Proving (A → -A) → -A In a Hilbert system

Question

Trying to prove (A → -A) → -A with the following axioms:

A → (B → A)

(A → (B → C)) → ((A → B) → (A → C))

(-A → -B) → (B → A)

Not an axiom, but you can also use A → (-A → B), which I've already managed to prove it.

Intuitively the proposition makes sense, but I'm not sure how to prove it

@KennyLau Then it would be straight forward. $${1.~A\to(\neg A\to \bot):\textsf{theorem 1}\2.~ (A\to(\neg A\to \bot))\to((A\to\neg A)\to (A\to \bot)):\textsf{axiom 2}\3.~(A\to\neg A)\to(A\to \bot):\textsf{modus ponens}}$$ — Graham Kemp, May 29 '18 at 00:13
@KennyLau No, ¬A is not really defined in any way, it's just known to be a unary operator which satisfies the 3rd axiom. — user1636815, May 29 '18 at 00:25
This problem can get solved quickly with the use of William McCune's Prover9. — Doug Spoonwood, May 31 '18 at 00:33

score 1 · Answer 1 · answered May 29 '18 at 06:41

Hint

With Ax.1 and Ax.2 you have to derive some preliminary useful results :

$\vdash A \to A$ and the derived rule Syll : $A \to B, B \to C \vdash A \to C$,

and eventually the Deduction Theorem.

Then, Ax.3 is crucial in deriving double negation laws:

$\vdash \lnot \lnot A \to A$ and : $\vdash A \to \lnot \lnot A$.

Double negation, in turn, give us : $\vdash (A \to B) \to (\lnot B \to \lnot A)$.

Two further Lemma :

1) $\vdash \lnot A \to (\lnot B \to \lnot A)$ --- Ax.1

2) $\vdash (\lnot B \to \lnot A) \to (A \to B)$ --- Ax.3

(A) $\vdash \lnot A \to (A \to B)$ --- from 1) and 2) by Syll

1) $\vdash \lnot A \to (A \to \lnot B)$ --- Lemma (A)

2) $\vdash (\lnot A \to (A \to \lnot B)) \to ((\lnot A \to A) \to (\lnot A \to B))$ --- Ax.2

3) $\vdash (\lnot A \to A) \to (\lnot A \to B)$ --- from 1) and 2) by MP

4) $\vdash (\lnot A \to \lnot B) \to (B \to A)$ --- Ax.3

(B) $\vdash (\lnot A \to A) \to (B \to A)$ --- from 3) and 4) by Syll

1) $\vdash (\lnot A \to A) \to ((\lnot A \to A) \to A)$ --- Lemma (B)

2) $\vdash ((\lnot A \to A) \to ((\lnot A \to A) \to A)) \to (((\lnot A \to A) \to (\lnot A \to A)) \to ((\lnot A \to A) \to A))$ --- Ax.2

3) $\vdash ((\lnot A \to A) \to (\lnot A \to A)) \to ((\lnot A \to A) \to A)$ --- from 1) and 2) by MP

4) $\vdash (\lnot A \to A) \to (\lnot A \to A)$ --- from $\vdash A \to A$ above

(C) $\vdash (\lnot A \to A) \to A$ --- from 3) and 4) by MP.

Now the proof :

1) $\vdash (\lnot \lnot A \to \lnot A) \to \lnot A$ --- Lemma (C)

2) $A \to \lnot A$ --- assumed [a]

3) $\lnot \lnot A \to \lnot A$ --- from 2)

4) $\lnot A$ --- from 1) and 3) by MP

5) $\vdash (A \to \lnot A) \to \lnot A$ --- from 2) and 4) by DT, discharging [a].

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