So I posted the same problem elsewhere and a user named Melanie Turner responded with a solution. Here's the proof, I've edited it slightly, but only in order to make the notation more consistent with the previous posts in this thread—everything else is untouched.
Axioms:
P1. a → (b → a)
P2. (a → (b → c)) → ((a → b) → (a → c))
P3. (~a → ~b) → ((~a → b) → a)
Lemma 1: P → P
1. P → ((P → P) → P) [P1]
2. (P → ((P → P) → P)) → ((P → (P → P)) → (P → P)) [P2]
3. (P → (P → P)) → (P → P) [1, 2, MP]
4. P → (P → P) [P1]
5. P → P [3, 4, MP]
Lemma 2: (P → Q) → ((R → P) → (R → Q))
1. (R → (P → Q)) → ((R → P) → (R → Q)) [P2]
2. ((R → (P → Q)) → ((R → P) → (R → Q))) → ((P → Q) → ((R → (P → Q)) → ((R → P) → (R → Q)))) [P1]
3. (P → Q) → ((R → (P → Q)) → ((R → P) → (R → Q))) [1, 2, MP]
4. ((P → Q) → ((R → (P → Q)) → ((R → P) → (R → Q)))) → (((P → Q) → (R → (P → Q))) → ((P → Q) → ((R → P) → (R → Q)))) [P2]
5. ((P → Q) → (R → (P → Q))) → ((P → Q) → ((R → P) → (R → Q))) [3, 4, MP]
6. (P → Q) → (R → (P → Q)) [P1]
7. (P → Q) → ((R → P) → (R → Q)) [5, 6, MP]
Lemma 3: P → ((P → Q) → Q)
1. (P → Q) → (P → Q) [Lemma 1]
2. ((P → Q) → (P → Q)) → (((P → Q) → P) → ((P → Q) → Q)) [P2]
3. ((P → Q) → P) → ((P → Q) → Q) [1, 2, MP]
4. (((P → Q) → P) → ((P → Q) → Q)) → ((P → ((P → Q) → P)) → (P → ((P → Q) → Q))) [Lemma 2]
5. (P → ((P → Q) → P)) → (P → ((P → Q) → Q)) [3, 4, MP]
6. P → ((P → Q) → P) [P1]
7. P → ((P → Q) → Q) [5, 6, MP]
Lemma 4: (~P → Q) → (~Q → P)
1. (~P → ~Q) → ((~P → Q) → P) [P3]
2. ((~P → ~Q) → ((~P → Q) → P)) → ((~Q → (~P → ~Q)) → (~Q → ((~P → Q) → P))) [Lemma 2]
3. (~Q → (~P → ~Q)) → (~Q → ((~P → Q) → P)) [1, 2, MP]
4. ~Q → (~P → ~Q) [P1]
5. ~Q → ((~P → Q) → P) [3, 4, MP]
6. (~Q → ((~P → Q) → P)) → ((~Q → (~P → Q)) → (~Q → P)) [P2]
7. (~Q → (~P → Q)) → (~Q → P) [5, 6, MP]
8. ((~Q → (~P → Q)) → (~Q → P)) → (((~P → Q) → (~Q → (~P → Q))) → ((~P → Q) → (~Q → P))) [Lemma 2]
9. ((~P → Q) → (~Q → (~P → Q))) → ((~P → Q) → (~Q → P)) [7, 8, MP]
10. (~P → Q) → (~Q → (~P → Q)) [P1]
11. (~P → Q) → (~Q → P) [9, 10, MP]
⊢ ~((~P → Q) → Q) → P
1. ~P → ((~P → Q) → Q) [Lemma 3]
2. (~P → ((~P → Q) → Q)) → (~((~P → Q) → Q) → P) [Lemma 4]
3. ~((~P → Q) → Q) → P [1, 2, MP]
As for Hex, I stopped paying much attention after the first post (the one with the rhetorical question about using logical tautologies as a means to determine whether or not a criminal is lying). However, from what I did pick up after lightly skimming over the other posts, Hex seems to be attempting to give an
a priori argument for some variety of radical empiricism. The argument is so incoherent that to engage oneself in any attempt to refute it is to waste one's time—and time is a precious thing.
Otherwise, thanks to Arising_uk for going to the effort to try to solve my problem.
—egg3000