How to prove that a statement is a theorem using Hilbert's system?

Question

I'm looking for an actual step-by-step way of proving that a statement is a theorem using Hilbert's system. For instance:

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As can be seen from the above picture, the solution consists in a series of steps starting from one of the three axioms (I can only use P2, P3 and P4 and Modus ponens from this https://en.wikipedia.org/wiki/Hilbert_system#Logical_axioms). However, how do I know which axiom to start with? Should I go backwards, starting from the actual statement provided and work my way until I find an axiom? I've also seen ways where you use Modus Ponens (a, a->b/b), but it's unclear exactly the process.

Can anybody help me with the actual step-by-step process on how to do this?

Modus ponens is used in most of the steps of the derivation above. The third step is derived with the tautology $\vDash [\lnot C \to (B \to (A \to B))] \to [B \to (\lnot C \to (A \to B))]$ that, by Completenss, gives : $\vdash \ldots$ and the second step with mp. — Mauro ALLEGRANZA, Jun 28 '15 at 19:11
Hilbert-style proof system is not easy to find proof; you have to build-up a basic "catalogue" of theorems (tautologies) and meta-theorems (like the Deduction Th) and re-use them. You can see Geoffrey Hunter, Metalogic : An Introduction to the Metatheory of Standard First Order Logic (1971), page 71-on. — Mauro ALLEGRANZA, Jun 28 '15 at 19:31
But how do I proceed with the actual solving process? Do I start from the bottom, or from the top? — Luca Matteis, Jun 28 '15 at 20:43
You have to use other proof systems, like Tableaux or Sequent calculus; they provide bottom-up procedures for proving theorems. — Mauro ALLEGRANZA, Jun 28 '15 at 20:47
But Tableaux will only tell me if it's valid/invalid. Here I actually have to show whether it's a theorem — Luca Matteis, Jun 28 '15 at 21:10
Maybe this answer and also this one might help to shed you some light on proofs in a Hilbert-style system. — Bruno Bentzen, Jun 29 '15 at 02:13
By completeness, if it is valid, it is provable and Tableaux is a proof procedure, like Natural Deduction and Sequent calculus. — Mauro ALLEGRANZA, Jun 29 '15 at 10:28

score 1 · Accepted Answer · answered Jun 29 '15 at 12:17

I figured out a way of doing these, from the bottom-up, starting from the statement provided without having to induce anything upfront. First off you should look into some algebra for this system (which is deduced from the main 3 axioms) here: https://it.wikibooks.org/wiki/Logica/Calcolo_delle_proposizioni/Sistema_di_Hilbert#Regole_derivate - it's in italian but what matters are the actual properties.

Now, let's say you're given a statement and you have to show it's a theorem in Hilbert's system. First thing is you write it down your page so you can write stuff on top of it:

$¬(A → B) → (B → C)$

Whenever you're confronted with something like $¬(A → B)$ - a NOT of two terms - you have to find a way to get rid of the NOT in front of it either by using the contrast property where $A → B$ is equal to $¬B → ¬A$, or by using the so called exchange of premises property where $A → (B → C)$ is equal to $B → (A → C)$. So from the above we go ahead and apply the exchange of premises property and exchange the $¬(A → B)$ with the $B$:

$B → (¬(A → B) → C)$
$¬(A → B) → (B → C)$

We moved it inside so we can get rid of the NOT, which as I said earlier, should be our main goal. In fact, now that $¬(A → B)$ is inside, we can apply the contrast property to help us negate it and therefore get rid of the NOT as such:

$B → (¬C → (A → B))$
$B → (¬(A → B) → C)$
$¬(A → B) → (B → C)$

Great so remember that our goal is to continue this process until we find a statement which is an instance of one of the 3 axioms of Hilbert (linked in the question). It seems we're getting close to match Axiom P2 $A → (B → A)$, if we just move the B inside:

$¬C → (B → (A → B))$
$B → (¬C → (A → B))$
$B → (¬(A → B) → C)$
$¬(A → B) → (B → C)$

Now we simply need to move ¬C out and we matched our P2 Axiom and therefore showed that the initial statement was in fact a theorem.

$¬C \vdash B → (A → B))$
$¬C → (B → (A → B))$
$B → (¬C → (A → B))$
$B → (¬(A → B) → C)$
$¬(A → B) → (B → C)$

Interesting approach. Did you prove each of the rules that you used here first using just substitution and detachment? Also, (¬C→(B→(A→B))) is a special case of a more general theorem. — Doug Spoonwood, Jul 04 '15 at 17:03

How to prove that a statement is a theorem using Hilbert's system?

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