The conditional statement is
$[(p \rightarrow q) \land (q \rightarrow r)] \rightarrow (p \rightarrow r)$
Here are the steps I took in an attempt to prove the above statement a tautology, but I can't seem to understand what kind of reasoning I have to use to complete this problem.
LINK TO A TABLE OF LOGICAL EQUIVALENCES FOR YOUR CONVENIENCE : http://en.wikipedia.org/wiki/Logical_equivalence
$\lnot[(p \rightarrow q) \land (q \rightarrow r)] \lor (p \rightarrow r)$ (Because $p \rightarrow q = \lnot p \lor q$ )
$\lnot (p \rightarrow q) \lor \lnot ( q \rightarrow r) \lor (p \rightarrow r)$ (Because $ \lnot(p \land q) = \lnot p \lor \lnot q $ )
At this point I just isolate and focus on the portion of the above that is:
$\lnot ( q \rightarrow r) \lor (p \rightarrow r)$
Then that can become:
$ (p \rightarrow r)\lor \lnot ( q \rightarrow r)$ (Because of the commutative law)
$(p \rightarrow r) \lor q \land \lnot r$ (Because $\lnot(p \rightarrow q) = p \land \lnot q$)
$(p \rightarrow r) \lor \lnot r \land q$ (Because of the commutative law)
$(\lnot p \lor r) \lor \lnot r \land q$ (Because $p \rightarrow q = \lnot p \lor q$)
$\lnot p \lor ( r \lor \lnot r) \land q$ (Because of the associative law)
$\lnot p \lor T \land q$ (Because of negation)
$T \land q$ (Because $\lnot p \lor T$ is true no matter what)
I don't know how to get from this point to "and this is why the original conditional statement is true no matter what".
$(p \rightarrow r) \lor (q \land \lnot r)$ is not equivalent to $((p \rightarrow r) \lor \lnot r) \land q$.
$\lnot ( q \rightarrow r) \lor (p \rightarrow r)=\text{(...as you did...)}=(p \rightarrow r) \lor (q \land \lnot r)$
So $\lnot (p \rightarrow q) \lor \lnot ( q \rightarrow r) \lor (p \rightarrow r)$
I used the distributive law $p \lor (q \land r)=(p \lor q) \land (p \lor r)$ but the fact we needed is actually $p \lor (\lnot p \land q)=p \lor q$, which I believe is the absorption law.