Complicated Logic Proof involving Tautology and Law of Excluded Middle

first of all the statement is true (checking truth values for $A$ and $B$), even though ive now given two incorrect answers... here we go again!

$\begin{align} & \neg A \lor \neg (\neg B \land (\neg A \lor B))\\ & =\neg A \lor \neg [(\neg B \land \neg A) \lor (\neg B \land B)]\\ & =\neg A \lor \neg [(\neg B \land \neg A) \lor FALSE]\\ & =\neg A \lor \neg(\neg B\land\neg A)\\ & =\neg A\lor(B\lor A) \text{ using } \neg(X\land Y)=\neg X\lor\neg Y\\ & =TRUE \end{align}$

formalize this in whatever system youre using, you might have to prove a few things beforehand...