TITLE: Proof of $(\neg A \supset A) \supset A$
QUESTION [5 upvotes]: As a (total) beginner in logic, I  read this introduction : http://www.loria.fr/~roegel/cours/logique-pdf.pdf (in french). They give an exercise I couldn't achieve. Could someone help me (give an answer or just a clue)? 
Using substitution, modus ponens and these axioms :
A1 : $(A\lor A)\supset A$ 
A2 : $B \supset (A \lor B)$ 
A3 : $(A\lor B) \supset (B \lor A)$
A4 : $(A\lor (B\lor C)) \supset (B\lor(A\lor C))$
A5 : $(B \supset C) \supset ((A\lor B) \supset (A \lor C))$
Prove : $(\neg A \supset A) \supset A$
I tried many combinations of these axioms and rules of inference but not the good one(s). 
Thank you
Edit : Here, logical implication $P \supset Q$ is an abbreviation for $\neg P \lor Q$ and $\neg$ is a primitive. 
Exercise is left undone on page 19 : "Nous laissons à titre d’exercice la preuve du troisième axiome de Lukasiewicz." $\Rightarrow$ "Prove the third Lukasiewicz axiom using Whitehead and Russell axioms" (page 18)

REPLY [3 votes]: Double-negation introduction: $B\supset \neg\neg B$. You know how to derive $\neg B\supset \neg B$, which is an abbreviation for $\neg\neg B\lor \neg B$. Then A3 gives you $\neg B\lor \neg\neg B$, which is the same as $B\supset \neg\neg B$.
Double-negation elimination: $\neg\neg A \supset A$. You know how to derive $A\supset A$, which is just $\neg A\lor A$. Now apply double-negation introduction (with $B=\neg A$) to the left-hand side of that (using A3-A5-A3), giving $\neg\neg\neg A\lor A$, which is the same as $\neg\neg A\supset A$.
Proof by contradiction: $(\neg A \supset A)\supset A$. The premise $\neg A\supset A$ is the same as $\neg\neg A\lor A$. By A3-A5-A3 with double-negation elimination, this implies $A \lor A$, and A1 then gives you $A$.