Complicated Logic Proof involving Tautology and Law of Excluded Middle I'm having great difficulty solving the following problem, and even figuring out where to start with the proof.
$$
\neg A\lor\neg(\neg B\land(\neg A\lor B))
$$
Please see the following examples of how to do proofs, I would appreciate it if you could attempt to give me guidance using the tools and the line numbers that it cites similar to those below:
This is a sample proof:

This is another sample proof (law of excluded middles):


The "complicated" formula :

$¬A∨¬(¬B∧(¬A∨B))$

can be re-written, due to the equivalence between $P \rightarrow Q$ and $\lnot P \lor Q$, as :

$A \rightarrow \lnot ((A \rightarrow B) \land \lnot B)$.

But $P \rightarrow Q$ is also equivalent to $\lnot (P \land \lnot Q)$; so the formula it is simply :


$A \rightarrow ((A \rightarrow B) \rightarrow B)$.


Now, a Natural Deduction proof is quite easy :
(1) $A$ - assumed
(2) $A \rightarrow B$ --- assumed
(3) $B$ --- from (1) and (2) by $\rightarrow$-elimination
(4) $(A \rightarrow B) \rightarrow B$ --- from (2) and (3) by $\rightarrow$-introduction
(5) $A \rightarrow ((A \rightarrow B) \rightarrow B)$ --- from (1) and (4) by $\rightarrow$-introduction.
