Hilbert Calculus statement proof

Question

I want to prove in $HPC$ that

$$ \vdash_{HPC} (A\rightarrow B)\rightarrow ((B\rightarrow C) \rightarrow (A\rightarrow C ))$$

I tried using different combinations of the $A\rightarrow (B \rightarrow A)$ and $ (A \rightarrow (B \rightarrow C)) \rightarrow ((A \rightarrow B) \rightarrow (A \rightarrow C)) $ axioms but didn't reach the result. What am I missing?

thanks.

Are you interested in a proof in HPC without using deduction theorem? — Taroccoesbrocco, Jul 29 '18 at 12:31

score 0 · Accepted Answer · answered Jul 29 '18 at 11:17

0

Solution:

Using the deduction theorem for $HPC$, this is correct iff: $$ A\rightarrow B, B \rightarrow C, A \vdash_{HPC} C$$

which is easily proven by using the $MP$ rule 2 times.

answered Jul 29 '18 at 11:17

Eloo

473

And if you really want to avoid the deduction theorem, then take this "MP twice" proof and run it through the proof of the deduction theorem. – Andreas Blass Jul 29 '18 at 16:14
@AndreasBlass As often holds, I don't think that's very efficient for producing an axiomatic proof of a relatively small size. If you do that, is the proof, including the axioms, more than 16 steps? – Doug Spoonwood Jul 29 '18 at 19:42
1

@DougSpoonwood The length of the proof would depend on the details of the axiom system. For example, $X\to X$ would be a reasonable axiom, but it can also be proved in 5 lines from the axiom schemes explicitly mentioned in the question. With just these two axiom schemes, I'd expect a direct application of the proof of the deduction theorem to take somewhat more than 16 lines, because the deduction theorem is applied 3 times. The advantage of this method would not be in the brevity of the resulting proof but in not needing any intelligence to find it. – Andreas Blass Jul 29 '18 at 20:36
@AndreasBlass I've studied some different axiom sets for what might get called the intuitionistic implicational propositional calculus (in significant measure by reading Dolph Ulrich's work here and associated work: https://web.ics.purdue.edu/~dulrich/C-pure-intuitionism-page.htm.) I have NOT seen any set of independent axioms where (X$\rightarrow$X) is an axiom as opposed to being a theorem. If you know of any such axiom sets that perhaps might be of interest. I'm not sure that the literature has such an axiom set in it. – Doug Spoonwood Jul 29 '18 at 21:21

score 0 · Answer 2 · answered Jul 29 '18 at 19:41

I use Polish notation. This means that well-formed formulas get re-defined such that every wff of the form ($\alpha$$\rightarrow$$\beta$) gets translated to C$\alpha$$\beta$. I don't spell out the substitutions made in the below, but I use spacing to show that a substitution follows the form of the axiom. The basic idea runs as follows: we prove CCbcCCabCac and then change the positions of 'Cbc' and 'Cab'.

1. CaCba axiom 
2. CCaCbcCCabCac axiom 
3. C CCaCbcCCabCac C Cbc CCaCbcCCabCac 1
4. CCbcCCaCbcCCabCac 3, 2
5. CC Cbc C CaCbc CCabCac CC Cbc CaCbc C Cbc CCabCac 2
6. CCCbcCaCbcCCbcCCabCac 5, 4
7. C Cbc C a Cbc 1
8. CCbcCCabCac 6, 7
9. CC Cbc C Cab Cac CC Cbc Cab C Cbc Cac 2
10. CCCbcCabCCbcCac 9, 8
11. C CCCbcCabCCbcCac C Cab CCCbcCabCCbcCac 2
12. CCabCCCbcCabCCbcCac 11, 10
13. CC Cab C CCbcCab CCbcCac CC Cab CCbcCab C Cab CCbcCac 2
14. CCCabCCbcCabCCabCCbcCac 13, 12
15. C Cab C Cbc Cab 1
16. CCabCCbcCac 14, 15

Hilbert Calculus statement proof

2 Answers2