Showing a conditional statement is a tautology without using a truth table. I wanted to show that [(p→q)∧(q→r)]→(p→r) is a tautology without using a truth table. This is what I got so far:
[(p→q)∧(q→r)] → (p→r)
=> ¬[(¬p v q) ∧ (¬q v r)] v (¬pvr)  (logical equivalence)
=> [¬(¬p v q) v ¬(¬qvr)] v (¬pvr)  (demorgan's law)
=> [(p ∧ ¬q) v (q∧¬r)] v (¬pvr)  (demogran's law)
I can't seem to figure out what comes after this step. Can someone help me?
 A: Expand the expression $(p \wedge \neg q) \vee (q \wedge \neg r)$ by distributing the $\vee$ over the $\wedge$:
\begin{align*}
&(p \wedge \neg q) \vee (q \wedge \neg r)\\
&[(p \wedge \neg q) \vee q] \wedge [(p \wedge \neg q) \vee \neg r]\\
&[(p \vee q ) \wedge (\neg q \vee q)] \wedge [(p \vee \neg r ) \wedge (\neg q \vee \neg r)]\\
&[(p \vee q) \wedge \top ] \wedge[(p \vee \neg r) \wedge (\neg q \vee \neg r)]\\
&(p \vee q) \wedge [(p \vee \neg r) \wedge (\neg q \vee \neg r)].
\end{align*}
Now overall we have $\{(p \vee q) \wedge [(p \vee \neg r) \wedge (\neg q \vee \neg r)]\}\vee (\neg p \vee r)$. If we distribute the $\vee$ over the $\wedge$, we get $[(p \vee q) \vee ( \neg p \vee r)]\wedge\{[(p \vee \neg r) \wedge (\neg q \vee \neg r)]\vee(\neg p \vee r)\}$. Focusing on the first half, you can manipulate $(p \vee q) \vee ( \neg p \vee r)$ to get $\top$ by shuffling parentheses around to get $(p \vee \neg p) \vee (q \vee r)$ (I'll leave that to you). 
So we are left with $[(p \vee \neg r) \wedge (\neg q \vee \neg r)] \vee (\neg p \vee r)$. Again let's distribute the $\vee$ over the $\wedge$:
\begin{align*}
&[(p \vee \neg r) \vee (\neg p \vee r)] \wedge [(\neg q \vee \neg r) \vee (\neg p \vee r)].
\end{align*}
Again both halves of this can be manipulated to get $\top$.
A: 
I can't seem to figure out what comes after this step. Can someone help me?

Yes.
$$\begin{align}\vdots\quad\\\iff&~\big((p \land \lnot q) \lor (q\land\lnot r)\big) \lor (\lnot p\lor r)
\\[1ex]\iff &~\big(\lnot p\lor (p\land \lnot q)\big)\lor \big(r\lor (\lnot r\land q)\big)&\quad\textsf{(Commutation and Association)}\\\vdots\quad\end{align}$$
A: You can use Double Distribution to get 
$$[(p \lor q)\land(q \lor \lnot q)\land(\lnot q \lor \lnot r)\land(p \lor \lnot r)]\lor (\lnot p \lor r)$$
$q \lor \lnot q$ is a Tautology so this becomes
$$[(p \lor q)\land(\lnot q \lor \lnot r)\land(p \lor \lnot r)]\lor (\lnot p \lor r)$$
which by distribution is
$$[(p \lor q)\land[\lnot r\land(p \lor \lnot q)]]\lor (\lnot p \lor r)$$
Association gives
$$[\lnot r \land [(p \lor q)\land(p \lor \lnot q)]]\lor (\lnot p \lor r)$$
Distribution again gives
$$[\lnot r \land [p \lor(q \land \lnot q)]]\lor (\lnot p \lor r)$$
$q \land \lnot q$ is a contradiction so this becomes
$$(\lnot r \land p)\lor (\lnot p \lor r)$$
Which by DeMorgan's Law is
$$(\lnot r \land p)\lor \lnot(\lnot r \land p)$$
Which is a tautology.
A: Notice that
$ [(p \land \neg q) \lor (q\land \neg r)] \lor (\neg p\lor r)$
is one big disjunction, so you can drop parentheses:
$ (p \land \neg q) \lor (q\land \neg r) \lor \neg p\lor r$
Now, if you have:
Reduction
$p \lor (\neg p \land q) \equiv p \lor q$
then you can apply that:
$ (p \land \neg q) \lor (q\land \neg r) \lor \neg p\lor r \equiv$
$\neg q \lor q \lor \neg p \lor r \equiv$
$\top \lor \neg p \lor r \equiv$ 
$\top$
But if you don't have Reduction:
$ (p \land \neg q) \lor (q\land \neg r) \lor \neg p \lor r \equiv$
$((p \lor \neg p) \land (\neg q \lor \neg p)) \lor ((q \lor r) \land (\neg r \lor r)) \equiv$
$(\top \land (\neg q \lor \neg p)) \lor ((q \lor r) \land \top) \equiv$
$\neg q \lor \neg p \lor q \lor r$
$\top \lor \neg p \lor r \equiv$ 
$\top$
