How to verify if a compound logical statement is a tautology using substitution

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?

The substitution leading to $(s \to r) \leftrightarrow (\neg r \to \neg s)$ is correct, and this formula is a well-known tautology. You are right in saying that it's not enough to consider the case $r = s$; maybe there's an error in your truth table somewhere? It should yield true in every row.
• By substitution, no further simplification seems to be possible. You could use some other equivalences, though (e.g., $P \to Q \equiv \neg P \vee Q$), if you know about them. – Johannes Kloos Jan 27 '13 at 10:56