TITLE: Help proving $ \sim \sim (p \to q),\sim \sim p \vdash \sim \sim q $ with intuitionistic axioms
QUESTION [6 upvotes]: Rule
modus ponens: $ p, p \to q \vdash q $
Axioms
A1  $ p \to (q \to p) $
A2  $ (p \to (q \to r)) \to ((p \to q) \to (p \to r)) $
A3  $ (p \land q) \to p $
A4  $ (p \land q) \to q $
A5  $ p \to (q \to (p \land q)) $
A6  $ p \to (p \lor q) $
A7  $ q \to (p \lor q) $
A8  $ (p \to r) \to ((q \to r) \to ((p \lor q) \to r)) $
A9  $ (p \to q) \to ((p \to \sim q) \to \sim p) $
A10 $ \sim p \to (p \to q) $
Rule I have derived: $ p \to q, q \to r \vdash p \to r $
Derived meta-rule (deduction theorem): $ \Gamma,p \vdash q $ implies $ \Gamma \vdash p \to q $
Theorems I have proven
T1  $ p \to \sim \sim p $
T2  $ (p \to q) \to (\sim q \to \sim p) $
T3  $ p \to (\sim p \to q) $
T4  $ \sim \sim (\sim \sim p \to p) $
T5  $ \sim \sim \sim p \to \sim p $
I am guessing axioms 3,4,5,6,7 and 8 are not needed for the proof.
On the other hand I believe A10 and T3 are critical.
Thanks for any help on this.

REPLY [2 votes]: The about 40 lines long derivation seems to me a little bit longer than needed ...
1) $\lnot \lnot (p→q)$ --- premise
2) $\lnot \lnot p$ --- premise
3) $\vdash \lnot \lnot (p→q) → [\lnot q \rightarrow \lnot \lnot (p→q)]$ --- Axiom 1
4) $\lnot q \rightarrow \lnot \lnot (p→q)$ --- from 1) and 3) by modus ponens
5) $\vdash \lnot \lnot p → [(p \rightarrow q) \rightarrow \lnot \lnot p]$ --- Axiom 1
6) $(p \rightarrow q) \rightarrow \lnot \lnot p$ --- from 2) and 5) by modus ponens
7) $\lnot q$ ---- assumed [a]
8) $p→q$ --- assumed [b]
9) $\lnot q \rightarrow \lnot p$ --- from 8) by T2 and modus ponens
10) $\lnot p$ --- from 7) and 9) by modus ponens
11) $(p \rightarrow q) \rightarrow \lnot p$ --- from 8) and 10) by Deduction Th, discharging [b]
12) $\lnot (p \rightarrow q)$ --- from 11) and 6) and Axiom 9 by modus ponens twice
13) $\lnot q \rightarrow \lnot (p \rightarrow q)$ --- from 7) and 12) and Deduction Th, discharging [a]
14) $\lnot \lnot q$ --- from 4) and 13) and Axiom 9 by modus ponens twice.
Thus, form 1), 2) and 14) :


$\lnot \lnot (p→q), \lnot \lnot p \vdash \lnot \lnot q$.