I am looking at the CNF of $((A \to B) \to (B \to A)) \to A$. For this I tried the following:
$${\quad((A \to B ) \to (B \to A)) \to A\\ \equiv ((\neg A \vee B) \to (\neg B \vee A) \to A \\\equiv (\neg (\neg A \vee B) \vee (\neg B \vee A )) \to A \\\equiv (( A \wedge \neg B) \vee (\neg B \vee A) \to A \\\equiv \neg ((A \wedge \neg B) \vee (\neg B \vee A)) \vee A \\\equiv \neg (A \wedge \neg B) \wedge \neg (\neg B \vee A) \vee A \\ \equiv \neg A \vee B \wedge B \wedge \neg A \vee A \\\equiv \neg A \vee B}$$ Now the solution is wrong at one little point, it should be $A$ and not $\neg A$ but I don't see where I did my mistake... Can someone please helps me out? Many thanks!
You are correct up until the indicated line. You dropped the brackets and so lost track of the associations.$$\require{cancel}{\quad((A \to B ) \to (B \to A)) \to A\\ \equiv ((\neg A \vee B) \to (\neg B \vee A) \to A \\\equiv (\neg (\neg A \vee B) \vee (\neg B \vee A )) \to A \\\equiv (( A \wedge \neg B) \vee (\neg B \vee A) \to A \\\equiv \neg ((A \wedge \neg B) \vee (\neg B \vee A)) \vee A \\\cancel{\equiv \neg (A \wedge \neg B) \wedge \neg (\neg B \vee A) \vee A \qquad\star \\ \equiv \neg A \vee B \wedge B \wedge \neg A \vee A \\\equiv \neg A \vee B}}$$ Keep the bracketing until you are concatenating like operators: $${\equiv \big(\lnot (A\land\lnot B)\land\lnot(\lnot B\lor A)\big)\lor A\\\equiv \big((\lnot A\lor B)\land(B\land\lnot A)\big)\lor A\\\equiv \big((\lnot A\lor B)\land B\land\lnot A\big)\lor A\\~\vdots\\\equiv B\lor A}\qquad$$
Now the remainder should be easy to fill in.