Determine whether $((p \Rightarrow q) \Rightarrow r) \Leftrightarrow (p \Rightarrow (q \Rightarrow r))$ is a tautology, a contradiction, or neither.
If $p,q,r = (0,0,0)$ then $((p \Rightarrow q)\Rightarrow r)$ $\Leftrightarrow$ $(p\Rightarrow(q\Rightarrow r)) = 0$, but if $p,q,r = (0,0,1)$ then $((p\Rightarrow q)\Rightarrow r) \Leftrightarrow (p \Rightarrow (q \Rightarrow r)) = 1$ . Therefore it is neither a tautology nor a contradiction.
Is this a good proof? Any feedback would be greatly appreciated. Thanks!
The truth table gives $$\begin{array}{c|c|c|c|c|c|c} p&q&r&p\implies q&(p\implies q)\implies r&q\implies r&p\implies (q\implies r)\\ \hline T&T&T&T&T&T&T\\ T&T&F&T&F&F&F\\ T&F&T&F&T&T&T\\ T&F&F&F&T&T&T\\ F&T&T&T&T&T&T\\ F&T&F&T&F&F&T\\ F&F&T&T&T&T&T\\ F&F&F&T&F&T&T \end{array}$$
Since there are two rows such that $(p\implies q)\implies r\iff p\implies(q\implies r)$ is not true, then it is not a tautology. Also since there is at least one row such that those columns are equivalent means that this is not a contradiction. Therefore, it's neither a contradiction nor a tautology.