Boolean function proving contradiction ,tautology or neither

Question

Determine whether $((p \Rightarrow q) \Rightarrow r)\Leftrightarrow (p \Rightarrow(q \Rightarrow r))$ is a tautology, a contradiction, or neither. $$\begin{array}{cccc} \underline{p}&\underline{q}&\underline{r}& \underline{((p \Rightarrow q) \Rightarrow r)\Leftrightarrow (p \Rightarrow(q \Rightarrow r))}\\ 0&0&0& 0\\ 0&0&1& 1\\ 0&1&0& 0\\ 0&1&1& 1\\ 1&0&0& 1\\ 1&0&1& 1\\ 1&1&0& 1\\ 1&1&1& 1\\ \end{array}$$

So it is neither contradiction or tautology since the last column has both 1 and 0 values. Is this correct? I need your feedback please Thank you.

Answer 1

Assuming you computed the right values, you are correct. It would be neither a tautology (always true) nor a contradiction (never true).