make sense; given the time you've spent thinking about this, you're probably my #1 target audience member for making the more intuitive proof (assuming it's possible, which I think it is).
ETA: You might have been asking if there should be a copy of Line 3 of the cartoon proof inside Line 1 of the cartoon proof. The goal is, yes, to basically to have a compressed copy of line 3 inside line 1, like how the strings inside this Java quine are basically a 2x compressed copy of the whole program:
https://en.wikipedia.org/wiki/Quine_(computing)#Constructive_quines#:~:text=java
The last four lines of the Java quine are essentially instructions for duplicating the strings in a form that reconstructs the whole program.
If we end up with a compressed proof like this, you might complain that the compression is being depicted as magical/non-reductive, and ask for cartoon that breaks down into further detail showing how the Quinean compression works. However, I'll note that your cartoon guide did not break down the self-referential sentence L into a detailed cartoon form; you used natural language instead, and just illustrated vaguely that the sentence can unpack itself with this picture:
I would say the same picture could work for the proof, but with "sentence" replaced by "proof".
Forgive me if I’m missing something but as far as I can tell, this line of inquiry was already pursued to its logical conclusion (namely, an Agda-verified quinean proof of Löb’s theorem) in this paper by Jason Gross, Jack Gallagher and Benya Fallenstein.
Thanks for raising this! I assume you're talking about this part?
They explore a pretty interesting set-up, but they don't avoid the narrowly-self-referential sentence Ψ:
So, I don't think their motivation was the same as mine. For me, the point of trying to use a quine is to try to get away from that sentence, to create a different perspective on the foundations for people that find that kind of sentence confusing, but who find self-referential documents less confusing. I added a section "Further meta-motivation (added Nov 26)" about this in the OP.
Motivation
Löb's theorem is pretty counterintuitive. Speaking informally about one or more agents engaging in logically valid reasoning, the theorem says something like this:
Formally, the theorem just says □(□C→C)→□C , where □X means "X is provable". This is most weird if X is a false statement, because it means you can't prove that you can't prove X (Gödel's theorem).
(Also: the box symbol above is not an unrendered symbol; it's supposed to be a box.)
Meta-motivation
I'd like to make Löb's theorem more intuitive for humans, because it shows how agents can sidestep the need to mentally simulate each other in games, by instead using reasoning/arguments that happen to self-fulfill in a good and defensible way. Basically, Löbian reflection helps agents to avoid metacognitive stack overflows when thinking about themselves and each other, and I want more human beings to understand how that can work, and sometimes already does work, in real world agents.
State of the art
The best attempt I've seen to make Löb's theorem more intuitive is Eliezer Yudkowsky's Cartoon Guide to Löb's Theorem, which is still quite confusing. The confusingness comes from thinking about a self-referential statement (Ψ on Wikipedia; S on Arbital) that's used to carry out the proof. The statement basically says "If this statement is provable, then C." Dealing with that sentence is pretty cumbersome, and requires a lot of fiddling around with nested statements and implications.
Doing better
I think we can make a new proof of Löb's theorem that doesn't use that weird self-referential sentence, by instead making the proof of Löb's theorem itself self-referential. Page 15 of the following paper poses an open problem on how to do this, which I think is possible to resolve affirmatively:
Here's a screenshot from it:
If we can make a proof like that work, we could then use the following much shorter and simpler cartoon guide to explain it:
In other words, I want to write a proof of Löb's theorem that is structured much like a Quine. I'd like the details to (eventually) be really crisp and readable, so it can be peer reviewed for correctness, at a level of rigor comparable to this earlier paper on Löb's theorem for proof systems with bounded length:
As a possible start to writing such a proof, I think some of the same machinery (e.g., the diagonal lemma) from Boolos's textbook "The Logic of Provability" can be used to cook up self-referential proofs fitting the cartoon template above... thereby making Löb's theorem less mysterious and more intuitive forever.
Further meta-motivation (added Nov 26)
A key reason I'm interested in having a self-referential self-validating proof, rather than a normal-ish proof about a sentence that is self-referential (like Ψ on Wikipedia), is that human documents often refer to themselves, but human sentences rarely refer directly to themselves in isolation. This sentence is an exception, but such utterances are generally rare. So, making Löb more intuitive to humans either means
The aim of this post is to try for the second approach. Notice how you didn't think the previous sentence was weirdly self-referential? That's because it referred to the whole document that it's part of, rather than just narrowly referring directly to itself. I'm not quite sure why humans are more comfortable with that than with directly-and-narrowly self-referential sentences. Maybe it's because I thing needs to have enough non-self-referential content before humans will care about it enough to tolerate it talking about itself. I'm not sure.
Countering naive impossibility arguments (added Nov 26)
It's in fact possible to use Löb's theorem to construct a self-referential proof of the kind illustrated above. So, naive arguments that "Proofs can't be self-referential like that" are not going to hold up. The hard thing I'm asking for is to make such a proof — of Löb's theorem specifically — without going through the traditional proof of Löb, i.e., without using a narrowly-and-directly-self-referential sentence (like Ψ on Wikipedia).
Difficulty assessment
I think getting a Qunean proof to work here is probably a bit easier than the JSL paper linked directly above. Like, I think if I spent 3 months on it (which is how long the JSL paper took), then I could do it. I do think getting it right involves mentally syncing with some fundamental and standard machinery in provability logic, which can take a bit of time if you've never worked in the area before (as I hadn't).
Anyway, if you can figure out how to write this proof, please post your solution or solution attempts here :)