# Proof of transitivity in Hilbert Style

We can use the following axioms: \begin{align} &A\to(B\to A)&\tag{A1}\\ &[A\to(B\to C)]\to[(A\to B)\to(A\to C)]&\tag{A2}\\ &(\lnot A\to\lnot B)\to(B\to A)&\tag{A3} \end{align}

We need to prove: $$A\to B, B\to C\vdash A\to C$$

The hint is to use The Deduction Theorem.

• Whose handwriting is it? – Asaf Karagila Jan 30 '13 at 18:40
• What deductive rules do you have available? – Carl Mummert Jan 30 '13 at 18:49

I assume you can use modus ponens as a deductive rule. Here is a Hilbert-style proof. As you can see, there is no reason to use the deduction theorem.

1. $A \to B$ [assumption]
2. $B \to C$ [assumption]
3. $(B \to C) \to (A \to (B \to C))$ [by A1]
4. $A \to (B \to C)$ [modus ponens, 2 and 3]
5. $(A \to (B \to C)) \to ((A \to B) \to (A \to C))$ [by A2]
6. $(A \to B) \to (A \to C)$ [modus ponens, 4 and 5]
7. $A \to C$ [modus ponens, 1 and 6]
• Thanks! and in the way we write it in the course we have to use deduction to get the assumptions.. – Rachel Bernoulli Jan 30 '13 at 19:12

$$A \to B, B \to C \vdash A \to C$$

by deduction theorem

$$A \to B, B \to C, A \vdash C$$

applying $B \to C$

$$A \to B, A \vdash B$$

applying $A \to B$

$$A \vdash A$$

tautology.

• you don't need to throw away hypothesis when you use them, but I'm doing so here because I knew in advance we only use each one once.. and it is simpler for the reader to see what's happening. – user58512 Jan 30 '13 at 18:37
• @CarlMummert, H1 means first hypothesis.. H2 is the second. can you follow it now? – user58512 Jan 30 '13 at 19:23
• I apologize; it may be that I am missing something, but I do not find it easy to read. I believe what you are saying in the first line is: "by the deduction theorem, it is sufficient to show ...". Certainly you must not be assuming the statement in the first line, which is what you are proving after all. Then "applying $B \to C$" must mean "to show this, because we would be able to use modus ponens and $B \to C$, it is sufficient to show". Then $A \to B, A \vdash B$ is just modus ponens, which is already rule of inference, so we already know $A \to B, A \vdash B$ without "applying $A \to B$". – Carl Mummert Jan 30 '13 at 19:42

If the deduction meta-theorem has an "if and only if" statement, then the deduction meta-theorem tells us that from "Cab, Cbc |-Cbc" we can obtain "Cab, Cbc, a|-c" by detachment and conversely (making the meta-logic into the object logic). I call the part which enables us to move from "Cab, Cbc |-Cbc" to "Cab, Cbc, a|-c" The Detachment Meta-Theorem, and the other part The Deduction Meta-Theorem. So far as I can tell, any metalogical proof of the deduction meta-theorem will necessarily tell us that we'll have modus ponens, so the detachment meta-theorem immediately holds even if not stated. So, now, here's another proof:

 1 Cab premise
2 Cbc premise
3 a premise
4 b 1, 3 modus ponens
5 c 2, 4 modus ponens
6 Cac 3-5 by Conditional Introduction.


Conditional Introduction comes as one of the rules of inference that The Deduction Meta-Theorem entails.

The Detachment Theorem, at least in my opinion, comes as MUCH more powerful and important than the Deduction Theorem, and it also comes as easier to metalogically argue for. There do exist logical systems where the part of The Deduction Theorem that enables you to move from "{$\Gamma$, $\alpha$} $\vdash$ $\beta$" to "$\Gamma$ $\vdash$ C$\alpha$$\beta" does not work, but you still have The Detachment Theorem. On the other hand there do not exist logical systems where the part of The Deduction Theorem that enable you to move from "\Gamma \vdash C\alpha$$\beta$" to "{$\Gamma$, $\alpha$} $\vdash$ $\beta$" does not work, but you still have the other half of The Deduction Theorem. The only logical systems where The Detachment Theorem does not work consist of those systems where detachment is not a primitive or derivable rule of inference of the logical system. In some sense, there exist "more" logical systems with The Detachment Meta-Theorem than the Deduction Meta-Theorem.