Prove that: ( → ( → )) → ( → ( → )) can be derived from 1 − 2 using .

Question

Zero-order logic

(1) → ( → );

(2) ( → ( → )) → (( → ) → ( → ));

Prove that: ( → ( → )) → ( → ( → )) can be derived from 1 − 2 using .

So far I have done up to here:

(1) → ( → ) (Premise)

(2) (Premise)

(3) → ( 1,2)

(4) ( → ( → )) → (( → ) → ( → )) (Premise)

(5) → ( → ) (Premise)

Now I don't understand what should I do next....Am I doing things correctly or I should approach it differently?

Answer 1

If you’re allowed the Deduction Meta-Theorem (DT), then this follows most easily by three instances of DT. For reference, DT states that if $\Gamma, A \vdash B$, then $\Gamma \vdash A \to B$.

Here’s an sketch of how it works:

1. $p \to (q \to r)$ [Premise]
2. $q$ [Premise]
3. $p$ [Premise]
4. $.$ [?]
5. $r$ [?]
6. $.$ [?]
7. $q \to(p \to r)$ [?]
8. $(p\to (q \to r)) \to (q \to (p \to r))$ [DT 1-7, discharging 1]

If you aren’t allowed to use the deduction theorem, it will take considerably longer. Note that this proof does not use either of the axioms, so to formally show that the axioms are enough to prove the formula in question, it is vital to show that DT follows from them.