Proving $[(P\lor Q)\land(P\to R)\land(Q\to R)]\to R$ is a tautology without using a truth table? 
$$[(P\lor Q)\land(P\to R)\land(Q\to R)]\to R\tag{1}$$

How can I prove that $(1)$ is a tautology without using a truth table?
I used the identity $$(P\to R)\land(Q\to R)\equiv(P\lor Q)\to R$$ but from there I get stuck and can't figure out where to go.

I'll outline one fairly easy way of solving this problem without a truth table; however, the solution is not one that you obtain deduction by deduction (which is what it seems you want). 
Using DeMorgan's law and the fact that $y\to z\equiv\neg y\lor z$, we may write your implication as follows:
$$
\underbrace{(\neg p\land\neg q)}_{(1)}\lor\underbrace{(p\land\neg r)}_{(2)}\lor\underbrace{(q\land\neg r)}_{(3)}\lor r.
$$
If $r$ is true, then you're done. Thus, suppose $r$ is not true. If either $p$ or $q$ (or both) is true, then either $(2)$ or $(3)$ is true (or both). If, however, neither $p$ nor $q$ is true, then $(1)$ is true. Thus, regardless of the truth values for $p,q,r$, we must have a tautology. $\blacksquare$
