How to prove $((A \to B) \to A) \to A$ using Lukasiewicz's axioms, MP and deduction theorem?

Question

This is an exercise from A.G. Hamilton's Logic for Mathematicians, section 2.1, p. 36. I have tried to do this for 10 long years, since 2010. Unsuccessful.

Exercise 3: Using the deduction theorem for $L$, show that the following wfs. are theorems of $L$, where $\mathcal{A}$ and $\mathcal{B}$ are any wfs of $L$.

(c) $((\mathcal{A} \to \mathcal{B}) \to \mathcal{A}) \to \mathcal{A}$

The axiom schemes of $L$ are:

1. $\mathcal{A} \to (\mathcal{B} \to \mathcal{A})$.
2. $\mathcal{A} \to (\mathcal{B} \to \mathcal{C}) \to ((\mathcal{A} \to \mathcal{B}) \to (\mathcal{A} \to \mathcal{C}))$.
3. $((\sim \mathcal{A}) \to (\sim \mathcal{B})) \to (\mathcal{B} \to \mathcal{A})$.

The only rule of inference of $L$ is modus ponens (MP): from $\mathcal{A}$ and $\mathcal{A} \to \mathcal{B}$, deduce $\mathcal{B}$.

The deduction theorem for $L$ says: if $\Gamma \cup \{\mathcal{A}\} \vdash \mathcal{B}$, then $\Gamma \vdash (\mathcal{A} \to \mathcal{B})$.

Thanks for helping me.

Answer 1

The statement is known as Peirce's Law, and the proof is pretty nasty. I can believe someone can spend $10$ years on it without cracking it!

The proof uses some helpful Lemma's.

First, let's prove: $\phi \to \psi, \psi \to \chi, \phi \vdash \chi$:

1. $\phi \to \psi$ Premise

2. $\psi \to \chi$ Premise

3. $\phi$ Premise

4. $\psi$ MP 1,3

5. $\chi$ MP 2,4

By the Deduction Theorem, this gives us Hypothetical Syllogism (HS): $\phi \to \psi, \psi \to \chi \vdash \phi \to \chi$

Now let's prove the general principle that $\neg \phi \vdash (\phi \to \psi)$:

1. $\neg \phi$ Premise

2. $\neg \phi \to (\neg \psi \to \neg \phi)$ Axiom1

3. $\neg \psi \to \neg \phi$ MP 1,2

4. $(\neg \psi \to \neg \phi) \to (\phi \to \psi)$ Axiom2

5. $\phi \to \psi$ MP 3,4

With the Deduction Theorem, this means $\vdash \neg \phi \to (\phi \to \psi)$ (Duns Scotus Law)

Let's use Duns Scotus to show that $\neg \phi \to \phi \vdash \phi$ (Law of Clavius):

1. $\neg \phi \to \phi$ Premise

2. $\neg \phi \to (\phi \to \neg (\neg \phi \to \phi))$ (Duns Scotus Law)

3. $(\neg \phi \to (\phi \to \neg (\neg \phi \to \phi))) \to ((\neg \phi \to \phi) \to (\neg \phi \to \neg (\neg \phi \to \phi)))$ Axiom3

4. $(\neg \phi \to \phi) \to (\neg \phi \to \neg (\neg \phi \to \phi))$ MP 2,3

5. $\neg \phi \to \neg (\neg \phi \to \phi)$ MP 1,4

6. $(\neg \phi \to \neg (\neg \phi \to \phi)) \to ((\neg \phi \to \phi) \to \phi)$ Axiom2

7. $(\neg \phi \to \phi) \to \phi$ MP 5,6

8. $\phi$ MP 1,7

Using Duns Scotus and the Law of Clavius, we can now show that $ (\phi \to \psi) \to \phi \vdash \phi$:

1. $(\phi \to \psi) \to \phi$ Premise

2. $\neg \phi \to (\phi \to \psi)$ Duns Scotus

3. $\neg \phi \to \phi$ HS 1,2

4. $\phi$ Law of Clavius 3

And so finally, by the Deduction Theorem, we have: $\vdash ((\phi \to \psi) \to \phi) \to \phi$