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I am trying to demonstrate to students how useful the deduction theorem for Hilbert-Style axiomatic systems turns out to be, and to emphasize that using the deduction theorem shows us that an axiomatic proof does exist without providing us with any real idea of how that axiomatic proof would go. To this end, I wanted to give them a homework problem to prove something axiomatically where it is easy to use the deduction theorem to show that there is some axiomatic proof, but where the axiomatic proof is difficult to construct. I was thinking about having them prove that the conditional $\rightarrow$ is transitive, i.e: $(P \rightarrow Q) \rightarrow ((Q \rightarrow R) \rightarrow (P \rightarrow R))$ is a theorem Super easy to show using deduction theorem. But I just realized that I have no idea how to give a genuine axiomatic proof of this formula. I am using the following axiom system:

(A1) $X \rightarrow (Y \rightarrow X)$

(A2) $(X \rightarrow (Y \rightarrow Z)) \rightarrow ((X \rightarrow Y) \rightarrow (X \rightarrow Z))$

(A3) $(\neg X \rightarrow \neg Y) \rightarrow ((\neg X \rightarrow Y) \rightarrow X)$

Can anyone either

(a) give me an axiomatic proof of $(P \rightarrow Q) \rightarrow ((Q \rightarrow R) \rightarrow (P \rightarrow R))$, or

(b) think of a better example where the formula is relatively hard to prove axiomatically but where it is easy to use the deduction theorem to demonstrate that such a proof does exist.

Dkn557
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Keith McPartland
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    Welcome to MSE. For some basic information about writing mathematics at this site see, e.g., basic help on mathjax notation, mathjax tutorial and quick reference, main meta site math tutorial and equation editing how-to. – José Carlos Santos Aug 29 '24 at 17:19
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    You can find many examples on this site. Search for Hilbert-style and similar. – Mauro ALLEGRANZA Aug 29 '24 at 17:23
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    All the proofs of the deduction theorem that I know of do easily give a concrete procedure for taking a proof in "more natural deduction style" and converting it to a proof in the Hilbert-style axiomatic system. Sure, it tends to blow up exponentially in practice, but it's a straightforward procedure. – Daniel Schepler Aug 29 '24 at 17:28
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    Under Curry-Howard correspondence, the natural proof would be $\lambda f. \lambda g. \lambda x. g(fx)$. Then, abstracting to SKI-combinator calculus, that becomes $\lambda f. \lambda g. S(Kg)f = \lambda f. S(S(KS)K)(Kf) = S(K(S(S(KS)K)))K$ (where $K$ corresponds to A1 and $S$ corresponds to A2). So, that points to the existence of a Hilbert-style proof with 15 steps (8 axiom instantiations and 7 applications of modus ponens). – Daniel Schepler Aug 29 '24 at 17:35
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    Maybe a simpler example where it's hard to come up with an axiomatic proof other than using the construction from the deduction theorem: $(P\to(Q\to R))\to(Q\to(P\to R))$. Alternatively, the similar $(Q\to R)\to((P\to Q)\to(P\to R))$ has a shorter axiomatic proof corresponding to $S(KS)K$ (7 steps) but that would still be hard to come up with just from scratch. – Daniel Schepler Aug 29 '24 at 17:38
  • @DanielSchepler - Very nice explanation! You should write an answer with the same content as your comments. I'll upvote it! – Taroccoesbrocco Aug 29 '24 at 18:51
  • @Taroccoesbrocco I'm not convinced that this advanced method would be an appropriate answer to the original question. Other than the comment about the proof of the deduction theorem being easily translated into an algorithm, which would be a duplicate of lots of other answers on this site. – Daniel Schepler Aug 29 '24 at 19:27
  • An example of converting a proof with DT to one without: 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 – Bram28 Sep 01 '24 at 19:27

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