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Take a proof to be a sequence (I'm working in Hilbert style system for simplicity). I have heard of infinitely long proofs.

When we speak of infinitely long proofs, are we talking about proofs indexed by $\mathbb N$, or some other ordinal? Or perhaps uncountably long proofs?

What kind of use can you think of for such proofs (especially proofs indexed by something larger than $\mathbb N$)?

EDIT: If we regard computations as derivations in some formal system, then an endless loop could be an example of an infinitely long derivation? But I think that's not an example of a derivation that needs to be indexed by something bigger than $\mathbb N$.

stranger
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It's actually vastly easier to view proofs as trees in this context, so I'll do that.

Generally, an "infinitely long proof" is more accurately a well-founded infinitely-branching tree (or something morally equivalent) - the idea being that we're working in some infinitary logic (or some logic of similar nature) where the basic deductive steps might involve infinitely many hypotheses. "Well-founded" means that the tree has no infinite paths, that is, no infinite regress: a node is "justified" by its immediate successors, with the root being the conclusion of the proof.

Now the point is that there is a notion associated to a well-founded tree which serves as a decent proxy for proof length - namely, the rank of the tree. Given a well-founded tree $T$, there is a unique function $\nu$ from the nodes of $T$ to the ordinals such that for each node $\alpha$ we have $$\nu(\alpha)=\sup\{\nu(\beta)+1: \alpha\prec \beta\}$$ (we take the supremum of the emptyset to be $0$, which means that $\nu(\alpha)=0$ for every terminal node $\alpha$). When we speak of the "length" of a proof in an infinitary system, this is what we (usually) mean. There are well-founded trees of arbitrary ordinal rank - while we generally are interested in at-most-countably-branching trees, which always have rank $<\omega_1$, we're not a priori limited to such.

As for their application, they show up in the study of infinitary logic in several ways (and Barwise's book has a good section on them) but they're also used to study classical first-order logic by replacing a complicated finitary system by an infinitary system which nonetheless has some better features; this happens with reasonable frequency in ordinal analysis, see e.g. here).

Noah Schweber
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For starters, you could prove Peano's Fifth Axiom, a.k.a. the Axiom of Induction.

subrosar
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    This is a good thought but could use some elaboration. – spaceisdarkgreen Aug 22 '19 at 04:10
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    Ah so something like a proof by cases? $\omega$ cases for induction on $\omega$ and a higher ordinal number of cases for induction on a higher ordinal? – stranger Aug 22 '19 at 04:26
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    You'd need an appropriate rule in the infinitary logic. But the obvious rule is to derive $\forall,n;P(n)$ from $P(0)$, $P(1)$, $P(2)$, etc, and then each instance of the induction axiom schema can be proved (without using that schema). And in a second-order logic, the induction axiom can be proved as a single statement, by universalizing $P$. So yes. – Toby Bartels Aug 22 '19 at 16:48
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    @stranger The infinitary rule Toby Bartels is referring to is the $\omega$-rule: similar to the finite $\wedge$-introduction rule, "From $A$ and $B$ infer $A\wedge B$," the $\omega$-rule says "From $A(n)$ for each $n$ infer $\forall x A(x)$." There are actually many different kinds of $\omega$-rule, since it's usually more interesting to look at weaker versions of the full $\omega$-rule, and these come up a lot in proof theory. – Noah Schweber Aug 22 '19 at 21:11
  • I don't know much about this topic myself, so if one of you (@TobyBartels or @NoahSchweber) wanted to expand this into a full length post I would be happy to delete mine. – subrosar Aug 22 '19 at 22:56
  • Well, I only know about the full $\omega$-rule, so @Noah Schweber knows more than I do. – Toby Bartels Aug 25 '19 at 05:25