Proving $A \rightarrow B \therefore \neg A \lor B$

Question

In forall x: Calgary, by P. D. Magnus, section 16 p.154, D. 3, appears this exercise (Fitch-style natural deduction):

I am trying to derive a contradiction on line 4. Is this approach correct? How can I continue ?

Answer 1

The text says that all of the exercises require the IP strategy. To review, the IP strategy means that when you want to prove a formula F, you need to assume the negation of that formula.

Answer 2

Hint

Assume $A$ and $\lnot (\lnot A \lor B)$.

With $A$ derive a contradiction and from it $\lnot A$.

Then, use it for a new contradiction.

Answer 3

We know that $\lnot A\lor A$ is true. By the hypothesis, whenever $A$ is true then $B$ is true. Therefore, $\lnot A\lor B$ is true.