Proving $(A \lor ¬B) \rightarrow (A \rightarrow B)$ is a tautology with Fitch-style deduction

Question

In forall x: Calgary, by P. D. Magnus, chapter 16, appears this exercise:

I do not know how to derive a contradiction in line 4. I came up with this basic skeleton. How can I continue the proof ?

Answer 1

You can't, because this statement is not valid!

Consider: $A$ is True and $B$ is False. Then the whole statement evaluates to:

$$(T \lor \neg F) \to (T \to F) = (T \lor T) \to F = T \to F = F$$

So (assuming the proof system is sound), this statement is just not provable.

Are you sure the statement is not $(\neg A \lor B) \to (A \to B)$? Because that one is valid, and so (assuming the proof system is complete) you should be able to prove that one.

BTW: If the statement were provable, your skeleton is good! Good job