Examine $\def\getsto{{\;\leftrightarrow\;}}\def\false{\underset{\tiny\text{false}}\bot}\def\true{\underset{\tiny\text{true}}\top} p \getsto (q \to r)$ and $(p \getsto q) \to r$
$\begin{align}p \getsto (q \to r) ~& \iff (\neg p\vee (q\to r)\wedge(p\vee\neg(q\to r))
\\ &\iff (\neg p\vee\neg q\vee r)\wedge(p\vee\neg(\neg q\vee r)))
\\ &\iff (\neg p\vee\neg q\vee r)\wedge (p\vee(q\wedge\neg r))
\\[3ex](p \getsto q) \to r ~&\iff \neg(p\getsto q)\vee r
\\ & \iff (p\oplus q)\vee r
\\ & \iff ((\neg p\vee \neg q)\wedge(p\vee q))\vee r
\\ & \iff (\neg p\vee \neg q\vee r)\wedge(p\vee q\vee r)
\end{align}$
Test when $r$ true but $p,q$ both false.
$ \false\getsto(\false\to\true)$ is false but $(\false\getsto\false)\to\true$ is true.