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In these systems of deduction, we assume some hypotheses $A_1, \dots, A_n$ and apply axioms and rules such as modus ponens to derive a conclusion $B$. But we also have the deduction theorem, which implies $A_1, \dots, A_n \vdash B$ iff $\emptyset \vdash A_1 \to \dots \to A_n \to B$. (Here $\to$ is right-associative.) So why allow hypotheses at all? Wouldn't it simplify the proofs of many (meta)theorems if we always forced ourselves to prove $\emptyset \vdash A \to B$ instead of $A \vdash B$? Are there any real issues with this, like a super-polynomial blowup in proof size, or anything like that?

Glenn Sun
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  • Usual math practice is $A_1, \ldots, A_N \vdash B$ where $B$ is a theorem of our mathematical theory and the $A_i$'s are the axioms. – Mauro ALLEGRANZA May 03 '23 at 07:46
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    What do you do if you have infinitely many hypotheses $\Sigma = { A_1, A_2, \dots }$? We can make sense of $\Sigma \vdash B$, but $A_1 \rightarrow A_2 \rightarrow \dots \rightarrow B$ isn't a well-formed sentence because it is infinitely long, so $\emptyset$ can't possibly prove it. – Nate Eldredge May 03 '23 at 13:44
  • Anyway, how would "tying your hands" like this simplify proofs? If you are trying to prove some meta-theorem whose conclusion is $A \vdash B$, then if it's more convenient to prove $\emptyset \vdash A \rightarrow B$, you are already perfectly free to do that and then add one more line that says "By the deduction theorem". – Nate Eldredge May 03 '23 at 13:49

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First note that when doing meta-theory, we are often not interested in creating an actual formal proof at all. We just want to know that something can be proven. And for doing that, the Deduction Theorem is super useful.

Also, if you do try and create a proof in a Hilbert system, it turns out that it is typically far easier to prove $A_1, \dots, A_n \vdash B$ than to prove $\emptyset \vdash A_1 \to (\dots \to (A_n \to B)..)$ ... the latter is often an absolutely heinous proof and next to impossible to figure out from scratch. As such, if we really want a proof, it is common practice to first prove $A_1, \dots, A_n \vdash B$ and then, using the constructive nature of the proof of the Deduction Theorem, systematically transform that proof into a proof of $\emptyset \vdash A_1 \to (\dots \to (A_n \to B)..)$. Here is an example conversion, which also shows how the former proof is so much easier than the latter.

Bram28
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  • "next to impossible to figure out from scratch" - I ofen see this, but it's not that hard. One can learn a bag of tricks and get really very good at writing Hilbert proofs directly. There's just no point in putting in the required practice (and rightfuly, people don't). The most I get out of it is a few karma points every few months when somebody asks an "is my insane random axiom system complete for classical logic?" question here. And I can earn a few pounds while tutoring philosophy majors who are forced to sign up for the world's toughest modal logic courses. – Z. A. K. May 03 '23 at 13:57
  • @Z.A.K. Well, I'm pretty good with proofs, but I really don;t know if I could have produced the eventual proof at the bottom of this post from scratch. And this is just a simple contraposition: https://math.stackexchange.com/questions/4607540/axiomatic-proof-of-%e2%8a%a2a%e2%86%92b%e2%86%92%c2%acb%e2%86%92%c2%aca-without-using-the-deduction-theorem/4615857#4615857 – Bram28 May 03 '23 at 14:01
  • Well, they're not gonna be short for sure - especially proofs constructed by the algorithm implicit in the deduction theorem, which cqn triple the size. But one can construct proofs of similar length by hand: with good intuition for which kinds of helpful tautologies do have shorter proofs than your goal, it often suffices to sketch out a proof using such tautologies, then fill in their proofs later (e.g. what I do here along with another trick: https://math.stackexchange.com/questions/4567579/axiomatic-derivation-of-conditional-from-biconditional/4569638#4569638). – Z. A. K. May 03 '23 at 14:38
  • There's of course no point doing this for propositional logic. The only point of Hilbert systems in 2022 is that certain proof-theoretic results are much easier to prove for axioms than for rules of inference, and Hilbert systems onoy have one rule of inference. If you actually need a Hilbert proof, the easiest way of getting one is finding and translating a sequent calculus or natural deduction proof - as you did in your excellent linked answer. The exception is the occasional modal logic which only has a Hilbert presentation. But finding direct Hilbert proofs isn't nigh-impossible at all. – Z. A. K. May 03 '23 at 14:46
  • @Z.A.K. Yes, proofs can be shortened, you're right. Thanks for the link too! – Bram28 May 03 '23 at 15:39
  • I'm aware that it's generally easier to prove $A_1, \dots, A_n \vdash B$ than $\emptyset \vdash A_1 \to \dots \to A_n \to B$, but is there an exponential sized gap in proof length, or is it just polynomial? From a computer science perspective, this is the important point that would make allowing hypotheses interesting. – Glenn Sun May 03 '23 at 22:57
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    @GlennSun If you read my linked post you'll see that the systematic method of transforming one into the other roughly multiplies the number of lines in the proof by a factor of 3 for each hypothesis being discharged. So with that method it will become exponential. And while proofs can be shorted, my strong suspicion is that you can't get around this kind of method .. my strong guess is that one can probably come up with some worst case scenario that proves it's going to be exponential indeed. – Bram28 May 04 '23 at 13:16
  • @Z.A.K. could you provide an example of a modal logic that only has an axiomatic presentation? – PW_246 Jul 17 '23 at 00:42
  • @PW_246 Take e.g. any of the systems discussed in arXiv:2307.05040. – Z. A. K. Jul 17 '23 at 00:56