I have two examples to figure out, and I've verified the first. The second one is giving me trouble, though. Here is the statement:
$[(p \lor q)\to r] \leftrightarrow [\lnot r \to \lnot(p \lor q)]$
All I've done is substitute $(p \lor q)$ with s, giving me:
$[s\to r] \leftrightarrow [\lnot r \to \lnot s]$
Since I couldn't figure out a way to simplify further, I made a truth table. When I made up a truth table based on this simplified statement, it doesn't seem to be a tautology. When s and r are the same value ($s = 1$ and $r = 1$, for example) then the biconditional ends up being true, but that's not enough for this to be a tautology, is it?