# Proof of a theorem in Hilbert's system

I have been trying to prove that the propositional formula $\big( \alpha \rightarrow \lnot \beta \big) \rightarrow \big((\alpha \rightarrow \beta) \rightarrow \lnot \alpha \big)$ is a theorem in Hilbert's system, i.e., to prove $\vdash \big( \alpha \rightarrow \lnot \beta \big) \rightarrow \big((\alpha \rightarrow \beta) \rightarrow \lnot \alpha \big)$using the following three axioms and Modus Ponens.

1. $\alpha \rightarrow \big(\beta \rightarrow\alpha \big)$
2. $\big({\alpha \rightarrow \big(\beta \rightarrow\gamma \big)}\big) \rightarrow \big({\big(\alpha \rightarrow\beta \big) \rightarrow \big(\alpha \rightarrow\gamma \big)} \big)$
3. $\big( \lnot \beta \rightarrow \lnot \alpha \big) \rightarrow \big( (\lnot \beta \rightarrow\alpha) \rightarrow \beta \big)$

Deduction theorem can be used as well. I am getting into long winded derivations which don't lead anywhere. Hints will be highly appreciated.

I'm able to "manufacture" a quite long proof, with ref to :

• Elliott Mendelson, Introduction to mathematical logic (4th ed - 1997); see page 35 for the axioms [those that you are requested to use] and the only rule of inference : Modus Ponens.

With Ax 1 and Ax 2 we can prove [see Mendelson, Lemma 1.8, page 36] : $\vdash \alpha \rightarrow \alpha$.

(1) $\alpha → ((\alpha \rightarrow \alpha) → \alpha)$ --- from Ax 1

(2) $(\alpha → ((\alpha \rightarrow \alpha) → \alpha)) → ((\alpha → (\alpha \rightarrow \alpha)) → (\alpha \rightarrow \alpha))$ --- from Ax 2

(3) $(\alpha → (\alpha \rightarrow \alpha)) → (\alpha \rightarrow \alpha)$ --- from (1) and (2) by MP

(4) $\alpha → (\alpha \rightarrow \alpha)$ --- from Ax 1

(5) $\alpha \rightarrow \alpha$ --- from (3) and (4) by MP

With it, Ax 1 and Ax 2, we can prove the Deduction Theorem [see Mendelson, page 37].

In addition, with Ax 1 and Ax 2, we can prove Hypothetical Syllogism : $\alpha \rightarrow \beta, \beta \rightarrow \gamma \vdash \alpha \rightarrow \gamma$; in this post we can find a proof of it.

We need two additional Lemmas :

Lemma 1 [see Mendelson, Lemma 1.11(a), page 38] : $\vdash \lnot \lnot \beta \rightarrow \beta$

(1) $\vdash (\lnot \beta \rightarrow \lnot \lnot \beta) \rightarrow ((\lnot \beta \rightarrow \lnot \beta) \rightarrow \beta)$ from Ax 3 with $\lnot \beta$ in place of $\alpha$

(2) $\lnot \beta \rightarrow \lnot \lnot \beta$ --- assumed [a]

(3) $\vdash \lnot \beta \rightarrow \lnot \beta$ --- from $\vdash \alpha \rightarrow \alpha$ : see above

(4) $\beta$ --- from (1), (2) and (3) by MP twice

(5) $\vdash (\lnot \beta \rightarrow \lnot \lnot \beta) \rightarrow \beta$ --- from (2) and (4) by Deduction Th, discharging [a]

(6) $\vdash \lnot \lnot \beta \rightarrow (\lnot \beta \rightarrow \lnot \lnot \beta)$ --- Ax 1

(7) $\vdash \lnot \lnot \beta \rightarrow \beta$ --- from (5) and (6) by HS

Lemma 2 [see Mendelson, Lemma 1.11(e), page 38] : $\alpha \rightarrow \beta \vdash \lnot \beta \rightarrow \lnot \alpha$

(1) $\alpha \rightarrow \beta$ --- assumed [a]

(2) $\lnot \beta$ --- assumed [b]

(3) $\vdash \lnot \beta \rightarrow (\lnot \lnot \alpha \rightarrow \lnot \beta)$ --- Ax 1

(4) $\lnot \lnot \alpha \rightarrow \lnot \beta$ --- from (2) and (3) by MP

(5) $\vdash \lnot \lnot \alpha \rightarrow \alpha$ --- Lemma 1

(6) $\lnot \lnot \alpha \rightarrow \beta$ --- from (1) and (5) by HS

(7) $\vdash (\lnot \lnot \alpha \rightarrow \lnot \beta) \rightarrow ((\lnot \lnot \alpha \rightarrow \beta) \rightarrow \lnot \alpha)$ from Ax 3 with $\lnot \alpha$ in place of $\beta$

(8) $\lnot \alpha$ --- from (4), (6) and (7) by MP twice

(9) $\lnot \beta \rightarrow \lnot \alpha$ --- from (2) and (8) by DT, discharging [b]

Now for the proof of : $\vdash (\alpha \rightarrow \lnot \beta) \rightarrow ((\alpha \rightarrow \beta) \rightarrow \lnot \alpha)$

(1) $\alpha \rightarrow \lnot \beta$ --- assumed [a]

(2) $\alpha \rightarrow \beta$ --- assumed [b]

(3) $\lnot \beta \rightarrow \lnot \alpha$ --- from (2) by Lemma 2

(4) $\alpha \rightarrow \lnot \alpha$ --- from (1) and (3) by HS

(5) $\lnot \lnot \alpha \rightarrow \lnot \alpha$ --- from (4) by Lemma 2

(6) $\vdash (\lnot \lnot \alpha → \lnot \lnot \alpha) → ((\lnot \lnot \alpha \rightarrow \lnot \alpha) → \lnot \alpha)$ --- from Ax 3

(7) $\vdash \lnot \lnot \alpha \rightarrow \lnot \lnot \alpha$ --- from $\vdash \alpha \rightarrow \alpha$

(8) $\lnot \alpha$ --- from (5), (6) and (7) by MP twice

(9) $\vdash (\alpha \rightarrow \lnot \beta) \rightarrow ((\alpha \rightarrow \beta) \rightarrow \lnot \alpha)$ --- from (1), (2) and (8), by two applications of DT, discharging assumptions [a] and [b].

Comment

It can be "istructive" to compare the above proof (I hope it can be simplified) with the Natural Deduction 9-lines proof.

• In the last part of the proof, in (3) $\lnot \beta \rightarrow \lnot \alpha$ is obtained from "(2)" by Lemma 2, right? – user183763 Nov 12 '14 at 13:23

This is just a re-labelling of $3$. It's 3s $\beta = \neg\alpha$ in theorem and $\alpha$ in 3. is $\beta$ in your theorem.

$$(\neg(\neg\alpha) \Rightarrow \neg\beta) \Rightarrow ((\neg(\neg\alpha) \Rightarrow \beta) \Rightarrow \neg\alpha)$$

For a proof of $\neg\neg\alpha \leftrightarrow\alpha$, you can refer to Thm. 6 on this site, to reduce to the proof of $(\neg\beta\Rightarrow\neg\alpha)\Rightarrow(\alpha\Rightarrow\beta)$ (Wich is their axiom 3)

• Yeah that's the crucial part. But how? – user183763 Nov 12 '14 at 8:22
• @jayakrishnan That's LEM (law of excluded middle / tertium non datur), it should be an axiom. – AlexR Nov 12 '14 at 8:58
• @AlexR This system does not have LEM explicitly. You must prove that LEM holds in this system. – Hanul Jeon Nov 12 '14 at 9:50