m | m <--> m | m --> m | (m <-> m) <--> (m --> m)
T | T | T | T
F | T | T | T
Note that each of $m \rightarrow m$ and $m \leftrightarrow m$ is a tautology (always true, regardless of the truth value of $m$), and hence, $$(m \leftrightarrow m) \leftrightarrow (m \rightarrow m)$$
is necessarily a tautology, as well, which means the following equivalence necessarily holds: $$(m \leftrightarrow m) \leftrightarrow (m \rightarrow m) \equiv T$$