Thank you for all the reponses.
I have focused on Mr. Makholm's helpful workup. Although it is true that I don't have access to $ \bot $ It seems to me that his steps 3-5 are not necessary:
Derive rule A: $ \sim q \vdash (p \to q) \to \sim p $
1 $ \sim q $ hypothesis
2 $ p \to q $ hypothesis
3 $ \sim q \to (p \to \sim q) $ A1
4 $ p \to \sim q $ 1,3 modus ponens
5 $ (p \to q) \to ((p \to \sim q) \to \sim p) $ A9
6 $ (p \to q) \to ~p $ 2,5 modus ponens
7 $ \sim p $ 4,6 modus ponens
8 $ \sim q, p \to q \vdash \sim p $ 1-7
9 $ \sim q \vdash (p \to q) \to \sim p $ 8 deduction theorem
Derive $ \sim \sim (p \to q),\sim \sim p \vdash \sim \sim q $
1 $ \sim \sim (p \to q) $ hypothesis
2 $ \sim \sim p $ hypothesis
3 $ \sim q $ hypothesis
4 $ \sim q \vdash \sim \sim (p \to q) $ 1
5 $ p \to q \vdash \sim \sim p $ 2
6 $ \sim q \to \sim \sim (p \to q) $ 4 deduction theorem
7 $ (p \to q) \to \sim \sim p $ 5 deduction theorem
8 $ (p \to q) \to \sim p $ 3 rule A
9 $ ((p \to q) \to \sim p) \to ((p \to q) \to \sim \sim p) \to \sim (p \to q)) $ A9
10 $ (p \to q) \to \sim \sim p) \to \sim (p \to q) $ 8,9 modus ponens
11 $ \sim (p \to q) $ 7,10 modus ponens
12 $ \sim q \vdash \sim (p \to q) $ 3-11
13 $ \sim q \to \sim (p \to q) $ 12 deduction theorem
14 $ ( \sim q \to \sim (p \to q)) \to (( \sim q \to \sim \sim (p \to q)) \to \sim \sim q) $ A9
15 $ ( \sim q \to \sim \sim (p \to q)) \to \sim \sim q $ 13,14 modus ponens
16 $ \sim \sim q $ 6,15 modus ponens
17 $ \sim \sim (p \to q),\sim \sim p \vdash \sim \sim q $ 1-16
My only points of discomfort are steps 4 and 5 of my second derivation (Mr. Makholm's "general assumptions" of his steps 6 and 7). They presume the meta-rule $ \varnothing \vdash p $ implies $ \Gamma \vdash p $. I suppose this should be easy enough to prove however.
I think I found a way to avoid using the meta rule $ \varnothing \vdash p $ implies $ \Gamma \vdash p $
To obtain step 6 of my second derivation without using step 4:
Prove $ \sim q \to \sim \sim (p \to q) $
1 $ \sim \sim (p \to q) $ hypothesis
2 $ \sim \sim (p \to q) \to ( \sim (p \to q) \to q) $ A10
3 $ \sim (p \to q) \to q $ 1,2 modus ponens
4 $ (\sim (p \to q) \to q) \to ( \sim q \to \sim \sim (p \to q)) $ T2
5 $ \sim q \to \sim \sim (p \to q) $ 3,4 modus ponens
To obtain step 7 of my second derivation without using step 5:
Prove $ (p \to q) \to \sim \sim p $
1 $ \sim \sim (p \to q) $ hypothesis
2 $ \sim \sim p $ hypothesis
3 $ p \to q $ hypothesis
4 $ \sim \sim p \to ( \sim p \to \sim (p \to q)) $ A10
5 $ \sim p \to \sim (p \to q) $ 2,4 modus ponens
6 $ (\sim p \to \sim (p \to q)) \to ( \sim \sim (p \to q) \to \sim \sim p) $ T2
7 $ \sim \sim (p \to q) \to \sim \sim p $ 5,6 modus ponens
8 $ \sim \sim p $ 1,7 modus ponens
9 $ p \to q \vdash \sim \sim p $ 3-8
10 $ (p \to q) \to \sim \sim p $ 9 deduction theorem