Proving $((A \to B) \to B) \to B$ is a tautology

Question

Working on P.D. Magnus. forallX: an Introduction to Formal Logic (pp. 154, exercise D. 7). It is the last exercise of Chapter 16. This is what I got so far.

Is my $IP$ strategy correct ?

$\def\fitch#1#2{\quad\begin{array}{|l}#1\\\hline#2\end{array}} \fitch{}{ \fitch{¬(((A \to B) \to B) \to B)}{ \fitch{¬((A \to B) \to B) }{ \fitch{¬(A \to B)}{ \fitch{B}{ \fitch{A}{ B } \\ A \to B \\ \bot } } } } }$

P.S: Here is a link to the open-source book: forallX

Answer 1

Hint: $(A \to B) \to B$ is logically equivalent to $A \lor B$. So $((A \to B) \to B) \to B$ is logically equivalent to $(A \lor B) \to B$, which is not a tautology: if $A$ is true and $B$ is false, $(A \lor B) \to B$ is false.

Answer 2

Use a truth table.

$$\begin{array}{|c|c|c|c|c|} \hline A & B & A \to B & (A \to B) \to B & ((A \to B) \to B) \to B\\ \hline 0 & 0 & 1 & 0 & 1\\ \hline 0 & 1 & 1 & 1 & 1\\ \hline 1 & 0 & 0 & 1 & 0\\ \hline 1 & 1 & 1 & 1 & 1\\ \hline \end{array}$$

Since the last column is not identically true (i.e., "$1$"), then you can conclude that $ ((A \to B) \to B) \to B$ is not a tautology.